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General Organic Chem



sp3

hybridizationGeometry is tetrahedron.

All the valencies of Carbon are satised byσ

-bonds

Carbon.otation about C C is free Conformational !somers.

Sa"-#orsepro$ections



%clipsed Stagge

red

&e"man'ro$ections

3 (cal )torsionalenergy*



'otential %nergy Changesduring rotation

' o t e n t i a l

%

n e r g y

otati

3(cal



Me

Me

HH

H

H

Me

Me

H

HH

H

%clipsed

Me

Me

H

H

H

H

Gauche

M

e

Me

H

HH

H

'artially%clipsed

Me

H

H

MeStag

red



'otential %nergy Changesduring rotation

' o t e n t i a l

%

n e r g y

otati

H

Me

H

H

Me

H

3.+

(cal

+.+-,.(cal



i/erent arrangements of atoms that can be converone another by rotation about single bonds arconformations.

Conformers cannot be isolated

Al"ays inter-converted. 'ositions not frozen.

sp3 carbon is least electronegative of all other hyatoms.

sp0

hybridizationGeometry is 1riangular 'lanar.

All the valencies of Carbon are not satised by2nsaturated Carbon.

otation about CC or C-C )in a ring* is restricted isomers are possible.



C=C

Me

H

Me

H

C=C

H

Me

Me

H

H

Me

H

Me

H

Me

Me

H



i/erent arrangements of atoms in space arisinghindered rotation about C-C or CC are called geoisomers.

sp0 carbon has higher electronegativity than sp3 carbo

sp

hybridizationGeometry is 4inear.

All the valencies of Carbon are not satised by2nsaturated Carbon.

#as highest electronegativity among carbon atomhybridizations.



!nductive%/ect

'olarization ofσ

- bonded electrons to"ards the atoelectronegativity.

1he standard of reference is # atom.

'ushing electrons a"ay is 5I e/ect "hile pulling electis 6I e/ect.

4ets chec( the inductive e/ect of C#3

H X+

δ

C XH

H

H

δ

-

δ

∴

C#3 has 5I

e/ect



4ets replace one # of C#3 "ith another C#3 and then c

C XH

H

H

δ-

δ

CH2CH3

X

δ

!ncreasing Cchain increases5I e/ect butthe rate ofincreasedecreases.4ets replace one # of C#3 "ith another C#3 and then

C#0 "ith a C#3.

CH2 CH2

X

δ

CH3

CH Xδ

CH3

CH3

I e/ectdecreases "ithdistance.



4ets replace the # "ith a sp0 carbon.

H X+ δ-

H2C=CH

∴

sp0 carbon has 6I e/ect. So does

sp carbon. !nductive e/ect is a particle e/ect. 1his meanmanifestation of the particle nature of electrons.

!nductive e/ect dissipates"ith distance.

!nductive e/ect is apermanent e/ect.

!nductive e/ect is a "ea(e/ect.

Order of decreasinginductive e/ects7

!nductive e/ect is an8al"ays9 e/ect.

5I groups7 O- : CO-0 : C3 : C#0 : C#0 :

-I groups7 &35 : S0

5 : &#35 : &O0 : SO0 : C& :

: Cl : <r : I : OAr : COO : O : CO :



esonance %/ect

elocalization of π electrons.

!t arises out of the "ave nature of electrons.

Since "ave nature cannot be dra"n= "e represent can"hich e>hibit particle nature and "hose hybridstructure can be assumed to be.

1he actual structure of the molecule?ion is the "eighthe canonical or parent structures. <ut the energy of is lo"er than all the canonical forms.

@hile dra"ing the structures of canonical forms posit

are to remain >ed "hile that of electrons changes.olecules sho"ing resonance must be con$ugated and

appropriately oriented.esonance e/ect is also called esomeric %/ect.

esonance e/ect is a permanent e/ect.esonance e/ect is a strong e/ect= generally st

!nductive %/ect.esonance is a relative e/ect. !t is a need based e/ect



<asic movement of electrons7

A=

<

C

5A

<-

C

A=< C

5A <=C-

A 7 A57

A 7 5A=

A -A=

A -A

A <7 5A=<-



. ouble bond con$ugated "ith 5ve charge7

5C

C=

C

#0 C#=C# C#=C#0C#0=C# C# C#=C#0

5

C#0=C# C#=C# C#0

5

! !! !!!

Stability order of resonating structures having 5 cha

. Of all such resonating structures the one in "hich octet complete is most stable and therefore most contributing.

0. &e>t is that resonating structure in "hich 5ve charge electropositive element. And then in that order.

