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ANSWERS TO EXAM-STYLE QUESTIONS 10 1CHEMISTRY FOR THE IB DIPLOMA CAMBRIDGE UNIVERSITY PRESS 2011
Answers to exam-style questions
Chapter 10 1 A
2 C
3 B
4 C
5 C
6 D
7 A
8 C
9 C
10 B
11 a Alcohols contain the OH functional group.
butan-1-ol
butan-2-ol
2-methylpropan-2-ol
2-methylpropan-1-ol
b Only primary alcohols can be oxidised to a carboxylic acid. Primary alcohols have only one C attached to the carbon with the OH group on it. Butan-1-ol and 2-methylpropan-1-ol are primary alcohols. The only change that occurs is that the CH2OH group of the primary alcohol becomes a COOH group in the carboxylic acid the rest of the carbon skeleton remains unchanged.
butan-1-ol becomes:
H C
H
OH
C
H
C
H
C
H
H
H H H
H C
H
OH
C
H
C
H
C
H
H
HH H
H C
H
OH
C
CH3
C
H
H
HH
H C
H
OH
C
CH3
C
H
H
HH
H C
H H
C C
H O H
H HH
C
O
H
H
C C
O H
H H
C
O
CH3
2-methylpropan-1-ol becomes:
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2 CHEMISTRY FOR THE IB DIPLOMA CAMBRIDGE UNIVERSITY PRESS 2011ANSWERS TO EXAM-STYLE QUESTIONS 10
Suitable oxidising agents are: acidi ed potassium dichromate(VI), K2Cr2O7/H+ or just Cr2O72/H+;
colour change orange to green or acidi ed potassium manganate(VII), KMnO4/H+ or MnO4/H+;
colour change purple to colourless.c The isomer must have the same molecular formula but no OH
group. Possible isomers are:
H C
HO
C
Cl
H
HH
H + ClH C+ C H
HHHO
HCl
H
H
C
HO
C H
H
H
H C
H H
Br
C C
H
H H
C
H
H
H
13 a Both compounds must be shaken with bromine water. But-2-ene decolorises the bromine water (orange to colourless), but there is no change with butane.
b
C C O C C
H
H
H
H
H
H
H
H
H
H
H C O C
H
H
H
H
C
H
H
C
H
H
H
H C O C
H
H
H
CH3
C
H
H
H
12 a C2H6 + Cl2 C2H5Cl + HClb i free radical substitution ii Homolytic ssion occurs in the initiation step. In the presence
of ultraviolet light the ClCl bond breaks so that one electron goes back to each Cl atom:
UV Cl2 2Cl A free radical is an atom/group of atoms with an unpaired
electron. In this case, Cl is a free radical. A CH3CH2 free radical is also formed in this reaction.
A termination step occurs when two free radicals react together this ends a chain reaction. There are three possible termination steps in this mechanism:
Cl + Cl Cl2 Termination Cl + CH2CH3 CH3CH2Cl Termination CH2CH3 + CH2CH3 C4H10 Terminationc i CH3CH2Cl + NaOH CH3CH2OH + NaCl or, as an ionic equation: CH3CH2Cl + OH CH3CH2OH + Cl
ii Chloroethane is a primary halogenoalkane and therefore reacts by an SN2 mechanism:
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ANSWERS TO EXAM-STYLE QUESTIONS 10 3CHEMISTRY FOR THE IB DIPLOMA CAMBRIDGE UNIVERSITY PRESS 2011
14 aC H O
Percentage 48.6 8.2 43.2
Divide by Ar 48.6 / 12.01 8.2 / 1.01 43.2 / 16.00
Moles 4.05 8.12 2.7
Divide by smallest 4.05 / 2.7 8.12 / 2.7 2.7 / 2.7
Ratio 1.5 3.01 1
Multiplying by 2 to get whole numbers we get C3H6O2, which is the empirical formula. The empirical formula mass is (3 12.01 + 6 1.01 + 2 16) = 74.09. The empirical formula mass is the same as the relative molecular mass, so the molecular formula is the same as the empirical formula. The molecular formula is C3H6O2.
b i Esters contain the COOC functional group:
ethyl methanoate
methyl ethanoate
ii
c Butanone can be obtained by oxidation of a secondary alcohol, butan-2-ol. To convert but-2-ene to an alcohol we must add water across the double bond:
c Acids react with magnesium to form hydrogen, so this isomer must be a carboxylic acid:
conc. H2SO4
heatC
H
H
OH +H C
H
H
H H C
H
H
C
H
H
H
OC
O
+O
H HO
H
C
O
H
H H
C C
O H
H H
C
O
H
H H
C C
C HH H
O
O
H
H C
CH
H
C H
H
O
O
conc. H2SO4
heatC
H
H
C
H
H
OH + +H C
H
H
H C
H
H
OHC
O
OH H
O
H
HC
O
Cr2O72/H+
heat
conc. H2SO4
heatC
O
C
H
H
C
H
H
HC
H
H
C
H
H
C
HH
C C
OH
H
C
H
H
HH+ H2O
H3C CH3
C
H
H
H
C
H
CH3
C
H
CH3
C
H
CH3
C
H
CH3
C
H
CH3
C
H
CH3
d Only the C=C part of the molecule reacts all other groups remain unchanged:
or
2CH3CH2COOH + Mg (CH3CH2COO)2Mg + H2
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4 CHEMISTRY FOR THE IB DIPLOMA CAMBRIDGE UNIVERSITY PRESS 2011ANSWERS TO EXAM-STYLE QUESTIONS 10
15 a 2-Bromobutane has a chiral centre, i.e. a C atom with four di erent groups attached.
