topic chemistry of carbon compounds -...

Chemistry of Carbon Compounds

Unit 29 An introduction to the chemistry of carbon compounds

Unit 30 Isomerism

Unit 31 Typical reactions of selected functional groups

Unit 32 Synthesis of carbon compounds

Unit 33 Important organic substances

Topic 8

KeyC o ncepts

An introduction to the chemistry of carbon compounds

• Homologous series• Systematic naming• Effects of functional groups and chain

lengthonphysical properties

Isomerism• Structural isomerism—chain

isomerism,position isomerismandfunctional group isomerism

• Stereoisomerism —geometricalisomerismand enantiomerism

Synthesis of carbon compounds• Synthetic routes for carbon

compounds• Preparation of simple carbon

compounds

Important organic substances• Aspirin• Soaps and soaplessdetergents• Nylons andpolyesters• Carbohydrates• Lipids• Proteins

Chemistry ofCarbon Compounds

Typical reactions of selected functional groups

• Reactions of alkanes, alkenes,haloalkanes, alcohols, aldehydes,ketones, carboxylic acids, esters andamides

Topic 8� Chemistry of Carbon Compounds �Unit 29 An introduction to the chemistry of carbon compounds

29.1 – 29.12 & 29.21

Summary

1 The following table summarizes the nomenclature of compounds in varioushomologous series.

29.1 The valueofmedicines: longer andhealthier lives

29.2 Functional groups: centreof reactivity

29.3 Naming alkanes and alkenes

29.4 Naming carbon compounds with one type of functionalgroup

29.5 Naming haloalkanes

29.6 Naming alcohols

29.7 Naming aldehydes andketones

29.8 Naming carboxylic acids

29.9 Naming esters

29.10 Naming amides

29.11 Naming amines

29.12 Naming compounds with more than one type of functionalgroup

29.13 Intermolecular forces and physical properties of carboncompounds

29.14 Physical properties ofhaloalkanes

29.15 Physical properties of alcohols

29.16 Physical properties of aldehydes andketones

29.17 Physical properties of carboxylic acids

29.18 Physical properties of esters

29.19 Physical properties of amides

29.20 Physical properties of amines

29.21 Common namesof carbon compounds

Homologous series

General formula

Functional group it contains

NomenclatureExample

Structural formula IUPAC name

Alkanes CnH2n+2 — addappropriateprefix to –ane CH3CHCH2CH3

CH3

2-methylbutane

Alkenes CnH2n C Creplace ‘ane’ ofthe correspondingalkanewith –ene

CH3CH CH2 propene

Haloalkanes RX–X

(X= F,Cl,Br or I)

add thenameofhalogenofunctional groupasprefix to thecorrespondingalkane

CH

Cl

CH3 CH3

2-chloropropane

Alcohols ROH –OH

replace the lastletter ‘e’ of thecorrespondingalkanewith –ol

CH3CH2OH ethanol

Aldehydes RCHOC H

O replace the lastletter ‘e’ of thecorrespondingalkanewith –al CH3CH2 C H

Opropanal

Ketones RCOR1

C

O replace the lastletter ‘e’ of thecorrespondingalkanewith –one

CH3 CH3C

Opropanone

Carboxylicacids RCOOH

OHC

Oreplace the lastletter ‘e’ of thecorrespondingalkanewith –oicacid

CH3CH2CH2 C OH

O

butanoic acid

Esters RCOOR1

C

O

O

thename consistsof two separatewords, the firstword comes fromthe alcohol, thesecondwordcomes from theacid

CH3CH3

O

OC

methyl

ethanoate

Amides RCONH2

(unsubstituted) C

O

N

replace the ‘oicacid’ endingofthe correspondingacidby –amide CH3 NH2C

Oethanamide

Amines RNH2

(primary)N

replace the lastletter ‘e’ of thecorrespondingalkanewith–amine

CH3NH2 methanamine

An introduction to the chemistry of carbon compoundsUnit 29


2 Sometimes there is more than one functional group in one compound. This kindof compound should be named according to the following order of precedence offunctional groups:

–COOH> –COO–> –CONH2> –CHO> –CO–> –OH> –NH2> –C=C–

Example

Give the IUPACnamesof the following compounds.

a)

(1mark)

CH2COOH

b)

(1mark)CH3 CH2CH(CH3)2C O

O

c) CH3CH=CHCH2OH (1mark)

d)CH3CH(NH2)CH2COOH (1mark)

Answer

a) phenylethanoic acid (1)

b)2-methylpropyl ethanoate (1)

c) but-2-en-1-ol (1)

d)3-aminobutanoic acid (1)

29.13 – 29.20

Summary

1 The physical properties (e.g. the boiling point and water solubility) of a carboncompoundare affectedby

a) the functional group it contains;

b)the lengthof the carbon chain inmolecules.

ExamtipsExamtipsExamtipsExamtips ♦ The longest continuous chain containing the carbon bearing the –OHgroup may NOT appear in a straight line. Questions often ask aboutthenamesof such structures.

3,4C2H5

H3C2C 1CH3

OH

2-methylbutan-2-ol

♦ DONOT spell ‘amine’ as ‘ammine’. ✔ ✘

♦ DONOT confuse ‘amine’ and ‘amide’.

Thegeneral formulaof primary amine is RNH2,while that of amide is(Hor R)CONH2.

