General Organic Chem

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Chemistry / General Organic Chemistry

General OrganicChemistry

This chapter is about an overview of organic chemistry. Here we’ll discuss the entire framework within which theconcepts of this subject will be built. A good feel of general organic chemistry is very necessary if you wish that themyriad of information available in this field stays coherently connected in your mind. As we progress along thestudy of organic chemistry, you’ll begin to understand that grasping a small number of guiding principles can beused to explain a vast array of existing disparate information and also to make predictions about previouslyunknown reactions and compounds.

Keeping this in mind, you should strive hard to really understand the underlying basic principles of organicchemistry.

Since organic chemistry is all about the study of carbon compounds we should have a good understanding of thechemistry of carbon atom.

Consider the ground state electronic configuration of the carbon atom.

We see that there are two half-filled orbitals in the 2p level. Thus, only two unpaired electrons might indicate thatcarbon atom is divalent.

The following visual will help you form a mental picture of what the second level orbitals of carbon look like.(Recall from the chapter on Atomic Structure that orbitals actually represent probability density distributions).

Section - 1 THE CARBON ATOM : HYBRIDISATION



2px

z

x

2py

z

2pz

z

x

yy

x

y

x

z

The four second-level orbitals (observe their orientations)

The three p-orbitals shown on the same axes

2s

z

x

y

Fig -01

Coming back to the question of carbon’s valency, although the ground state configuration of carbon indicates thatcarbon should be divalent, carbon is actually tetravalent in an overwhelming majority of its compounds (In fact,compounds in which carbon is divalent exist but they are very unstable; an example of such a compound is CCl2).

[ METHANE ]

Consider one of the simplest possible compounds of carbon as an example : methane (CH4). The carbon atomexhibits tetravalency in this compound. Not only that, each carbon-hydrogen bond length is precisely the same;whereas if we have one 2s and three 2p orbitals in the valence shell of carbon, we should expect the four bondlengths to be different. How do we explain all these facts about carbon’s bond formation ? The answer ishybridisation.

It is clear that for the carbon atom to exhibit tetravalency, we can’t just do with two unpaired electrons; we needfour. This can happen if one of the 2s electrons can be excited to the empty 2p orbitals, which would result in fourunpaired electrons.

2s2 2p 2p 2s1

Ground state Excited state

Excite one of

the 2s electrons1x

1y 2p1

x 2p 1y 2p1

z

This is how the tetravalency is explained : carbon atom undergoes bond formation with other atoms not in itsground state (where it has only two unpaired electrons) but in its excited state (where it has four unpaired electrons).



But even now a problem remains. A carbon atom still cannot bond in the configuration 1 1 1 12 2 2 2x y zs p p p of theexcited state even though it has four unpaired electrons. This is because if the carbon atom were to do so, it wouldlead to the formation of three directed bonds with the three 2p orbitals (which would be mutually at right angles)and one different non-directed bond with the spherical 2s orbital. However, as already stated earlier for the caseof methane, the four carbon-hydrogen bonds are of identical lengths and lie symmetrically; the angle between anytwo bonds being 109°28' (the four hydrogen atoms lie at the four vertices of a regular tetrahedron). It is thissymmetry that hybridisation explains.What hybridisation says is that before undergoing bond formation (in methane, for example) the 2s and the three2p orbitals are hybridised (mixed) to yield four new identical orbitals which are capable of forming stronger bonds.These four orbitals point towards the four vertices of a regular tetrahedron. These orbitals are referred to as sp3

hybrid orbitals.

2s1Four new sp3 hybrid

orbitals

Hybridisation

2p1x 2p 1

y2p1

z

To form a visual picture in your mind, refer to the next figure which shows on the right hand side the four new sp3

hybrid orbitals that are identical in all respects. The angle between any two hybrid bonds in 109°28'.

Fig - 02



The next figure shows these four hybrid orbitals in the same figure, which should make it clear how these bonds aresymmetrically oriented:

C

The four symmetrically oriented sp3 hybrid orbitals

Fig - 03

Hybridisation is actually a mathematical process that is carried out on the electron - wave functions(which represent the orbitals) and is not something that physically occurs like we have put it here. It isa tool used to understand the behaviour of different atoms in bond formation. The full significance of these concepts will become clear in advanced courses at university level.

Once we have four sp3 hybrid orbitals at our disposal, we can easily explain bond formation in a molecule likemethane:

+ 4 H

H

C

H

HH

H

C add 4

H-atomsC

H

H

H

mix

1s and sp3 C

H

H

H

H

H

H

H

H

H

HC

HH

C

σ-bonded methaneFig - 04

From the figure above, observe that each sp3 hybrid orbital of carbon overlap with the 1s orbital of a hydrogenatom to form a sp3 – s carbon - hydrogen sigma bond (σ bond), These four σ bonds, being tetrahedrally placedwith respect to each other, explain the symmetric structure of methane. (The significance of the term “σ - bond”is as follows : whenever the overlap of two atomic orbitals takes place along their major axis, the resulting (bondingmolecules) orbital is the σ - bonding orbital. Any such bond will be referred to as a σ - bond .)

[ ETHANE ]

Let us consider another simple compound of carbon : C2H6 called as ethane

Each carbon in ethane is sp3 - hybridised. The following figure depicts in detail the bond formation process inethane:



C

H

H

CC

H

H

H

HC

C

C

H

H

H

H

H

H

H

H

H

H

H

H

a C — C σ−bond (wedge-shaped three dimensional representation will be learnt soon)

Fig - 05

As depicted in the figure above, two sp3 hybrid orbitals, one from each carbon, undergo a “head - on” overlap toform a carbon-carbon σ - bond. The other six sp3 hybrid orbitals, three on each carbon, undergo σ - bondformation with the 1s orbitals of the hydrogen atoms. Note that both the carbon atoms are still in the tetrahedralconfiguration.It has been found that free rotation is possible about a carbon–carbon single bond. Thus, an infinite variety ofstructures are possible according to the relative positions of the groups attached on the two carbon atoms. Suchstructures which are obtainable by free rotation about a carbon–carbon σ–bond are called conformations. (Inother words, they are different arrangements of the same group of atoms that can be converted into one anotherwithout the breaking of any bonds.) Consider ethane. We will discuss two extreme conformations of ethane: Onein which all the hydrogen atoms are aligned and one in which they are oppositely aligned.These two conformations are known by the names “Eclipsed” and “Staggered” conformations.There are two ways that we can use to represent conformations, as done in the following figure:

Fig - 06The two extreme conformations of ethane



Observe that in the staggered conformation, the hydrogen atoms are as far apart from each other as they canpossibly be, so this would be a minimum repulsion configuration. On the other hand, the eclipsed form is theconformation with the maximum possible ‘crowding’ and hence the maximum repulsion. Hence the staggered formshould be the most stable while the eclipsed form is the least stable.However, although there is a difference in stability since the hydrogen atoms are very small, this difference instability is negligible to an extent that even at room temperature conversion from one conformation to the othereasily takes place with the help of ordinary thermal motions.Later on we will encounter molecules where such differences in stability of the conformations are more pronounced.We’ll discuss the properties of σ - bonds soon enough. Right now, let us see organic compounds containingcarbon atoms in a state of hybridisation different than sp3.

[ ETHENE ]

Ethene has the molecular formula C2H4. Each carbon is bonded to only three atoms: two hydrogens and the othercarbon atom. How can we use hybridisation to explain the structure of ethene?

This is how hybridisation takes place : only three orbitals undergo hybridisation (the 2s and two of the three 2porbitals) to yield three new identical sp2 hybrid orbitals placed symmetrically in a plane at 120° to each other, whilethe remaining 2p orbital stays unhybridised.

2s1 2p1x 2p1

y 2p1z

sp2 hybridisation

2p1z

Three new identical hybrid orbitalssp2

x

zy

x

zy

x

zy

x

zy

x

zy

x

zy

2s

2px 2py

Three new identical sp2 hybridised orbitals (They have a planar configuration)

Hybridisation

along the y-axis

in xyplane

in x-yplane

Fig - 07



The next figure shows two views of how the three sp2 hybrid orbitals and the fourth unhybrid 2p orbital are placed

Hybrid orbital

z

y

x

Top-view

y

x

The unshaded lobes, lying in the same plane correspond to the three sp2 hybrid orbitals while the fourth (shaded) orbital along the z-axis is the unhybridised 2p orbital.(Minor lobes have not been shown)

The three hybridised sp2 orbitalssymmetrically placed at 120º to each other. The fourth unhybridised2p orbital is perpendicular to the plane of paper (not shown here). z axisis coming out of the plane of paper

120º

120º

120º

unhybridised orbital

Side-view

Fig - 08

The bond formation/structure of ethene can now be explained using two such sp2 hybridized carbon atoms

H

z

C

Head-on overlap

Side-way overlap z

CH

H

H+

CH

HC

H

H

Fig - 09

Observe the following:• Head-on overlap occurs between

two sp2 hybrid orbitals (one fromeach of the two C-atoms) orientedalong the internuclear axis, thus,forming a C—C σ−bond (referFig 09 below.

• Two sp2 hybrid orbitals of eachC-atom overlap with two 1s Horbitals forming a total of 4 C—Hσ–bonds.

• The remaining p orbital on eachC-atom mutually overlapside-ways to form a molecularorbital representing the π bond.

π-bond

σ-bondπ-bond



We thus see that two sp2 hybrid orbitals, one from each carbon, overlap head-on to form a C – C σ - bond.Similarly, the remaining four sp2 hybrid orbitals are used to form σ - bonds with the four hydrogen atoms.

However, notice how the unhybridised 2p orbitals overlap. Since these two 2p orbitals are parallel to each other,they overlap “side-ways” resulting in an (bonding) orbital that spreads over both the carbon atoms and is situatedabove and below the plane containing the two carbon and the four hydorgen atoms. Such a bond is referred to asa π -bond (pi - bond).

Thus, from now on, we can visualise the ethene molecule as containing an electron cloud (due to the π - bond)above and below the plane containing the six atoms.

We can represent an ethene molecule conveniently as

π-cloud (bond)

H

CC

H H

H H

CC

H H

H

or

σ-bond The double-line indicates a σ-bond and a π-bond

Fig - 10

From Fig - 8, it should be intuitively obvious that this sideways overlap will be the maximum when the twounhybridised 2p orbitals are exactly parallel to each other (or equivalently, when the six atoms of ethene arecoplanar)

If one tries to cause rotation about the carbon — carbon double bond, it is obvious that ethene will deviate fromits state of maximum overlap leading to a lesser stable configuration (a decrease in strength of the bond). Thus, anysuch attempt to cause rotation about the double bond will be resisted.

Thus, in the ethene molecule (and any other molecule containing a double bond), free rotation about a carbon —carbon double bond is not possible and thus the question of conformations as in ethane, does not arise.

Observe that due to the cloudπ − , the carbon — carbon double bond is a spatial region of high negative chargedensity so it will favour an attack on itself by electron-seeking reagents. We’ll have to say more about this later.

[ ETHYNE ]

Ethyne has the molecular formula C2H2. Each carbon is bonded to only two atoms : one hydrogen and the othercarbon atom.

Hybridisation in ethyne takes place as follows : only two orbitals undergo hybridisation ( the 2s and one of thethree 2p orbitals) resulting in two new sp hybrid orbitals placed symmetrically in a linear fashion, while the remainingtwo 2p orbitals stay unhybridised



2s1

sp hybridisationTwo new identical sp hybrid orbitals

2p 1y

2p1x 2p1

z 2p 1y 2p 1

y

x

zy

x

zy

x

zy

Two new identical sp hybridised orbitals(They have a linear configuration)

Hybridisationx

zy

Fig - 11

The following figure shows two views of how the two sp hybrid orbitals and the remaining two unhybrid2p orbitals are placed.

sp hybrid orbital

z

y

x

Top-view

y

x

Side-view

Unhybridisedorbitals

180°

CC

The unshaded lobes, lying along the x-axis Two sp hybrid orbitals lie in a linearcorrespond to the two sp hybrid orbitals fashion, at 180º to each other. Thewhile the two orbitals (shaded) lying along two unhybrid orbitals lie along the ythe y-axis and the z-axis are the two unhybrid and z axis (not shown here)2p orbitals. (Minor lobes have not been shown)

Fig - 12



The bond formation in ethyne takes place using two sp-hybridised carbon atoms:

Sideways overlapz

y

C

Head-on overlap

z

y

C HH +

Fig - 13Observe carefully that two sp hybrid orbitals, one from each carbon, overlap head-on to form a C—Cσ - bond. The remaining two sp hybrid orbitals are used to form σ - bonds with the two hydrogen atoms.There are two pairs of unhybridised 2p orbitals, one pair parallel to the y-axis and one to the z-axis. These pairsoverlap sideways resulting in two π - bonds (these π - bonds lie in planes at right angles to each other)Due to these two π- bonds, the ethyne molecule can be effectively considered as having a cylindrical region ofnegative charge along the carbon-carbon σ - bond.

bondσ −

CH H

-bond

cloud of negative charge

An approximaterepresentation of ethyneC

Fig - 14The ethyne molecule can be represented as

H C C H

A triple line indicates oneσ - bond and two π-bonds

We have thus seen three valid ways in which the orbitals of carbon can be hybridised, resulting in the followinghybrid orbitals: sp3, sp2 or sp. (Other valid modes of hybridisation are also possible but we shall not discuss themhere.) The following figure summarizes the structures, bond lengths and bond angles in ethane, ethene and ethyne.

C HH C

π-cloud (bond) due tooverlap of pz orbitals

π-cloud (bond) due tooverlap of py orbitals

σ-bonds



C

C

H

109 .28’°

H1.10Å

H

1.54Å

HH

H

ethane

C

C

180°

1.20Å

H

H

1.06Å

ethyneethene

C

C

HH

HH

121°1.34Å

1.09Å

118°

Fig - 15

You must understand why hybridisation takes place at all and what is the advantage gained out of it.It is a universal law of nature that any natural system will tend to take up a configuration where its total instrinsicenergy is minimum. Hybridisation lets the molecule take up a minimum energy configuration because hybrid orbitalscan form stronger bonds; also, hybrid orbitals ensure that the other atoms bonded to the hybridised atom remainas far apart from each other as possible. Thus, the process of hybridisation is energetically favoured.

BONDING CHARCTERISITICS

We know why atoms hybridize. Let us now see how hybridization affects various bond characters of molecule.We will be discussing the following three characteristics:

(i) Bond length(ii) Bond angle(iii) Bond strength (energy)

Before that, let us first calculate the percentage s (or p) character in a hybrid orbital and after this calculation, wewill see its effect on the above characteristics of a bond.

Percent of s(or p) character = No. of s(or p) orbitals mixed 100Total no. of orbitals mixed

×

Thus, % s character in:

3 1sp 100 25%4

= × =

2 1sp 100 33.33%3

= × =

1sp 100 50%2

= × =

Percentage s-character has two effects on the bond characteristics:I. Orbital containing more s-character is closer to the nucleus and hence, forms shorter bonds. We know

that, the shorter the bond, the stronger it is.II. s-character also affects the electronegativity of carbon. Carbon atoms having hybrid orbitals which posses

high s-character are more electronegative.Thus, increasing order of electronegativity of the hybrid orbitals is:

3 2sp sp sp< <% s-character: 25 33.33 50



Thus, we can summarize:More s character

Closer to nucleus + More electronegative

Shorter bond

Stronger bond

High bond energy

−↓

↓

↓

↓

Now let us study the characters one by one.

(i) BOND LENGTH: Experimental measurements indicate that similar bonds have fairly constant bondlengths. eg. carbon-carbon bond length in ethane, propane, cyclohexane, isobutane etc. is approximatelythe same.However, the bond length decreases as % s-character increases.

3 2Hybridization : spsp sp% s character : 5025 33.33

1.18Bond length(Å) : 1.53 1.31

≡≡

−

C – C C == C C C

Apart from increase in s-character, there is one more reason for shortening of bond as we move fromC— C to C== C and finally to C ≡≡C. The reason is:

• C— C contains only one sigma bond.

• C== C contains one sigma bond and an additional π bond which shortens the distance betweenthe two carbon atoms.

• C ≡≡C contains one sigma bond and two additional π bonds which further shorten the distancebetween carbon atoms.

In molecules like C CH CH C — — — , it has been observed that the C— C single bond length is

only 1.48Å. The shortening of bond length is because this single bond is formed between two sp² hybridcarbon atoms (having more s-character than sp³).

Thus, for a molecule like C C C C ,− −≡ ≡— the carbon-carbon single bond length is even shorter, 1.38Åbecause here, the single bond is formed between two sp hybrid carbon atoms (50% s-character).In addition to C—C bond, it has also been observed that C–H bond length depends on the hybridizationof carbon atom.