C
H3C CH3
HH
C
H H
C C
H H
H
C
H H
H
C
C
H
HH3C
C
CH3
iii 1-Bromobutane will react more quickly than 2-bromobutane via an SN2 mechanism. There is less steric hindrance around the C with the Br on it in 1-bromobutane, as there is only one alkyl group attached to this C. It is therefore easier for the nucleophile to get in to attack the C in 1-bromobutane.
An alternative explanation is in terms of the transition state. The transition state will be more crowded in 2-bromobutane, as there are two alkyl groups around the ve co-ordinate C. This makes the transition state less stable for 2-bromobutane, and the activation energy is thus higher. This means that this reaction will occur more slowly.
c i When a halogenoalkane is heated with KOH in ethanol, an elimination reaction occurs and an alkene is formed. H and Br are eliminated from adjacent C atoms and a C=C forms where they were removed.
H C C
Br
HH
H
C
+NH3
H
C
H
HHH C C
HH
HH
H + Br
H + NH4+
C+ C H
HH
NH3
H
C
H
H
C
H
H
H
C
HH
H H
H
C C
+N
H
C
H
H
H
H
H
H C
H
H
H C
H
H
C
NH2
H
C
H
H
Br
H
C
NH3
C H
H
H
+
NH3+
Br
C H
CH3
H5C2Br
CH
CH3
C2H5
b i N stands for nucleophile. A nucleophile is a molecule/negatively charged ion, possessing a lone pair of electrons, which is attracted to a more positively charged region in a molecule (region with lower electron density) and donates a lone pair of electrons to form a covalent bond.
ii
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ANSWERS TO EXAM-STYLE QUESTIONS 10 5CHEMISTRY FOR THE IB DIPLOMA CAMBRIDGE UNIVERSITY PRESS 2011
c We need to form a compound containing three carbons from bromoethane, so we need to add an extra carbon to the carbon chain. The only way that we have seen to do this in this part of the course is to use CN (if you study Option G, this can also be done with a Grignard reagent).
The only reaction of nitriles that we have encountered in this part of the course is the reduction of the nitrile group to form an amine. This means that it is the amine with three C atoms that can be formed from bromoethane in two steps. P is thus propylamine/propan-1-amine/1-aminopropane. The reaction pathway for its formation is:
heat+ C
H
H
C
H
H
H
C
H
H
C
H
H
H
C C
H
H
C
H
H
C
H
H
H
H
NC
H
H
C
H
H
C
H
H
H N
O
+ H2OO
H
CH
H
O
C
CH3
H
C
H
H
C
H
CH3
C
H
H
C
C
H
H
C
H
H
C
H
H
C
H
H
N
H H
N
C
O OCH3
C
H
CH3
C
H
H
C
H
H
C
H
C
O
C
H
H
C
H
H
C
H
H
C
H
H
N
H H
N
C
O
H2 Ni catalyst
heat, high pressure
KCN(aq) methanol
heat under refluxC
H
H
C
H
H
C
H
H
H NC
H
H
CC
H
H
HC
Br
H
HC
H
H
H N
H
H
If you have studied Option G, you will realise that there is another way of answering this question. The nitrile could be converted to a carboxylic acid by heating with H+/H2O, so P could have also been the carboxylic acid.
d i This polymer is a polyamide, as it is the amide group that links the monomers. This will be formed in a condensation polymerisation reaction.
ii To work out the structures of the monomers, we must split apart the polymer at the amide groups and add H2O at each split.
ii A and B/C are structural isomers of each other. They have the same molecular formula but di erent structural formulas. The atoms are joined together di erently.
B and C are geometrical isomers. These have the same structural formula, but the atoms are arranged di erently in space. B and C are di erent, as the rotation of the groups are restricted by the C=C group.
16 a This molecule contains the amide functional group.
b This is a condensation reaction in which an amine reacts with a carboxylic acid to form the amide.
H H
C N
CH
H
C
H
H
C
H
H H H
C C
H H
H
O
C
CH3
H
C
H
H
C
H
CH3
C
H
H
CCC
H
H
C
H
H
C
H
H
C
H
H
N N
O
O O
H HO
H
H
H
H
H is added to the N and OH to the C=O group.The monomers are:
