Worked solutions: Chapter 11 Organic chemistry · ... Chapter 11 Organic chemistry This page from...

Worked solutions: Chapter 11 Organic chemistry

© Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2008.

This page from the Chemistry: For use with the IB Diploma Programme SL Teacher’s Resource may be reproduced for classroom use.

Section 11.1 Exercises

1 Members of the same homologous series have:

• the same general formula, formulas of successive members differ by CH2

• similar chemical properties

• a gradual change in physical properties.

2 a C4H10

b C5H10

c C6H14

d C4H8

a c

b d

3 a alkene

b alkene

c alkane

d alkane




4

5 a 2-methylpentane

b but-1-ene

c but-2-ene

d pent-2-ene

6 a 2-methylbutane

b pent-2-ene

c 3-methylpentane

d propene

7 Pentane (C5H12) has straight chain molecules. These molecules pack closely together and are bonded together by van der Waals’ forces. 2-Methybutane (C5H12) has branched molecules. These branched molecules do not pack together so closely, leading to weaker van der Waals’ forces and hence a lower boiling point.

8

9 m(H2O) = 1.442 g

n(H2O) = 02.18

442.1 = 0.08002 mol ∴ n(H) = 0.1600 mol

m(CO2) = 3.521 g

n(CO2) = 01.44

521.3 = 0.08000 mol = n(C)




C : H 0.08000 : 0.1600

1 : 2 The empirical formula is CH2.

n(vapour) = mV

V = 4.22

797.0 = 3.558 × 10–2 mol

M = nm = 210558.3

000.2−×

= 56.21 g mol–1

Empirical formula mass = 12.01 + 2 × 1.01 = 14.03

03.1421.56 = 4

The molecular formula is C4H8.

10 m(H2O) = 0.736 g

n(H2O) = 02.18

736.0 = 0.04084 mol ∴ n(H) = 0.08169 mol

m(CO2) = 1.80

n(CO2) = 01.44

80.1 = 0.04090 mol = n(C)

C : H

0.04090 : 0.08169

1 : 2

The empirical formula is CH2.

PV = nRT

n(vapour) = RTPV =

37331.800.184

×× = 0.0271 mol

M = nm =

0271.0279.2 = 84.10 g mol–1

Empirical formula mass = 12.01 + 2 × 1.01 = 14.03

03.1410.84 = 6

The molecular formula is C6H12.





1 a alcohol

b ester

c chloroalkane

d carboxylic acid

e amine

2

a c

b

d

3 a propanal

b butanone

c 1,2-dichloroethane

d hex-2-ene

e methanoic acid

f pentan-3-ol

4 a 2-methylbutane

b 2,3-dichlorobutane

c butanone

d pentanal

e propanoic acid

5 a It is a homologous series.

b Boiling point would increase as the size of the molecules increases, due to increasing van der Waals’ forces between molecules.




c i They are more soluble, as the NH2 group is polar, allowing for some hydrogen bonding with water.

ii They are less volatile, as they have stronger intermolecular bonding and hence higher boiling points.

6 a b c

Names: a pentan-1-ol; b pentan-2-ol; c pentan-3-ol

7 a Alcohols have hydrogen bonding between their molecules (in addition to van der Waals’ forces). These bonds are stronger than the van der Waals’ forces found between the alkane molecules; hence their boiling points are higher.

b Van der Waals’ forces become the more significant intermolecular forces as the hydrocarbon chain increases in length. The chain is similar in the alcohols and alkanes with the same number of carbon atoms, hence the boiling points are similar.

8

Semi-structural formulas are CH3CH2CH2OH, CH3CH(OH)CH3, CH3OCH2CH3

9 a

ethanol hexan-1-ol

Semi-structural formulas are CH3CH2OH and CH3(CH2)4CH2OH.

b The –OH group in ethanol has only one possible position, hence no number is needed. In hexanol, the –OH could be placed in a number of positions (e.g. on C1, C2 or C3), hence a number is required to show its position.




10 a Van der Waals’ forces between octane (C8H18) molecules are stronger than van der Waals’ forces between propane (C3H8) molecules due to the larger molecule size. The stronger the van der Waals’ forces, the higher the boiling point.

b Van der Waals’ forces for ethane (C2H6) are weaker than those for butane (C4H10). Weaker forces and smaller molecules lead to greater volatility.

11 a tertiary

b secondary

c primary

d tertiary

12 a secondary

b primary

c tertiary

d secondary

13

a

c

amine ester

b

d

1: amine; 2: carboxyl group 1: amine; 2: carboxyl group; 3: benzene ring

14 a butanoic acid

b hexan-2-one

c 2-methylbutan-2-ol

d butanal





1 a C5H12 + 8O2 → 5CO2 + 6H2O

b 2C4H10 + 13O2 → 8CO2 + 10H2O

c 2C6H14 + 13O2 → 12CO + 14H2O

d 2C6H14 + 7O2 → 12C + 14H2O

2 Carbon monoxide is produced and this is poisonous as it combines with haemoglobin in preference to oxygen, starving the body of oxygen. Also, if carbon is produced it causes respiratory problems due to the minute particles in the air.