23Hybridization of Carbon spsp% s-character 33.3325C–H bond length 1.08Å1.09Å

3 3 2 2CH – CH CH = CH

Similar is the case for other bonds: C O, C N, C S, C X− − − − etc.



Example 1Which has a shorter C = O bond length: formaldehyde or carbondioxide?

Solution: C

O

HH O C O — —

Hybridization of carbon : sp2 sp% s-character: 33.33 50

Thus, CO2 will have a shorter C = O and length than formaldehyde.

2Experimental values: Formaldehyde COC O bond length 1.21Å 1.16Å=

(ii) BOND STRENGTH: The strength of a bond can be measured by measuring the energy of the bond. Itcan be expressed in two ways:

• Dissociation energy (D).• Bond energy (E).

D is the energy needed to cleave a bond to constituent radicals. Let us consider the example of a watermolecule. For the process,

2H O H OH, D 118 kcal/mol. → + =However, for the process,

OH H O, D 100 kcal/mol. → + =

We can see that it has two different values of D for the same bond.

Thus, the bond energy of O–H bond is taken to be the average of the two (D) values, which is called thebond energy, E and is equal to 109 kcal/mol for H2O.Obviously, for diatomic molecule, D = E.We know, the shorter the bond, the stronger it is. Or we can say, larger amount of energy is required tocleave a shorter bond.Let us summarize it:

short bond

strong bond

High bond energy

↓

↓

Always keep in mind that in a given group of the periodic table, the bond weakens on moving down thegroup. Thus, C–S bond is weaker than C–O bond.Similarly, the carbon-halogen bond strength decreases in the order: C F C Cl C Br C I− > − > − > −



We have seen that C C≡ is a shorter and thus, a stronger bond than C=C and even more stronger thanC–C. But we should keep in mind that C=C is not twice as strong as C–C because C=C consists of 1σbond and 1π bond. We know that extent of overlap in π bond is less than that in σ bond. Thus σ bondsstronger than π bonds. Hence, C=C is not twice as strong as C–C.Hence, C—C bond dissociation energy will be maximum for ethyne.

(iii) BOND ANGLE: Bond angle depends upon the hybridization of the central atom.When the central atom is sp³ hybridized, the bond angle is normally 109°28', where all the four groupsattached to the central atom are identical.Similarly, when the central atom is sp² hybridized, the bond angle is 120° and for sp hybridized centralatoms, the bond angle is 180°.However, the measure of angle deviates when different groups are attached to the central atom. eg.

Although carbon is sp³ hybridized, the C–C–Br bond angle in 3 3CH CH CH|

Br

− − is 114.2°

Hetero BondsCarbon-oxygen and carbon-nitrogen bonds: Carbon can bond to oxygen through single or double bonds.Consider the electronic configuration of oxygen atom:

Hybridisation in the oxygen atom can take place in two ways: sp3 or sp2

Thus, observe that two of the four hybridised orbitals will be used to accomodate the two lone pairs on oxygenwhile the other two can be used in bond formation with other groups. For example, consider a class of organic

compounds called ethers, which have a C — O — C —

::— linkage. The oxygen atom here is sp3 hybridised.

There are two lone pairs accomodated in two of the four sp3 orbitals of oxygen while the other two hybrid orbitalsare used to form bondsσ − with the carbon atoms which are themselves sp3 hybridised. Oxygen atom can also

be sp2 hybridised. For example, we consider a class of organic compounds called ketones, which have a C O

::

linkage. Observe that the carbon atom is sp2 hybrdised. The oxygen atom is also sp2 hybridised:

sp2 hybridisation

1s2 2s2 1s2

Three new sp2

hybridised orbitalsUnhybridisedp orbital

2p2x

2p1y

2p1z 2p1

z



Observe that two of the three sp2 hybrid orbitals are used to accomodate the two lone pairs on oxygen. The thirdhalf-filled hybrid orbital forms a bondσ − with the sp2 hybridised carbon. The unhybridised p orbital on oxygenoverlaps sideways with the unhybridised p-orbital on carbon to form a carbon-oxygen bondπ−

Sideways overlap forming a π bond

C O OC

σ-bond Carbon-oxygen π-bondFig - 16

An approximate sketch showing the carbon-oxygen double bond. is depicted in Fig 16 .Carbon can bond tonitrogen using single, double and triple bonds. Consider the valence shell electronic configuration of nitrogen

2s2 2p1x 2p1

y 2p1z

Nitrogen can undergo hybridisation in three ways: sp3, sp2 or sp:

(i)

2s2 2p1x 2p1

y 2p1z

sp3 hybridisation Four new hybridised orbitalssp3

The lone pair is accomodated into one of the sp3 hybrid orbitals while the remaining three orbitals can beused in bondσ − formation. An example would be the following compound:

CH — N — CH3 3

:

CH3

The nitrogen atom is hybridised and

has a lone pairsp3

(ii)

2s2 2p1x 2p1

y 2p1z

sp2 hybridisation

2p1z

Three new hybridised orbitals

sp2 Unhybridised orbitalp

The lone pair is accomodated into one of the three sp2 hybrid orbitals. The remaining two sp2 hybrid orbitalscan be used in the formation of bondsσ − . The unhybridised 2p orbital will form a bondπ− . An exampleof this case is the following compound:



C C

CH3

CH3 CH3

N

CH3

N

HH

The carbon and nitrogen atoms are both Carbon and nitrogen are bonded sp2 hybridised. The lone pair on nitrogen by a double bond. is accomodated into one of the sp2 hybrid orbitals. The unhybridised 2p orbitals on carbon and nitrogen form a p-bond.

Fig -17

(iii) We finally consider the case when nitrogen is sp hybridised.

2s2 2p1x 2p1

y 2p1z

sp hybridisation

2p1z

Two new hybridised orbitals

sp

2p1y

The lone pair is accomodated into one of the sp hybridised orbitals. The other hybrid orbital can be used toform a bondσ − . The two unhybridised 2p orbitals can form two bonds.π− Let us consider an example:

C N NCH3

CH3

Lone pair

C

The carbon and nitrogen atoms are both sp The drawing depicts an approximatehybridised. The lone pair on nitrogen is electronic charge distribution in the spatialaccomodated into one of the sp hybrid region of the carbon nitrogen bond. As inorbitals. The other hybrid orbital forms a ethyne, here also a cylindrical region ofσ-bond with an sp hybrid orbital of carbon. negative charge density will envelop theThe two unhybridised 2p orbitals on carbon-nitrogen bond regioncarbon and nitrogen each form two π-bonds

Fig - 18



On the basis of structure and the type of atoms present, different terms have been designated to organic compounds:hydrocarbons, saturated, unsaturated, alicyclic, aromatic etc.Let us see what these terms mean.

HYDROCARBONS:The term hydrocarbon refers to compounds containing only carbon and hydrogen atoms. These include alkane,alkene, alkynes and aromatic hydrocarbons.Alkanes: are hydrocarbons that contain only single bond and no multiple bonds. They are also referred to assaturated compounds because they contain maximum number of hydrogen atoms that the carbon compound canpossess.

eg. 3

3 3 3 3 2

3 3

CH|

CH ,CH CH ,CH C CH| |

CH CH

− − −

Alkenes are the hydrocarbons that contain carbon-carbon double bond. eg. 2 2 2 2,CH CH CH CH CH= =—

Alkynes are the carbon-carbon triple bond containing hydrocarbons. eg. 3,HC CH CH C CH≡ ≡—

Aromatic compounds are organic compounds containing multiple bonds and possessing special ring structure.Benzene and its derivations are called benzenoid.

eg, ,

CH3

, hydrocarbons which do not contain a benzene ring are called non-

benzenoid compounds.

e.g.

In aromatic compounds , if atoms, other than carbon (eg O, N, S etc.) constitute the ring, then aromatic compoundsare not classified as hydrocarbons. They are called heterocylic aromatic compounds.

e.g.

N NH

O, ,

Alkenes, alkynes and aromatic compounds are referred to as unsaturated compounds (compounds containingmultiple bonds) because they have fewer number of hydrogen atoms than the maximum number the carbon compoundcan possess. Thus, they can react with hydrogen under suitable conditions.

S

Section - 02 CLASSIFICATION OF ORGANIC COMPOUNDS



ACYLIC COMPOUNDS:Organic compounds may exist as long chain or branched chain. These compounds are referred to as acyclic oropen chain or aliphatic compounds.

e.g. 3 3CH CH— , CH CH CH3 2 — — — CH3

CH3

, CH C OH3 — —

O

etc.

ALICYCLIC COMPOUNDS:Sometimes the long chain organic compounds may form a loop and arrange to form a ring-like structure. Thesecompounds are referred to as alicyclic or closed chain or ring compounds.If only carbon atoms constitute the ring, it is called homocyclic.

eg.,

If atoms other than carbon (heteroatoms eg. O, S, N etc.) form the ring, the cyclic compounds are called heterocyclic.[Heteroatoms are atoms that form covalent bonds and have unshared electron pairs].

eg. ,O N

HWe will study each of these class of compounds in later chapter. Let us now try to summarize.

ORGANIC COMPOUND

Acyclic Cyclicor or

Open chain or Aliphatic compounds Closed chain or Ring compounds_

Homocyclic Heterocyclic

↓ ↓

↓ ↓

_______________|______________

____________|_____________

______|_______ ______|

Alicyclic Aromatic Alicyclic Aromatic

Benzenoid Non Benzenoid

↓ ↓ ↓ ↓

↓ ↓−

_______

_________|________

MAJOR FUNCTIONAL GROUPSMost of the organic compounds concerning us also contain atoms such as oxygen, nitrogen, etc. Such an atom orgroup of atoms or an exclusive kind of chemical bond gives exclusive properties to the carbon compound. In otherwords, it determines the structure and properties of a family of organic compounds. Such an atom or group ofatoms or special kind of bond is called a functional group as it determines the overall function of the organiccompound.



Now, let us consider an example. In the following compound:CH3 — Cl

the halogen atom, here chlorine, is the functional group as it determines the overall chemical properties of thecompound. It should be noted here that the carbon portion, i.e. CH3— , of the molecule, under suitable conditions,undergoes the reactions typical of the class of compounds with C — C single bonds ( called as alkanes - detailswould be covered later). However, the set of reactions that characteristically determine the family of compoundsR — X (here X is a halogen atom and R is an alkyl group) are the reactions that occur at the halogen atom. Let ustake another example:

CH3 — OHHere, — OH (read as ‘hydroxyl’ group) determines the characteristics of the family of compounds represented asR — OH. Hence ‘— OH’ is the functional group.Let us now go on to examine a few very important functional groups that we will frequently encounter in the rest oforganic chemistry. What you are expected to do while going through the following discussion is to learn to identifythe functional groups wherever they appear in molecular or structural formulae of organic compounds. Also, yourjob would be to learn their names. This would ease your understanding of assigning names to organic compoundswhich we would discuss later. Apart from the names of functional groups, sometimes special names are assignedto classes of compounds containing particular functional groups. For example, functional group —OR (where R isa hydrocarbon framework or can be called as an alkyl group) called as alkoxy functional group when present in aclass of compounds R1—O—R2 (R1, R2 may be two different alkyl groups) gives it the general class name ‘ether’.Now, we move on to discuss the major functional groups as follows:

(1) C — C— double bond in alkenesAlkenes, sometimes also called as ‘olefins’, contain carbon-carbon double bonds imparting exclusive reactivity tothe alkene molecule. Hence, the double bonds are treated as functional groups. In case you are finding it awkwardto classify a kind of bond (here , C — C— double bond) as a function group, please recall the definition of ‘functionalgroup’. An atom or group of atoms or any special feature of an organic molecule that imparts it a characteristicreactivity has been discussed as a functional group. Since C — C— double bond too attributes special reactivenature to the organic compound in which it is present, hence this too is treated as a functional group.

Example 2

H C — H C — HC — CH — CH3 2 3— C C double bond is a functional group and the class of

compounds in which it is present is called as alkenes——

(2) C — C—— triple bonds in alkynesCarbon-carbon triple bonds also impart special kind of reactivity to the compounds in which they are present.Hence, such triple bonds are considered as functional groups and the corresponding class of compounds arecalled as alkynes.

Example 3

H C — C — C — CH 3 3—— triple bond as a functional groups.



Note : Alkanes (simplest class of organic compounds) do not contain any functional group. They simply consistsof C — C single bonds and attached hydrogen atoms as shown below. Hence, they are extremely unreactive.

Example 4

CH3 — CH2 — CH2 — CH3 } no functional group, hence very less reactive

(3) Hydroxyl (—OH) group in alcoholsThe functional group represented as ‘—OH’ attached to a carbon atom is known as ‘hydroxyl’ group. The classof organic compounds containing this group is called as ‘alcohols’. They are usually represented as R — OHwhere R is the hydrocarbon (alkyl) group.

Example 5

CH —OH3 'hydroxyl' functional group

(4) Alkoxy (—OR) group in ethersEther is a class of compound consisting of two alkyl groups (usually denoted as R', R2, etc.) mutually connectedthrough an oxygen atom. The functional group here is —OR (called as alkoxy) where R is smaller of the two alkylgroups.

Example 6

CH — O — C H3 2 5alkoxy functional group (here of — is ' —' group)R OR CH3

(5) Nitro group (—NO2) in nitro compoundsThe functional group nitro (—NO2) is present in a class of compounds represented as R—NO2 and called as nitrocompounds.

Example 7

CH —NO3 2 nitro functional group

Note: Observe the following structural representation of the nitro group. It will assume major significance as wemove into organic reactions later.

RN

O

O

correct structural formula of group —NO 2

RN

O

O

incorrect structural formula asnitrogen cannot have five bonds



(6) Halogen functional groups present in alkyl halidesAtoms such as fluoro, chloro, bromo and iodo are present as functional groups in R — F, R — Cl, R — Br andR — I respectively. The class of such compounds R — X (where X denotes a halogen atom) is called as alkylhalide.Example 8

CH CH — Cl3 2

functional group: chloro

CH — F3

functional group: fluoro

CH — Br3CH2

functional group: bromo

(7) Carbonyl group ( C O— ) in aldehydes and ketones

The presence of carbonyl functional group, structurally represented as C O— , gives rise to the following twoclasses of compounds:

(i) aldehydes, structurally represented as C O—H

R

(ii) ketones, structurally represented as C O—R1

R2

Note: The only difference between the above two structural representations is that aldehyde always has at leastone hydrogen atom attached to the carbonyl carbon atom, while ketone has alkyl groups (R1 and R2, say) as thetwo substituents on the carbonyl carbon.Example 9

C O—CH3

CH3

C O—C H2 5

CH3

two alkyl groups (and no atom) on carbonyl carbon, hence ketone

H

C O—H

HC O—

H

CH3

at least one atom on carboxyl carbon, hence aldehyde

H

Carbonyl functional group

(8) Carboxyl group (—COOH) in carboxylic acidsThe carboxyl functional group imparts to its class of compounds, carboxylic acids the special reactivity of acids asthe name ‘carboxylic acids’ itself suggests. Structurally, the functional group is represented as follows:

R C—O H—

Carboxyl functional group

is the acidic proton that imparts acidic nature to the functional group and hence to the compound (note that ' ' could be atom too).R H—

O



Example 10

H — COH

OCH — C3

Carboxyl functional group

OH

O

(9) Carboxyl group with an additional alkyl group (—COOR) in esters.This functional groups is written by replacing H atom ( of carboxylic —OH) with alkyl group ‘R’. Structurally, thefunctional group being discussed here is represented as (note that R1 is not a part of the ester functional group):

functional group (obtained from carboxyl group) characteristic of esters

alkyl group replacing the acidic atom of the corresponding carboxyl functional group.

HR C O R1 — — —

O

Example 11

functional group characterising esters

CH C O 3 — — — CH3

O

alkyl group replacing the acidic atom of the carboxyl functional group.

H

(10) Functional group (—COCl) present in acyl chlorides.Structurally, the functional group is represented as

functional group in acyl chlorides.R C —O

Cl

It is written by replacing the ‘—OH’ in carboxylic acid with the chloro group‘—Cl’.

Example 12

carboxyl functional group

CH C 3 —O

OH

replace

by

—

—

OH

Cl

functional group in acyl chlorides

CH C 3 —O

Cl



(11) —CONH2, —CONHR or —CONR1R2 functional groups in amides.The class of compounds called as amides may contain any of the following functional groups.