♦ Studentsmayneedtoidentifythefunctionalgroupspresentinunfamiliarcompounds.

e.g.

CH2CH3

CH3

CH3CH2

CH2CH3

CHNH2

C

NH

O

OC

OO

oseltamivir

Besides the ether linkage, functional groups present in oseltamivirinclude:

– C=Cbond;

– amide functional group;

– amine functional group; and

– ester functional group.

➤for (a),the group is a phenyl group. It is attached to the ethanoic

acid.

➤ (b)

CH3 CH2CH(CH3)2C O

O

this part comes from ethanoic acid

this part comes from 2-methylpropan-1-ol

➤ (c) Thedouble bond takes the form -en-.

➤ (d) –COOHistheprinciplefunctionalgroup.The–NH2groupisnamedas a prefix.

RemarksRemarks*


Homologousseries Intermolecular forces Physical properties

Haloalkanes

•permanent dipole-permanent dipoleattractionsbetweenmolecules

Clδ+ δ–CH3

Clδ+ δ–CH3

Clδ+ δ–CH3

key:

permanent dipole-permanent dipole attraction

•boiling points higher than those ofalkanes of similar relative molecularmasses

•polar molecules can interact withwatermolecules

•slightly soluble inwater

Alcohols

•h y d r o g e n b o n d i n g b e t w e e nmolecules

CH3CH2 H

O

H

CH2CH3

O

key:

hydrogen bond

•boiling points much higher thanthose of alkanes of similar relativemolecularmasses

•hydrogen bonding between alcoholmolecules andwatermolecules

CH3CH2 H

CH3CH2 H

O

O

H

H

O

key:

hydrogenbond

H

H

O

•alcohols with less carbon atomsare miscible with water in allproportions

•alcoholswithalongcarbonchainintheirmoleculesaremuchlesssolubleinwater

2 The following table summarizes the physical properties of members of somehomologous series.


Aldehydesand

ketones


C Oδ+ δ–

CH3

CH3

key:

permanent dipole-permanent dipole attraction

C Oδ+ δ–

CH3

CH3

C Oδ+ δ–

CH3

CH3

•boiling points higher than those ofalkanes of similar relative molecularmasses

•hydrogenbondingbetweenaldehyde/ ketone molecules and watermolecules

CH3

HH

H3C

OO

C

key:

hydrogen bond

HH

O

•aldehydes and ketones with lesscarbonatomsshowappreciablewatersolubility

Carboxylicacids

•h y d r o g e n b o n d i n g b e t w e e nmolecules;moreextensive than thatin alcohols

CH3

key:

hydrogen bond

O

HO

O

C

H

O

CCH3

•boiling points higher than those ofalcoholsofsimilarrelativemolecularmasses

•hydrogen bonding between acidmolecules andwatermolecules

CH3

key:

hydrogen bond

H

C

O

O

O

H

H

O

H

H

O H

H

•the first fouracids aremisciblewithwater in all proportions

Topic 810 Chemistry of Carbon Compounds 11Unit 29 An introduction to the chemistry of carbon compounds


Esters


•boilingpointsareaboutthesameasthose of aldehydes and ketones ofsimilarmolecularmasses

•hydrogen bonding between estermolecules andwatermolecules

CH2CH3

CH3 C

O

O

key:

hydrogen bond

HH

O

•simple esters are very soluble inwater

Amides

•h y d r o g e n b o n d i n g b e t w e e nmolecules

H

CH3 C

O

N

H

key:

hydrogen bond

H

CH3 C

O

N

H

•boiling points arehigh

•hydrogen bonding between amidemolecules andwatermolecules

H

CH3 C

O

N

H

key:

hydrogen bond

HH

O

HH

OH

HO

•simple amides are very soluble inwater


Amines

•hydrogenbondingbetweenmoleculesofprimaryamines;hydrogenbondingless strong than that in alcohols

CH2CH3

CH3CH2

CH2CH3

key:

hydrogen bond

N

HH

N

HH

N

HH

•boiling points of primary amineshigher than those of alkanes butgenerallylowerthanthoseofalcoholsof similar relativemolecularmasses

•h y d r o g e n b o n d i n g b e t w e e nprimary amine molecules and watermolecules

key:

hydrogen bond

CH2CH

3

NH

H

HH

O

H

H

O

•primary amines with less carbonatoms are very soluble inwater

ExamtipsExamtipsExamtipsExamtips ♦ Questionsmayask students toarrangecarboncompounds inorderofincreasing boiling point.

e.g.

CH3(CH2)2CH3<CH3(CH2)3CH3<CH3(CH2)3Cl <CH3(CH2)3OH

Attractions weak instantaneous dipole- permanent dipole- hydrogen between induceddipole attractions permanent dipole bonds molecules attractions

♦ The following carbon compounds aremiscible withwater:

– methanol, ethanol and propan-1-ol;

– ethanal and propanone;

– methanoic acid, ethanoic acid, propanoic acid and butanoic acid.

Topic 812 Chemistry of Carbon Compounds 13Unit 30 Isomerism

Example

CompoundsW,X,YandZareallcolourlessliquids.Suggesthowyouwoulddistinguishthe four compounds fromeachother.