3. esonating structures "hich are similar have eBual stabili

+. Among resonating structures in "hich charge is on simila

structures are di/erent= the stability is decided by inductive e

! !!! :!!



0. ouble bond con$ugated "ith double bond7

C=C C=C

CH2=CH CH=CH CH=CH2CH2=CH CH=CH CH+ CH2

- CH2=CH CH+

C

+CH2 CH=CH CH=CH CH2-

!! !!!

!

!

Stability order of resonating structures having net cha

. Of all such resonating structures the one in "hich nocharge is most stable.

0. &e>t is that resonating structure in "hich charge separaand then in that order.

! : !! : !!! :!

%vidences of resonance7

. All bonds are bet"een a single and a double bond.

0. On treating the molecule "ith <r0 in CCl+ "e get + produc

and =,.



C#0=C# C#=C# C#

C#0 <r <r

C#0=C# C# C#=C#

C#0<r <r=0

product

=+productC#0=C# C# C# C#

=C#0<r <r

3=+product

C#0 C#=C#

C#0<r=,

prouct

=, product : =0 product : =+ produc: 3=+ product

=, product is most stable because the double

con$ugated and non-terminal. =0 is ne>t since doubcon$ugated but one of them is terminal. =+ is ne>t adouble bonds is non-terminal= "hile 3=+ is the leastbonds are terminal.



3. ouble bond con$ugated "ith 6ve charge7-

C C=

C

O C#=C# C#=C#0!

O=C# C# C#=C#0

-

!

!

O=C# C#=

!!!

Stability order of resonating structures having - char

. Of all such resonating structures the one in "hich the 6the most electronegative atom is most stable and then in that


3. Among resonating structures in "hich charge is on similastructures are di/erent= the stability is decided by inductive e

! : !! :!!!



+. ouble bond con$ugated "ith lone pair7

:C C=C

#0& C#=C# C#=

C#0

:

!

#0&=C# C# C#=

C#0

-

!!

5

#0&=C# C#

C#0

5

!!!

Stability order of resonating structures having net cha

. Of all such resonating structures the one in "hich nocharge is most stable.

0. &e>t is that resonating structure in "hich charge separaand then in that order.

! : !! :!!!



E. ouble bond con$ugated "ith unpaired electron7

C C=C

C#0

C#=

C#

C

#=O! C#

0

=

C#

C#

C#

=O!! C#

0

=

C#

C#

O!!!

Stability order of resonating structures having nohaving unpaired electron7. Of all such resonating structures the one in "hich the un

is on the least electronegative atom is most stable and the in


3. Among resonating structures in "hich the unpaired electratoms but the structures are di/erent= the stability is decide/ects.

! : !! :!!!



,. 4one pair con$ugated "ith 5ve charge7

7&

C5

#0& C#=C# C

#0

5

7

!

#0& C# C#=C

#0

7

5

!!

#0&=C# C#=C

#0

5

!!!

Stability order of resonating structures having 5 cha

. Of all such resonating structures the one in "hich octetcomplete is most stable and therefore most contributing.

0. &e>t is that resonating structure in "hich 5ve charge electropositive element. And then in that order.


+. Among resonating structures in "hich charge is on similastructures are di/erent= the stability is decided by inductive e

!!! : ! :!!

i $ d i h i d l



F. 4one pair con$ugated "ith unpaired electron7

:&

C

7&#0

C#0

&#0 C

#0

+

! !!

Stability order of resonating structures having nohaving unpaired electron7. Of all such resonating structures the one in "hich the un

is on the least electronegative atom is most stable and the in


3. Among resonating structures in "hich the unpaired electatoms but the structures are di/erent= the stability is decide/ects.

! : !!

1he con$ugation need not be necessarily via a singlebe via a double bond as "ell.

C#0CCC#0

@herever "e have studied that a double bond is con

something= the double bond could also be a triple bon



Aromaticity

olecules that satisfy the follo"ing conditions arearomatic. Aromatic molecules have e>tra stability non-aromatic or anti-aromatic molecules.

. 1he molecule should be cyclic.

0. !t must have )+n50* π electrons in the periphery of the mo

3. 1he molecule must sho" cyclic delocalization.

+. All the atoms must be sp0 hybridized in at least structure.

!n order to chec( "hether a molecule is aromatic or nrst learn ho" to ma(e a molecule aromatic.

1o do this "e "ould rst ma(e an empty ring structure

4ets start "ith the smallest ring structure= a 3 membe

@ 9ll t dd l t t h f th



@e9ll rst add one π electron to each of theperipheral atoms.