R C —O

NH

H

R C 2 —O

NH

R1

possible functional groups

R C 3 —O

NR1

R2

or or

These are written by replacing the ‘—OH’ of the carboxyl group by the following respective groups:

R C—O

OH

replace by

' OH'' NH '

—— 2

or R C2 —O

OH

replace by

' OH'' NHR '

—— 1

or R C3 —O

OH

replace by

' OH'' NR '

—— 1R2

R C —O

NH2

R C 2 —O

NH R1

R C 3 —O

N R1R2

The dashed region shows the possible functional groups present in amides.

Example 13

CH C 3 —O functional group

alkyl group on atom is not a part of the functional group.

N

(12) Amino (—NH2) group in amines.Here, the amino functional group ‘—NH2’ is present on a carbon atom of the hydrocarbon framework, R. It isrepresented as R — NH2 and the ‘—NH2’ group imparts a basic (contrary to acidic nature of carboxyl functionalgroups) character to the compound.



Example 14

CH3CH NH2 2— amino functional group

hydrocarbon framework,represented above as ' 'R—

Note: Compare the slight difference in the names of the two classes of compounds ‘amides’ (in discussion (11)above) and amines (in discussion (12) above). The structural difference between the two classes of compounds,as you must have noticed by now, is that the amide class of compounds have carbonyl functional group present inthem, while amines do not have any carbonyl group.

(13) Cyano group (— C N)——— in cyanides.

Cyanide compounds are also called as nitriles or nitrile compounds. This class of compounds is represented asR — CN.

Example 15

CH C N3 — — cyano functional group.

(14) Acetal class of compounds

Acetals are those compounds that have two oxygen atoms singly bonded to the same carbon atom of the organiccompound. Structurally, we can represent it as follows:

Cfunctional group characterising the class of compounds called as acetals

CH3

C2H5

O

O

R1

R2

Example 16

Ccharacteristic functional group.CH3

CH3

O

O

CH3

CH3

the carbon atom to which the two oxygen atoms are singly bonded

The following table enlists a general formula for various classes of compounds and corresponding functionalgroups. Note that subscript ‘n’ stands for number of carbon atoms present in the compound.



Table – 1 : Functional groups and general formula

Class of compound Functional group General formula ( n = no. of C–atoms)

Alkenes C C—— double bond 2n nC H

Alkynes C C——— triple bond 2 2n nC H −

Alcohols — OH (hydroxyl) 2 2n nC H O+

Ethers — OR (alkoxy) 2 2n nC H O+

Aldehydes ——O (carbonyl) 2n nC H O

Ketones ——O (carbonyl) 2n nC H O

Carboxylic acids — COOH (carboxyl) 2 2n nC H O

Esters — — — C O O

(carboxyl with an additional 2 2n nC H O

alkyl group on saturated oxygen atom)

Let us now devise a method to conveniently represent organic molecules on paper. This would improve ourapproach to deal with those organic molecules whose structures appear complex. We will here focus upon thecentral chain of carbon atoms that runs through the entire molecule. This may even form a ring structure. Observethe following compounds:

C C C— —

H

H

HHH

CH CH CH C H3 2 3 8or 3

(1)

C C C C H H

| |—

| |

C

H

HH

HH

HH

H

(2)

C H5 10

H

H

H

Note that with increasing number of carbon atoms, it becomes increasingly inconvenient to draw the structures.Also, the C-atoms forming the skeleton practically do not exist in straight lines. On the other hand, let us drawthem in a zig - zag fashion to realise their practical way of existence and also increase our comfort at drawing themolecules. Study the following example.

Section - 3 REPRESENTATION OF ORGANIC COMPOUNDS



CC

CC

HH

H

HH

HH

simple

(4)(3)

HH

H

representation

Here, the zig - zag pattern (4) is very clear to comprehend and it eliminates any confusion between the hydrogenatoms attached to the various carbon atoms. However, the following aspects of arriving at diagram (4) fromdiagram (3) should be kept in mind.

(i) Every kink in representation (4) and the two ends of the line show a C-atom.

(ii) Hydrogen atoms bonded to C-atoms are not shown; not even the bonds between hydrogen and carbonatoms. Even without drawing them their presence is implied. Any carbon atom in representation (4) whosefour bonds are not explicitly shown is assumed to have as many hydrogen atoms as needed to satisfy thevalency of carbon which is 4.

(iii) Most important is the rule that C atoms are not shown explicitly by writing ‘C’.

Study the following figure for complete understanding of the above stated rules:

this atom must also carry 3 because only 1 bondis shown attached to it

C-H-atoms

these 2 atomsshould also carry 1 atom because only 3 bonds are shown for each

atom

C-

H-

C-

each of these 2atoms have two

bonds shown, hence each should also carry 2 atoms

C-

H-

this atom has allits 4 bonds shown,hence no atomsare implied

C-

H-

OHthis atom is shown because it is bonded to oxygen: a hetero atom (atom other than or )

H-

C H

end of line represents a atomC-

O

every kink in the chain (bond-line diagram)represents a atomC-

Fig - 19

Here, it is noteworthy that all hetero atoms are explicitly shown; even H atom(s) bonded to such hetero atoms areshown. We will now consider some examples.

Example 17

H H H H H C C C C H H H

| | | |— — — — —

| |

representedas

these atoms carry1 atom each

C-H

—



Example 18

CH CH CH COOH3 — — —— representedas

OH

OC-

Hatom has

3 atoms

CH

atom has no atom

Example 19

OH CH|

C O||

OHCH3

representedas

OH

COOH

sometimes, atom, being part of the functional group, may be clearly shown

C-

Example 20

representedas

O

OHCH3

O

OH

Example 21

representedasCH3 NH2

CH2NH2

H atoms bonded to hetero (here, nitrogen)atom is shown

Example 22

representedas

CH2

CH2CH2

CHCH2N N

lone pair of electrons on atom is shown hereN-

From the above examples we can conclude that the basic purpose of highlighting the functional group(s) should beserved. At times, we may even show hydrogen atom(s) bonded to the carbon skeleton or we may even show alone pair of electrons (a pair of electrons not involved in a chemical bond). The core idea is that any molecular sitewhich is participating in a given reaction should be explicitly shown.



Also, you can draw the bond-line diagrams in any other convenient orientation. For example:

Example 23

represented

asCH3 COOHCH

OH |

COOH

OH

OH

O

OHor

For further convenience and to focus our attention only on the reactive part of the organic compound, we put acurvy line which is nothing but a pictorial indication of an incomplete structure. Observe the following representation:

represented

as

O H

O

curvy (wiggy) line to indicate rest of the compound

O

O

H

The curvy line within the dashed space above simply denotes that we are, for the time being, not concerned withthe non-carboxylic part of the molecule. This approach not only reduces our efforts to repeatedly draw irrelevantmolecular portion but also prevents us from unnecessarily thinking about it.Note: If we want to represent the three dimensional (spatial) orientation of organic molecules on paper, weproceed as follows:(i) Show one of the substituents on the carbon atom as coming out of the plane of the paper towards us, and

the other substituent going away from us, into the plane of the paper.(ii) The ‘coming-out-of-the-paper’ bond can be shown using a bold-wedged diagram: COOH

(iii) The ‘going-into-the-paper’ bond can be shown using a dashed-wedged (or non-wedged) diagram fadingaway from us: dashed-wedged dashed (not wedged)or

(iv) The other bonds attached to the carbon atom can be depicted using straight lines as if in the plane of thepaper.

Observe the following representation:

CO H2

NH2

three-dimensionallyH

CO H2H2N

other two bonds in the plane of the paper

bond coming out of the paper

bond going into the paper

represented as

While drawing such diagrams, prefer showing the functional groups as ‘coming-out-from’ or ‘going-into’ the planeof the paper but retain the basic carbon skeleton in the plane of the paper.



Example 24

IOH

H

— and — are respectivelyshown 'coming-out' and 'going-into' the plane of the paper

I OH

OH Idrawn

as

Let us now study the various types of carbon atoms on the basis of number of neighbouring carbon atoms to whichit is bonded. Following five types of carbon atoms can be arrived at:

(1) Methyl carbon CH OH3 —

Here, the carbon atom is not bonded to any other carbon atom.

(2)OCH3

neighbouring-atomC

Primary carbon or 1º carbon.

Here, the carbon atom is attached to only one neighbouring carbon atom.

(3) Secondary carbon or 2° carbon O

OH

two neighbouring -atoms.C

The carbon atom is attached to two neighbouring carbon atoms.

(4) Tertiary carbon or 3° carbon Cl

3 neighbouringC-atoms.

Here, the carbon atom is attached to three neighbouring carbon atoms.

(5) Quaternary carbon or 4° carbon OH

The carbon atom is attached to four neighbouring carbon atoms.

Certain organic compounds exhibit conjugation which we’ll discuss in this section. Conjugated systems are of somuch importance to organic chemistry that they merit a separate section for their discussion.

Section - 4 CONJUGATED SYSTEMS



A system is said to be conjugated if it contains alternate single and multiple bonds. For example, the followingcompounds are conjugated:

CH — CH — CH — CH2 2— — CH — CH — CH — CH — CH — CH — CH — CH3 3

— — —

Such compounds in which the bonds are conjugated are found to be slightly more stable than the compounds inwhich these bonds are isolated. For example, consider the following two compounds which have the same molecularformula.

CH — CH — CH — CH — CH3 2

— —(i)

CH — CH — CH — CH — CH2 22— —

(ii)Even though these two compounds appear to be the same except for the positioning of one of the double bonds,it is found that compound (i) is thermodynamically more stable than compound (ii).The fact that conjugated compounds are slightly more stable than their corresponding non-conjugated counterpartsis experimentally provable:

(i) Conjugated molecules exhibit a lower heat of combustion(ii) They exhibit a lower heat of hydrogenation(iii) It is found that isolated double bonds can be made to migrate under certain conditions quite easily to form

conjugated system. This shows that conjugated systems are more stable.As an example, consider the following reaction which takes place easily under alkaline conditions. Observe howthe double bond migrates to form a conjugated system which is stabler:

|— CH — CH — CH — C — O2

— —Non-conjugated system

Base |— CH — CH — CH — C — O2

— —Conjugated system

Now the question that arises is: Why should conjugated systems be more stable?

In organic chemistry, a very powerful rule of thumb can be used to explain a lot of phenomenon regarding stabilityof molecules : the greater the delocalisation (spread) of charge on the molecule’s surface (or surroundings), themore stable that molecule tends to be. This is intuitively appealing: if we try to localise charge on some region ofthe molecule, that region would tend to get destablised because of the high concentration of charge in a smallregion. However, spreading the charge will result in lower local charge density and hence more stability. A completejustification of this rule has its basis in Molecular Orbital Theory which we will not delve into here.

We can now use this rule to explain why conjugated systems are stabler. In particular, we’ll discuss why compound(i) above is stabler than compound (ii). First visualise the orbital picture for compound (ii), and observe that it hastwo ‘localised’ bondsπ − , these being at the two ends of the carbon chain:

H C — CH — CH — CH—CH2 2 2

In this figure, only the uhybridised 2 orbitals used in the -bondformation are being emphasisedp π

Orbital picture of compound (ii)Fig - 20



Now visualise the orbital picture for compound (i) :

CH CH CH CH CH3 2— — — — Here too, only the unhybridised2 orbitals are being emphasisedp

Orbital picture of compound (i)

Fig - 21Can you notice the difference between the orbital pictures of the two compounds?If we number the carbon atoms from left to right, notice that in compound - (i), a sideways overlap is possible evenfor the 2p orbitals on the third and fourth carbon atoms. This overlap is depicted using the dashed curve in thefigure. This means that sideways overlap can take place between all the four 2p orbitals on the adjacent carbonatoms. This overlap will result in a (molecular) orbital delocalised over all the four carbon atoms as shown below:

CH — CH — CH — CH — CH3 2

A very approximate pictureshowing the delocalised molecularorbital due to adjacent overlap offour 2 orbitals in compound (i)p

Fig - 22Such an extended delocalisation, which is not possible with compound (ii), is the reason that compound (i) is morestable.

Benzene: A highly conjugated systemAs another example of conjugation, we’ll discuss in brief here the orbital picture of benzene, which exhibits a highdegree of conjugation.The molecular formula for benzene is C6 H6 and the structure of benzene is known to be planar. The six carbonatoms are connected in a hexagonal ring. This implies that each carbon atom must be sp2 hybridised: two of thethree hybrid orbitals will be used in bonding with two other adjacent carbon atoms in the benzene ring while thethird hybrid orbital will be used to bond with the hydrogen atom. The fourth unhybridised 2p orbital on eachcarbon remains perpendicular to the plane of the ring

In this approximate orbitalpicture of benzene, the unhybridised 2p orbitals have been emphasised.Note that the carbon atoms have been numbered

1

2 3

56

4

Fig - 23Sideways overlap of the 2p orbitals could take place in one of the two possible ways, leading to two possiblestructures which are represented below:



1 2, 3 4, 5 6↔ ↔ ↔ 1 6, 2 3, 4 5↔ ↔ ↔

⇓ ⇓

⇓ ⇓

Fig - 24

The two possible structures so formed are called the Kekule structures of benzene.However, no such Kekule structures exist as such. The fact that overlap does not take place has been describedin the figure above. Note that in the benzene ring, the bonds are conjugated. Therefore, what actually happens isthat all the six adjacent p–orbitals overlap. We’ll not discuss in detail what exactly happens when these sixp–orbitals overlap (actually six new Molecular Orbitals will be formed; this is part of Molecular Orbital Theory).It will suffice to say for now that the overlap of all these six p-orbitals will result in the formation of an annularelectron cloud, above and below the plane of the benzene ring.

Fig - 25

This figure should make it clear to you that the benzene molecule is a highly conjugated system (the 2p electronsare delocalised all over the ring). This explains the fact that benzene exhibits a very high thermodynamic stability.We’ll discuss more about the stability of benzene in a question later on.

In the last section, we discussed about the two possible (theoretical) structures of benzene, called the Kekulestructures. However, as mentioned there only, these Kekule structures actually do not exist. The actual structure ofbenzene involves a molecular orbital delocalised over the entire ring.

We can say that the actual structure of benzene is a hybrid of the two Kekule structures. The figure below depictsthis relation between the two thoretical Kekule structures and the ‘real’ benzene molecule which is a hybrid ofthese two structures.

Section - 5 RESONANCE



The two (theoretical) Kekulestructures of benzene

The real benzene moleculein which the molecularorbital is delocalised overthe entire ring

On the left hand side above, we have used curved arrows in the first Kekule structure. You must understand verycarefully that these curved arrows do not represent an actual movement of electronsπ− . (In fact, as alreadymentioned, this Kekule structure does not even really exist). The curved arrows are just a representation tool thatwe use to link one Kekule structure to the other (in other words, the curved arrows in the Kekule structure tell ushow to obtain the other Kekule structure; they do not in any way represent a real movement of electrons).Such a phenomenon in which different theoretical structures can be written for a molecule, and the actual structureof the molecule is a hybrid of all the theoretical structures, is called resonance. The various theoretical structuresare called the canonical forms of the molecule under consideration while the actual structure is the resonancehybrid of these canonical forms.

Thus, for example, benzene exhibits resonance. It has two canonical forms given by the two Kekule structureswhile the actual benzene molecule is a resonance hybrid of these two Kekule structures.

You should understand this carefully that resonance is a way of representing molecules, that cannot be representedadequately by a single structure, by a combination of two or more classical structures (the canonical forms). Thesestructures can be related to one another through curved arrows as already used for benzene: the tail of the curvedarrow indicates where an electron pair moves from and the head of the curved arrow indicates where it moves toin the molecule. To emphasize again there is no such movement of electron pairs in reality. You should not visualisethe canonical forms as interconvertible structures that are rapidly oscillating from one form to another. What existsin reality is a single, stable structure that can be said to be a hybrid of the various (non–real) canonical forms.

Let us consider another example of a molecule which exhibits resonance. Consider the acetate anion, 3CH COO− ,

whose theoretical structure is represented below:

OCH — C3

O –———

This figure indicates that the two carbon–oxygen bonds should turn out to be of different lengths since one is asingle bond while the other is a double bond. However, this is not the case. Experiments prove that the twocarbon-oxygen bonds in the acetate anion are of exactly the same length; in other words, it is impossible todistinguish between the two oxygen atoms.This is explained as follows: the actual structure of the acetate anion is a hybrid of the two canonical formspossible.

——— CH — C 3

O

O

The real acetate anion is a resonance hybrid of its twocanonical forms

CH — C 3

O—

OCH — C 3

O——O

The two (theoretical) canonical forms of the acetate anion

-

-



Observe that in the real (hybrid) acetate anion, the negative charge is delocalised over both the oxygen atoms (andnot just one). This explains the fact that the two oxygen atoms are indistinguishable.