CH3CH2OH (CH3)3COH CH3(CH2)2Br CH3(CH2)7OH

W X Y Z (5marks)

Answer

Distinguishing the liquidsbywater solubility

Addwater to the liquids. BothCH3CH2OHand (CH3)3COHaremisciblewithwater. (1)

Distinguishing the two liquidswhich aremisciblewithwater

Warmeachof these two liquidswith acidifiedpotassiumdichromate solution. (1)

OnlyCH3CH2OH turns thedichromate solution fromorange to green. (1)

Distinguishing the two liquidswhich arenotmisciblewithwater

Warm each of the two liquids which are not miscible with water with AgNO3(aq) inethanol. (1)

OnlyCH3(CH2)2Br gives a creamyprecipitate slowly. (1)

➤In questions of this type, it is common to distinguish the compounds bytheir water solubility before other chemical tests.

➤Hydrocarbons such as cyclohexane and cyclohexene often appear in thistypeof questions. They are insoluble inwater.

RemarksRemarks*

30.1 Isomerism

30.2 Structural isomerism

30.3 Geometrical isomerism

30.4 Physical properties of geometrical isomers

30.5 Chirality

30.6 Enantiomers

30.7 Test for chirality—planeof symmetry

30.8 Properties of enantiomers

IsomerismUnit 30

Topic 81� Chemistry of Carbon Compounds 1�Unit 30 Isomerism

30.1 – 30.2

Summary

The following charts show the classificationof isomers.

Example

Consider the isomeric compoundsX andY shownbelow:

compound X

OH

CHO

compound Y

OH

CHO

a) Name the typeof isomerism involved. (1mark)

b)Whichof the above compoundshas ahighermeltingpoint? Explain. (3 marks)

Answer

a) Position isomerism (1)

b)Themeltingpointof compoundY ishigher than thatof compoundX. (1)

Only compoundX can form intramolecularhydrogenbonds. (1)

CompoundY formsmore intermolecularhydrogenbonds than compoundXdoes. (1)

CH3CHCH2CH3

CH3

isomersdifferent compounds that have the same molecular formula

structural isomersatoms are linked in different orders

stereoisomersatoms are linked in the same way but with different spatial arrangements

structural isomers

chain isomersisomers with the same functional group but different carbon skeletonse.g.

CH3CH2CH2CH2CH3

and

functional group isomers

isomers with the same molecular formula but different functional groupse.g.

and

position isomersisomers with the same carbon skeleton and functional group, but the position of the functional group is differente.g.

CH3CH2CH2CH2OH

and

CH3CHCH2CH3

OH

CH3C CH3

O

O

CH3CH2C OH

O

ExamtipsExamtipsExamtipsExamtips ♦ When ask about the type of isomerism, give the precise type, insteadof just stating structural isomerism.

e.g.

OH andH3C OCH3

The typeof isomerism involved is functional group isomerism.

♦ Questionsoftenaskaboutmethodsfordistinguishingbetweenisomericcompounds.

e.g.

OH andH3C OCH3

canbedistinguished by

– a physical method

comparingtheirboilingpoints/meltingpoints; OHH3Chas a higher boiling point /melting point.

– a spectroscopic method

comparing their IR spectra; OHH3C has a broad and

strong absorption at 3230 – 3670cm–1.


Example

Consider the melting points and boiling points of cis-1,2-dichloroethene and trans-1,2-dichloroethene.

Compound Melting point (°C) Boiling point (°C)

cis-1,2-dichloroethene –80 60

trans-1,2-dichloroethene –50 48

Explainwhy

a) cis-1,2-dichloroethenehas ahigherboilingpoint; (3marks)

b)trans-1,2-dichloroethenehas ahighermeltingpoint. (2marks)

Answer

a) Theboilingpointof a compounddependson its intermolecular attractions. (1)

In a molecule of the cis isomer, the dipole moments of the two polar C–Cl bondsreinforce each other. Thus, the molecule has a net dipole moment. Thus, thesemolecules areheld togetherbypermanentdipole-permanentdipole attractions. (1)

Inamoleculeofthetrans isomer,thedipolemomentsofthetwopolarC–Clbondscanceleachother.Thus,themoleculehasnodipolemoment.Thus,thesemoleculesareheld togetherbyweaker instantaneousdipole-induceddipole attractions. (1)

b)Inadditiontointermolecularattractions,themeltingpointofacompounddependsalsoon thedegreeof compactnessofmolecules in the solid state. (1)

The cis isomer has a lower degree of symmetry. It fits into a crystalline latticerelativelypoor and thushas a lowermeltingpoint. (1)

30.3 – 30.4

Summary

1 Stereoisomers thathave adifferent arrangementof their atoms in spacedue to therestricted rotation about a carbon-carbondoublebond are geometrical isomers.

2 Compoundswith twodifferentgroupsattached toeachcarbonof thedoublebondhave two alternative structures,which are geometrical isomers.

andCC

CH3CH3

HH

cis-isomer

CC

HCH3

CH3H

trans-isomer

3 Geometrical isomers normally have similar chemical properties. However, theirphysical properties areoftenquitedifferent.

➤Questions often ask students to compare the melting points and boilingpoints of geometrical isomers.

e.g.

m.p. 102 °C m.p. –19 °C

CCCH3OOC

H

H

COOCH3

CCH

CH3OOC

H

COOCH3

Their intermolecular attractions are van der Waals’ forces of comparablestrength.

