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Chemistry 3.5Chemistry 3.5
Describe the structure and Describe the structure and reactions of organic compounds reactions of organic compounds
containing selected organic containing selected organic groupsgroups
You will notice the first thing the 3.5 standard says is :
You will be expected to know the principles of Organic Chemistry you covered in AS 2.5
How well do you know your stuff from last year?
The standard then goes on to say….
Organic compounds described are limited to those containing no more than 8 carbon atoms
Don’t think you get off that easy
It then goes on to say ………
Larger organic molecules may be used in questions involving the application of organic principles (e.g. the identification of functional groups)
So lets start with Alkanes
1. What’s the general formula for an alkane?
CnH2n+2
Can you draw the structural formula and Can you draw the structural formula and molecular formula for the following?molecular formula for the following?
Methane
Molecular Formula -
Structural Formula
CH4
How about drawing the 3 dimensional structure for methane?
Use the molymods to make a 3 D model
Any ideas on how to draw it?
This wedge represents the bond coming out from the paper
These lines represent the bond going behind the paper
These two lines represent bonds on the same plain as the paper
Drawing a 3 dimensional Structure
Describe the shape of this molecule
Tetrahedral
Is CH4 polar or non polar?
Non polar
Can you give a reason why?
What is the bond angle between each of the atoms?
109 o
109 o
Alkane Nomenclature - Give the names, molecular and condensed structural formula for the first ten alkanes
NameName MolecularMolecular Condensed Structural FormulaCondensed Structural Formulamethanemethane CHCH44 CHCH44
CC77HH1616
CC66HH1414
CC44HH1010
CC33HH88
CC22HH66
CC55HH1212
CC88HH1818
CC99HH2020
CC1010HH2222
ethane
propane
butane
pentane
hexane
heptane
octane
nonane
decane
(CH(CH33))22
CHCH33(CH(CH22))22CHCH33
CHCH33(CH(CH22))55CHCH33
CHCH33(CH(CH22))66CHCH33
CHCH33(CH(CH22))77CHCH33
CHCH33(CH(CH22))88CHCH33
CHCH33(CH(CH22))33CHCH33
CHCH33(CH(CH22))44CHCH33
CHCH33CHCH22CHCH33
Can you remember the alkane order? A simple mnemonic is-ManyElderlyPeopleBuyPentHousesHighOverNorthDakota
MethaneEthanePropaneButanePentaneHexaneHeptaneOctaneNonaneDecane
Cyclic Alkanes
These are Alkanes with at least 1 ring of carbons eg cyclohexane
Draw the structure of cyclohexane in your book?
Cyclohexane
Draw cyclopropane
The general formula for an alkane with one ring is:
CnH2n
Do you remember how to name a branched alkane?
Steps
1. Find the longest chain of continuous carbons (called the parent chain) and name it:
ie 5 carbons – name pentane
2. Identify any side branches or functional groups
ie methyl if there are more than one of the same type use prefixes
di (2) , tri (3) etc. 2 methyls = dimethyl.
