organic chemistry hl only 20.1 introduction more functional groups esters contain functional group...

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20.1 Introduction

More functional groups

Esters

Contain functional group COOR, where R = an alkyl group such as CH3 (methyl group).

COR

O

Naming esters

CO

OR

R

name of this alkyl group forms first part

of name.

second part of name is derived from the name of the carboxylic

acid salt.

Name the following ester

CO

OCH3CH2

CH2CH3

this is an ethyl group.

this is a propanoate group.

ethyl propanoate

Name the following ester

CO

OH

CH3

this is a methyl group.

this is a methanoate group.

methyl methanoate

Draw out the structure of

CO

OC6H5

CH2CH3

ethyl benzenecarboxylate

Draw out the structure of

CO

OCH3CH2CH2

CH2CH3

ethyl butanoate

Amidesuffix –amide

For example

CH3CONH2 ethanamide

CNH2

O

CH3CO

NHCH3

We will also come across amides where one of the hydrogen atoms attached to the N atom is replaced by an alkyl group these are referred to as N-substituted amides.

The structure shown is N-methylethanamide.

Nitrilessuffix -nitrile

For example

CH3CH2CN propanenitrile

C N

Remember to include the nitrile group as part of the C chain.

C NH2 Aminessuffix –amine or prefix amino-

For example

CH3NH2 methylamine

H2NCH2COOH aminoethanoic acid

-amine suffix tends to be only used for short chain amines e.g. propylamine.

Amino- prefix is used for longer names so

CH3CH2CH2CH2CH2CH2NH2 is 1-aminohexane

What would the structure of 2-aminohexane look like?

For secondary and tertiary amines, the longest alkyl chain attached to the N is identified and we then name in a similar fashion to N-substituted amides. So for example:

N

H

CH3 C2H5

N-methylethanamine

N

H

C2H5 C3H7

N-ethylpropanamine

N

C2H5

CH3

CH3

N,N-dimethylethanamine

20.2 Nucleophilic Substitution ReactionsThe examples of nucleophilic substitution

looked at earlier are not the only examples. Other nucleophiles which will react with

halogenoalkanes include H2O, NH3 and CN-. Using water as the nucleophile would produce

an alcohol but the reaction is much slower than with hydroxide ions as the hydroxide ions have a negative charge so are attracted more strongly to the + on the C atom.

The order of reactivity with these nucleophiles is:

C – I > C – Br > C – Cl

As explained earlier, this is due to increasing bond strengths.

From experimentation we also found:

tertiary > secondary > primary

This is harder to explain but it is thought that the activation energy to form the tertiary carbocation intermediate in SN1 is less than that required to form the transition state in SN2.

propanenitrileCH3CH2I (ethanol) + CN-(aq) CH3CH2CN + I-

1. With cyanide ions

This is an important reaction for organic chemists as it increases the length of the carbon chain. The cyanide group can then be converted to other functional groups:

CH3CH2CN

CH3CH2CH2NH2

CH3CH2COOHH+

H2O

H2

Ni

propanoic acid

1-aminopropane or propylamine

The mechanism for the nucleophilic substitution reaction is:

-CN

NC

H

CH3C

H

Br-

BrC

CH3

H

H

C

+

H

CH3 H

NC Br

-

SN2

Transition state

2. With ammonia

aminoethaneCH3CH2Br + NH3 CH3CH2NH2 + HBr

NH3

H2N

H

CH3C

H

Br-

BrC

CH3

H

H

+

C

H

CH3 H

Br

SN2

Transition state

H2N

H

There is still a lone pair of electrons on the N atom in ethylamine so this can react with a further molecule of bromoethane to form a secondary amine.

C2H5NH2 + C2H5Br (C2H5)2NH + HBr

N-ethylethanamine

The secondary amine still contains a lone pair so further reaction occurs forming a tertiary amine.

(C2H5)2NH + C2H5Br (C2H5)3N + HBr

N,N-diethylethanamine

The tertiary amine still contains a lone pair so further reaction occurs forming a quaternary ammonium salt.

(C2H5)3N + C2H5Br (C2H5)4N+Br-

tetraethylammonium bromide

A better method for making a primary amine involves catalytic (Ni catalyst) hydrogenation of a nitrile. The yield is larger and there is no further reaction possible.

CH3CH2CN + 2H2 CH3CH2CH2NH2

propanenitrile propylamine

20.3 Elimination Reactions

When warm aqueous sodium hydroxide reacts with a halogenoalkane, an alcohol is formed via a nucleophilic substitution reaction. The hydroxide ions behave as a nucleophile so:

C2H5Br + OH-(aq) C2H5OH + Br-

However, if the sodium hydroxide is dissolved in hot ethanol and the mixture heated under reflux a different product is formed, an alkene.

This is an elimination reaction and the sodium hydroxide is behaving as a base (proton acceptor) rather than a nucleophile.

C2H5Br + OH-(alc) C2H4 + H2O + Br-

Overall HBr is eliminated (removed) from the halogenoalkane.

There are two possible mechanisms known as E1 and E2.

CH3

H H

H

CC

OH-

CH3

H H

HCC

Br H

propene

H OHBr -

acting as a base

CH3

H H

HCC

Br H

CH3

H H

HCC

H+

OH-acting as a base

E1

E2

elimination

+ OH-

RCH=CH2 + H2O + X-

(ethanol)

nucleophilic substitutionalcohol

+ OH-

RCH3CH2OH + X-

(aqueous)

RCH2CH2X

alkene

hydroxide acts as a base

hydroxide acts as a nucleophile

20.4 Condensation Reactions

A condensation reaction involves two molecules reacting together to form a larger molecule with the elimination of a small molecule such as water or hydrogen chloride.

Esterification is an example of a condensation reaction. Esterification involves the formation of an ester from a carboxylic acid and an alcohol in the presence of concentrated sulphuric acid.

For example, ethyl ethanoate is formed when ethanoic acid and ethanol are mixed in the presence of a few drops of concentrated sulphuric acid.

CH3COOH + C2H5OH Ý CH3COOC2H5 + H2O

The conc sulphuric acid acts as a catalyst and also shifts the equilibrium to the right hand side by removing water.

The water is formed from the –OH of the carboxylic acid combining with the H from the alcohol.

Uses of Esters

Solvents – esters are volatile and polar. Polarity means that they act as solvents for many polar organic compounds. Low b.p. means that they evaporate from less volatile solutes. e.g. ethyl ethanoate is used as the solvent in glues such as polystyrene cement.

Plasticisers – plastics are often not flexible as chains cannot move over each other easily. The addition of plasticisers allows chain movement. Over time these additives escape so the plastic becomes brittle and stiff.

Food flavourings – many esters have a sweet, often fruity smell and are used as artificial food flavouring.

For example –

pentyl ethanoate – pear

2,2,dimethylpropyl ethanoate – banana

octyl ethanoate – orange

ethyl butanoate – pineapple

pentyl pentanoate – apple

Amide Formation

Carboxylic acids also undergo a condensation reaction with amines (or ammonia) to form an amide. The –OH group from the carboxylic acid reacts with one of the hydrogen atoms attached to the N atom to form water.

RCOOH + R’NH2 RCONHR’ + H2O

R’ = an H atom amide

= an alkyl group N-substituted amide

CH3CO

OH

+ NH3 CH3CO

NH2

+ H2O

CH3CO

OH

+ CH3NH2 CH3CO

NHCH3

+ H2O

Why is this reaction of biological importance?

C COOHR

H

NH2

primary amino group

carboxylic acid group

Two amino acid molecules can react together to form an amide.

What is the correct IUPAC name of the two amino acids on the next slide?

Deduce the structure of two possible condensation products.

C COOHH

H

NH2

C COOHCH3

H

NH2

glycine

alanine

The two amino acid units are held together by what is known as the peptide bond (peptide link).

C N

H

O

The two condensation products are known as dipeptides.

peptide link

H2NCHCOOH

R1

H2NCHCOOH

R2

+

H2NCH – C – N – R1

CHCOOH

R2H

O

+ H2O

On one end of the dipeptide is an amine group and on the other is a carboxylic acid group so further reaction at each end is possible. The result is a CONDENSATION POLYMER. In this instance a polyamide.

Condensation Polymers

Condensation polymers are formed by the reaction between molecules having two functional groups, involving the loss of small molecules such as H2O, CH3OH or HCl.

There are two types of condensation polymer:

• polyamides

• polyesters

Polyamides

The reaction between a dicarboxylic acid and a diamine leads to the formation of a polyamide.For exampleNylon-6,6 is formed from hexanedioic acid and hexane-1,6-diamine.Nylon-6,6 is so called as it is made from a 6 C diacid and a 6 C diamine.

nH2N(CH2)6NH2+nHOC(CH2)4COH

O O

HN(CH2)6NHC(CH2)4C

O O

- C(CH2)4C

O O

HN(CH2)6NH -n

repeating unit

+ 2n-1H2O

As with naturally occurring polyamides (polypeptides and proteins), synthetic polyamides are susceptible to hydrolysis and can be broken down into component monomer units. Consequently, polyamides are biodegradable.

Polyesters

The reaction between a dicarboxylic acid and a diol leads to the formation of a polyester.

The most important example is terylene which can be formed from benzene-1,4-dicarboxylic acid and ethane-1,2-diol.