The trans isomer has a higher melting point because it is moresymmetrical.

RemarksRemarks*

➤Questions often ask students to compare the melting point / volatility ofisomeric compounds.

➤CompoundXformsintramolecularhydrogenbondsduetothecloseproximityof the –OHgroup and –CHOgroup.

O

C

H

O

H

hydrogen bond

RemarksRemarks*

ExamtipsExamtipsExamtipsExamtips

♦ andCCH3C

H3C

Br

ClCC

H3C

H3C

Cl

BrareNOT

geometrical isomers. They are identical molecules.

Oneof the carbon atomsof thedouble bondhas twomethyl groupsattached to it.


ExamtipsExamtipsExamtipsExamtips ♦ Questions often ask about enantiomers.

♦ Questions may ask students to mark chiral carbons on the structuresof unfamiliar compounds.

e.g.

Tamiflu Aspartame

♦ DONOT confuse the terms ‘chiral’ and ‘achiral’.

30.5 – 30.8

Summary

1 Achiralmolecule isdefinedasone that isnot superposableon itsmirror image.Achiralmolecule and itsmirror image are called apair of enantiomers.

2 Most simple chiral molecules contain one carbon atom bonded to four differentatomsor groupsof atoms. Such a carbonatom is called a chiral carbon.

3 A pair of enantiomers have the same physical property and chemical property,except

a) their behaviour towardsplane-polarized light; and

b)their reactionswith chiral reagents.

Example

CompoundXhas the following structural formula:

CH(CH3)COOH

The above structural formula can represent two stereoisomers.

a) Draw three-dimensional structuresof the two stereoisomers. (2marks)

b)State aphysical propertywhich is different for the two stereoisomers. (1 mark)

CH2CH3

CH3

CH3CH2

CH2CH3

CHNH2

C

NH

O

OC

OO

* **

NH

CH2

CH

CH2

CH

CO2H

H2N OCH3

O

C

O

C* *

➤Students need to give good drawings of three-dimensional structures. Usetheconventionscommonlyused in the representationof three-dimensionalstructures.

➤Questionsoftenaskstudentstosuggestadifferentphysicalpropertybetweena pair of enantiomers. One of them turns the plane of polarization of abeamof plane-polarized light clockwise,while theother anticlockwise.

RemarksRemarks*

Answer

a)

COOH

CH3

C

HHOOC

CH3

C

H(1) (1)

b)Theyrotate theplaneofpolarizationofabeamofplane-polarized light tooppositedirections. (1)

Topic 820 Chemistry of Carbon Compounds 21Unit 31 Typical reactions of selected functional groups

31.1 – 31.7

Summary

1 The figurebelow summarizes the addition reactionsof alkenes.

HBr(g) Br2(aq)

RCH2CH3

alkane

RCHCH2

diol

RCHBrCH2Br

dibromoalkaneOHOH

RCHCH3

major product minor product

bromoalkanes

Br

RCHCH2Br

bromoalcohol

RCH2CH2Br RCHBrCH2Br

dibromoalkane

+ +OH

RCH CH2

alkene

Br2

(in organic solvent)

cold alkalinedilute potassiumpermanganate

solution

H2(g)Pt catalyst

2 Markovnikov’s rule for addition reactionof an asymmetric alkene:

When a molecule HA adds to an asymmetric alkene, the major product is the oneinwhichthehydrogenatomattachesitselftothecarbonatomalreadycarryingthelargernumberofhydrogen atoms.

3 Substitution reactionsofhaloalkanes—alkalinehydrolysis ofhaloalkanes

R OHR XNaOH(aq)

reflux

31.1 Introduction

31.2 Important reactionsof alkanes

31.3 Addition reactionsof alkenes

31.4 Addition ofhydrogen to alkenes

31.5 Addition ofhalogens to alkenes

31.6 Addition ofhydrogenhalides to alkenes

31.7 Substitution reactionsofhaloalkanes

31.8 Reactions of alcohols

31.9 Reactions of aldehydes andketones

31.10 Reactions of carboxylic acids

31.11 Hydrolysis of esters

31.12 Hydrolysis of amides

Typical reactions of selected functional groupsUnit 31

Topic 822 Chemistry of Carbon Compounds 23Unit 31 Typical reactions of selected functional groups

Example

Describe, by giving reagent(s) and stating observations, how you could distinguishbetween the following two compoundsusing a simple test tube reaction.

(4marks)

CH3CH2CHCH3

compound X

and

Cl

CH3CH2CHCH3

compound Y

I

ExamtipsExamtipsExamtipsExamtips ♦ Questionsoftenaskstudentstopredictthemajorproductofanadditionreaction involving an asymmetrical alkene.

e.g.

CH3

CH3

CH3HBr

H

C C

C2H5

CH3

C2H5HBr

H

C C

HI

CH3

CH3 CH3

Br C

H

C H

CH3

C2H5 C2H5

Br C

H

C H

I

♦ Questions may ask students to deduce the structure of a compoundbasedon the typeof isomerism it can exhibit and its reactions.

Given information:

– Molecular formula C6H12

– It has a pair of enantiomers.

– It loses its chiral centre after hydrogenation.

Deductions:

– The compound should be an alkene as it can undergohydrogenation.

– The compound should have a chiral carbon.

– Structures of the compound:

H2C=CH

C2H5

CH3

C HHC=CH2

C2H5

H3C

CH

Answer

Put about 2cm3 of ethanol and 1cm3 of silver nitrate solution in each of two testtubes. (1)

Place the test tube in awater bath at 60°C. (1)

Add several dropsof compoundsX andY separately to each test tube.

Ayellowprecipitate forms rapidly in the test tube containing compoundY. (1)

Awhiteprecipitate forms slowly in the test tube containing compoundX. (1)

➤Questions often ask about chemical tests for distinguishing haloalkanes.RemarksRemarks*

31.8 – 31.9

Summary

CH3CH CH2

propene

CH3CH2CHO

propanal

C3H7Cl

1-chloropropane

C3H7OH

propan-1-ol

(1° alcohol)

C3H7Br

1-bromopropane

CH3COOC3H7

propyl ethanoate

CH3CH2COOH

propanoic acid

C3H7I

1-iodopropane

• reflux with conc. HCl +

ZnCl2 catalyst; or

• mix with PCl5 ; or

• reflux with SOCl2• reflux with NaBr + conc. H2SO4; or• reflux with red P + Br2

• re

flux

with N

aI +

co

nc. H 3P

O 4; or

• re

flux

with re

d P

+ I 2

CH3COOH +conc. H2SO4

• K2Cr

2O7 / H

3O +, reflux; or

• KMnO

4 / H3O +, reflux

1 LiAlH4 / ethoxyethane

2 H3O +

K 2Cr 2O

7/ H

3O+

disti

l off

the p

ropa

nal

1 Li

AlH4 /

ethox

yeth

ane

2 H 3O

+

• excess conc. H2SO4, 180 °C; or • Al2O3, 300 °C

K2Cr2O7 /H3O

+, heat

CH3CH(OH)CH3

propan-2-ol

CH3COCH3

propanone

(2° alcohol)1 LiAlH4 / ethoxyethane2 H3O

+

K2Cr2O7 / H3O+

reflux

(CH3)3COH

methylpropan-2-olno reaction

(3° alcohol)

K2Cr2O7 / H3O+

reflux

Topic 82� Chemistry of Carbon Compounds 2�Unit 31 Typical reactions of selected functional groups

Example

Describe, by giving reagent(s) and stating observations, how you could distinguishbetween the following two compoundsusing a simple test tube reaction.

(2marks)

OH

compound X

andCH3

CH2OH

compound Y

Answer

Warmeach compoundwith acidifiedpotassiumdichromate solution. (1)

Only compoundY turns thedichromate solution fromorange to green. (1)

➤Theoxidation reaction is commonly used to distinguish

– aprimary / secondary alcohol froma tertiary alcohol;

– an aldehyde froma ketone.

RemarksRemarks*

TestCompound

Alcohols Aldehydes Ketones

WarmwithK2Cr2O7 /H3O

+

1° and2° alcohols— clear orangesolution turnsgreen almostimmediately

clearorangesolution

turnsgreen

noobservable

change

3° alcohols —noobservable change

ExamtipsExamtipsExamtipsExamtips ♦ Theformationofhydrogenchloridefumesuponreactionwithphosphoruspentachloride is a test for the presence of a hydroxyl group in acompound.

♦ The relative ease with which alcohols undergo dehydration shows thefollowing order:

Ease of dehydration: 3°>2°>1° alcohol

♦ Questions often ask about theoxidation of alcohols:

– the oxidizing agent required;

– the name(s) of theproduct(s).

♦ Remember that LiAlH4will NOT act uponC=Cbonds.

♦ The transformation of –COOH to –CH2OH involves 2 steps: reductionby LiAlH4, followedby treatmentwithH3O

+.

31.10 – 31.12

Summary

CH3CH2OH + conc. H2SO4 CH3CH2COOH

propanoic acid

CH3CH2COO– NH4+

ammonium propanoate

CH3CH2CONH2

propanamide

CH3CH2CH2OH

propan-1-ol

aqueousNH3

heat

1 LiAlH4 / ethoxyethane2 H3O

+

CH3CH2COO– Na+

sodium propanoate

CH3CH2COOH

propanoic acid

CH3CH2COO– Na+

sodium propanoate

CH3CH2COOH

propanoic acid

++

CH3CH2OH

ethanol

CH3CH2OH

ethanol

NaOHsolution

H2O /H3O

+

NaOHsolution

H2O /H3O

+

CH3CH2COOCH2CH3

ethyl propanoate

ExamtipsExamtipsExamtipsExamtips ♦ When propan-1-ol and propanoic acid are heated under reflux toproduceanester,theestercanbeseparatedfromthereactionmixtureby fractional distillation or using a separating funnel.

♦ Giventhestructureofanester,studentsshouldbeabletoanalyzetheester and deduce the alcohol and carboxylic acid forming the ester.

e.g.

The active ingredient of a superglue has the following structure:

H

H CN

C C

C

O

OCH3

It is an ester formed from

H

H CN

C

COOH

C and CH3OH.