So the name so far is dimethylpentane
3. Number the parent chain from the end that gives the side branch groups the lowest number
Pentane (parent chain)
methyl
methyl
Steps continued
3. Number the parent chain from the end that gives the side branch groups the lowest number ie from the left
methyl
methyl
1 2
3 4 5
4. Separate numbers by comma’s and separate numbers from words with a dash – now you can name it
2,3-dimethylpentane
Exercise – Draw the structures for the following
a) 3 – ethylheptane
b) 2,2,4-trimethylpentane
Lets look at making some structureLets look at making some structure
Turn to Expt 1 in your bookletTurn to Expt 1 in your booklet
In pairs make and draw the structures for In pairs make and draw the structures for Q 1 to Q 3Q 1 to Q 3
Answer the questionsAnswer the questions
Structure Classification Name
alkyne pent-2-yne
CH3CH2CHCH2CH2CH2CH2CH3
OH
alcohol
(secondary)octan-3-ol
CH3CH2CH C O
CH3 OH
carboxylic
acid
2-methylbutanoic acid
C C H
H3C CH2CH3
H
alkene cis-pent-2-ene
CH3CHCH2CH3
Brhaloalkane 2-bromobutane
CH3CH2CH2CH2CHCH2CH3
CH3
alkane 3-methylheptane
CH3CH2C CCH3
Structure Classification Name
alkene 2,3-dimethylpent-2-ene
CH3CH2CH2C
CH3
OH alcohol
(tertiary)2-methylpentan-2-ol
C C
H3C CH3
ester butyl methanoate
CH3CH2CHCH2CHCH3
CH3
alkane 2,4-dimethylhexane
CH3CH2 CH3
CH3
H C O
O CH2CH2CH2CH3
CH3
Structure Classification Name
alkyne 3-methylbut-1-yneCH C CH CH3
CH3
CH3CHCH2CHCH3
CH3 Clhaloalkane 2-chloro-4-methylpentane
CH3CH2CH2CH2CH2CH2COOHcarboxylic
acid heptanoic acid
CH3CH2 C O
O CH2CH2CH3
ester propyl propanoate
HO
CH3
CH3C8H17
Cholesterol is a major component of gallstones. From the following structure of the compound predict its reaction with ..
cholesterol(a) Br2
(b) H 2 with a Pt catalyst(c) CH3COOH
HO
CH3
CH3C8H17
The reaction with Br2 results in addition of bromine to the double bonded carbons forming a single carbon bond. The solution would change colour from orange to colourless.
BrBr
With H2 and a Pt catalyst a hydrogenation reaction would occur and H atoms would be added across the double bond forming a single C – C bond.
HO
CH3
CH3C8C17
H H
CH3
CH3C8H17
Ethanoic acid reacts with the hydoxy group to form an ester and water
OH3C C
O
+ H2O
We have some new functional groups to learn this year Title your page Functional groups
Use the photocopied sheet and copy the complete the table neatly into your exercise book
Yes the whole table!We must learn these!
Complete the task on the handout , glue into your lab book
Use your chart to help you classify and name the listed compounds (complete in pencil)
CH3CH2CH2CH2CHCH2CH3
Cl
Class haloalkane Name 3-chloroheptane
CH3CH2CH2 C O
Cl
Class acyl chloride Name butanoyl chloride
C CH2CHCH3
HO CH3
OClass carboxylic acid Name 3-methylbutanoic acid
Answers to the left hand column on handout
CH3CH2CH2CH2NH2
Class amine Name 1- aminobutane
CH3CH2CH2CH2CH2C O
NH2
Class amide Name hexanamide
Class ester Name butyl pentanoate
Answers to the left hand column continued
CH3CH2CH2CH2 O
O C CH2CH2CH2CH3
CH3CH2CH2COCH2CH3
Class ketone Name hexan-3-one
CH3CH2CH2CH2CH2OH
Class alcohol Name pentan-1-ol
Class aldehyde Name hexanal
Answers to the Right hand column
CH3CH2CH2CH2CH2C O
H
CH3CH2CH2CH2CH2COCl
Class acyl chloride Name hexanoyl chloride
CH3CH2CHCH3
NH2
Class amine Name 2-aminobutane
Class amide Name octanamide
Answers to the right hand column continued
CH3CH2CH2CH2CH2CH2CH2CONH2
Structural Isomers (are also called constitutional isomers)
These are molecules with the same molecular formula but different structural formula.
The isomers of a particular molecule will have different physical properties e.g. melting and boiling points. They may also have different chemical properties.
Draw and name the structural isomers of C4H10
Name: 2-methylpropaneName: butane
Boiling point 36.1o CBoiling point -11.7o C
Task – in pairs use the models to make hexane
Draw the structural formula for hexane
Then make as many structural isomers of hexane as you canName and draw each one
There should be 5?
C6H14
H
C H
H
H
CH
H
H
C
H
H
C
H
H
C
H
H
C
H
Hexane
CH3
2-methylpentane
H
C H
H
H
CH
H
C
H
H
C
H
H
C
H
Structural Isomers of Hexane
H
CH
H
C
H
C
H
H
C
H
CH3
CH3
H
H
CH
H
C
H
C
H
H
C
H
CH2
H
H
CH3
Structural Isomers of Hexane
2,2-dimethylbutane
3-methylpentane
H
CH
H
C
H
C
H
C
H
CH3
H
H
CH3
Structural Isomers of Hexane
2,3-dimethylbutane
Geometric Isomers will only occur if….
The compound has a double or triple bond where there can be no rotation around the
C C bond
Remember alkanes don’t exhibit geometric isomerism because there is rotation around the single C C bond
Geometric (cis and trans) Isomers
e.g. The Geometric Isomers of but-2-ene
cis-but-2-ene trans-but-2-ene
Bpt 3.7o C Bpt 0.88 o C
* Geometric isomers have similar chemical properties but different physical properties
To exist as geometrical isomers the C atoms at both ends of the double bond must each have two different groups (or atoms attached).
Geometric isomerisim does not occur if one of the carbon atoms in the fixed (ie the double bond) has two identical atoms or groups of atoms attached
But-1-ene does not have geometric isomers, because it has two groups, H atoms, attached to the C atoms on either side of the double bond
Therefore but-1- ene does not have geometric isomers
flip it over and it’s the same as
Geometric isomers are a form of stereoisomerisim
Stereoisomerism – are where the atoms are bonded in the same sequence but are arranged differently in space in a molecule.
e.g. but-2-ene
cis-but-2-ene trans-but-2-ene
Same sequence of atoms
Two different geometric isomers exist where atoms are arranged differently in space
Bpt 3.7o C Bpt 0.88 o C
Exercise
Draw the structures of the following alkenes and decide which of them can exist as cis-trans isomers
a) 2-methylbut-2-ene
b) 3 – methylpent-2-ene
forms no cis/trans isomers
Occurs as cis trans isomers
cis-methylpent-2-ene trans-methylpent-2-ene
Identify whether cis trans isomers occur with in the following molecules
HO
C
H
Cl
C
CH3
C
H
Cl
C
CH3
H3C
C
Cl
C
CH3
H3C CH3
No Yes Yes
C
H
C
H
Geometric isomers have the same chemical properties, but different physical properties. Why?
Because cis isomers have bulky side groups and cannot be packed closely together, this causes weaker intermolecular forces between molecules and therefore lower melting points than the trans version
H3C
CH3
C
H
C
H
H3C CH3
C
H Cl
C
HCl
C
H
Cl
C
H
Cl
However cis forms are sometimes polar (as above) and therefore have stronger intermolecular forces between molecules causing higher melting points.
Testing for Cis - Trans Isomers Testing for Cis - Trans Isomers
Weigh out 2 grams of maleic acid into a 50ml beaker
Add 4mls of water and warm slightly to dissolve the acid
Pour this into a pear shaped flask
Carefully add 5mls of concentrated HCl
Place a condenser on top of the flask and secure it in a retort stand with a water bath. Then warm the solution until a solid forms.
Cool the solution to room temperature by placing the flask in a cold water bath ie a 250ml beaker of cold water
Pour the solution into an evaporating dish, pouring off the excess liquid then carefully rinse with water – Then complete task B on page 159
C
H H
C
CC OO H
O
O H
C
H H
C
COOHCOOH
Maleic acid (cis isomer)
Give the names and structural formula for the following substances from their condensed structural formulae.
(c) CH3CH2CHCHCH2CH2CH3
(d) (CH3)3COH
H H
C
H
H
CH
H
C
H
H
C
H
C
CH3
C OHCH3
hept-3-ene
2-methylpropan-2-ol
H
C H
H
H
C
H
CH3
Give the names and structural formula for the following substances from their condensed structural formulae.
(a) CH3CH2CHClCH2CH3
(b) (CH3)2CHCH2CH2CHCH2
H H
C H
H
H
CH
H
C
H
Cl
C
H
H
C
H
CH3 H H
CH
C
H
C
H
H
C
H
H
CCH3
3-chloropentane
5 -methylhex-1-ene
Another form of Stereoisomerism is Optical isomerism
E,Z Nomenclature of bond geometryIn cis and trans – nomenclature the ‘like” groups are identified and used to specify the type of isomer.With E,Z rules the pair of substituent's at each end of the double bond are given a priority.Highest priority = atom of highest atomic no attached directly to the double bondEg 2-chloro-3-methylpent-2-ene
C
Cl CH3
C
CH2 -CH3CH3
1st pair Cl highest priority
2nd pair CH2CH3 highest priority
This isomer has the highest priority groups on opposite sides of double bond therefore is the E isomer
The Z isomer has the highest priority groups on the same side of the double bond
Optical Isomerism
Optical isomers involve an asymmetric carbon – a carbon bonded to four different atoms or groups of atoms such as:
H, OH, CH3, C2H5, C3H7 etc
A molecule with an asymmetric carbon is known as a chiral molecule. The two forms of the chiral molecule are known as enantiomers or optical isomers
C
OHH
H3C
C2H5
C OH
H
CH3
C2H5
Structural formula of
butan-2-ol
3-D diagram
of butan-2-ol
*
C* = asymmetric carbon
*
C
OHH
H3C
C2H5
C
HOH
CH3
C2H5
* *
3-D optical isomers of butan-2-ol
Mirror line
Optical isomers are:
mirror images of each other
Are not superimposable on each other
Glucose is an optically active compound.
On the straight-chain form of glucose shown here, all four of the carbons in the middle of the chain are chiral centres – they each have four different groups attached to them.
Optical activity
Four different groups can be arranged around a central atom in two different ways. These are optical isomers and are mirror images of each other and cannot be superimposed — just as your left and right hands are mirror images and cannot be superimposed.
The structure for glucose shows four chiral centres, which means a total of 16 different forms are possible. Glucose is just one of those forms.
Molecules which have one or more chiral centres
rotate plane-polarised light.
Molecules which are mirror images of each other
rotate plane polarised light in opposite directions.
Optical isomers are
If we add a second grill at right angles to the first, it will block the red wave.
Two polarising filters placed at right angles will block all light.
Two sheets of polarising film are placed on an overhead projector stage. The arrows on the sheets show their orientation.
On the left we see that when the two sheets are orientated in the same direction light passes through them, but when the sheets are at right angles (right) the light is blocked.
Beakers of water and sucrose solution (which is cheaper than glucose and also optically active) are placed on a sheet of polarising film sitting on the overhead stage. A second sheet of polarising film is on top of the beakers, at right angles to the first.
Water Sucrose
Although the film above the water beaker is dark, light shines through the film above the sucrose solution.
The sucrose has rotated the light waves sufficiently so that they are able to pass through the second film.
When we rotate the top sheet, the film above the sucrose solution is now dark, while light passes through everywhere else.
Water Sucrose
Different wavelengths (colours) of light are rotated by different amounts, so that as the polarising film is rotated by different angles, we see these different colours.
Remember:
• Optically-active solutions rotate plane-polarised light
• optical isomers rotate plane-polarised light by equal angles in opposite directions.
Optical Isomer Properties
Many organic compounds have optical isomers including hormones and enzymes involved in biochemical reactions. The shape of a molecule is very important in these reactions and this means that the mirror image (ie the optical isomer) of an enzyme will not work properly in the body.
Optical Isomer Properties
Optical isomers have identical chemical and physical properties except that they rotate the plane of polarised light in opposite directions.
Optical isomers react differently with other optically active compounds.
an equal mixture of both enantiomers is called a racemic mixture
In fact, only one of the two optical isomers of thalidomide appear to cause these birth defects, although it could be that once ingested each isomer readily changes into the other form. Today thalidomide is being used successfully as a treatment for leprosy (although not for pregnant women).
The drug thalidomide, prescribed as a treatment for morning sickness in early pregnancy in the 1960s, tragically caused the development of serious birth defects (badly deformed limbs, or none at all).
Exercise
Draw the structure of each of the following molecules and then decide which ones are optically active. Mark the chiral carbon with an asterisk.