As another example, observe carefully how resonance takes place in the following ion (which is allyl carbocation(discussed later)):

CH — CH — CH2 2—

+CH — CH — CH2 2

—+

——— CH — CH — CH2 2 +

The actual structure of this ion shows how the positive charge is spread (delocalised) over the entire moleculerather than on just one carbon atom as in the theoretical canonical forms.

What physical significance does resonance have? Observe carefully that what resonance actually implies isdelocalisation of charge. Thus resonance in a compound implies increased stability due to delocalisation. In general,the more the canonical structures that can be written for a compound, the greater should be the delocalisation ofelectrons, and the more stable the compound will turn out to be.

Let us now list down some general rules pertaining to the writing of canonical forms for a given compound. Thejustifications of these rules are straight forward:

(a) In each canonical form, the constituent atoms of the compound must occupy the same relative positions.(b) All the canonical forms must contain the same number of paired electrons.(c) The various canonical forms should not differ too much in their energy content. If such a case occurs, the

canonical forms with higher energy content will contribute very little to the hybrid; their contribution to thehybrid will be irrelevant.

(d) For canonical forms that have the same energy content (for example, the two forms for the acetate anion),the stabilising effect of resonance is more pronounced. Thus, we should expect benzene molecule, acetateanion and allyl carbocation to be stable entities.

(e) Structures which have charge separation will be of a higher energy content than structures which don’t, andhence the former will contribute lesser to the stability of the hybrid

CH — CH — CH2 2—+

CH — CH — CH2 2

+ + –

(i) (ii)In the figure above, the contribution of carbocation (ii) in the stability of the hybrid allyl carbocation willbe negligible since it has high degree of charge separation.

(f) Delocalisation (resonance) occurs most effectively when the atoms attached to the carbon atoms involvedin multiple bonds lie the same plane (or nearly in the same plane). In such a case, the lateral overlap of theunhybridised 2p orbitals is the most effective.As a concluding example, consider the stabilising effect of resonance in the following anion:

O O O O OH–

Canonical forms Actual Structure

-

-

-

-

In the actual structure, the negative charge is not concentrated only on the oxygen atom. It is also delocalisedover the entire ring due to resonance.



Example 25The C — O double bond energy is significantly less than twice the C—O single bond energy. What could be thepossible reason?

Solution: The oxygen atom in C — O is sp2 hybridised with its two lone pairs of electrons present in sp2 hybridorbitals (120° apart). On the other hand, the oxygen atom in C—O is sp3 hybridised (as in

3 3— —CH O CH , for example) implying presence of lone pairs of electrons in tetrahedral (109°apart) hybrid orbitals. Hence, lone pairs on oxygen are initially more distant in than in C—Odue to reduced electrostatic repulsion amongst the lone pairs.

Further, no such lone pair is available on carbon atom. Hence, the double bond energy (of1 and 1σ π bond) is significantly lower than twice the C—O single bond energy (one σ bond).

Example 26In the spectroscopic analysis of dienes, it is observed that simple dienes absorb in the ultra-violet region of thespectrum while as the extent of conjugation increases, the absorption moves towards the visible range of thespectrum. Thus, highly conjugated dienes are colored. Justify this observation.

Solution: As the extent of conjugation increases, the delocalisation of π bonding electrons also increases, thusimparting extra stability to the excited states of dienes. Their extra stability implies lowering the energylevel of the dienes’ excited states. It leads to reduction in energy gap between ground and excitedstates of conjugated compounds, in comparison to compounds containing isolated double bonds. Theoverall effect is that with increasing conjugation, energy gap goes on decreasing, implying lower amountof energy (and hence higher wavelength radiation) for exciting a ground state electron. Thus, dieneswith increasing extent of conjugation increasingly absorb in the visible range (higher wavelength)compared to simple dienes absorbing in the ultra-violet region (lower wavelength). Hence, highlyconjugated dienes are coloured.

Example 27Draw the possible canonical forms of the following compounds and also represent their respective resonancehybrid.

(a) CH — C — CH3 2

O

(b) CH — CH — CH — CH — O3

Solution: Canonical forms and corresponding resonance hybrids are drawn as below:

(a) CH C CH3 2 — —

–O

CH C CH3 2 — —

–O

canonical forms

CH C3 — CH2

O –

reronance hybrid



(b) CH3

CHCH

CHO CH3

CHCH

CHO

⊕

CH3

CHCH

δ+

CH δ–

O

-

Example 28

Is delocalization possible in allene, CH — C — CH2 2? Explain with reasoning.

Solution: CH — C — CH2 2 (called as allene) has its central C-atom doubly bonded to two other C-atoms(of methylene groups, CH2). There is no single bond separating the two double bonds. Hence, noconjugation is possible, leading to a complete absence of delocalisation of π electrons.

Let us understand with the help of orbital picture as to why delocalisation cannot occur. In the followingfigure, it is easy to see that 2C is sp hybridised while 1C and 3C are sp2 hybridised. Further, the twoπ bonds are mutually perpendicular. Absence of their coplanarity prevents delocalisation of π electrons.

H

HCCC

H

H

sidewaysoverlap

1 2 2 3C — C and C — C bonds (formed as a result of overlap of orbitals along mutually perpendicular axes) are at right angles to each other

πp

1 2 3

H

HCCC

in horizontal plane (or plane of C-skeleton)

2 3

in vertical plane (or plane perpendicular to C-skeleton)

Here, along with the two -bonds, the two terminal CH (methylene) groups are perpendicularly oriented (as shown by the horizontal and vertical planes)

π2

H

H1

Fig - 26



Q. 1 Consider the compound C H — C H — C H — C H — C H3 2

1 2 3 4 5 explain the following observations.

(i) The 3 4—C C bond length is shorter than a C — C single bond.

(ii) There is significant restriction on rotation about the 3 4—C C single bond.

Q. 2 Consider the following enthalpies of hydrogenation:

(i) H2+ 1120 /H kJ mol−∆ = −

(ii) 2H2+ 1232 /H kJ mol−∆ = −

(iii) 3H2+ 1208 /H kJ mol−∆ = −

Observe that enthalpy of hydrogenation of the cyclic diene in (ii) is almost twice of cyclohexene in (i). Thus, ifbenzene were to exist as a kekule structure (with three double bonds), we could expect the enthalpy of hydrogenationof this hypothetical structure to be three times that of cyclohexene, i.e., approximately 3x (– 120 kJmol–1)= – 360kJ mol–1.However, the actual enthalpy of hydrogenation, for the ‘real’ benzene, is only – 208 kJ mol–1, as given in (iii).Thus, real benzene is thermodynamically more stable than the hypothetical ‘cyclohexatriene’ by – 208 – (– 360)= 152 kJ mol–1.On the other hand, a conjugated diene is more stable than a non-conjugated diene by only around 17 kJ mol–1.Explain why benzene exhibits such high stability.Q. 3 Mention the possible reason(s) that phenol, PhOH, exhibits acidic behavior, while aliphatic alcohol, of the

general form ROH, does not.Q. 4 Draw possible canonical forms and corresponding resonance hybrids of the following compounds:

(i) CH CH CH — — 22 2

+(ii) CH CH CH — — — 2 2CH

(iii) ⊕

(iv) CH CH CH CH NH3 2— — — —+

Q. 5 Given below are pairs of resonating structures Predict which one would contribute most to the hybrid andexplain your answers.

(i) CH C2 —O

CH3

CH C CH2 3— —

O

(ii) H C—O

OHH C—

O

OH⊕

(iii) CH CH OH3 — —⊕

CH CH OH3 — —⊕

Q. 6 Though cyanic acid ( — — )H O C N≡ and isocyanic acid ( — )H N C O= = yield the same anionon losing a proton, the two acids are not regarded as resonance structures of each other. Explain

Q. 7 Which has a shorter carbon-nitrogen bond length: methylamine or formamide?

TRY YOURSELF - I



A general organic reaction may be depicted as:

Substrate Attackingreagent → [Intermediate] → Products(s) + By-products(s)

As we can see, there are two reactants: substrate and reagent. Substrate is the reactant that supplies carbon to thenewly formed bond. The other reactant is called the reagent.

In case, both reactants supply carbon to the new bond, the molecule on which attention is focussed is arbitrarilychosen as the substrate.

Mechanism is the actual process by which the reaction takes place. It tells us which bonds are broken, which newbonds are formed, the order in which bond fission-formation takes place, the steps involved etc.In most of the organic reactions that we will encounter, bonds are broken. This can occur in either of two ways:

(i) Homolytic fission(ii) Heterolytic fission

Bond fission leads to formation of reactive intermediates and the natue of these intermediates depends on themode of bond cleavage. In other words, what type of intermediate species (carbocation, carbanion etc.) areformed will depend upon the way a particular bond breaks.

Futher, reagents and intermediates, together, decide the type of reaction a given substrate will undergo e.g. addition,substitution etc.

(a) BOND CLEAVAGEA knowledge of how bond cleavage takes place in molecules is important to understand the various transientintermediate structures formed during reactions. There are three ways in which a covalent bond can be broken:

(i) — • •A B A B → +This sort of cleavage is termed homolytic fission because each group separates with one electron. Theentities so formed are very reactive and are known as radicals. Such cleavage generally occurs in the gasphase or in non-polar solvents. It can be catalysed by light (as in the substitution of methane which will bediscussed suitably under free radical reactions) or by the addition of other radicals

(ii) A — B A + B:

(iii) A + B:A — B

In the process of cleavage, it could happen that one group separates with both the electrons of the covalentbond. This will result in the formation of an ion pair. Note that it can happen in two ways as in (ii) and (iii)above. Such cleavage is termed heterolytic fission. This generally occurs in solution in polar solvents.There are two reasons for this.

(a) a polar solvent aids in the separation of charge (by favouring the bond cleavage process)

(b) the resultant ion pair so formed can be stabilised through solvation (by the molecules of the polarsolvent.)

Section - 6 MECHANISM IN ORGANIC REACTIONS



In the context of carbon compounds, a carbocation is a positive ion in which a carbon carries the positive chargewhile a carbanion is a negative ion in which a carbon atom carries the negative charge.

The three transient intermediates discussed here may be formed in an actual reaction for only very short durationsand in very low concentrations, but they are the prime controlling factors that determine the outcome of thereaction. Finally, other reaction intermediates also exist. We’ll discuss about them in later chapters. However, freeradicals, carbocations and carbanions are the intermediates mostly encountered in organic reactions.

(b) REACTION INTERMEDIATESCarbocations: As we have already understood that carbocation is simply a positively charged organic ionwhere the positive charge is present on the C-atom. In other words, the C-atom is electron-deficient as it isbonded to three other atoms, thus amounting to a total of only 6 electrons (2 electrons less than octet) in thevalence shell. The positively charged C-atom is sp2 hybridised yielding three sp2 hybrid orbitals lying in the sameplane as that of the C-atom, and are directed towards the corners of an equilateral triangle. Let us analyse thefollowing diagram of the simplest possible carbocation - the methyl cation:

(a)structural representation

HH

H

planar trigonal shape due to 3 bondsσ

sp s 2– bondσ

empty orbitalp

HH

H

(b)orbital representation

orH

CH

H orH

HH

Fig - 27Notice here that the planar trigonal shape due to 3σ bonds (involving 3 sp2 orbitals of the positively chargedC- atom and three 1s orbital of the H atom) is flat. However. Fig. (b) shows an empty (and unhybrid) p orbitallying perpendicular to the plane of the three σ bonds. This empty p orbital imparts, to the entire carbocation,reasons to be unusually stable or unstable. There are various factors that contribute to increase or decrease incarbocation stability. The degree of stability attained by a carbocation is mainly determined by how well itaccommodates the positive charge. Electrostatically, a charged system is more stable if the charge is dispersedover more atoms. Thus, it can be concluded that any phenomenon which encourages the spread of positivecharge and disperses it over the rest of the ion stabilizes the carbocation.Consider a case when the positively charged C-atom has a substituent, say A, in place of a H-atom. Relative to theH-atom, this substituent A may either withdraw electrons or release electrons. An electron-releasing A wouldreduce the positive charge at the C-atom; A’s electron-releasing tendency would impart slight positive charge to Aitself, thus overall a dispersal of the positive charge would occur imparting stability to the carbocation. Observe thefollowing figure:

electron-releasing substituent :dispersal of charge; stabilizes cation

C

HH

A

(a)

electron-withdrawing substituent : intensification of charge; destabilizes cation

C

HH

A

(b)Fig - 28



Further, an electron-withdrawing substituent intensifies the positive charge on the already electron-deficientC-atom, making it less stable. Refer Fig.... (b) above.Hence, we can say that the higher is the number of electron-releasing substituents, more stable is the carbocation.Such a phenomenon of availability (release or withdrawal) of electrons at the reaction site is called as polareffect.Carbanions: We shall now move onto another kind of transient intermediates carbanions. As the name itselfsuggests, this kind of intermediate has a negative charge being carried by the C-atom. This negatively chargedC-atom could be either sp3 hybridised (resulting into a pyramidal configuration Fig... (a)) or sp2 hybridised (yieldingplanar trigonal arrangement, Fig.... (b)). Examine the following possible configurations.

CR3

R2R1

smaller lobe of the orbital is not shown here

or CR3

R2

R1

unhybrid orbital containing the –ve charge

p

(a) (b)Fig - 29

In the above figure, ‘R’ (R1, R2, R3) is any hydrocarbon group. Now, of the two possible configurations shown inFig...., which one actually exists in a given organic reaction depends on the nature of R1, R2 and/or R3. For the timebeing, it is sufficient for you to understand that a carbanion may have any of the two possible configurations. Infact, the actual configuration may be intermediate between the two.

The stability of a carbanion is judged on grounds similar to those applicable to carbocations. The higher thenumber of electron-withdrawing substituents on the negatively charged C-atom, the more is the dispersal of thenegative charge and hence, increased stability of the carbanion.Free Radicals: We now move onto understand free radicals one of the most reactive transient intermediates. Bydefinition, an atom or group of atoms having an odd electron (an electron which is not paired; it does not have apartner of opposite spin) is called a free radical. Its symbolic representations includes a dot (•) to indicate an oddelectron. Further, when a free radical participates in a reaction, the movement of the single electron (or even incases where no free radical is innolved, but we need to exclusively depict the movement of one single electron andnot one pair of electrons) is indicated by a single - barbed curved arrow, as shown below:

single-barbed curvedarrow denotes movement of a single electron

homolysisA B: A• + B •

Note: Movement of a pair of electrons in an organic reaction is shown by the normal double - barbed curvedarrow as drawn below:

normal double-barbed curved arrow denotes movement of an electron pair

heterolysisA B: A B: +ion pair



Structurally, the case of a free radical is somewhat similar to that of carbanions as far as location of the oddelectron is concerned. Two configurations are possible: one, the unpaired electron is present in an unhybridp orbital (with rest of the sp2 hybrid orbitals forming planar trigonal configuration); and second, the odd electroncould be present in a sp3 hybrid orbital (thus giving a pyramidal shape to the radical). The two possibilities are asshown below:

C120°120°R1

R2

R3 120°

unhybrid orbital containing the odd electron

p

sp -2 hybrid C atom

(a)

CR2

R1

R3

sp3 hybrid orbital containing the odd electron

sp -3 hybrid C atom

(b)

or

Fig - 30

Experimentally, it has been found that most simple alkyl radicals are planar trigonal at the carbon having theunpaired electron. The simplest alkyl radical is CH3• called as methyl radical. However, an experimentally observedexception is that of CF3• which has essentially sp3 hybrid C-atom. Hence CF3• has a pyramidal geometricalshape.Let us now discuss the stability aspects relating to a free radical. Although alkyl radicals are uncharged, the carbonthat bears the odd electron is electron-deficient (as its octet in the valence shell is not complete). Hence,electron-releasing substituents (recall ‘polar effects’) bonded to such a C-atom would stabilize it, and the moresuch substituents are, the more stable the radical would be.

Carbenes: Carbenes are neutral species in which carbon atom forms only two bonds (C is bonded to only 2atoms) and has only 6 valence electrons. Out of these 6 electrons, 2 are present in each bond and the remaining2 are non-bonding electrons.A carbene may be represented as:

1R

C:

2RCarbenes are very reactive species and have a lifetime of less than 1 second. They can only be isolated by trappingthem in matrices at low temperature.Generation of carbenes:(i) α-elimination: In these types of eliminations, both leaving groups are from the same atom. e.g. When

chloroform is treated with a base, it leads to the formation of dichlorocarbene.

Cl C H + OH— — Cl C— — Cl C

Cl

Cl

Cl

ClCl

Cl

very acidic protondue to presence ofthree electronwithdrawing Cl atoms

loss of Clleads to formationof carbene

— dichloro carben



e.g. Decarboxylation of sodium trichloro acetate.