Topic 82� Chemistry of Carbon Compounds 2�Unit 32 Synthesis of carbon compounds

Example

Oil of wintergreen is a common ester. Salicylic acid can be obtained from it in twosteps.

oil of wintergreen

Step 1a sodium salt of

salicylic acid+

CH3OH

COOCH3

OH

salicylic acid

COOH

OH

Step 2

a) Give the reagent and conditionused in Step 1. (2marks)

b)Suggest a structure for the sodium salt of salicylic acid formed in Step 1. (1mark)

c) Suggest a reagent that canbeused in Step 2. (1mark)

Answer

a) Sodiumhydroxide solution (1)

Heatunder reflux (1)

b)

(1)

COO– Na+

OH

c) Dilute sulphuric acid / dilutehydrochloric acid (1)

♦ Questions often give carbon compounds with similar structures (such

as

O

H C O CH3 and OHCH3

O

C ) and compare them.

– Whether theyhave the samemolecular formula / relativemolecularmass.

– Whether they are soluble inwater.

– Whether they have the sameodour.

– Whether they have the sameboiling point.

– Whether they have the same chemical properties.

Totacklethesequestions,first identifythehomologousseriestowhicheachcompoundbelongs.Thenansweraccordingtothegeneralpropertiesof the series concerned.

➤To draw the structure of sodium salt of salicyclic acid in (b), just use thegeneral equation for thehydrolysis of an ester for deduction.

O– Na+

O

R C O NaOH+ R1 O HR1

O

R C +

➤Students may also work backwards from the structure of salicyclic acid todeduce the structure of its sodium salt.

RemarksRemarks*

32.1 Introduction

32.2 Planning a synthesis

32.3 Problems indevising a synthesis

32.4 Laboratory preparationof simple carbon compounds

32.5 Preparing 1-bromobutane in the laboratory

Synthesis of carbon compoundsUnit 32

Topic 82� Chemistry of Carbon Compounds 2�Unit 32 Synthesis of carbon compounds

c)OH–(aq)

heat

(1) (0.5)

Cl OH

conc. H2SO4

heat

or conc. H3PO4

heat(1) (0.5)

Br2 (in organic

solvent)

(1)

Br

Br

➤Inpart(a),theimmediateprecursorofthetargetmolecule(analcohol)maybe a carboxylic acidor a haloalkane.As a carboxylic acid canbeobtainedfrom the starting molecule (an amide) via hydrolysis, so the synthesis canbedone in the two steps shown.

➤Many conversions involve

– thehydrolysis of haloalkanes to alcohols;

– theoxidation of alcohols to ketones and carboxylic acids;

– the reduction of aldehydes, ketones and carboxylic acids to alcohols.

RemarksRemarks*

32.4 – 32.5

Summary

The laboratorypreparationof a carbon compound involves fivemain stages:

Stage1—planning thepreparation;

Stage2—carryingout the reaction toproduce thedesiredproduct;

Stage3— separating the crudeproduct from the reactionmixture;

Stage4—purifying anddrying theproduct;

Stage5—calculating thepercentage yieldof theproduct.

Common separation and purification methods for products

Type of product Separation and purification method to employ

Liquidproduct

•simple distillation

•fractional distillation

•liquid-liquid extraction

Solidproduct •re-crystallization

32.1 – 32.3

Summary

Steps fordevising a synthesis:

a) identify an immediateprecursor to the targetmolecule;

b)continueuntil the startingmolecule is reached.

targetmolecule 1st precursor 2ndprecurso startingmolecule

ExamtipsExamtipsExamtipsExamtips ♦ The synthesesdiscussedwouldNOT involvea change in the lengthofthe carbon chain.

♦ Studentsshouldbefamiliarwiththereactionsoforganicfunctionalgroupsin order to suggest workable synthetic routes for the transformations.

Example

Outlineasyntheticcroute, inNOTMORETHANTHREESTEPS, toaccomplisheachofthe following conversions. For each step, give the reagent(s), the conditions and thestructureof theproduct.

a) CH3CH2CH2CONH2 CH3CH2CH2CH2OH (3marks)

b)

(4marks)

CH2CH3 COCH3

c)

(4marks)

Cl

Br

Br

Answer

a) CH3CH2CH2CONH2H2O /H3O

+

heat (1)CH3CH2CH2COOH (1)

1 LiAlH4 / ethoxyethane2 H3O

+ (1)CH3CH2CH2CH2OH

b) CH2CH3

(1) (0.5)

CHBrCH3

Br2

UV light or heat

(1) (0.5)

CH(OH)CH3

OH–(aq)

heat

(1)

COCH3

K2Cr2O7 / H3O+

reflux

Topic 830 Chemistry of Carbon Compounds 31Unit 32 Synthesis of carbon compounds

Example

2-chloro-2-methylpropane can be prepared by reacting methylpropan-2-ol withconcentratedhydrochloric acid.

(CH3)3COH(l) +HCl(aq) (CH3)3CCl(l) +H2O(l)

Thispreparation follows the stepsoutlinedbelow:

Step 1 Shake excess concentrated hydrochloric acid with methylpropan-2-ol in aseparating funnel.

Step 2 Separate theorganic layer from the aqueous layer.

Step 3 Wash the crudeproductwith10% sodiumhydrogencarbonate solution.

Step 4 Dry theproductwith a suitable reagent.

Step 5 Purify thedriedproduct to remove the remainingmethylpropan-2-ol.