3 a substitution

b free-radical mechanism

c Homolytic fission occurs in which each atom of the halogen molecule receives one electron and two free radicals are produced.

4 a C2H6 + Br2 → C2H5Br + HBr

b chloroethane, dichloroethane, trichloroethane, tetrachloroethane, pentachloroethane, hexachloroethane

5 a C : H : Br

01.123.29 :

01.17.5 :

90.790.65

2.44 : 5.7 : 0.81

3 : 7 : 1

Empirical formula is C3H7Br.

Empirical formula mass = 123, Mr = 120 (approx)

Molecular formula is (C3H7Br)x where x = 123120 = 1

Molecular formula is C3H7Br.

b

Semi-structural formulas: CH3CH2CH2Br and CH3CHBrCH3

c 1-bromopropane and 2-bromopropane

d CH3CH2CH3 ⎯⎯⎯ →⎯ /catalystBr2 CH3CH2CH2Br + HBr

CH3CHCH2 ⎯⎯⎯ →⎯ stHBr/cataly CH3CH(Br)CH3





1 a i CH3CHCHCH3 ⎯⎯⎯ →⎯ /catalystH2 CH3CH2CH2CH3

ii CH2CH2 ⎯→⎯ 2Br CH2BrCH2Br

iii CH3CH2OH(1) + 3O2(g) → 2CO2(g) + 3H2O(1)

iv CH3CH3 ⎯⎯ →⎯ /UVCl2 CH3CH2Cl + HCl

b UV light

c Decolourization of the bromine solution would occur.

2 a CH3CH3 ⎯⎯ →⎯ /UVCl2 CH3CH2Cl + HCl

b CH2CH2 ⎯⎯⎯ →⎯ stHCl/cataly CH3CH2Cl

3 Add bromine to the hydrocarbon. If it is a saturated hydrocarbon there will be no effect and the liquid will retain the reddish colour of the bromine. If it is unsaturated, the bromine will be decolourized.

4 a addition

b n(Cl2) = n(alkene) and n = Mm

∴ n(Cl2) = 90.7091.5 = 0.0834

∴ n(alkene) = 0.0834 = M33.2

∴ M(alkene) = 27.95 = 28

∴ n = 2. The molecular formula is C2H4.

5 a CH3CHBr2 or CH2BrCH2Br (by substitution)

b CH2BrCH2Br only (by addition)

6 a CH2CH2 + H2O → CH3CH2OH

b Water is in the form of steam, and both steam and ethene are passed over a catalyst (phosphoric acid on silica) at 300°C and 70 atm pressure.

c CH2CH2 ⎯⎯⎯ →⎯ stHCl/cataly CH3CH2Cl

7 A polymer is a very large chain-like molecule that consists of many smaller units called monomers which are linked together to give the chain-like structure.

8 a nCH2CH2 → (CH2CH2)n

b nCH2CHCH3 → (CH2CHCH3)n

c nCH2CHCl → (CH2CHCl)n




9

a b

c

d

10


1 a CH3COOH, ethanoic acid

b CH3CH2OH, ethanol

c CH3CH2COCH3, butanone

d Tertiary alcohols resist oxidation so no product is formed.

2 a An aldehyde can form.

b The reaction is carried out at room temperature and the aldehyde is distilled from the mixture.

c It is heated under reflux.


1 A nucleophile may be any substance with an electron pair that can form a dative (coordinate) bond and is attracted to a positive charge centre.

2 They are nucleophilic substitution reactions.

3 The iodoalkane would react more rapidly than the chloroalkane as the strength of the C–I bond is less than that of the C–Cl bond.

4 Tertiary halogenoalkanes undergo SN1 mechanisms.




5 This is an SN2 mechanism.

6 As this is a tertiary halogenoalkane, it undergoes an SN1 mechanism.


1 a addition reaction

b addition polymerization

c addition reaction (hydrogenation)

d oxidation

2 a CH3CH3 ⎯⎯ →⎯ /UVCl2 CH3CH2Cl ⎯⎯ →⎯NaOH CH3CH2OH ⎯⎯⎯ →⎯+−/HOC 2

72r CH3COOH

ethane

Heat under reflux

ethanoic acid

b CH3CHCHCH3 ⎯⎯→⎯ OH 2 CH3CH2CH(OH)CH3 ⎯⎯⎯⎯ →⎯+− /HOC 2

72r CH3CH2COCH3

but-2-ene

Heat under reflux

butanone

c CH2CH2 ⎯⎯→⎯HCl CH3CH2Cl ⎯⎯ →⎯NaOH CH3CH2OH ⎯⎯⎯⎯⎯⎯⎯ →⎯+− /HOC 2

72r CH3CHO

ethene

Heat at room temperature

ethanal

Distill off product




Chapter 11 Review questions

1 a Homologous series: Molecules in a series in which each member has a CH2 more than the previous member.

b Saturated molecule: A molecule in which the carbon–carbon bonds are all single bonds.

c Structural isomers: Molecules with the same molecular formula, but different structures.

d Functional group: An atom or group of atoms that influences the chemical properties of a compound.