Cl C C— — Cl C— — Cl C

Cl

Cl

Cl

ClCl

Cl

Loss of CO , leavingbehined the electronpair

2

O

O Na⊕

> 80ºC—CO2

Cl leaves with bondingpair of electron, leading to formation of carbon

dichloro carbene

(ii) Disintegration of compounds containing certain types of double bonds.

R R

C Z C Z= → +: !

R R

e.g. Decomposition of diazomethane.

CH2 — — N N⊕ hv CH + N N2 —

loss of leads toformation of carbene

N 2

Types of carbenes:There are two types of carbenes: (i) Singlet (ii) Triplet.Both singlet and triplet carbenes are sp2 hybridised and the two differ only in the way the non-bonded electronsoccupy the orbitals.(i) Singlet carbene: In case of singlet carbenes, the two non-bonded electrons are present in the sp2 hybrid-

ized orbital. The unhybridized p-orbital remains vacant.

C

H

H

smallerbond angle(100-110º)

Sp2 hybridized orbitalcontains non bondingelectrons

2 electrons cause more repulsions

unhybridized-orbital emptyp

(ii). Triplet carbene: The two non-bonding electrons in triplet carbenes occupy different orbitals. One electronis present in sp2 hybrid orbital and the other is present in unhybridized p-orbital



C

H

H

Largerbond angle(130-150º)

sp2 hybridized orbitalcontains one non-bondingelectrons

only 1 electrons less repulsions

unhybridized -orbital contains one nonbondingelectron

p

All carbenes can exist either as singlet or triplet. However, most carbenes are stable as triplet statesbecause the energy that can be gained by bringing the electron to more stable sp2 orbital is insufficient toovercome the repulsion that arises due to pairing of electron.

A number of factors operate in molecules that govern the charge density available at different sites in the molecule.We will discuss these effects in the context of organic molecules in particular.

(a) INDUCTIVE EFFECT:This effect takes place whenever two atoms/groups of different electronegativities are bonded to each other.The more electronegative atom/group of the two will tend to draw the shared electron pair more towardsitself, resulting in a skewed rather than a symmetrical electron charge distribution.Consider two atoms A and B bonded to each other, with B being more electronegative. B will tend to pull theelectron pair of the covalent bond towards itself. This means that overall, the electron pair of the covalentbond will spend more time near B than A, resulting in a slight negative charge ( –)δ on B and a slight positivecharge ( )δ+ on A:

Aδ+

Bδ–

Aδ+

Bδ–

B is more electronegative than A, so its pull on the electron pair is greater than A

An approximate orbital picture showing the skewed electron charge density between A and B

Fig - 31Consider the compound 3CH Cl (methyl chloride). Chlorine is more electronegative than carbon. Thus,methyl chloride shows the inductive effect : electron density is greater near the chlorine atom

C

H

H

H

δ+Clδ–

H

H

H

Cδ+

Clδ–

Inductive effect in methyl chloride

An approximate orbital picture showing the skewed charge distribution between carbon and chlorine

Fig - 32

Section - 7 ELECTRONIC EFFECTS IN MOLECULE



Thus, it should be noted that inductive effect in a molecule is a state of permanent polarisation of the molecule.This will show up in the physical properties of the molecule like its dipole moment.

Inductive effect can be propagated down a chain. Consider the following carbon chain, with chlorine at oneend. Observe how the inductive effect propagates down the chain:

C C C Clδδδ+ δδ+ δ+ δ−

123It should be clear that the strength of the inductive effect falls as we proceed down the chain. For allpurposes, the transmission of the inductive effect after the second carbon atom is negligible.

C3

C2

δδ+C1

δ+Clδ– A crude orbital picture showing how the

inductive effect dies down along the chain

Decreasing skewness of charge distribution along the C-skeleton

Fig - 33Inductive effect is abbreviated as the I-effect

Further, any atom or group which attracts electrons more strongly than hydrogen, is said to be showing –I(negative-inductive / electron - withdrawing) effect. On the other hand, any atom or group attracting electrons lessstrongly than hydrogen is said to be showing +I (positive inductive / electron-releasing) effect. Following are twosequences in decreasing order of their effects:

–I groups : 3 3 2 3 3 6 5( )CH N NO CN F COOH Cl Br I CF OH OCH C H H

+> > > > > > > > > > > >

+I groups : 3 3 2 3 2 2 3 2 3( ) ( )CH C CH CH CH CH CH CH CH CH H> > > > >

(b) MESOMERIC EFFECTS:Consider an organic compound in which the linkage C = O is present. We can write two canonical formsfor this linkage so that the actual linkage will be a hybrid of these two canonical forms :

C — O— C — O ——— C Oδ+ δ−

(i) (ii) (iii)You might say that (ii) is not a stable canonical form since it involves the separation of charge and thereforewill be of a higher energy content. However, the negative charge being on electronegative oxygen atom willstabilise this form to an extent. Thus, (ii) will not have a negligible contribution to the hybrid (iii).

Thus, we should actually expect that the linkage will be in a state of permanent polarisation due to theresonance possible. There will be an electron withdrawing inductive effect of the oxygen atom too, but sincethe σ electrons are much less polarisable than the π electrons, we can expect the polarisation in (iii) to bemostly due to the π electrons.

Lets see another example in which such redistribution is possible. We consider a conjugated system in whichthe C = O linkage is conjugated to a carbon - carbon double bond. For this compound, we can writetwo canonical forms, and the actual structure will be a hybrid of these two forms:



———

H |— CH — CH — C — O

δ+ δ−

*

H |— CH — CH — C — O* —

H |— CH — CH — C — O— —

Thus, the actual structure is in a state of permanent polarisation, with the oxygen atom being slightly negativelycharged and the (starred) carbon atom being slightly positively charged. Notice that this polarisation hasbeen transmitted through the π electrons only.Having seen these two examples, we can now properly define the mesomeric effect : it is essentially anelectron redistribution that can take place in unsaturated systems (especially in conjugated systems) via theπ orbitals.Since transmission of this effect occurs via the π orbitals while inductive effect operates through the sigmaorbitals (which are much less polarisable than the π orbitals), the mesomeric effect suffers much lessattenuation in its transmission, with the result that this effect can operate across long carbon chains(the polarity of adjacent carbon atoms will alternate along the chain due to this effect). Electron-donatingmesomeric effect is represented as the +M effect while the electron withdrawing mesomeric effect isrepresented as –M effect.

Consider the following two examples:

C C C C NH CH — — — — — 3

shows an electron-donating(+M) mesomeric effect

C H — C — C — C — O2 5

carbonyl group shows -M mesomeric effect

Following is a list of +M and –M atom or group of atom:

+M groups: — 2 2 3,— ,— ,— ,— ,— ,—Cl Br I NH NR OH OCH

–M groups:

Like the inductive effect, mesomeric effect is a permanent polarisation state of the molecule and would thus,show up in physical properties of the molecule like its dipole moment. For the sake of emphasis, inductiveand mesomeric effects have been compared in the table below:

Table-2: Comparison between inductive and mesomeric effects

1. Operate in both saturated and Operate only in unsaturated compounds unsaturated compounds (and especially in conjugated compounds)

2. Involve bond electron σ

Inductive effects Mesomeric effects

s Involve electrons and orbitals 3. Can be transmitted only over Can be transmitted across large carbon chai

very short distances due to lowpolarisability of the orbitals

π

σ

ns(provided, of course, that conjugation is presentacross the carbon chain)

(c) HYPERCONJUGATION:In part (a), we discussed the inductive effect as an electron donating or withdrawing effect of a groupdepending on the relative electronegativity of the other group with which it is bonded.



It is known that alkyl groups have the following order of electron donating inductive effect (which we’lldenote as +I effect):

CH C 3

CH3

CH3

(i)

CH

(ii)

CH — CH 3 2

(iii)

CH 3

(iv)

CH3

CH3

Thus, tertiary alkyl groups as in (i) above have the highest +I effect while methyl group (in (iv)above) willhave the lowest +I effect.

However, this order of electron donating effect does not hold true when the alkyl group is attached to anunsaturated system like a double bond or a benzene ring. In some cases, this order might even be reversed,so that the methyl group will have the highest and the tertiary alkyl group the lowest electron releasing effect.

Obviously, some other mechanism of electron donation is at work when alkyl groups are connected tounsaturated systems. This mechanism is termed hyperconjugation and is a sort of an extension of the mesomericeffect. In hyperconjugation, the delocalisation of electrons occurs through σ - π overlap of orbitals.

Let us discuss this effect in detail in the case of CH — CH CH3 2—— . First, let us understand the physicalpicture of how the σ - π overlap occurs. Then we’ll understand how to represent the process ofhyperconjugation on paper, through canonical forms.Consider the following approximate orbital picture of CH — CH CH3 2—— .

H

HCCC

HH

H Only one of the three carbon-hydrogen molecular orbital has been drawn. Note that in addition to the 2p-2p unhybridised orbital overlap, an overlap can also take place of the carbon-hydrogen orbital with the orbital system of the double bond. This is r ep re s e n t ed b y t he t h r ee d o t t e d l i n es

σ

σ

Fig - 34As is evident in the figure above, (each of) the carbon hydrogen sigma molecular orbital can overlap with theπ-orbital system of the carbon-carbon double bond, resulting in a sharing of electron charge density of thecarbon-hydrogen single bonds with the carbon-carbon double bond. This explains the electron donatingeffect of the methyl group when attached to a double bond (or any unsaturated system). We can representhyperconjugation in this case using canonical forms as follows:

H |H — C — CH — CH | H

2—

H H — C — CH — CH | H

2—

H |H C — CH — CH | H

2—

H |H — C — CH — CH

H

2—

H H C C — CH

H

2

In the actual structure, the carbon- hydrogen orbitals charge density is shared with the carbon-carbon double bond.

σ



We can now explain the observed reversal of the electron donating effects of alkyl groups when attached tounsaturated systems using hyperconjugation

H |H — C — CH — CH | H

2—

(i)

CH |CH — C — CH — CH | CH

3

3 2

3

—

(iv)

H |H — C — CH — CH | CH

2

3

—

(ii)

H |CH — C — CH — CH | CH

3 2

3

—

(iii)

In (i), we have three ‘hyperconjugable’ hydrogen atoms, because all three of the carbon-hydrogen σ orbitalscan overlap with the π molecular orbital system. In (iv), for example, there is no hyperconjugable H-atomand thus electron-donating effect of the alkyl group, here, through hyper conjugation is the lowest.

Hyperconjuation explains a lot of phenomenon which could not otherwise be explained. It tells us, forexample, why (A) is thermodynamically more stable than (B) below:

CH|

CH — C — CH — CH

3

3 3

— —

(A)

CH|

CH — CH — C — CH

3

3 2 2

(B)

(A) has nine hyperconjugable H-atoms while (B) has only five. Thus, the effective delocalisation of charge ismore in (A) than in (B) (due to hyperconjugation) and hence (A) is more stable.

(d) TIME VARIABLE EFFECTS (ELECTROMERIC EFFECT) :The three effects discussed above are permanent in the sense that they represent permanent polarisations inthe compound. They would thus show up in the physical properties of the molecule, like its dipole moment.Molecules can also show time-variable effects, which occur only on some external stimuli, like the approachof a reagent, for example.We will discuss here one of the more important time variable effects, the electromeric effect. This effectoccurs via π - electrons. On the approach of some external reagent, the π - electron pair shifts entirely to themore electronegative atom. Consider the following example

C — O

—

—(A nucleophile ;we'll discuss moreabout these soon)

C — O + Nu–

δ+ δ−—

——

The approach of Nu– above causes the π - electron pair to shift entirely to oxygen. Electromeric effect is atemporary effect. If the attacking reagent is removed, the molecule will revert back to its ground state. Thus,this effect will not show up in the physical properties of the molecule.

Electromeric effect is the time-variable counterpart of the mesomeric effect. Like wise, there’s atime-variable inductomeric effect which corresponds to the (permanent) inductive effect



Example 29

Draw the hyperconjugation possible in toluene (Ph — CH3) by sketching various possible canonical forms.

Solution: Following canonical forms depict hyperconjugation in toluene:

H — C — H

H

H — C — H

H⊕ H — C H

H⊕ H — C H

H⊕

.... so on

for the other two H atoms on the C atom bonded to the phenyl group. Hence, there would be a totalof 10 such canonical forms.

Note: While writing such canonical forms, try to explicitly show all hyperconjugable H atoms(s) so that you don’tmiss out drawing any possible canonical form.

Example 30

A student represents the hyperconjugation possible in propane as follows:

H C CH CH — — — 2

H

H

H C CH CH — — — 2

H

H

⊕

−

Can we infer from the second canonical form that a proton separates from the main carbon chain? What is thesignificance of this second structure?

Solution: No, the second canonical form does not imply that a proton becomes free. Separation from the maincarbon chain would mean movement of the proton from its original position - a situation inhibitingdelocalisation whose basic pre-requisite is that all the constituent atoms must occupy essentially thesame relative positions in each canonical form.

(The significance of the second structure lies in clarifying how alkyl groups here), CH3 – are capableof giving rise to electron release, thus, enabling delocalisation.

Q. 1 Explain why reactions involving ionic intermediates take place more easily in solution in polar solvents thanin non-polar solvents.

Q. 2 Which carbocation is expected to be more stable:

3 2 2— —CH CH C H⊕

or 2 2CH CH C H

⊕−=

Q. 3 What should be the direction of the inductive effect exerted on carbon by magnesium in the compound3H C MgX− where X is a halogen atom? (This compound is an example of a Grignard reagent which we

will encounter later in organic chemistry).

TRY YOURSELF - II



In this section we will formally categorise the types of reagents that we’’’ encounter in the course of our study oforganic chemistry.

(a) Electrophiles (electron-seeking reagents)

These are species which tend to attack electron rich sites. For example, a site like the oxygen atom in theC = O linkage will tend to be attacked by an electrophile. Substitution on the benzene ring will be almost

always by an electrophile because of the high electron charge density on the ring.

Electrophiles can be positively charged cations like NO2 . They might also be species that are actually notcations but posess an atom or centre that is electron deficient and will tend to react with an electron rich site.For example the sulphur atom SO3 will act as an electrophile, due to the electron deficient sulphur atom,(even though SO3 is not a full blown cation):

Oδ− S δ+++Oδ−

Oδ−

(b) Nucleophiles (“nucleus” - seeking reagents)These are reagents which will tend to attack electron deficient sites. For example, in the following compound

|— C Cl |

δ+ δ−

the carbon atom acquires a slightly positive charge due to the –I effect of the chlorine atom. This carbonatom will tend to be attacked by nucleophiles.

As in electrophiles, nucleophiles can either be negatively charged anions like OH–, Cl– etc, or they can bespecies which are not actually anions but which posess an atom or a centre that is electron rich. For example,the nitrogen atom in ammonia , NH3, will tend to act as a nucleophile.

(c) Free radicalsThese are highly reactive reagents (intermediates) which are not charged, and thus they are much less proneto variations in electron density in the substrate (which they attack) than reactions which involve polarintermediates ( electrophiles or nucleophiles). An example of a free radical is the chlorine free radical, whichcan be formed by the action of sunlight on the chlorine molecule :

Cl2sunlight Cl Cl• + •

We will discuss more about the properties of these reagents (like their hybridisation states etc) as and whenwe encounter them. For the time being, this introduction should suffice.

Example 31

Classify the following as electrophiles or nucleophiles.(i) 2NO+ (ii) RO– (iii) OH–

(iv) AlCl3 (v) SO3 (vi) CO2

(Note that electrophilicity of Al, S and C atoms are being refered to wherever present)

Section - 8 TYPES OF REAGENTS



Solution: Electron-seeking reagents or electrophiles are as follows:

2NO+ , AlCl3, SO3, CO2

Nucleophiles are: RO– and OH–.

The reactions of organic compounds can be classified into four major categories:(a) Substitution (b) Addition(c) Elimination (d) Rearrangement

(a) Substitution reactions are those in which a group attached to a carbon atom is substituted by anothergroup.

A classic example of a substitution reaction is the nucleophilic substitution in alkyl halides :

| | — C — X + Nu — C — Nu + X | |

The Nu (a nucleophile)anion displaces the halide ion. (This reaction will be discussed in some othercontext ).Substitution is, of course, not limited to nucleophilic reagents. It can be electrophilic as well as radicalinduced.