The table below lists some information of methylpropan-2-ol and 2-chloro-2-methylpropane:

Compound methylpropan-2-ol 2-chloro-2-methylpropane

Density (gcm–3) 0.78 0.84

Boilingpoint (°C) 82 51

Water solubility miscible very slightly soluble

a) Draw a diagram of the separating funnel after the reaction has taken place in Step 1, labelling clearly the aqueous layer andorganic layer. (2marks)

b)SuggestONEadvantageofusing a separating funnel to carryout Step 1. (1 mark)

c) OutlinetheexperimentalprocedureforwashingthecrudeproductinStep 3.Includethenecessary safetyprecautions. (4marks)

d)Suggest a suitable reagent fordrying in Step 4. (1mark)

e) Nameamethod forpurifying thedriedproduct in Step 5. (1mark)

ExamtipsExamtipsExamtipsExamtips ♦ DONOT confuse an experimental set-up for fractional distillationwiththat for simple distillation.

♦ Questionsoftenaskstudentstodescribethere-crystallizationprocedurefor thepurification of a crude solid product.

Answer

a)

(2)

aqueous layer

organic layer

b)Anyoneof the following:

• Allow better mixing of the reactants by vigorous shaking of the separatingfunnel. (1)

• Alloweasyisolationoftheproductbydrainingouttheloweraqueouslayerfromthe separating funnel. (1)

c) Add slowly 10% sodium hydrogencarbonate solution. Wait until no effervescenceoccurs. (1)

Stopper and shake the separating funnel vigorously. (1)

Duringshaking,openthetapoftheseparatingfunnelregularlytoreleasethepressurebuilt inside the funnel. (1)

Remove the aqueous layer. Repeat the washing procedure until no effervescenceoccursupon the additionof sodiumhydrogencarbonate solution. (1)

d)Anhydrous calciumchloride / sodium sulphate (1)

e) Simpledistillation / fractionaldistillation (1)

➤Questionsoftenaskaboutthepurposesofdifferentstepsinthepreparationsof carbon compounds.

➤When giving a reagent for drying, remember to include the word‘anhydrous’.

RemarksRemarks*

Topic 832 Chemistry of Carbon Compounds 33Unit 33 Important organic substances

33.1 – 33.9 & 33.13

Summary

1 a) The structure of the active ingredient of aspirin tablets, acetylsalicylic acid, isshownbelow. It contains two functional groups.

CH3O C

O

OHC

Ocarboxyl group

ester group

b)Thepercentagebymassofacetylsalicylicacidinaspirintabletscanbedeterminedbyback titration.

2 a) There are two typesofdetergents:

i) soap—made fromnatural fats andoils; and

ii)soaplessdetergents—made fromchemicals derived frompetroleum.

b)The structureof a typical anionicdetergent is shownbelow:

hydrophobichydrocarbon ‘tail’

hydrophilic anionic ‘head’

c) Adetergenthelpswater to removedirt by

i) the ability to act as awetting agent; and

ii)the emulsifying action.

3 Fats andoils aremixed triglycerides.

4 Soap is made by boiling natural oils or fats with a strong alkali, usually sodiumhydroxideorpotassiumhydroxide solution.

33.1 Introduction

33.2 Aspirin — fromherbal remedy tomoderndrug

33.3 Analyzing aspirin tablets byback titration

33.4 Detergents

33.5 How dodetergentshelpwater to clean?

33.6 Thewettingandemulsifyingpropertiesofdetergentsinrelationto their structures

33.7 The cleaning actionofdetergents

33.8 Manufacture of soaps and soaplessdetergents

33.9 Thecleaningabilitiesofsoapsandsoaplessdetergentsinhardwater

33.10 Nylons

33.11 Polyesters

33.12 Carbohydrates

33.13 Lipids

33.14 Proteins andpolypeptides

Important organic substancesUnit 33

Topic 83� Chemistry of Carbon Compounds 3�Unit 33 Important organic substances

Example

The structureof themain chemical constituentof an animal fat is shownbelow:

(CH2)nCH3

(CH2)nCH3

(CH2)nCH3

O C

O

O C

O

O C

O

C

C

H C

H

H

H

H

ExamtipsExamtipsExamtipsExamtips ♦ Studentsshouldbeabletodrawthecorrectstructureoftheesterlinkagein a triglyceride (see the structure shown below). The three carboxylicacids bearing long chains are attached to a glycerol backbone.

♦ Whenfatsandoilsareheatedwithsodiumhydroxidesolution,theyarehydrolyzed first to form glycerol and carboxylic acids. The acids thenreact with the alkali to form sodium carboxylates, which are soaps.Such reactions are called saponification.

♦ Questions often give the structure of a soap / soapless detergent andask about its properties.

– Whether it is made from natural fats and oil or from hydrocarbonsobtained frompetroleum.

– Whether it forms scum in hardwater / seawater.

– Whether it is biodegradable.

♦ Hydrogenation of vegetable oils producesmargarines.

