chemistry 3.5 describe the structure and reactions of organic compounds containing selected organic...

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







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


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


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


2,2-dimethylbutane

3-methylpentane

H

CH

H

C

H

C

H

C

H

CH3

H

H

CH3


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


C2H5OH C2H5ClConc HCl/ZnCl2


• 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


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


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.

chemistry 3.5 describe the structure and reactions of organic compounds containing selected organic...

Documents

n slide

dimethylpentane slide

trimethylpentane slide

o slide

dimensional structure

general formula

selected organic groups

branch groups