@e "ould add ? remove electrons if reBuired inorder to ma(e the π electrons eBual to )+n50*. !nthe present case the closest value of )+n50* to 3 is0. So "e remove one electron from any of the atom. !f "e remove then "e put a 5ve charge= if "e adput a 6ve charge.

&o" "e $oin the otherπ

electrons byπ

bonds.

1he structure made this "ay "ill be aromatic. conditions are any"ay met. 1he last condition is alchec( the 3rd condition by dra"ing all the resonating the compound.1o dra" the resonating structures in a cyclic molecchoose a direction of the arro" )cloc("ise or anticlomaintain this direction for moving electrons in astructures. &e>t "e choose "hether "e put the arroutside the ring.

+

+ +

1his ion

1roponium

@ th t ll b d h b d bl



@e can see that all bonds have become double onatom has had the positive charge once. 1herefmolecule sho"s cyclic delocalization.1he closest value of )+n50* to + is either

"e have already seen the removal of eadd electrons this time.

@e add or remove from ad$acent atomsneed to $oin the leftover π electrons.

4ets dra" all the resonating structures and then chec(



<enzene

&aphthalene

Anthracne

'henanthrene

+

+

+

'y

ne

& l t d t h t h th i



&o" let us nd out ho" to assess "hether a given moion is aromatic or not.

Count the no. of atoms that ma(e up the ring aanother ring of the same no. using only carbon atoms.

&o" ma(e this ring aromatic as done before.

!f all that is present in your structure is present structure then it is aromatic other"ise not. emembcharge is eBuivalent to a lone pair.

N

H

'yrrole

N

N

H !midazole

N

'yridine

N

N

N

N

'yrimid

NN

'yridazi

+

'yrazi



#ypercon$ugation

C CH=CH

H

H

H

H

C CH=CH

CH3

CH3

CH3

H

<

esonance involving single bonds.

i/erent as compared to resonance as it invo)generally C # bond*.

Also (no"n as no bond resonance.

#ypercon$ugation is a permanent e/ect.@ea(er than esonance but stronger than !nductive %

Single bond con$ gated ith do ble bond



. Single bond con$ugated "ith double bond7

C C#=C#

#

#

#

#

↔

C=C# C#

#

#

#

#

5

↔

C=C# C#

#

#

#

#

+

↔

C=

#

#

#+



C CHH

Hδ+

δ+

δ

+

H

CH2

δ

0 Single bond con$ugated "ith 5ve charge7



0. Single bond con$ugated "ith 5ve charge7

C

C#

0

#

#

#5

↔

C=

C#

0

#

#

#

5

↔

C=

C#

0

#

#

#5

↔

C=C

#

#

#5

3. Single bond con$ugated "ith unpaired electron7

C

C#

0

#

#

#

↔

C=

C#

0

#

#

#

↔

C=

C#

0

#

#

#

↔

C=

C

0

#

#

#

+. Single bond con$ugated "ith -ve charge

C C#

0Cl

C

l

Cl-↔

C=C#

0Cl

C

l

Cl

-

↔

C=C#

0Cl

C

l

Cl

C=

C

C

l

C

↔

%l t i



%lectromeric%/ect

A sudden e/ect )temporary* due to the presenmolecules or groups in the same molecule.

Causes permanent change in the nature of the molecu

esponsible for reactions bet"een a neutral moleculeion or bet"een t"o neutral molecules.

1he movement of electrons due to the presencemolecule or ion leading to the reaction is electromeric

C#0

%

<

-

5

5

! t di t 1 iti



!ntermediates 1ransitionStates

Substances formed during a reaction, which are formed in sare consumed in the subsequent steps are called intermedia

Intermediates are real. Their presence can be detected.

Transition states are imaginary species. In each step an imagbe imagined between the reactant of that step & the produ Transition states may resemble the reactant or product other.

Eery step has a transition state.

If there are n steps in a reaction then there would be n trans!n"#$ intermediates.

%et us tae the S'# reaction of !($)**l with +H".



!($)*−*l

+H−

()*−+H

*l−

C

C

Cl

!($ * !($ * *l"Step-



!($)*−

*l

!($)* *l"

C

Cl

Step-#

C

Cl

5

δ

55

(eactant

Intermediate

TransitionState

!($ * ( *+HStep-



!($)*

+H−

()*+HStep-

C

5

C

5 5δ

Intermediate

/roduct

Transition

ti



#

eactive!ntermediates

*arbon with a positie

charge.

#. *arboniumIon

+ctet incomplete. Short by twoelectrons.

*H)+H

H5

*H)−+H

5

*H)5 H+

s p

sp

#

C

Triangular /lanar with an empty p"orbital perpendicular to th

0ree



*arbon without anycharge.