(b) Addition reactions occur in unsaturated systems. As the name suggests, an addition reaction is one inwhich two molecules ‘add’; a good example is the addition of a hydrogen halide, HX, to the carbon-carbondouble bond

| |C — C + H — Br — C — C — | | H Br

——

———

Addition reactions too can be electrophilic, nucleophilic or radical-induced, depending on what entity initiatesthe addition reaction. In general, addition to carbon-carbon double bonds is electrophilic or radical induced,while addition to, say carbon oxygen double bond would be usually nucleophilic. (More on this later)

(c) Elimination reactions are the reverse of addition reactions; a molecule is lost/eliminated/ “subtracted”from the original molecule. The most common elimination reactions are those in which there is a loss of ahydrogen atom and another atom or group from adjacent carbon atoms. The dehydrohalogenation of alkylhalides in the presence of a base is a good example of this class of reactions:

–HXC C

H

X

C C

(d) Rearrangement reactions under certain reaction conditions, a molecule may rearrange its structure toanother form which might be thermodynamically more stable (in having a lower energy content). Carbocationrearrangement is an example of this class of reactions ( the reason why it happens will be discussed here)

Section - 9 TYPES OF REACTIONS



H H | |CH — C — CH — OH CH — C — CH — OH | | CH CH

3 2 3 2 2

3 3

H

–H O2

H | CH — C — CH CH — C — CH | | CH CH

3 3 3 2

3 3

hybrideshift

1° carbocation

(a rearrangement)

3° carbocation

This example illustrates that rearrangements do take place. Details of this reaction will be dealt with later.Rearrangements can also proceed through intermediates that can be cations (electrophiles), anions(nucleophiles) or radicals. The most common intermediates in rearrangements are carbocations.

(A) DEFINITIONSWe will, in our course of the study of organic chemistry, frequently encounter acids and bases. Questionsmight be posed as to which of a given set of organic compounds is the most acidic or most basic, and so on.It is thus of significant importance to study the concept of acidity and basicity, the various theories involvedand the factors influencing acidity and basicity.The three theories of acidity / basicity that we’ll discuss are as follows in increasing order of generality :(i) Arrhenius definition

Acids are compounds that yield hydrogen ions (H +) in solution. Bases are compounds that yield hydroxideions (OH –) in solution. This was one of the first definitions proposed and it sufficiently covers reactionsin which water is the reaction medium.As an example of an Arrhenius acid and base, you can think of HCl which would be an acid since it canyield H + ion in solution ; NaOH is a possible base since it can yield OH – in solution. However, thisdefinition cannot account for acidity and basicity in certain compounds and thus it has become moregeneralised.

(ii) Bronsted definitionAcids are proton (H +) donors while bases are proton acceptors. For example, consider the followingreaction: HCl + H O2 : Cl + H O–

3+

Acid Base Conjugatebase

Conjugateacid

Observe that HCl acts as a Bronsted acid since it can denote an H + ion to the water molecule; it therebygets converted into Cl – which is termed its conjugate base. H2O acts as a Bronsted base since it canaccept the H + ion from HCl; doing so converts H2O into H3O + which is its conjugate acid.The Bronsted definition is an improvement over the Arrhenius definition in terms of more generality, butit can be further generalised to include even more compounds. This is what the Lewis definition does.

Section - 10 ACIDS AND BASES



(iii) Lewis definition

Acids are molecules or ions which can co-ordinate with unshared electron pairs provided to them byother molecules or ions. Bases are molecules or ions which have unshared electron pairs available forco-ordination with other molecules or ions.

To understand how this definition is more generalised, consider the following reactions

Me N + BF Me N BF3 3 3 3: :

Cl + AlCl Cl + AlCl2 3 4

In these reactions, there is no involvement of protons (H+ ions)/ hydroxide ions (OH– ions). Even then,we can use the Lewis definition to say that BF3 and AlCl3, for example, are Lewis acids because eachof them are capable of co-ordinating with (accepting) an unshared electron pair.

(B) ACID - BASE REACTIONSCan we predict whether a given pair of acid and base will react to give appreciable amounts of products ornot?Let us see how to do so. It is said that initially acid-base reactions exist in equilibrium .

StrongAcid

StrongBase

WeakBase

WeakAcid+ +

Conjugate

Conjugate Pairs

The products of the reaction are determined by the position of equilibrium. We know that equilibriumreactions always favour formation of most stable species. Since weaker acids and bases are more stablethan stronger acids and bases, the equilibrium shifts towards formation of weaker acid and base.

Thus, we should always keep in mind the following rule: “Acid-base reactions always favour the formationof weaker acid and weaker base”.

(C) MEASURE OF ACIDITY/ BASICITYAs has been discussed in the chapter on equilibrium, the strength of acids and bases is measured in terms ofpKa and pKb respectively (We will be concerned with only proton acids and bases for the time being)Ka and Kb are equilibrium constants for the following equilibrium (considered in an aqueous solution)

–2 3

–3

10

Acids : ...(i)

[ ][ ] [ ]

– log

a

a a

HA H O H O AH O AK

HApK K

+

++ +

=

=

!!"#!!



–2

–

10

Bases : ...(ii)

[ ][ ] [ :]

– log

b

b b

B H O BH OHBH OHK

BpK K

+

++ +

=

=

!!"#!!

It is evident from these relations that stronger acids/ bases imply higher corresponding values of Ka/Kb andtherefore lower corresponding values of pKa/pKb. Thus, for example, an acid with a pKa of 1 will bestronger than an acid with a pKa of 2. A similar analogy would hold for such a pair of bases.

Conventionally, the strength of bases is also described in terms of Ka (and pKa) instead of Kb and (pKb).Doing so will result in a single scale/measure for the strength of both acids and bases and will thus be moreconvenient. To do so, we consider, instead of the reaction (ii), the following reaction as our referencereaction for bases :

2 3 BH H O B H O+ ++ +: :!!"#!!

For this reaction, we can write the Ka (and pKa) as :

3[ :] [ ][ ]a

B H OKBH

+

+

⋅=

10loga apK K= −

Thus, Ka (pKa) here would be a measure of the strength of the conjugate acid BH+ of the base B : . A lowerpKa value will imply a stronger conjugate acid BH + and thus a weaker conjugate base B:. Similarly, a higherpKa value implies a stronger base B :

⇒ +Lower value Stronger acid Weaker base :apK BH B⇒ ⇒

⇒ +Higher value Weaker acid Stronger base :apK BH B⇒ ⇒

For example, NH3 has a pKa value of 9.25 while CH NH3 2 has a pKa value of 10.64. Thus, CH NH3 2

is astronger base.One point must be emphasised here : the pKa scale is logarithmic and you must appreciate this. Thus, an acidwith a pKa value of 2 will be 10 times weaker than an acid with a pKa value of 1. In fact, acids with pKa >16 will not even be detectable as acids at all in water, because their ionisation will be less than the autolysisof water itself.

For very stronger acids, full ionisation will be achieved in water, and thus all strong acids will appear to be ofthe same strength in water. This levelling effect of water will be reduced to an extent in case the solvent isother than water and is a weaker base than water ( because for a weaker base, ionisation of the acid will be100% complete).

This means that relative strengths of strong acids can be compared in a solvent that is a weaker base thanwater. Similarly, relative strengths of very weak acids can be compared in bases stronger than water.



(D) FACTORS AFFECTING ACIDITY

In this section, we will discuss the factors which affect the strength of acids in the form of HA. If HA gives upits proton easily, it will be a stronger acid than in the case when it is reluctant to give up the proton.

Also, the acid which gives up its proton easily results in formation of a weaker have than when it is reluctantto give up the proton.The factors which can affect acidity/basicity are:

(i) Strength of H — A bond : The general rule is: The stronger the H — A bond, the weaker the acid.”Amongst the compound, of the same group of the periodic table, bond strength to proton decreases onmoving down the group.

H F—

H Cl—

H Br—H I—

Increases

Decreases

pKa Acidity

This is because, as we move down the group, the effective overlap between hydrogen and the halogen atomdecreases. Thus, overlap is minimum in HI which results in formation of weaker H — I bond. Weaker bondimplies stronger acid.Now let us compare the basic strength of the conjugate bases of these acids. Since HI is the strongest acidamongst them, its conjugate base (I—) will be the weakest one. On the other hand, F— will be a strongerbase because it is the conjugate base of a weaker acid (H — F).The same holds true for compounds of other groups of periodic table.

H O2 Increases

Decreases

pKa Acidity

H S2

H Se2

Here, H2O is the weakest acid and H2Se is the strongest one.(ii) Electronegativity: On moving left to right across a periodic table, bond strength varies but the predominant

factor is the increase in electronegatvity. Recall that electronegativity is a measure of an atom’s attraction forelectrons. The higher the electronegativity, the greater the attraction.Electronegativity has the following two effects on the acidity



(a) Polarity of the bond.(b) Stability of the anion formed (conjugate base) after the loss of proton

High electronegativity increases the polarity of the bond. Let us examine the compounds of any one period.e.g. CH4, NH3, H2O, HF.

electronegativity C N O FIncreases →

Thus, we can say, F being most electronegative causes most polarization of the H — F bond. In otherwords, H acquires maximum positive charge in HF and least in CH4. Thus, HF can readily lose a proton toa base and hence, is most acidic.

H C H3 —δ— δ+

H N H2 —δ— δ+

HO H—δ— δ+

F H—δ— δ+

IncreasesPolarity

DecreasespKa

Acidity Increases

Now let us compare the strength of their conjugate bases.HF being the strongest acid forms the weakest have F—. This is because fluorine is most electronegative andcan most readily accommodate the negative charge. Thus, F— is most stable and least reactive and hence,least basic.

On the other hand, CH4 is the weakest acid so leads to formation of strongest base —CH3. Carbon being theleast electronegative atom is least ready to accept the negative charge and is most unstable and reactive.Thus, it is most basic.

(iii) Hybridization: We have already studied about the effect of hybridization on the acidity of the molecule.Recall that more s-character increases the electronegativity. Thus,

H C C H — — — C C—H

H

H

HH — — — C C H

H

H H

H

Increases

DecreasesAcidity

pK4

(iv) Inductive effect: We studied about inductive effect and the types of inductive effects in section-7. Weknow that a + I group (electron releasing group) increases the electron density around the atoms to which itis attached. If it is attached to a base, it will increase the electron density around it making it more unstableand henc, more reactive or more basic.



On the other hand, –I group (electron withdrawing group) decreases the electron density around the atomto which it is attached. –I group attached to a base decreases electron density of the base making it lessreactive and less basic.–I groups attached to an acid facilitate release of proton by withdrawing electrons. On the other hand, +Igroups have an opposite effect on acidity ie. they decrease acidic character.

Acidic nature Basic nature+I effect Decreases Increases–I effect Increases Decreasese.g. CH C OH 3 — —

O(I)

CH C OH 2 — —

O(II)

Cl

Amongst the two compounds, (II) will be more acidic because of presence of a –I group at α-carbon.

e.g .

NH2

(I)

NH2 NH2

CH3 NO2

(II) (III)

The increasing order of basicity of the above compounds is:

NH2

(I)

NH2NH2

CH3NO2

(II)(III)

<<

(III) is least basic because it has an electron withdrawing (–I group) attached which decreases basicity. Onthe other hand, (II) has an electron-donating (+I group) which increases basicity.

Inductive effect: We studied about inductive effect in Section -7. The application of inductive effect and itsinfluence on acidity of a molecule is explained using example 37.



(v) Effect of stability of A– with respect to HA: This factor is very important in determining the strengths ofacids. If the conditions are such that A– (the conjugate base of HA) can be stabilised with respect to HAitself, we will find that HA will tend to be more acidic.The reason for this is clear. A stable conjugate base A– because of its stability, will be reluctant to gain aproton to form back the acid HA. Thus, a stable A– will be a weak base and consequently, HA will be astrong acid. You can think of it this way : if A– can be stabilised somehow, it will be ‘energetically favorable’for HA to give up a proton to form A– can ; thus, HA will lose its proton with ease, and will therefore be astrong acid.

All factors that can possibly stabilise A– will have something to do with dispersion/delocalisation of charge.Let us consider the various factors through examples :

Example 32The order of acidity of the hydrogen halides in aqueous solution is HF < HCl < HBr < HI . Explain.Solution: Observe this order carefully. Do you realise that this order is contrary to what we should expect due

to the affect of electronegativity described earlier ? For example, since fluorine is the most electronegativeof all the halides, we should expect HF to be the strongest acid. However, as it turns out, it is theweakest. (Before proceeding on, one point requires attention here. The order of acidity is beingtalked about in water as a solvent, since this is the most common solvent used for organic reactions).One factor to which we can attribute the reversal of the expected acidity order is the relative stabilityorder of the halide anions (the conjugate bases) formed on the loss of a proton. This stability order isas follows:

– – – Order of stabilityF Cl Br I−< < <

The reason for this stability order is that the larger the atomic radius of an anion, the more the volumeof the region over which the negative charge can spread and thus more stable that anion will be. Sinceatomic radii of the halide atoms are in the order F < Cl < Br < I, stability of the anions will alsofollows this order. Consequently, as described earlier, acidity of the hydrogen halides will also followthis order.

Example 33

The compound HCOOH is by organic standards, a moderately strong acid. It has a pKa of of 3.77 (The name ofthis compound is methanoic acid). Explain.Solution: We must again look into the possibility of the conjugate base of this acid being somehow stabilised.

A minor reason for this compound showing acidity is the fact that the

carbonyl group is electron withdrawing, so it increases the

electron affinity of the oxygen atom of the OH group (which contains the acidic hydrogen atom.)

C — O

The major reason for methanoic acid showing strong acidity is that the conjugate base is resonancestabilised.

H C O

OH C

O

OH C

O

Oδ-

δ-

methanoate anion



The delocalisation of the negative charge is in fact extremely effective since the two canonical formsare of identical energy content. Thus, the methanoate anion is quite stable.Observe that although delocalisation through resonance can take place in methanoic acid itself, it willnot be really effective since it will involve charge separation in the canonical forms.

H C O

O HThis structure has a high energy content

H C O

O H

-

+

Thus, resonance will tend to stabilise the methanoate anion with respect to the methanoic acid, whichresults in the acidity of the latter.

Example 34The alcohol ROH (where R is an alkyl group) is a much weaker acid than phenol, PhOH, altough both contain thesame functional group (– OH). Explain.

Solution: Again, we look into factors stabilising the conjugate base of phenol and which do not work for thealcohol

Consider the conjugate base of phenol, PhO–. Observe that there is a possibility of resonancestabilisation in this anion :

O O O Oδ−

The 'actual' phenoxide anion

Observe that in the phenoxide anion, the negative charge is delocalised over the entire benzene ringand thus this anion is stabilised.

Resonance can occur in the undissociated phenol molecule also but that will involve charge separationand will therefore be less effective than resonance in the conjugate base.This discussion justifies why phenol shows an acidity stronger than alcohols (where no such resonancestabilisation is possible in the conjugate base RO –).

As an example, the pKa of CH3OH is 16 while that of phenol is around 10, indicating that phenol is aconsiderably stronger acid than CH3OH.

Example 35Phenol (PhOH) is a considerably weaker acid that the carboxylic acids (in which the acidic part is the functional

group C O

OH) . Explain.



Solution: As discussed in the previous case study, the phenoxide anion PhO– is stabilised by resonance and thusPhOH is acidic. Here we need to look into how carboxylic acids should be even stronger acids.Consider the conjugate base of a general carboxylic acid (RCOOH) and observe that it will be resonancestabilised too:

R — C

O

OR — C

O

O——— R — C

O

O

δ−

δ−

Carboxylate anion

-

-

However, resonance stability in the carboxylate anion will be much more than that in the phenoxideanion. This is due to two reasons:

(i) For the carboxylate anion, both the canonical forms are of identical energy content

(ii) In the canonical forms of the carboxylate anions, the negative charge is on the highly electronegativeoxygen atoms and therefore we can expect these forms to be stable.

In the canonical forms of the phenoxide anion drawn earlier, there is a negative charge on a carbonatom in three of the four structures, and hence these forms are likely to be of a higher energy content(and therefore less stable) than the structure in which the negative charge is on oxygen.

These two reasons justify which carboxylic acids are stronger acids than phenol. As an example, the

pKa of phenol is around 10, while that of methanoic acid, CH — C 3

O

OH is 4.76.

Note:The inductive effect of the groups connected to the acidic group in a molecule can have a significant effecton the acidic properties of that molecule. We first discuss as an example, inductive effects in carboxylic acids.Consider a general carboxylic acid of the form G - COOH, where G could be any group. If G has an electronwithdrawing inductive effect, it will tend to stabilise the carboxylate anion by aiding in the disperson of charge onthe —COO group. On the other hand, a + I effect of G would tend to distabilise the carboxylate anion.