2 a Five isomers

b

Three isomers: but-1-ene, but-2-ene, 2-methylprop-1-ene

3 a There are too many hydrogens on the middle carbons.

b There is one hydrogen too many on the final carbon.




4

5 a C2H5OH

b C6H12

c C6H12

d CH3CH2COOCH3

6 a chloro group, carboxyl group, benzene ring

b carbonyl group (C=O), bromo group, hydroxyl group

c ester group (COO), chloro group, C=C double bond

7 a CH3(CH2)4CH3. It is the larger alkane, and hence has stronger van der Waals’ forces.

b NaCl. Ionic bonding produces much stronger bonds than the dipole–dipole intermolecular forces in CH3Cl.

c CH3CH2OH. Hydrogen bonding between molecules is stronger than the dipole–dipole bonds between CH3OCH3 molecules (similar van der Waals’ forces for both).

8 a 2-methylpentane

b 2,2-dimethylbutane

c 2,3-dimethylbutane

9 a Pentan-1-ol has hydrogen bonding and van der Waals’ forces between its molecules. These bonds are stronger than the van der Waals’ forces between pentane molecules, so the boiling point is higher.

b CH3CH2CH2CH2COOH (pentanoic acid)

10 A = ethanol, B = ethanoic acid, C = bromoethane, D = 1,2-dichloroethane

11 a C : H : O

01.1260 :

01.13.13 :

00.167.26

4.996 : 13.17 : 1.669




3 : 8 : 1

Empirical formula is C3H8O.

Mr(C3H8O) = 60 ∴ Molecular formula is C3H8O.

b

CH3CH2CH2OH CH3CH(OH)CH3

c propan-1-ol and propan-2-ol

12 a b

13

14 nCH2CHCH3 → (CH2CHCH3)n

15 butan-1-ol: hydrogen bonding; butanal: dipole–dipole attraction; butanoic acid: hydrogen bonding

butanal < butan-1-ol < butanoic acid

16 a

b The reaction between 1-chloropropane and sodium hydroxide would be slower than the

reaction between 1-bromopropane and sodium hydroxide.

17 In an SN1 reaction, only one species is required to form the reactive intermediate; in an SN2 reaction, two species are required to form the reactive intermediate. Any tertiary halogenoalkane will undergo an SN1 reaction; any primary halogenoalkane will undergo an SN2 reaction. Secondary halogenoalkanes may undergo SN1 or SN2 reactions.




18 a C3H8 + 5O2 → 3CO2 + 4H2O

b C4H8 + 4O2 → 4CO + 4H2O or C4H8 + 2O2 → 4C + 4H2O

c 2C3H7OH + 9O2 → 6CO2 + 8H2O

19 a SN2 mechanism

b SN1 mechanism

20 a A butane; B butan-1-ol; C butanal

b i free-radical substitution

ii nucleophilic substitution

iii oxidation

c SN2

d Reaction is carried out at room temperature and the aldehyde can be distilled off from the mixture.




Chapter 11 Test

Part A: Multiple-choice questions

Question Answer Explanation 1 C Each hydrocarbon is named by first choosing the longest carbon chain and

numbering them such that the substituents have the lowest possible number. Thus I is 2-methylpentane; II is 2,2-dimethylbutane and III is 2,3-dimethylbutane. Therefore, both I and III are correct.

2 C Double bonds are stronger than single bonds and hydrocarbons containing a carbon to carbon double bond are more reactive than those hydrocarbons containing carbon to carbon single bonds only. Statements I and II are therefore both correct.

3 D The combustion of a hydrocarbon produces either CO2, CO or C, depending on the amount of O2, and combustion reactions are always exothermic. C2H4 + 3O2 → 2CO2 + 2H2O. C2H4 + 2O2 → 2CO + 2H2O. Thus 2 mol of O2 will produce CO.

4 D Ethene undergoes an addition reaction with Br2 in which the carbon to carbon double bond is broken and each bromine is added to a carbon atom, therefore forming CH2BrCH2Br.

5 D M(C2H4O) = 44. Therefore the molecular formula is twice the empirical formula; C4H8O2. The only formula which corresponds to this is CH3CH2CH2COOH.