. 0ree(adical

+ctet incomplete. Short by one

electron.*H1

*l

*H) H*l

s p

sp

Triangular /lanar with a p"orbital containing one electron pthe plane of the radical.

#

#

C

) *arbanion



*arbon with a negatiecharge.

). *arbanion

+ctet

complete.*H)'+

+H−

+' *H−

H+s p

sp3

Tetrahedral geometry. Shape is/yramidal.

#C

#

2ny pyramidal molecule has a tendency to 3ip its lone pbellow the plane of the *arbon atom. This is due to t

frequency of the lone pair dipole ector. This e4ect is

1



Cl

Cl

*arbon with no charge.

1.*arbene

+ctet incomplete. Short by two

electron.**l) +H **l)

H+H **l

*l

**l

*arbenes are of two types. !i$ Singlet and!ii$ Triplet. !i$ Singlet-

sp

C

Molecular shape is 8ent. +ne loneplane of the three atoms and one em

perpendicular to this plane.

!i$



!i$ Triplet-

sp

CCl

y

Molecular shape is %inear. Twop"orbitals haing one electroneach are orientedperpendicular to the moleculara9is and to each other.

In Triplet carbene the electrons are in separateorbitals as compared to the same orbital inSinglet. This reduces the electron"electronrepulsion in Triplet maing it more stable

compared to the Singlet.

*arbenes show insertion reaction in alenes and alynes



*arbenes show insertion reaction in alenes and alynes.

The stereochemistry of addition is di4erent for the two carben

If the alene is cis, then a singlet adds to yield a cis product.

This is because the reaction occurs in one step, where both

electrons moe simultaneously and form the two bonds with 8y the time the π bond of the alene breas completely the r

formed ensuring the two groups on the alene are restricted t

##

C

C C

CC

I T i l t b b th th l t f th b f



In a Triplet carbene both the electrons of the carbene are ofspin !up spin$.

8ut the bond electrons of the alene are of opposite spins !oone down spin$.

So one bond, between the electrons of up spin of the carbene

spin of the alene forms immediately. 8ut the other bond cthe spin of the carbene !or alene$ changes to down spin.

:uring this spin change the bond between the carbon atomis single and the groups on the carbons can rotate freely.

If this happens a cis alene can change to trans by the time carbene changes and the second bond forms.

This implies that this would be a two step process. 0irst breaing of alene bond and maing of one bond betweealene. Second step would be the spin turning of the second

carbene and then second bond formation with the alene.



##

C

Cl

Cl

CC



##

C

Cl

Cl

CC



##

C

Cl

Cl

CC



##

C

Cl

Cl

CC



##

C

Cl

Cl

CC



##

C

Cl

Cl

CC



##

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC



#

#

C

Cl

Cl

CC

2 cis alene will gie a mi9ture of cis and trans cyclic productis triplet



#

#

C

Cl

Cl

CC

is triplet.

;. 'itrene



'itrogen with no charge.

+ctet incomplete. Short by electrons.

+

H

8

( * ' 8r

+

( * '

+

( * ' 8r

'itrenes are of two types. !i$ Singlet and!ii$ Triplet.

!i$ Singlet-

sp

C

&

!i$Triplet-

y



Triplet-

sp

y

&

Application of%/ects

;



%/ectsStability of!ntermediates

0

3

;

4

;

E

0 : :3

0 is more stable than because 0 has t"o 5I e/ecte/ects "hile has only one 5I e/ect and three 5# e/

3 is more stable than 0 because 3 has three 5I e/ectse/ects "hile 0 has only t"o 5I e/ect and si> 5# e/ect

: +

+ is less stable than even though + sho"s lone pairresonance due to "hich octet of carbon gets complcompromising the octet of ;luorine. 1his is becauhalogens= their 6I e/ect dominates the 5 e/ect. 1he

Application of%/ects

;



%/ectsStability of!ntermediates

0

3

;

4

;

E

0 : :3

E is more stable than + because it has one additional three additional 5# e/ects due to a methyl group.

stable than because the net e/ect of the presence E is electron "ithdra"ing e/ect "hich destabilizes thion E compared to .

: + :E

, is less stable than + because in , only the 6I e/ect operative "hile in + a ma$or part of the 6I e/ect ocancelled by its 5 e/ect. So the net "ithdra"ing e/e

than that in ,.

CH2 CH2 CH2 CH2



OMe Cl NO

2 3 4 5

O

0:

is more stable than 0 because three 5I e/ects e/ects in out"eigh the 5 e/ect of benzene in 0. benzene loses its aromatic character in resonance iinHuencing reason.