(a)2-chlorobutane (b) 2-methylpropanoic acid
(c) 3-methylpentanal (d) 2-aminobutanoic acid
CH3 CH CH2
Cl
CH3 C
OH
O
CHCH3
CH3
C CH2 CH CH3CH2
O
H CH3
C
OH
O
CHCH2
NH2
CH3
*
* *
No optical isomers
Homework
Read unit 28 page 111 in pathfinder
Complete Q’s 4 and 5 page 114
Complete Enantiomers on BestChoice before next Friday please
Group work exercise Each group is to work on problems giving great detailed answers. But all people in the group must have the answers written in their lab books
Random people from each group will be asked for their groups answer
Answer questions on alkanes and alkenes page 172 -174
Turn to page 6 in your booklet
Properties of alkanes and alkenes
Demo Alkene addition
Turn to page 145 in the year 13 lab Turn to page 145 in the year 13 lab bookbook
Comparison: Alkanes vs AlkenesComparison: Alkanes vs Alkenes
Complete Structural isomer starter
Complete Worksheet two Q’s 1 and 2 in organic
booklet
Alkane Reactions
Alkanes are used as fuels and undergo combustion reactions
In excess air (oxygen) they form the products H2O and CO2
In limited air (oxygen) they form the products H2O and C or CO2
Because the carbon atoms in alkane molecules are saturated with hydrogens (ie they don’t have any double or triple bonds) they are called saturated hydrocarbons.
Properties of AlkanesProperties of Alkanes
Insoluble in waterInsoluble in water
Soluble in non polar solventsSoluble in non polar solvents
Don’t conduct – no free electronsDon’t conduct – no free electrons
Float on water because HFloat on water because H22O is relatively denserO is relatively denser
Boiling/melting point increases with chain length Boiling/melting point increases with chain length because as molecular mass increases the because as molecular mass increases the intermolecular forces between molecules intermolecular forces between molecules increasesincreases
Questions
1. Name the type of intramolecular bonding and the type of intermolecular bonding in:
a) Methane b) Liquid pentane
2. Explain in terms of bonding why:
a) Methane gas can be collected over water
b) Petrol floats on water
c) Oil dissolves in petrol
Alkane Reactions
Because alkanes have no double bonds they react slowly with halogens in the presence of UV light.
This reaction is called a substitution reaction in which an H atom is replaced by a halogen atom (eg Cl or Br)
butane + bromine
bromobutane + hydrogen bromide
What’s missing in this reaction?
The bromine solution decolourises slowly and the HBr formed is an acidic gas that turns moist blue litmus red
UV
light
These are unsaturated hydrocarbons
This means they have at least one double or triple covalent bond
These types of bonds are called functional groups because it’s at these bonds that reactions occur
Alkenes (CnH2n) and Alkynes ( Cn H2n - 2 )
Like alkanes these unsaturated hydrocarbons are non polar and insoluble in water.
They undergo combustion the same as alkanes giving the same products
They have very similar physical and chemical properties to alkanes
Form addition reactions because of the reactive double or triple bond
Alkenes exhibit a different form of isomerism called geometric isomerisim
Alkenes and Alkynes
Starter ADraw and name the 2 possible products formed when HCl is added to 2-methylbut-2-ene.Name the products and identify which is the major product.
2-chloro-3-methylbutane
2-chloro-2-methylbutane – major product
H
CH
H
C
H
C
H
C
H
CH3
Cl
H
H
H
CH
H
C
H
C
H
C
HCH3Cl
H
H
Hydrogenation of an Alkene (Addition reaction)
Hydrogen can be added across the double C bond
The conditions for this reaction are:
• platinum catalyst
• 150 - 200 O C
• pressure of 4 atmospheres
The reaction is
C2H4(g) + H2(g) CH3CH3(g)
ethene ethane
The Good Oil
This hydrogenation process is
used to harden plant oils to commercially produce margarine.
Natural plant oils contain many double bonds per molecule, and because several double bonds exist after hydrogenation, the margarine is said to be polyunsaturated.
Bromination of an Alkene (Addition reaction)
Alkenes and alkynes undergo addition reactions with halogens to form a dihaloalkane (or tetrahaloalkane).