6 A Alcohols become less soluble in water the longer the hydrocarbon chain. The longer the hydrocarbon chain, the less hydrogen bonding can occur with water molecules because the non-polar part of the molecule is getting larger. Consequently, both statements are true.

7 B There is only 1 possible structure with 6 carbons in a chain; there are 2 isomers with 5 carbons in a chain and there are 2 isomers with 4 carbons in a chain. This gives a total of 5.

8 D The aldehyde functional group is –CHO. The only compound containing this is CH3CH2CHO.

9 A The products of incomplete combustion of a hydrocarbon are either CO and H2O or C and H2O.

10 A An alcohol undergoes oxidation when treated with acidified potassium dichromate (VI) to form an aldehyde, which can then be further oxidised to form a carboxylic acid. A is an aldehyde. Note that D is a tertiary alcohol which resists oxidation.

Part B: Short-answer questions

1 a Any two of the following would be satisfactory:

• They have the same general formula.

• Successive members differ by CH2.

• They have similar chemical properties.

• There is a gradual change in physical properties.

• They have same functional group. (2 marks)




b Add bromine to the hydrocarbon.

If the hydrocarbon is an alkane there is no change; the solution stays brown.

If the hydrocarbon is an alkene, the bromine decolourizes. (3 marks)

2 a Boiling point increases with an increase in the number of carbons. This is because the relative molecular mass increases, so the strength of intermolecular forces (van der Waals’ forces) increases and more energy is needed to separate the molecules.

(3 marks) b Methanol is polar (has an O–H group), therefore hydrogen bonding occurs between

molecules. This intermolecular attraction is stronger than the van der Waals’ forces in ethane and so it has a higher boiling point than ethane.

(2 marks)

3 a Acidified potassium dichromate(VI) is needed.

The reaction is an oxidation. (2 marks)

b During the reaction the colour changes from orange of dichromate ions to green of Cr3+ ions.

(1 mark)

4 a i C4H10 + Br2 → C4H9Br + HBr

CH3CH2CH2CH2Br or CH3CHBrCH2CH3 would represent the structure of the product.

(2 marks) ii Homolytic fission occurs in bromine.

The species that reacts with butane is a free radical (Br•). (2 marks)

b i C4H8 + Br2 → C4H8Br2

CH3CHBrCHBrCH3 would represent the structure of the product. (2 marks)

ii There is an addition reaction. (1 mark)




Part C: Data-based questions

1 The following points should be covered in your answer. Names of intermolecular forces:

• Van der Waals’ forces

• Dipole–dipole attractions

• Hydrogen bonding

Relative strengths: H-bonding > dipole–dipole > van der Waals’ Reasoning:

• Greater Mr means stronger van der Waals’ forces.

• Greater polarity means stronger dipole–dipole attractions.

• Stronger intermolecular forces means more energy needed to break these bonds.

In the parts of this question, the focus should be as follows:

a Ethane and butane: difference in van der Waals’ forces

b Ethane and bromoethane: van der Waals’ versus dipole–dipole

c Bromoethane and ethanol: dipole–dipole versus hydrogen bonding (8 marks)

2 a 2C2H5OH + O2 → 4CO2 + 6H2O (2 marks)

b i Mr(C5H9OH) = 88

∆Hөc = –3325 ± 25 kJ mol–1 (The graph should be extrapolated.)

(2 marks) ii You would expect that the value of the standard enthalpy of combustion of pentan-2-ol

should be about the same as that of pentan-1-ol because the same number and type of bonds are being broken and made in both compounds.

(2 marks)




Part D: Extended-response question

a i CH3CH2OCH2CH3 would have the lowest boiling point because it is the only one without hydrogen bonding (it has only dipole–dipole attractions).

(2 marks) ii T = (70 + 273) = 343 K

V = P

nRT

V = 51010.134331.80200.0

×××

= 0.518 dm3 or 518 cm3 (3 marks)

b The alcohol from which CH3COCH3 can be prepared is propan-2-ol. This occurs in an oxidation reaction and the other reagents required are potassium dichromate(VI) (or potassium manganate(VII)) and sulfuric acid.

The mixture should be heated under reflux. (5 marks)

c This is a secondary halogenoalkane.

i The intermediate in SN2 is

(2 marks)

ii SN1 mechanism: CH3CH2CH(CH3)Br → CH3CH2CH(CH3)+ + Br– CH3CH2CH(CH3)+ + OH– → CH3CH2CH(CH3)OH

(2 marks) d If 2-chlorobutane was used instead of 2-bromobutane the rate would decrease, as the C–Cl

bond is stronger than C–Br. (2 marks)

Worked solutions: Chapter 11 Organic chemistry · ... Chapter 11 Organic chemistry This page from...

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