CH2 CH2 CH2 CH2



OMe Cl NO

2 3 4 5

O

0:

3 is more stable than 0 because it has a fourth resonain "hich the octet of all atoms is complete. !t is alsothan as this could be a dominating reason.

:3 : 4

+ is less stable than 0 because its 6I e/ect is strone/ect.

: 5

E is less stable than + because of its strong 6! eunstable resonating structure in "hich positive char

CH2 CH2 CH2 CH2



OMe Cl NO

2 3 4 5

O

0:

, is more stable than + E because the resultant e/ec! "hile that in , is 5. !t is less stable than 0 becauresonating structures "hile , has only t"o.

:3 : 4 : 5 :

CH2 CH2 CH2 CH2



↔

↔

↔

OMe OMe OMe OMe

↔

NO

ON

O

ON

O

ON

O

O

CH2 CH2



0 3

NO

4 5

O

: 0 is more stable than 0 because it has less 5I e/ee/ect destabilizes carbanions.

: 3

0 is more stable than 3 because it has less 5I e/ee/ect destabilizes carbanions.

:+

+ is more stable than because it stabilizes the negat

the 6 e/ect of the phenyl group.

CH2 CH2



0 3

NO

4 5

O

: 0E is more stable than + because of the 6I e/ect and&O0 group. 1here is an additional resonating stru

"hich the negative charge resides on O atom. 1hoverall stability of carbanion E more stable.

: 3:+:5

CH2 CH2



0 3

NO

4 5

O

: 0, is less stable than E because of 6I e/ect and 6 e/eis almost eBual to + though + has more resonatinbecause carbon accommodates a positive charge bdoes a negative charge. 1his ma(es it diIcult for charge to spread to the phenyl ring brea(ing its aro

1his is "hy the benzylic carbonium ion is more stabl

: 3:+:5≈

CH2 CH2 CH2 CH2

C



↔

↔

↔

NO

ON

O

ON

O

ON

O

O

↔

O

CH2



0 3 4

0 : 0 is more stable than because 0 has t"o 5I e/ect

e/ects "hile has only one 5I e/ect and three 5# e/stability is generally similar to carbonium ion stability

:3

3 is more stable than 0 because 3 has three 5I e/ectse/ects "hile 0 has only t"o 5I e/ect and si> 5# e/ect

CH2



0 3 4

0 : #ypercon$ugation and resonance both decrease for

electron compared to a positive charge as positivegreater electron deciency. Since hypercon$ugation bond= it decreases more than resonance. 1herefore + iresonate the unpaired electron "hile the hypercon$ug

relatively poor.

:3:+

CH2 CH2 CH2 CH2



↔ ↔ ↔

C;0 CCl0 C<r CC!0



0 0 3 +

Singlet7

: 0 : 3 : +

!n singlet carbene there is an empty p-orbital on thehalogens "ould donate their lone pair to this orbitfrom ; to ! the size of the atoms become bigger com

of carbon. 1he di/erence bet"een the energy ohalogen and that of carbon increases conseBuently debond strength of the π-bond. 1herefore ; is more staCl and in that order.

1riplet7

+ : 3 : 0 :

!n triplet carbenes there is no empty orbital. So hsho" their 5 e/ect. So only 6I is operative.



J

J C

AcidicStrength

;. +H =.+

?.+

*



#.*H0)

.*H

*l)).*H)+

H1.*H+

H

<.+H

*H

)

H

*H)

>.+H

*H

)

H*

l

#@.+H

*l

#).+

#;.*++

#=. *++H #?.*++H

'+



H

'+

#1.+H

'+

H

#<.*++H

*H)

*H)

#>.*++H

*H)

'+

@. *++H

'+

) 1

+

;



8enAene Sulphonic 2cid !;$ is the strongest acid in thcompounds. This is because it is the only molecule, among twhose anion has three equialent resonating structuresnegatie charge is localised on the more electronegatie +all the gien molecules the negatie charge is either locateon +9ygen. +9ygen being more electronegatie of the twonegatie charge better than carbon.

0ormic 2cid !)$ is more acidic than 8enAoic 2cid !#;$ becau

bond of the carbo9yl group is in conBugation with the doubphenyl group. This resonance would decrease the double bothe carbo9yl group maing it less aailable for resonance wicharge in 8enAoate ion. Therefore the negatie charge dispersed in #; as it is in ).