The common test for an unsaturated hydrocarbon (ie a hydrocarbon with a C C or C C bond) is therefore the rapid decolourisation of an orange bromine solution in the absence of sunlight
The reaction is
CH2H4(g) + Br2(l) CH2 Br CH2 Br (l)
ethene 1,2-dibromoethane
Alkene molecules can create polymers (plastics) by addition reactions where many alkene monomer units are joined together in the presence of heat and a catalyst
The process involves the breaking of one of the bonds in the double bond in each alkene molecule.
Each of the two electrons from the bond go to each end of the molecule to create a bond with another molecule that has undergone the same process.
This creates long chains of joined monomers to create a polymer.
H
C
H
H
C
H
Can be drawn as
H
C
H
H
C
H
. .
H
C
H
H
C
H
. .
H
C
H
H
C
H
. .
H
C
H
H
C
H
. .
H
C
H
H
C
H
. .
H
C
H
H
C
H
. .
H
C
H
H
C
H
. .
Heat and a catalyst added
3 ethene monomers
Repeating monomer unit
Polyethylene
polymer
Representation of adddtion polymer reaction
H
C
H
H
C
H
H
C
H
H
C
H
H
C
H
H
C
H
Cl
C
H
H
C
H
. .
Cl
C
H
H
C
H
. .
Cl
C
H
H
C
H
. .
Cl
C
H
H
C
H
. .
Cl
C
H
H
C
H
. .
Cl
C
H
H
C
H
. .
Heat and a catalyst added
3 vinyl chloride monomers
Repeating monomer unit
Polyvinylchloride
Polymer
Aka PVC
Changing the side chain of the monomer in the reaction gives different polymers ie
C
H
H
C
H
Cl
C
H
H
C
H
Cl
C
H
H
C
H
Cl
This year we will also look at polymer reactions involving condensation reactions
•Addition polymers are formed when alkene monomers undergo addition to form a polymer eg. polythene from ethene, P.V.C. from vinyl chloride (chloroethene), polypropene from propene.
Haloalkanes (alkyl halides) RX
where X = F, Cl, Br, I
Named as a chloroalkane or bromoalkane etc, with the position of the halogen given by the appropriate number of the carbon that it is attached to in the chain.
Exist as primary, secondary, tertiary
The haloalkanes can be classified as: primary - the C atom to which X is attached is only attached to one other C atomeg
H
CH
H
H
C Br
HCarbon attached to Br is attached to one carbon
Secondary haloalkane - the C atom to which X is attached is attached to two other C atomseg
H
CH3
C Br
CH3Carbon attached to Br is attached to two other carbons
Tertiary haloalkane - the C atom to which X is attached is attached to three other C atoms.eg
CH3
C Br
CH3Carbon attached to Br is attached to three other carbons
CH3
The Lucas TestThe Lucas Test
The Lucas test is used to distinguish between The Lucas test is used to distinguish between the primary, secondary and tertiary alcoholsthe primary, secondary and tertiary alcohols
The Lucas reagent consists of ZnClThe Lucas reagent consists of ZnCl22 in in concentrated HClconcentrated HClThe zinc chloride catalyses a substitution The zinc chloride catalyses a substitution reaction between the alcohol and the reaction between the alcohol and the concentrated HClconcentrated HClChloroalkanes form and appear as a cloudy Chloroalkanes form and appear as a cloudy suspension in the water because they are suspension in the water because they are insolubleinsoluble
The Lucas testThe Lucas test
*Important*Important
The rates at which the different types of The rates at which the different types of alcohol react to form chloroalkanes enable alcohol react to form chloroalkanes enable them to be classified as follows: them to be classified as follows:
The Lucas TestThe Lucas Test
Type of Type of alcoholalcohol
ExampleExample ObservationObservation ProductProduct
PrimaryPrimary Butan-1-olButan-1-ol
No No cloudiness, cloudiness, very slow if very slow if at allat all
1-chlorobutane1-chlorobutane
SecondarySecondary Butan-2-olButan-2-ol
Cloudiness Cloudiness after 5-15 after 5-15 minutesminutes
2-chlorobutane2-chlorobutane
TertiaryTertiary 2-methylpropan-2-ol2-methylpropan-2-ol Cloudiness Cloudiness after 1-2 after 1-2 minutesminutes
2-methyl-2-2-methyl-2-chloropaopanechloropaopane
Starter: Write out and fill in the missing wordsThe s_____ of the C-OH b____i_______ from tertiary to secondary to primary alcohols (which have the strongest C-OH bond.)The test for the C-OH strength is called the l____ test which uses concentrated ____ and Z_________ as the catalyst. The speed of this s_________ reaction of the OH for a Cl indicates the type of Alcohol
trength ond ncreases
ucas HCl inc chloride
ubstitution
Properties of haloalkanesProperties of haloalkanes
Haloalkanes do not form hydrogen bonds, so they have lower boiling points than alcohols and are not miscible in water.