+ G

GCOOH

I effect of willdestabilise the conjugatebase of

G C O

O

δ−

δ−G C

O

O

δ−

δ−

– G

GCOOH

I effect of willstabilise the conjugatebase of

We can thus conclude that groups with –I effect will increase the acidity of the carboxylic acid while groups with+I effect will decrease the acidity. This is what is actually observed, as the data in the table below proves.

Methanoic acid, H—COOH (pKa = 3.77) has been used as a reference. On the left hand side are carboxylic acidsweaker than methanoic acid due to the +I effect of the group attached to the carboxylic group; on the right handside are carboxylic acids stronger that methanoic acid due to the –I effect of the group attached to the carboxylicgroup.



These acids have been arranged in increasing order of acidity.

Table 3 : Major acids in increasing order of acidity

Acids weaker than methanoic acid pKa

Acids stronger than methanoic acid pKa

(1) 5.05

CH C COOH3

CH3

CH3

CH CH COOH3 2(2) 4.88

CH — CH COOH | CH

3

3

4.86(3)

CH CH CH COOH3 2 2(4) 4.82

CH COOH3

CH CH CH — COOH |Cl

2 2 2

CH — CH — COOH2—

CH — CH — CH COOH | Cl

3 2

HO CH — COOH2

(5) 4.76

(6) 4.52

(7) 4.25

(8) 4.06

(9) 3.83

3.58(10)||O

(11) 3.53

(12)||O

(13) I CH COOH2

3.35

3.16

Br CH COOH2(14) 2.90

(15) Cl CH COOH2 2.86

(16) F CH COOH22.57

(17) NC CH COOH2 2.47

(18)

(19) NO CH COOH2 2

( )CH N CH COOH3 3 21.83

1.68

(20)Cl

ClCH COOH 1.25

(21)

Cl C COOH

Cl

Cl

0.65

CH — O — CH COOH3 2

*Data quoted from “Peter Sykes : A Guidebook...”



You are urged to go through this table and observe carefully how the pKa varies as the group attached to thecarboxylic group is varied.

It would in fact be a good exercise to compare two different compounds of the same (or similar) molecularformula or similar structure and try to reason out why their acidities differ. We will now do this in a few examples.

As mentioned at the end of casestudy 1, here also the justification of the acidity order based only on the inductive effect is only partially complete. The actual scenario requires a more detailed and complete justification. You can refer to the appendix for that prupose. However,for our current needs, the justification provided above would suffice.

Example 36

(a) Why is F — CH2 — COOH a stronger acid than I — CH2 — COOH ?

(b) Why is CH — COOHCl—

—Cl

a stronger acid than Cl — CH2 — COOH ?

Solution : You should find this question very straightforward since it requires simple reasoning based on theinductive effect.

(a) F is more electronegative than I ; the electron withdrawing inductive effect (–I effect) of F willtherefore be more than I, and thus stabilisation of conjugate bases will be greater in the case of Fthan in the case of I. This explains the observed greater acidity of F — CH2 — COOH.

(b) The –I effect of two chlorine atoms will obviously be more than that of just a single chlorine atom.This explains why the dichloro-substituted acid is stronger.

Example 37

Why is CH — CH — COOH2 — a stronger acid than CH3— CH2— COOH ?

Solution : The α - carbon (the C-atom immediately next to the carboxylic C-atom) in the first compound is sp2

hybridised. Electrons in an sp2 hybrid orbital will be drawn closer to the carbon nucleus than in an sp3

hybrid orbital. The reason for this has already been discussed. sp2 orbitals contain a greater s-character.Thus, an sp2 hybridised carbon atom will be less electron donating than an sp3 hybridised carbonatom.

Thus, the +I effect on the carboxylic group will be lesser in the case of the first compound than in thesecond in which the α -carbon is sp3 hybridised. This explains why CH — CH — COOH2

— is a strongeracid

Note: Let us now understand the effect of inductive/mesomeric electron withdrawal/donation on acidity in aromaticcompounds.

Inductive and mesomeric effects can have significant influence on acidity in aromatic compounds too, as in thecase of aliphatic compounds.



For a start, we’ll discuss acidity in substituted phenols. Our reference compound for discussion will be phenol(pKa = 9.95). The conjugate base of phenol is phenoxide anion which is stabilised by resonance :

O O O Oδ−O

Although resonance to an extent delocalises the negative charge over the benzene ring, any group which has anelectron withdrawing effect, when attached to the benzene ring, will further increase the stability of the phenoxideanion by aiding in further delocalisation of the negative charge. Such a substituted phenol will be more acidic thanphenol itself.

On the contray, any group which has an electron donating effect, when attached to the benzene ring, will tend todestabilize the phenoxide ion, and hence the substituted phenol will be less acidic than phenol itself.* If the substituted group has an inductive effect, we can expect the inductive influence of this group to fall

off with distance as we go from ortho meta para→ → substituted phenol.

* If the substituted group has a mesomeric effect, it will have an influence when the group is attached to theortho or para position but not at the meta position. This will become clear when we draw the canonicalstructures possible for each of the three positions and observe that only if the group is at the ortho or thepara position can it promote ionisation by stabilisation of the anion.

We do this as an example for NO2 attached at para position, i.e., for p-nitrophenoxide ion.

O

N–O O+

O

N–O O+

O

N–O +

O

N–O O+O–

(A) (B)Observe the form (A) – (B) and note that it was because of the fact that the nitro group is at the p-position that itcould exert –M effect and aid in the stabilisation of the phenoxide anion.

This would not have been possible had the nitro group been on the meta position, as should be evident from thefollowing canonical forms:

O O

NO2

-

-

NO2

O

NO2-

O

NO2

-



In the canonical forms above, the negative charge never appears at the meta position ; therefore, the nitro groupdoes not have a possibility of delocalising the negative charge over itself through its –M effect.

Let us now discuss other cases of acidity through examples.

Example 38

Justify the following observed values of pKa :

COOH COOH COOH COOH

Cl

4.20 2.94 3.83 3.99Cl

Cl

Solution: We must first of all focus our attention on the substituent which is Cl here. Cl has a power –I effectbecause of its high electronegativity. However, Cl can also exert a +M effect due to the lone pair whenattached at the ortho or para position. This means that there are two opposing effects here : Cl willtend to stabilise the system by its –I effect but at the same time tend to destabilise the system by its +Meffect

* o–chlorobenzonic acid is the strongest of the four given acids because of the extremely shortdistance over which the powerful –I effect of Cl operates (this more than compensates for the +Meffect of Cl). As we go from − → − → −o m p chlorobenzoic acid, the strength of this –I effectfalls off very quickly.

* p–chlorobenzoic acid is weaker than m-chlorobenzoic acid due to a weaker –I effect in theformer (because of the larger distance) and the possibility of the +M effect due to chlorine beingat the para position.

Example 39

Explain the observed values of pKa for the following nitro-substituted phenols (phenol itself has a pKa of 9.95) :

OHNO2

OH

NO2

OH

NO2

OHNO2NO2

OHNO2NO2

NO27.23 8.357.14

4.01 1.02

Solution: Here we present a point - by - point analysis of these observed values of pKa.

* All nitro-substituted phenols are stronger acids than phenols due to the –I effect of NO2 and –Meffect of NO2 if it is at the ortho or para position

* o – and p–nitrophenol are stronger acids than m-nitrophenol due to the –M effect the NO2canexert at the ortho and para positions



* Although in o-nitrophenol, the –I effect of NO2 operates over a shorter distance than in p-nitrophenol(and we should therefore expect o-nitrophenol to be stronger than p-nitrophenol), it actually turnsout that p-nitrophenol is the stronger acid (albeit only marginally). This happens because there area lot of (complicated) factors contributing to acidity, like the solvation patterns etc (and which wedo not consider here for the sake of simplicity).

* 2, 4 - dinitrophenol will obviously be a still stronger acid due to two reasons : the powerful –Ieffect of the two nitro groups and also their –M effect since they are at the ortho positions.

* 2,4,6 - trinitrophenol is a very strong acid due to the effect of the three nitro groups which willexert both –I effect and –M effect to stabilise the conjugate base to a large extent.

Example 40

2, 6-dihydroxybenzoic acid is found to be highly acidic by organic standards (pKa = 1.30)

(A) (B)

COOH COO–

–H+OH OH OH OH

Such high acidity cannot be accounted for by the –I effect of the OH groups alone. What could be the otherpossible factors ?

Solution : This is an example of a general trend : the behaviour of ortho - substituted acids many times turns outto be significantly greater than expected because of direct interaction possible between the adjacentgroups.

It turns out that here a direct interaction is possible between the carboxylic group (—COO–) and thetwo hydroxyl groups (–OH) in the form of hydrogen bonding, which stabilises the anion (B) to a verylarge extent.

This stability is the reason that the acid (A) readily gives up its acidic proton to form the anion (B). Itshould be pointed out here that intramolecular H–bonding will also be possible in the undissociatedacid (A) itself but it will not be as effective as in (B) in which there is a negative charge and hencestronger H-bonding.

(iv) Effect of solvent: A mention was made earlier about the levelling effect of water. All strong acids get100% dissociated in water and thus appear to be equally strong in water. To compare the relativestrengths of these strong acids, a base weaker than water can be used as the solvent. This is one of themany examples which show that the nature of the solvent can have a profound determining effect on theacidity of the solvent. As the example above shows, this is particularly the case when water is thesolvent.



We can divide organic solvents into two broad classes :

(a) polar solvent - like water

(b) non-polar solvent - like toluene (methylbenzene)

We might find cases where a particular molecule shows acidic properties in a particular solvent but istotally non-acidic in another. The next example considers this scenario.

Example 41

HCl is a very strong acid in water. In fact, it exists almost fully dissociated in water. In alcohols also, like methanol(CH3OH) for example, HCl is a moderately strong acid. However, in a non-polar solvent like toluene (methylbenzene), HCl is almost totally non-acidic i.e. it occurs almost totally undissociated in toluene. Justify theseobservations.

Solution: For any acid, the stronger the solvent as a base, the more the dissociation of that acid in the solventwill be observed and correspondingly, the weaker the solvent as a base, the lesser the dissociation ofthat acid will be observed. Water and alcohol are good solvents in this sense. In fact, water is a verygood polar solvent :

(a) it has a high dielectric constant which aids in the formation of the ion pairs from the acid

(b) it aids in the stabilisation of the ion pair because it has a very strong ion-solvating power due to itsability to form hydrogen bonds.

Alcohols share these properties to an extent and thus HCl is found to be a moderately strong acid inmethanol also.

However, toluene is non-polar (and non-basic) and has none of the ion-pair stabilisation abilities thatwater and alcohol have. Thus, it it energetically unfavorable for the HCl molecule to dissociate intoluene. This explains the observations.

(D) FACTORS AFFECTING BASICITY

As discussed earlier, the reference reaction for bases is the following equilibrium :

2 3BH H O B H O+ ++ +:!!"#!!

with Ka and pKa now being given by

310

[ :][ ] ; log[ ]a a a

B H OK pK KBH

+

+= = −

Thus, higher the pKa, stronger is the base B: In physical terms, the strength of a base is the measure of easewith which B: can accept a proton, or equivalently, the measure of the availability of the unshared electronpair on B. Factors which increase this availability of electrons will tend to increase the basicity of B: whilefactors which decrease this availibity of electrons will tend to reduce its basicity.

However, as in acidity, the solvent also has an effect on the basicity. We will now discuss the various factorsthrough examples.



Example 42The basicity order of ammonia (NH3) and alkyl-substituted ammonia (amines) in a non-polar solvent is found to bein the following order

1° amine

NH3 R — NH2 NH—R—R — N

R

RR

2° amine 3° amineExplain.

Solution: The basicity order being talked about has been measured in a non-polar solvent which means that wecan safely neglect any effects of solvent molecules on the basicity order (such as H-bonding) andconcentrate only on structural reasons that give rise to this order.

Recall that basicity will increase as the unshared electron pair availability on the nitrogen atom increases.Recall also that alkyl groups have a + I effect.This is the reason behind the observed basicity order. As we go from 1° amine to 3° amine, theincreasing +I effect of the substituted alkyl groups increases the electron pair availability on nitrogen,causing an increase in basicity :

NH3 R NH2 NHRR N

R

RR

Example 43The basicity order of NH3 and the amines in water follows the order:

NH3 R — NH2 NH—R—R — N

R

RR

This means that basicity increases up to the 2º amine but then decreases in the 3º amine! Explain.

Solution: Obviously, apart from the +I effect of the alkyl groups, there’s another factor at work now, which mustbe due to the solvent, water. This factor is that water due to its polar nature, has the ability of ionsolvation. The amine, after the uptake of a proton, can become solvated by water molecules throughhydrogen bonding and can thus get stabilised.

If you observe the structures of the amines carefully, you will realise that after protonation, a 1° amineis capable of forming 3 H-bonds, a 2° amine can form 2 H-bonds while a 3° amine can form only 1 H-bond :

H |R — N — H OH | H

2

H O2

OH2

H |R — N — H OH | R

2

OH2 H |R — N — R | R

OH2

Decreasing stabilisation due to solvationby water molecules

Thus, while progressing along the series

NH RNH R NH R N3 2 2 3→ → →



the inductive effect increases the basicity but lesser stabilisation of the cation formed after protonationby the water molecules through solvation decreases the basicity. The net result is that the effect ofintroducing alkyl groups becomes progressively smaller and from 2° → 3° amine, a reversal takesplace because the solvation factor becomes stronger than the inductive factor.This explains the observed basicity order. As an example, consider the following pKa values whichclearly confirm this trend :

NH3 CH NH3 2 NHCH3

CH3

N

CH3

CH3

CH3

9.25 10.64 10.77 9.80

Example 44

Aniline, 2 ( 4.62)aPhNH pK =$$ is a very weak base compared to ammonia, 3 ( 9.25)aNH pK =$$ . Explain.

Solution: We must look into the factors affecting the unshared electron pair availability on the nitrogen atom.Obviously, for aniline to be a weaker base, there must be a factor (s) at work which is reducing thee– - pairavailability on nitrogen. If we analyse the structure of aniline, we’ll be able to identify two such factors:(a) The unshared electron pair on nitrogen can interact with the π-molecular orbital system on the

benzene ring, and get delocalised over the benzene ring to an extent, reducing its availability forprotonation :

Thus, it would be energetically unfavorable for aniline to gain a proton since then the lone pair willno longer be available for delocalisation

(b) Nitrogen atom is attached to an sp2 - hybridised carbon atom, which also plays a role (howsignificant is difficult to say) in making the unshared pair lesser available for protonation.

Example 45Nitro-substituted anilines are extremely weak bases. This should be evident from the following pKa vallues

NH2

4.62

AnilineNH2

–0.28

o-nitroaniline

NO2

NH2

m-nitroanilineNH2

p-nitroaniline

NO2

2.45NO2

0.98

Explain.



Solution: As discussed earlier, the NO2 group will have two effects :

(a) A – I effect which decreases with distance as o– → m– → p–.

(b) A – M effect which operates if NO2 is at the o– or the p– positions.

The point (b) explains why o– and p–nitro aniline are weaker bases than m-nitro aniline. For the sakeof emphasis, let us draw the canonical forms for p-nitro aniline and see the –M effect of the NO2group:

N+–O O

N+–O O

N+–O O–

NH2

N+–O O

NH2

- -

+ NH2

NH2

++

Forms (A) – (B) should make it clear to you how NO2 exerts a – M effect being at the p–position.

The point (a) partially explains why o-nitroaniline is a much weaker base than p-nitroaniline. Anotherreason is the stabilising effect of the intra-molecular H-bonding possible between the NO2 and NH2groups when NO2 is at the o– position.

Example - 46

Amides, of the general structural form ||O

R — C — NH2 , are very weak bases. Justify.

Solution: One possible reason for this observation is that the nitrogen atom is bonded to a group with an overallelectron-withdrawing inductive effect through the sp2 - hybridised carboxyl carbon :

||O

R — C NH2

-

N-

I effect of the group will reduce the unsharede -pair availability on the atom–

||O

R — C —

A more significant reason is that the ||O

R — C — group can give rise to an electron-withdrawing mesomericeffect (–M effect ) :

||O

R — C — NH2

|O–

R — C — NH2— +

Thus, amides are weak bases for these two reasons. As an example, consider the pKa of acetamide :CH — C — NH3 2

||O

acetamide

: = 0.5pKa



Example - 47

Guanidine, with the structure: NH

| C

2

NH2HN is an extremely strong organic base, with a pKa of 13.6. Explain.