; F #? F #



ortho"'itro 8enAoic 2cid !#?$ is second most acidic in the gibecause of three reasons. 0irstly the strong "I e4ect of C'+.

anion more stable. Secondly the "( e4ect of C'+. This crecharge on the α"carbon. Therefore there is an additional CI "I$. Thirdly the presence of '+ ortho to 6*++H causes both

go aboe and below the plane of the phenyl group respesteric repulsion. This is nown as +rtho E4ect !also called Sof (esonance"SI( E4ect$. This reduces the resonance betw

bond and the double bond of the phenyl group. Therefore thcharacter of the *D+ increases which enables the anionstabiliAed. 2ny group ortho to C*++H causes the +rtho E4ortho substituted carbo9ylic acids are more acidic than theirmeta and para isomers irrespectie of the nature oIntramolecular hydrogen bonding does not occur because

membered rin is dicult

'e9t is para"'itro 8enAoic 2cid !#$. 8ecause C'+ shows it

creates a positie charge on the α"carbon This creates

; F #? F # F #< F @ F ) F #;



'e9t is ortho"Methyl 8enAoic 2cid !#<$. This is due to +rtho E

'e9t is meta"'itro 8enAoic 2cid !@$. This is weaer than #?

e4ect does not impact the acidic strength much since it iwhich gets the positie charge. It is more acidic than the rese4ect of '+. The CI e4ect would be stronger in @ than in

than in #?.0ormic 2cid !)$ is ne9t. This is because all the other acids taromatic acids in which the double bond of the carbconBugated with a double bond of the aromatic ring which d

acidic strength.'e9t is 8enAoic 2cid !#;$ because all the others that are carbo9ylic acids hae releasing groups in them which decreastrength.meta"Methyl 8enAoic 2cid !#=$ is ne9t. This is in spite of #= he4ect than #> !methyl group being closer to *++H gromethyl present in para position hyperconBugates and the α

negatie charge This decreases the acidic strength of #>

; F #? F # F #< F @ F ) F #;F F 1 F #1



para"Methyl 8enAoic 2cid !#>$ follows. This is more acidic thin the latter the *H) group is directly attached to the *++

decreases the acidic strength of .*arbonic 2cid !1$ is less acidic than all the carbo9ylic

because in 1 there is a lone pair double bond conBdiminishes the double bond character of *D+ bond decreasing the resonance of it with the negatie charge. Eealso the double bond character of *D+ decreased due to double bond resonance, it diminishes more in 1 because lobond resonance is stronger than double bond" double boSince it is the weaest of all carbo9ylic acids, all the cadissole in H*+)

solution giing *+. This is true for acids th

than H*+).

H*l *H)*++

*l *H)*++H

SA WA

2mong all the /henols that are gien para"'itro /henol !#1$as the C( e4ect of −'+ gies rise to an additional resonan

the /heno9ide ion. ortho"'itro /henol !#$ also shows

resonance along with greater CI e4ect It is less acidic than

; F #? F # F #< F @ F ) F #;F F 1 F #1 F # F ? F #) F #



ortho"*hloro /henol !?$ is more acidic than meta"'itro /hthough C'+ is a stronger electron withdrawing group th

because *l is closer to + in ? than 6'+ is in #).

meta"'itro /henol !#)$ is more acidic than meta"chloro phsame distance '+ is more electron withdrawing than *l.

It may appear that para"*hloro /henol !##$ is more acidicisomer !#@$ due to the fact that in one of the resonating/heno9ide ion the negatie charge gets located on the carb*l atom. This would mae that structure more stable as the

be able to withdraw the negatie charge better because of c8ut the e9perimental obseration shows that #@ is more a This is e9plained by the fact that in #@ the *l atom is resonating structure in which the negatie charge is locatatom. 'ow both #@ and ## stabiliAe one structure each bother. 8ut #@ stabiliAes the more contributing structure whi

; F #? F # F #< F @ F ) F #;F F 1 F #1 F # F ? F #) F #



/henol !;$ follows ne9t because in o"Methyl /henol !<$, m"Meand p"Methyl /henol !>$ there is an electron releasing grou

decreases their acidic strength compared to ;.

; F =

m"Methyl /henol !=$ is more acidic than o"Methyl /henol !</henol !>$. This is due to the fact that in both < and >becomes negatiely charged because of hyperconBugation group. In = there is hardly any hyperconBugation since talready negatiely charged because of resonance of the /hen

F >

p"Methyl /henol !>$ is more acidic than o"Methyl /henol !<methyl group is farther away from the acidic H in > whileMethyl shows I e4ect which decreases acidic nature.

F <

o"Methyl /henol !<$ is more acidic than 8enAyl 2lcohol delocaliAes the charge of the anion in the phenyl ring by reno resonance delocaliAation is seen in 1.

; F #? F # F #< F @ F ) F #;F F 1 F #1 F # F ? F #) F #



8enAyl 2lcohol !1$ is more acidic than Methyl 2lcohol !)phenyl group shows CI e4ect in 1 while the methyl shows I

; F = F > F < F 1 F )

Methyl 2lcohol !)$ is more acidic than 0luoroform !#$ andbecause the negatie charge on the anion is located on + located on * in # and . +9ygen being more electronegatienegatie charge better.

F F

*hloroform !$ is more acidic than 0luoroform !#$. Initiallymore acidic than due to the higher electronegatiity of 0 stronger CI e4ect. Though this e4ect does e9ist, thephenomenon that dominates. *hlorine being in the third perd"orbitals which 0 does not hae. Ghen H5 is lost by the locarbon forms a π"bond with the d"orbitals of chlorine. This bonding. :ue to this the negatie charge is more delocaliA3uorine does not hae any acant orbitals the negatie charg

reside only with carbon Therefore the anion of is more s

S ++

+H

S ++

+−

S +−+

+

S + −+

+



+ +

"H

↔ ↔

)

*

+H+

- H

*

+−+

↔

*

++−

*

+H+−

H−*−+

+

H−*−+−

+-H

H−*=+

+−

*

+H+

*

+ +

*

+−+ +



*

"H+

'+

*

'

+

+

↔

*

'

+

+−

*

↔

+

*

++

'

+−

+

H



*

+H

+

*

+−+

*

+ +

*

+ +

*

++



*

'+

H5

*

'

+

+

↔

*

'

+

+

↔

*

'

+

+

↔

*

'

++

*

+H

+

*

+H

+

*

+H+

*

+H

+



*

* HH

H

↔

*

* H

H

H

↔

*

* H H

H

↔

*

*H

H

H

H+−*−+H

+

↔ H+D*−+H

+

+H + + + +



'+

"H

'

+

+

↔

'

+

+

↔

'

+

+

↔

'

+

+

'

+

+

H

+H + + + +



*l

"H

*l

↔

↔

↔

*l *l

+H * H

+H

H

* H

H5 +H



* H

H

* H

H

* H

H

H H+ +

+

H

* H

H

H

+

H

*

H



Cl

CCl

Cl

In 8*l) as well as 80) there is a bac bonding- between the lo*l$ and the empty p"orbital of 8. 8ut since 0 is closer in siAe the bond with 0 is stronger So the p orbital of 8 is more



<

80

)

8*l

)

;

;

the bond with 0 is stronger. So the p orbital of 8 is more electrons in 80). So it is weaer %ewis acid.

<asicStrength



'H

)

#

*H)'H

!*H)$ 'H

)

#FF) F This is the order solent.

F F F

)#

This is the order ifis the solent.

In water, ammonia and its deriaties would form a lecomple9 with the lewis acid along with the formation of hywith water



with water.

'

H

H

H

%2

+

H H

+

H H

+

H

H

It is the stability of this cdetermines the basic strengthwords, the comple9 that

ma9imum energy when it is fostable and therefore most basic

%et us assume that each hreleases; cal and in 'H) the coordinate

' and %2 releases 1 cal.

Therefore in 'H) the comple9 formed in water release in total

To get *H)'



'

H

H

H

%2

+

H H

+

H H

+

H

H

; cal

;cal

;cal

1cal

'

*H

)

H

H

%2

+

H H

+

H

H

; cal

;cal

##cal

)

replaced # H of

This would resua hydrogen bo

strengthen thbond between '

is a releasing *H)'H is a stro

'H) it mean

strengthening ' .%2 bond energy than thdue to a hydrog

Therefore !*H)$'H releases # cal in w

To get !*H)$'H we hae replaced # H of !*H)$'H with a *result in the loss of a hydrogen bond but would strengthen bond between ' and %2 as *H is a releasing group Since



'

*H

)

H

%2

+

H H

; cal

#= cal

bond between ' and %2 as *H) is a releasing group. Since

stronger base than !*H)$'H it means that the strengthen

bond releases more energy than the loss incurred due to a hy Therefore in !*H)$'H the comple9 fo

release in total cal. It should also b

from 'H) to *H)'H the di4erence is ofrom *H)'H to !*H)$'H the di4erenc

This signiJes that the rate of increase is one can e9pect in the ne9t case that tthe coordinate bond may not een co

loss of a hydrogen bond.