However, they are polar compounds, so have higher boiling points than their parent alkanes. The lowest mass haloalkanes are gases at room temperature, but the rest are volatile liquids.
To make a haloalkane we can substitute the OH on an alcohol using eg. PCl3, PCl5,SOCl2 or conc
HCl/ZnCl2
C2H5OH C2H5ClPCl3 or PCl5
ethanol chloroethane
C2H5OH C2H5Cl + HCl +SO2
SOCl2
ethanol chloroethane
C2H5OH C2H5ClConc HCl/ZnCl2
ethanol chloroethane
• Haloalkanes are relatively nonpolar overall (despite the polarity of the C-X bond) and are insoluble in water.
A monohaloalkane eg. 2-bromopropane can be formed by:
a) addition of HBr to propene (forming only one product)
CH3
C
H
H
C
H
+ HBr
H
C CH
H
H
C Br
H
H
H
b) substitution of propane using Br2.
(forming two products, the bromoalkane and HBr)
H
C CH
H
C H
H
H
+ Br2
HH
H
C CH
H
C H
H
H BrH
+ HBr
Lab book Page 201 Preparation of a Haloalkane
Like tertiary alcohols, tertiary haloalkanes are easily substituted.
A tertiary haloalkane will react with cold water to form an alcohol:
R—X + H2O → R—OH + HX
We can tell whether this reaction has taken place by the presence of the X– ions, which will form precipitates with silver nitrate solution.
Substitution of haloalkanes to form alcohols
• Tertiary haloalkanes form alcohols in cold water
• Secondary haloalkanes react when the water is warm
• Primary haloalkanes do not react with water, but react to form alcohols with aqueous sodium hydroxide.
summary!
Nucleophiles
A nucleophile is any species which loves nuclei, that is anything attracted to a positive charge.
Nucleophiles are therefore species that carry a negative charge or a lone pair of electrons,
eg OH- , H2O and NH3 (in alcohol)
The C -X bond is polar and the slighty positive C atom is prone to attack from a negative nucleophile.
eg
R X(l) + OH- (aq) R OH + X- (aq)
Elimination to form an alkene
Haloalkanes can also undergo an elimination with hydroxide to form an alkene:
R—X + alcOH– → R’=C + HXUse the above reaction as a template to write your own formation of an alkene from a haloalkane
When deciding where to put the double bond in an elimination reaction, apply the rule ‘the poor get poorer’.
One carbon of the double bond will be the carbon that lost the halogen.
To decide whether the bond goes to the left or right of that carbon, look at the number of hydrogen atoms on each of those carbons.
The carbon to lose the hydrogen atom (and thus become the other half of the double bond) is that carbon which has the fewer hydrogen atoms already bonded to it.
H
CH
H
Cl
C
H
H
C
H
H
C H
H
H
CH
H
C
H
C
H
H
C H
H
+ HCl
(Alc)KOH
Draw the structural formula for the reaction of the tertiary haloalkane 2-chloro, 2-methylpropane with water
+ H2O
H
C CH
H
C H
CH3
Cl HH
+ HCl
H
C CH
H
C H
CH3
OH HH
2-methylpropan-2-ol
Tertiary alcohol
Amines (aminoalkanes)
Amines are named as substituents eg aminomethane, CH3NH2.