Solution: Observe that the guanidine molecule exhibits a certain sort of symmetry. This symmetry is more evidentwhen we consider the protonated form of guanidine. For the protonated cation, we find that threeidentical canonical forms (of equal energy content) can be drawn :

+

NH|

C

2

NH2H N2

NH||

C

2

NH2H N2

+ NH|

C

2

NH2H N2

+

The outcome of the fact that three identical canonical forms can be drawn is that this cation is extremelystable, with the positive charge being delocalised over the entire molecule. Thus, it is energeticallyfavorable for the guanidine molecule to accept a proton, which justifies its strong basicity.

Observe that the non-protonated guanidine molecule will also show resonance but not as effective asthat possible in the protonated cation.

Example 48

Compare to aniline, the pKa values of the OH– substituted anilines are found to be as follows:

NH2

OH

OHOH

:

4.62 4.72 4.17 5.30(An) (oAn) (mAn) (pAn)

NH2 NH2 NH2

: : :

Explain.Solution: Let us first discuss what effect(s) the substitutent OH can have :

(a) a base strengthening +M effect from the ortho and para positions

(b) a base weakening –I effect which falls off with distance as o– → m– → p–.

In mAn, there can be no base strengthening +M effect. However, the –I effect operates and thisexplains why mAn is a weaker base than An.

Both oAn and pAn are stronger bases than An due to the +M effect possible in these two molecules.However, between oAn and pAn, oAn is a weaker base because of

(a) powerful –I effect of OH which is a short distance away from NH2 in oAn.(b) other steric and polar effects possible in oAn due to direct interaction between NH2 and OH

group.



Q. 1 Explain the following observed pKa values :

CH CH COOH3 2— —4.88

CH CH COOH2 ———

4.25

CH C COOH————1.84

Q. 2 2, 4, 6-trinitrophenol (below) is a very strong acid. Explain the possible reason(s).

O N2

OHNO2

NO2

pKa = 1.02

Q. 3 Explain the following observed pKa values:

COOH

CH3 NO2 Cl OH

COOH COOH COOH COOH COOH

4.344.87 4.20 3.43 3.99 4.58Q. 4 The following compound is found to be non-basic even though it contains a nitrogen atom

N —CF3

CF3

CF3Explain.

Q. 5 Ethanamidine (drawn below) is an extremely basic compound, with pKa = 12.4. What is (are) the possiblereason(s)?

CNH2HN

CH3

Q. 6 Consider the following two compounds:

NO2

(A)NO2

(B)

NO2

N HH NCH3 CH3

O N2 NO2O N2

It is observed that (B) is about 40000 times more basic than (A). What could be the possible reason(s)?

TRY YOURSELF - III



(i) Polarity of bond: We know that a covalent bond is formed by sharing of electrons between two atoms.The electrons are shared equally when the two atoms forming the covalent bond have, approximately, thesame tendency to attract electrons. However, when one of the atoms, forming the covalent bond, is moreelectronegative than the other, the electron cloud shifts towards the atom with higher attracting power.This results in increase in electron density around one atom and a decrease in electron density around theother making the bond negative at the electronegative end and positive at the other. Or we can say, theredevelops a positive and a negative pole. In other words, the molecule constitutes a dipole due to separationof two opposite poles in space.It is symbolyzed as:

+veend

–veend

Such bonds are said to be polar and they possess polarity. Polarity is indicated using symbols δ + and δ −

which imply partial (or slight) positive and negative charge.

H F—δ+ δ—

H

Clδ—

δ+

Remember, the greater the difference in electronegativities of the atoms forming covalent bonds, the morepolar the bond will be.The polarity of bond affects various physical and chemical properties of the molecule like solubility, meltingpoint, boiling point, the type of reaction that it undergoes etc.Dipole moment is a physical property which can be measured experimentally. It may be defined as theproduct of magnitude of charge (in electrostatic units, esu) and the distance between the charges (in cm).

Dipole moment = charge × distancee dµ = ×

Thus, its unit will be esu cm.Since the charge of most molecules is of the order 10–10 esu and separation between them is approximately10–8 cm, (µ = 10–10 esu × 10–8 cm = 10–18 esu cm) for convenience 10–18 esu cm is taken to be one debye,D.So far we have only discussed diatomic molecules. We concluded that any diatomic molecule in whichtwo atoms have different electronegativities will be polar and will necessarily have a dipole moment.

What if a molecule consists of more than one polar bond? Will it necessarily be polar? We will see that itmay or may not be polar. Lets see how.

(ii) Polarity of molecule: As an example, we will discuss the polarity of the carbon dioxide molecule. Weknow carbon and oxygen have appreciable difference in electronegativity values. Thus, carbon-oxygenbond will be polar with slight positive charge on carbon and slight negative charge on oxygen. Hence, weshould expect CO2, having two polar bonds, to be polar.But actually, CO2 has zero dipole moment. If we look at the structure of CO2, we see that it is linear(carbon being sp hybridized). The two carbon-oxygen bonds are definitely polar but the two dipoles actin opposite direction.

O C O— — δ—δ+δ—

Section - 11 DIPOLE MOMENT



Since the individual dipoles are equal in magnitude, they cancel each other and result in no net dipolemoment.Let us take another molecule CCl4. Obviously, the C—Cl bond will be polar due to large difference inelectroegativities of carbon and chlorine. But just like CO2, CCl4 also doesn’t possess any dipole momentand is non polar.This is because CCl4 is tetrahedral and if we consider the direction of individual bond moments, we willsee that they cancel each other.

C

Cl

Clδ—

δ+

δ—

Clδ—

Clδ—

C

Cl

Cl

ClCl

If on the other hand, we consider another molecule CH3Cl, we will find that it possesses considerableamount of dipole moment. This is because it has three carbon–hydrogen bonds which have negligibleindividual dipole moments (C and H have nearly the same electronegativities). However, the carbon-chlorine bond is highly polar and accounts for the dipole moment of CHCl3

C

Clδ—

δ+

HH

H

Example 49Predict whether the given molecules will possess net dipole moment or mot:

3 4 2 2 3( ) , , ,C CH SO H O NH

Solution: (i) 3 4( )C CHThis case is similar to CCl4. Being a symmetric molecule, it doesn’t possess any overall dipolemoment.

CH3

CCH3CH3 CH3

(ii) SO2According to VSEPR theory, the shape of SO2 molecule is angular (Unlike CO2 which has linearas geometry).

δ—

S δ+

O O δ+

Resultant

Thus, it possesses dipole moment.



(iii) H2OWater also has angular shape with two lone pair of electrons on oxygen. Lone pairs contributemore to the dipole moment because they are not bonded to any atom.

O

HH

Resultant

(iv) NH3Like H2O, NH3 also possesses dipole moment. The shape of NH3 is pyramidal with one lone pairon nitrogen.

ResultantN

H

H

H

Example 50Predict whether the following molecules possess net dipole moment or not. Also give the direction of net dipolemoment for the molecule.(i) cis CHCl CHCl= (ii) trane CHCl CHCl= (iii) 2 2CH CBr= (iv) 2 2CBr CBr=Solution: (i) cis CHCl CHCl=

Resultant

C C —

Cl Cl

δ+δ—

δ— δ—

H H

(ii) trans CHCl CHCl= Resultant

C C —

Cl

Cl

δ+δ—

δ—

δ—

H

H Zero

(iii) C C — δ+δ—

δ—

Br

Resultant

H

Br

δ—

Br

(iv) C C — δ+δ—

δ—

ResultantZero

Br

δ—

Br

δ—

Br

δ—

Br



1. Indicate the direction of dipole moment if any.

(i) diethyl ether (v)

CH3

(ix)

CH3

NO2

(ii) N(CH3)3 (vi)

NO2

(x)

OH

(iii) ICl (vii)

CH3

CH3

(iv) Methanol (viii)

CH3

CH3

TRY YOURSELF - IV



Q. 1 Given that benzene contains an annular electron cloud above and below the plane of the ring. what type ofreagents are expected to attack benzene most readily?

Q. 2 What are the important criteria for delocalisation of electrons through orbital overlap to take place ?

Q. 3 Explain how conjugation takes place in the following compound (draw the possible canonical forms):

——CH3

CH— CH

CH— ——O

Q. 4 Classify the following as electrophiles or nucleophiles :

(i) 3O (ii) RC C−≡ (iii) H − (iv) H +

(A starred atom means electrophilicity / nucleophilicity is being talked about for that particular atom )

Q. 5 What do you expect should happen when the following compound is treated with an acid ?

CH3

—

CH C C CH3 3— — —

CH3

—

— —

OHOH

H +

Think as freely as you can, and remember that one of the most important factors driving a reaction is thestability of its end products.

Q. 6 Explain the following observed pKa values :

CH CH CH COOH3 2 2— — —4.82

—

CH CH CH COOH3 2— — —

Cl

2.84

—

CH CH CH COOH2 2— — —2

Cl4.52

—

CH CH CH COOH3 — — —2

Cl4.06

Q. 7 Which of the following compounds will be more acidic?

or

OH OH

CH3

EXERCISE



Q. 8 Explain why o-hydroxybenzoic acid is found to be much more acidic than p-hydroxybenzoic acid.

p-hydroxy benzoicacidOH

COOHCOOHOH

o-hydroxy benzoicacid


3 6 5 21.233.77 2.834.76 4.17— — —H COOH CH COOH C H COOH HOOC COOH HOOC CH COOH

Q. 10 The basicity of butyl amines in chlorobenzene is found to follow the following order:

BuNH Bu NH Bu N2 2 3 < < while in water, it follows the following order:

Bu N BuNH Bu NH < < 23 2

ExplainQ. 11 Consider the following compound (which is known as phthalimide):

O

O

NH

Do you expect this compound to be acidic or basic?

Q. 12 The compound diphenylamine, Ph2NH, is an extremely weak base, with pKa = 0.8. Explain the possiblereason.

Q. 13 Do you expect the following compound to be extremely basic

NO2

NO2

NH2

O N2


OCH3

4.62

NH2 NH2 NH2 NH2

4.49 4.20 5.29OCH3

OCH3



[ ANSWER ]

TRY YOURSELF - I1. (a) Hint: C3 — C4 bond has a partial double bond character

(b) Hint: delocalisation of electrons would require the four p atomic orbitals to be essentially parallel, thus,restricting rotation about the C3 — C4 single bond.

2. Benzene being flat has an aromatic character due to cyclic overlap of p orbitals and fulfillment of the4n + 2 π electrons as per the Hückel’s rule. Aromaticity imparts extra stability than the hypothetical‘cyclohexatriene’.

3. The conjugate base PhO– of PhOH is resonance stabilised while it is not so in case of an aliphatic alcoholof the general form ROH.

4. Resonance hybrids are as follows:

(i) CH — CH — CH2 2 2

δ+ δ+[ ]⊕` (ii) CH — CH — CH2 2 2 — CH2[ ]

δ+ δ+

(iii) δ+δ+

⊕

(iv) CH CH CH CH NH3 2— — — —δ+ δ+

5 (i) CCH2 CH3

O

because negative charge on highly electronegative atom makes the structure more

stable.

(ii) CH OH

O

because charge separation decreases stability. Also, more covalent bonds a structure

has, more stable it is

(iii) 3 —CH CH O H+= because all atoms have a complete octet of electrons.

6. Cyanic acid and isocyanic acid differ in position of hydrogen atomLoss of proton from.

Cyanic acid yield. –

—O C N≡and from isocyanic acid yields:

–—N C O=

They are resonating structures.

O C N — — O C N— —

7. (I) Methylamine CH3 — NH2 (II) Formamide C O—H

NH2

Carbon in (I) is sp3 hybridized and in (II) is sp2 hybridized. Thus, carbon in (II) has more s-character thanin (I) and hence forms a shorter and a stronger bond with nitrogen.



TRY YOURSELF - II

1. Hint: due to greater ease of separation of charge and stabilisation of the resultant ion pairs throughsolvation.

2. CH — CH — CH2 2

⊕ is more stable

3. H C Mgx3

TRY YOURSELF - III

1. Higher is the degree of s-character in the hybridization of the α-carbon atom, more closely electrons areheld to the carbon nucleus. Hence, sp hybridised α-carbon is less electron-donating (thus, correspondingcarboxylic acid is more acidic) than sp2 and similar order of decreasing electron-donation follows fromsp2 to sp3 hybridized α -carbon atom.

2. Due to the following reasons:• –M effect of nitro substituents in the ortho and para positions• –I effect of all the three nitro groups (though it falls off rapidly with distance on moving from

− → − → −o m p ).• the pattern of negative charge increases stability of the anion via solvation.

3. Use the following hints to explain the pKa values: • phenyl group has lower +I effect (than a saturated C-atom) due to the sp2 hybridized C-atom to which the

carboxyl group is attached. • +I substituents (e.g. alkyl groups) decrease acidity • – M groups at o– and p– position (e.g. – NO2) increase acidity • Groups with –I but +M effect when located at o– and p–positions (here, –OH, – Cl) have an overwhelming

electron-donating mesomeric effect, thus reducing the acidity.4. Hint: presence of three powerful electron-withdrawing CF3 groups.

5. Delocalisation stabilises both the neutral (given) molecule as well as the cation resulting from its protonation.Further unlike the given neutral molecule, the cation has two exactly equivalent canonical forms, thus,making protonation energetically profitable.

6. The –NMe2 group sterically interferes with the two very large –NO2 groups at o-position, thus, preventingthe p orbitals of the N-atom from being parallel to the p orbitlas of the ring-carbon atoms, Hence +Meffect of – NMe2 is inhibited. Also, –M effect of –NO2 groups does not take place and their base-weakening influence is restricted to their inductive effects. Thus, N-atom in – NMe2 is more electron-richthan that in – NH2.



TRY YOURSELF - IV1. (i) diethyl ether

O

C H2 5 C H2 5

Resultant

(ii) N(CH3)3

N

CH3

Resultant

CH3

CH3

(iii) I Cl—δ+ δ— Resultant

(iv) Methanol

CH3

Resultant

O

H

(v)

CH3

Resultant

(vi)

NO2

Resultant

(vii)

CH3

CH3

Resultant

Zero

(viii)

CH3

CH3

Resultant

(ix)

CH3

NO2

Resultant(x)

OH

Resultant



EXERCISES[ ANSWER ]

1. Electron-seeking or electrophilic reagents.

2. For delocalisation, all participating atomic orbitals should essentially be parallel by virtue of planarity of themolecule. This planarity may get effected due to steric factors. Further, a large increase or decrease incompound’s energy content and/or charge separation discourages delocalisation.

3.CH3

CHCH

CHO CH3

CHCH

CHO

⊕

CH3

CHCH

CHδ+

Oδ+

4. Electrophiles : O3, H⊕

Nucleophiles : RC C , H–

5. The product (mechanistic details would be learnt later under pinacol-pinacolone rearrangement) is asshown below:

C — C — CH3

O

(CH )33

6. Farther located is the halogen atom from the carboxyl group, its –I effect dies down more rapidly along asaturated chain. Hence, the negative charge in carboxylate anion becomes progressively spread.

7. C6H5OH is more acidic.8. Formation of intramolecular hydrogen bonding stabilises the anion from o-hydroxybenzoic acid by

delocalising its charge. This is not possible in p-hydroxybenzoic acid.

9. Ponder upon the following hints:• the –I effect of carboxylic group as a substituent falls off rapidly with saturated C-atom(s) separating thetwo –COOH groups• phenyl group has lower electron-donating nature than saturated α -carbon as in CH3— group.

10. In solvents such as water where stabilisation by solvation via the hydrogen-bonding can occur, introductionof a second alkyl group increases the basic strength. However, the introduction of a third alkyl groupdecreases the basic strength as it is also determined by the extent to which the cation, formed by uptakeof a proton, can be solvated and hence, stabilised.In a solvent such as chlorobenzene where there is no scope for hydrogen-bonding, the higher the numberof alkyl groups, the more is the basicity of the butylamines as here, basicity is determined exclusively byelectron availability on the nitrogen atom.

11. Compound is acidic as loss of proton would lead to the following resonance stabilised anion.

N δ–

δ–O

δ–O

–



12. Reasons are as follows:• N-atom is bonded to two sp2 hybridised C-atom• Unshared electron pair on nitrogen get protonated, the resultant cation does not have unshared electronpair for delocalisation. Thus, the neutral molecule is stabilised with respect to the cation (or protonation ofthe neutral molecule is energetically unfavorable).

13. The given compound has very low basicity due to effective +M effect of –NH2 group (hardly any stericrepulsion as –NH2 is a sufficiently small group) and hydrogen bonding between the oxygen atoms of the o-NO2 groups and the hydrogen atoms of the NH2 group (this hydrogen-bonding assists in holding thegroups in the required planar orientation).

14. Use the following hints:• – OMe group has +M effect (at o- and p- position) and –I effect.• o-substituent has both steric and polar effects apart from the usual +M and –I effect

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