COHHOC

O O

+

benzene-1,4-dicarboxylic acid

HOCH2CH2OH

ethane-1,2-diol

-H2O

C OCH2CH2OCCOC

O O

OCH2CH2O

O O

n

+ 2n-1H2O

Polyesters (like esters) are susceptible to hydrolysis and can be broken down into their component monomers. So polyesters (like polyamides) are biodegradeable.

20.6 Stereoisomerism

Stereoisomers are compounds that have the same structural formula but their atoms are arranged differently in space.

There are two types:

1.Geometrical isomerism and,

2.Optical isomerism.

Geometrical Isomerism

Arises due to lack of rotation at C = C in alkenes.

C = C

R R

H HC = C

H R

R H

cis isomer trans isomer

It is not possible to have geometrical isomerism when there are two identical groups attached to the same C atom in the double bond. So but-1-ene does not exist as geometric isomers whereas but-2-ene does.

The C atoms in the double bond are sp2 hybridised.

One of the bonds is a bond formed by overlap of two sp2 hybrid orbitals.

The other is a bond formed by sideways overlap of p orbitals. These must be in the same plane to overlap. Any attempt to rotate will mean that these will no longer be in the same plane which would mean the bond would be broken.

Consequences of Geometric Isomerism

Melting points are influenced by how closely molecules pack together.

cis but-2-ene melts at -139 °C, trans but-2-ene melts at -106 °C.

cis 1,2-dichloroethene melts at 60 °C, trans 1,2-dichloroethene melts at 48 °C.

Draw out the cis and trans isomers of but-2-ene-1,4-dioic acid.

Which of the two isomers will have the highest melting point? Why?

trans isomer has an m.p. of 286 °C, cis isomer has m.p. of 131 °C.

trans isomer has strong hydrogen bonding between molecules but cis isomer has strong hydrogen bonding within molecules.

Which of the two isomers will dehydrate more easily? Why? What will the product be?

Geometric isomerism is also found in cyclic compounds where the rigid structure of the ring prevents free rotation.

For example: 1,2-dichlorocyclopropane exists as cis and trans isomers.

Draw them!

How many isomers are there of dichlorocyclobutane? Draw them and name them.

Optical IsomerismArises when there are four different groups attached to a carbon atom.

This means that the molecule has no centre, plane or axis of symmetry.

The molecule is said to be CHIRAL and possesses an asymmetric carbon atom.

Two tetrahedral arrangements in space are possible so that one is the mirror image of the other.

For example 2-hydroxypropanenitrile

C

NC

CH3

HHO

CCN

H3C

HOH

mirror

Stereoisomers of this type are referred to as enantiomers.

Enantiomers have exactly the same physical properties except for their effect on the plane of plane-polarised light. As a result they are referred to as optically active.

Plane polarised light is made of waves vibrating in one plane only. When it is passed through a solution of a chiral molecule, the light emerges with its direction of polarisation changed.

One enantiomer rotates the light in a clockwise direction and is referred to as the (+) isomer or dextrorotatory (d).

Its mirror image rotates the light by exactly the same angle but in an anti-clockwise direction. It is referred to as the (-) isomer or laevorotatory (l).

A mixture of equal amounts of both enantiomers is optically inactive because the effects of each enantiomer is cancelled out. Such a mixture is called a racemic mixture or racemate.

Draw out the structures of the following molecules and state whether they are optically active or not. If they are optically active, highlight any chiral carbon atoms.

1. Butan-2-ol

2. 3-methylhexane

3. 3-methylpentan-3-ol

4. 2-aminopropanoic acid

5. 1-aminopropan-2-ol

6. 2-methylpropan-2-ol

CC

H

H

OH

H

C

H

H

C

H

H

H H

C

H

H

H C

H

H

C

H

H

C

CH3

H

C

H

H

C

H

H

H

butan-2-ol

3-methylhexane

3-methylpentan-3-ol

C

H

H

H C

H

H

C

CH3

OH

C

H

H

C

H

H

H

C C

NH2

H OH

OC

H

H

H2-aminopropanoic acid

CC

H

H

OH

H

C

H

H

NH2H1-aminopropan-2-ol

CC

H

H

OH

CH3

C

H

H

HH

2-methylpropan-2-ol

Many naturally occurring molecules exist as single enantiomers, for example most amino acids, such as 2-aminopropanoic acid (alanine).

The chemical properties of enantiomers are identical except in reactions with other optically active substances.

Enzymes are stereospecific. This means they will catalyse the reactions of only one of a pair of isomers.

Acts much faster when administered as the (+)-enantiomer.

(+)-carvone is found in spearmint oil, (-)-carvone is the main constituent of caraway seed oil.

thalidomide