These may be classed as primary, secondary or tertiary, but unlike the haloalkanes the classification depends on the number of C atoms attached to the N atom. Primary RNH2, secondary R2NH, tertiary R3N.
H
C NH H
HH
H
C NH CH3
HH
H
C NH CH3
H CH3
H
C
HAminomethanePrimary amine
N-methylaminomethaneSecondary amine
N,N-dimethyaminoethaneTertiary amine
Amines (aminoalkanes)
Amines have an unpleasant “fishy” smell. The smaller amines, up to C5, are soluble in water but larger
amines are insoluble, as the size of the non-polar hydrocarbon chain cancels out the effect of the polar amino functional group.
Like ammonia itself, water soluble amines form alkaline solutions. They react with water by proton transfer to form OH- ions. This means aqueous solutions of amines turn litmus blue.
RNH2 + H2O RNH3+ + OH
Amines also react with acids to form salts.
CH3NH2 + HCl CH3NH3+
Cl
aminomethane methyl ammonium chloride
The formation of an ionic salt increases the solubility of the amine in acidic solutions (compared to their solubility in water).
This is why we put lemon juice onour fish to get rid of the amine smell
Formation of amines
Another nucleophilic substitution reaction occurs between haloalkanes and alcoholic ammonia:
R—X + NH3(alc) → R—NH2 + HX
amineWhy do you think the ammonia has to be alcoholic?
It must be alcoholic ammonia: if water is present the alcohol could be formed instead.
CH3CH2CH2CH2CH2CH2COCl
Class acyl chloride Name heptanoyl chloride
CH3CH2CHCH2CH3
Class amine Name 2-aminopentane
Class amide Name propanamide
CH3CH2CONH2
NH2
Write these out give the compound class and name
CH3CH2CH2CH2COCH2CH3
Class ketone Name heptan-3-one
CH3CH2CH2CHOHCH2
Class secondary alcohol Name pentan-2-ol
Class aldehyde Name hexanal
CH3CH2CH2CH2CH2C O
H
Write these out give the compound class and name
Draw and name the structural isomers of C4H10O
CH3CH2CH2CH2OH
butan-1-ol
CH3CH
CH2CH3
OH
butan-2-ol
H
CH3 C
CH3
CH2 OH
2-methylpropan-1-ol
OH
CH3 C
CH3
CH3
2-methylpropan-2-ol
Alkene Reactions
Alkenes react readily by adding small molecules across the double C C bond.
These reactions are known as addition reactions because molecules add “across” the double bond
CH3CH2CH2CH2CH2NH2
Class amine Name 1- aminopentane or pentylamine
CH3CH2CH2C O
NH2
Class amide Name butanamide
Class ester Name ethyl pentanoate
Write these out give the compound class and name
CH3CH2 O
O C CH2CH2CH2CH3
When deciding where to put the double bond in an elimination reaction, apply the rule ‘the poor get poorer’.
H
CH
H
Cl
C
H
H
C
H
H
C H
H
H
CH
H
C
H
C
H
H
C H
H
+ HCl
One carbon of the double bond will be the carbon that lost the halogen.
To decide whether the bond goes to the left or right of that carbon, look at the number of hydrogen atoms on each of those carbons.
The carbon to lose the hydrogen atom (and thus become the other half of the double bond) is that carbon which has the fewer hydrogen atoms already bonded to it.
Haloalkanes are molecular substances, so they do not contain free X– ions.
When silver nitrate solution is added to 1-bromo-butane no cream precipitate of silver bromide forms.
When silver nitrate solution is added to 2-chloro, 2-methyl propane, the water in the solution reacts with the haloalkane, forming an alcohol and releasing chloride ions which then react with the silver nitrate to form a white precipitate.
Primary haloalkanes can also be converted into alcohols, but a stronger nucleophile is needed: OH–.
Dilute sodium hydroxide solution is added to 1-bromo butane and shaken.
The excess NaOH is neutralised with dilute nitric acid.
When silver nitrate is added the solution turns cloudy.