R

R

R

OH

OH

OH

O C

O

O C

O

O C

O

C

C

H C

H

H

H

H

3NaOH+

triglyceride in fat or oil

C

C

H C

H

H

H

H

+

glycerol

alkali

3RCOO– Na+

sodiumcarboxylate

(soap)

a) Animal fat is oneof the rawmaterials in theproductionofdetergentX.

i) NameanotherrawmaterialthatisneededintheproductionofdetergentXfroman animal fat. (1mark)

ii)Name the typeof reaction that takesplace in theproductionofdetergentX.(1mark)

iii)Write a chemical equation for the reaction involved in thepreparation. (1mark)

b)An oil tanker was wrecked and spilt a lot of crude oil in the sea. State whetherdetergentX is suitable for treating theoil spill. Explainyour answer. (3marks)

Answer

a) i) Sodiumhydroxide (1)

ii)Saponification / alkalinehydrolysis (1)

iii)

(1)

(CH2)nCH3

(CH2)nCH3

(CH2)nCH3

O C

O

O C

O

O C

O

C

C

H C

H

H

H

H

3NaOH+

C

C

H C

H

H

H

H

H

H

HO

O

O

+ 3CH3(CH2)nCOO– Na+

b)Not suitable (1)

Seawater contains a lot ofmetal ions, such as calcium ions andmagnesium ions.(1)

DetergentXwill reactwith thesemetal ions to form scum. (1)


33.10 – 33.12 & 33.14

Summary

1 a) Nylons arepolyamideswhich contain the amide linkage N

H

C

O

.

b)Nylons are formedby condensationpolymerization.

c) The repeatingunit ofnylon-6,6 is

from diamine

H

N

O

C (CH2)4 C

O

H

N (CH2)6

from dicarboxylic acid

2 a) Polyesters contain the ester linkage

O

C O .

b)Polyesters are formedby condensationpolymerization.

c) The repeatingunit ofpoly(ethylene terephthalate) is

from dicarboxylic acid from diol

O

C

O

C O CH2 OCH2

3 a) Glucose can exist inopen-chain and cyclic forms.

b)The open-chain form of glucose contains an aldehyde group and five hydroxylgroups.

4 a) Fructose can exist inopen-chain and cyclic forms.

b)The open-chain form of fructose contains a keto group and five hydroxylgroups.

5 a) Whentwoaminoacidmoleculesundergocondensationreaction,awatermoleculeis eliminated and apeptide link forms.Theproduct is adipeptide.

+H N

R

HH

C

O

C OH

H N

R

HH

C

O

C

H

+ H2O

N

R1

HH

C

O

C OH

N

R1

HH

C

O

C OH

amino acid 1 amino acid 2

peptide link (or amide group)

a dipeptide

b)The peptide links in peptides and proteins can be hydrolyzed to release theindividual amino acids.

ExamtipsExamtipsExamtipsExamtips ♦ Acondensation reaction isa reaction inwhich twoormoremoleculesreacttogethertoformalargermoleculewiththeeliminationofasmallmolecule such aswater.

♦ Nylon and poly(ethylene terephthalate) are thermoplastics as well ascondensation polymers. NOT all thermoplastics are addition polymers.

♦ Questions often give the repeating unit of a polymer and ask aboutinformation concerning thepolymer.

e.g.

– whether it is an addition polymer or a condensation polymer;

– whether it is formed from one monomer or two differentmonomers.

➤Questions often ask about the purpose of adding concentrated sodiumchloride solution in soappreparation.

Concentrated sodium chloride solution is added to salt out the soapproduced.

➤Questionsoftenaskstudentstogivetheproductsformedfromthecompletehydrolysis of triglycerides existing in natural fats andoils.

e.g.

OH

OH

OH

CH2

CH

CH2 (CH2)16CH3

(CH2)7CH

(CH2)16CH3

CH(CH2)7CH3

O C

O

O C

O

O C

OCH2

CH

CH2

+ CH3(CH2)16COOH +

+

complete

hydrolysis

CH3(CH2)7

H

H

(CH2)7COOHC C

CH3(CH2)7

H

(CH2)7COOH

HC C

RemarksRemarks*


Example

Twaron is a heat resistant, high strength fibre used in protective clothing. A shortsectionof the structureofTwaron is shownbelow.

O

CC

H

N

O N

N

O

CC

H

N

O N

N

O

CC

O

a) Draw the repeatingunit ofTwaron. (1mark)

b)Draw the structuresof the twomonomers that couldbeused toprepareTwaron. (2marks)

c) ExplainwhyTwaronhas a great strength. (1mark)

Answer

a)

(1)H

N

N

N

O

CC

O

b) NH2H2N (1)

COOHHOOC (1)

c) There are stronghydrogenbondsbetween chainsofTwaron. (1)

➤Given the structure of a polymer, students should be able to deduce thestructure(s) ofmonomer(s) used to produce thepolymer.

Look at thepolymer structure.

– If the polymer backbone contains only carbon atoms, then the polymeris an addition polymer produced from one monomer.

e.g.

C6H5

C

H

H

C

H

C6H5

C

H

H

C

H

C6H5

C

H

H

C

H

is an addition polymer formed

from the following monomer:

C6H5

C

H

H

C

H

– If the polymer backbone contains other atoms, then the polymer is acondensationpolymer,probablyproducedfromtwomonomers(seeTwaronshownabove).

➤Some condensation polymers are produced from one monomer.

e.g.

the repeating unit of polylactide is shown below:

CH3

O C

H

C

O

It is a condensation polymer formed from one monomer, lactic acid.

HO OH

lactic acid

CH3

C

H

C

O

RemarksRemarks*

topic chemistry of carbon compounds -...

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