*H

)

'

*H

)

*H

)%2

@ cal*H

)

The decrease in the rate of strengtcoordinate bond could be due to the fbond has a certain limit of ma9imum stre The approach of that limit may slow doincrease for eery further addition of

chlorobenAene there is no hydrogen bon

the strength of the coordinate bo

'H 'H

H ' Me

Me ' Me

H ' /h

Me ' /h



)

#

e

)

e

1

h

;

h

<

# F In 2niline !$ the lone pair on ' resonates with the π"bo

thereby getting de"localiAed in the ring. This decreases its adecreasing its basic strength compared to 2mmonia !#$.

F

)

+n replacing a H of C'H with *H) the steric repulsion tilts

orbital of ' which maes it less parallel with the p"orbital of reduced resonance of the lone pair with the ring. *oupled witof *H) basic strength of ) is more than and 1 more than )

) and 1 are less than # su estin that the resonance e4ec

1

F

'H 'H

H ' Me

Me ' Me

H ' /h

Me ' /h



)

#

e

)

e

1

h

;

h

<

# F '"/henyl 2niline !;$ is less basic than aniline as the lone pair

resonating with /h groups consequently less aailable for %e

F )F 1 F ;

<

F

'"Methyl"'"phenyl 2niline !<$ is more basic than ; because thgroup on ' would mae its lone pair orbital to go out of planebasic than as the oerall e4ect of a −/h group and a −*H) g

electron"withdrawing.

F =

',','"Triphhenyl 2niline !=$ is least basic in the gien set as i

phenyl griups which resonate with the lone pair on '

'H 'H



O#

O#



ee

&O0

ee &O0

>

#

,1":imethyl"1"nitro phenol !#$ is more acidic than ),;":imet

phenol !$ because due to the ortho e4ect of two methyl grouout of plane and this reduces it C( e4ect. Githdrawing e4ectsincrease the acidic strength. Ghile in !$ '+ is able to show





C#3C#0COO

#

*H*H*++

H*H≡**++H

0 3

<<



# H 0 3

/ropanoic 2cid !#$ is less acidic than /ropenoic 2cid !$ whicacidic than /ropynoic 2cid !)$ because from !#$ to !)$ elect

the α"carbon is increasing.H−*−*H *−H

+ +H

*H)−*−*H−*−*H)+ +#

Me+−*−*H−*−Me

+ +H

# )

> >

The *H in the middle of the two *D+ groups is called met

The hydrogens attached to the methelene carbon are the mthe molecule as the carbanion formed can be resonanceresonating with two *D+ groups. /ropandial !#$ is more acbutandione !$ because of the I e4ect of the two C*H) group

!$ is more acidic than dimethyl malonate !)$ because in ) t+ is conBugated with the *D+. This resonance would decre

bond characeter of *D+ bond maing it less aailable for r

>

In #,)":ithiane !#$ the carbaa bac bond with the d"orbiwhile d"orbitals are not pre



S

0

S + +while d"orbitals are not pre#,)":io9ane !$. So no such will occur in .

'

H

' '

H

'

H

&#3)C

# ) 1 ;

:::::

;+3

/yrole !#$ is least basic as in # the electrons are conBugdouble bond of the ring and would be in resonance. In resonance of the lone pair which maes # aromatic /yridi



resonance of the lone pair which maes # aromatic. /yridibasic than # because the lone pair lies in the same plane awould not be able to resonate with it. 8ut the lone pairs arewhich are shorter than sp) and the lone pairs would be held

/iperidine !)$, /yrrolidine !1$ and :iethylamine !<$ are secoso they are more basic than ;. 1 is more basic than < becauthe ring the carbon chain is more tightly held bac e9posingmore thereby allowing the lewis acid to form the bond easbasic than 1 because of greater number of carbon chain cause4ect. 2mmonia !;$ is more basic than # & because both t

withdrawing e4ects. It is, howeer, less basic than the resothers hae releasing e4ects.

'H 'H &#0 &#0



'+

*'

<

# &O0

ee

C&

e

:

3 +

p"*yano aniline !$ is more basic than p"nitro aniline !#$ beca

stronger withdrawing group due to CI and C(. 8ut in ),;":imaniline !)$ due to the presence of two methyls at ortho of C'

hindrance cause C'+ to go out of the plane !+rtho e4ect$ and

( e4ect decreases. C*' is not a4ected by the steric factor asgroup due to sp hybridiAed *. So the C( e4ect of *' is till ope:imethyl"1"cyano aniline.

'!*H)$ 'H &)C#3

*0



Me Me Me Me

:

0

e e::

3

'"),;":imethyl phenyl"','"dimethyl amine !#$ is more basi:imethyl phenyl amine !$ because the two C*H

)

groups on

out of plane due to steric factor. So the lone pairs are less dthe ring. In '",<"dimethyl phenyl"','"dimethyl amine !)$ theon the ortho position cause greater steric hindrance than the

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