448 ANSWERS TO EXERCISES AND REVISION QUESTIONS17Q
(i)Revision questions
17.1C 6 H 14 , CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 , CH 3 (CH 2 ) 4 CH 3
17.217.3 (i)pent-2-ene (ii)2,5-dimethylhex-2-ene
17.4 (i) CH 2 CHCH 2 CH 2 CH 3 (ii) CH 3 CHCHCH 2 CHCH 3
|
Cl
(iii) (iv)(v)17.5 (ii)and(iii)
17.6 (i) CH 2 Br–CHCH 2 CH 2 CH 3 (1,2-dibromopentane)
|
Br
(ii) CH 2 OHCHCH 2 CH 3 (butan-1,2-diol)
|
OH
(iii) CH 3 CH 2 CH 2 CH 3 (butane)
17.7
(i) (ii)(iii)17.8tetrafluoroethene: CF 2 CF 2
PTFE:
···CF 2 – CF 2 – CF 2 – CF 2 – CF 2 – CF 2 – CF 2 – CF 2 – CF 2 – CF 2 ···
a non-stick coating for frying pans.
17.9 (i)
CHno of moles 88.9 11.1 12 1
ratio 07.4 : 11.1
2 : 3The empirical formula is C 2 H 3.
(ii)C 4 H 6
(iii)CH 3 CH 2 CCH, but-1-yne
CH 3 CCCH 3 , but-2-yne
17.1017.11We used bond energies to find the standard
enthalpy change for the reaction:
6C(s)3H 2 (g)C 6 H 6 (l) H^ — 1 ? (equation 1)
We start by calculating the standard enthalpy change for
the reaction in which gaseous carbon makes gaseous
benzene:
6C(g)3H 2 (g)C 6 H 6 (g) H^ — 2 ? (equation 2)
From Table 13.4,
H—^ H–H436 kJ mol–^1
H—^ C–H412 kJ mol–^1
H—^ CC612 kJ mol–^1
H—^ C–C348 kJ mol–^1
Assuming that benzene has the Keluké molecular
structure, the heat absorbed in breaking bonds in
reactant molecules, i.e. three H–H bonds is
heat absorbed 3 436 1308 kJ
The heat released when bonds in product molecules are
assembled, which involves making three CC bonds,
three C–C bonds and six C–H bonds, is
heat given out (3612)(6412)(3348)
5352 kJ
The overall change in enthalpy is
H—^2 5352 1308 4044 kJ mol–^1
The difference between equations (2) and (1) is that we
have to allow for the production of gaseous carbon
atoms and formation of gaseous benzene being
endothermic:
6C(s)6C(g)H^ — 3 (298 K) (7176)4302 kJ mol–^1
C 6 H 6 (l)C 6 H 6 (g) H—^4 (298 K) 31 kJ mol–^1
Use of the enthalpy diagram confirms that:
H—^1 H—^4 H^ — 2 H^ — 3 or
H—^1 H—^3 H^ — 4 H^ — 2
H—^1 4302 31 (4044)227 kJ mol–^1
Therefore, the difference between the theoretical and
experimental values is (227 50)177 kJ mol–^1. We
explain this difference by assuming that the Keluké
structure is not the true structure of benzene – benzene
has a lower enthalpy than this structure suggests.Unit 18
Exercises
18A
(i)(a) CH 3 CH 2 Cl (b) CH 3 CHBrCH 3
(c) CH 2 ClCHClCHFCH 3
(d)(ii)(a) 1-bromo-1-chloroethane (b) 3-iodopentane
(c) trichlorofluoromethane
(d) 3-chloromethylbenzene (or 3-chlorotoluene)18B
(i)(ii)CH 2 OHCHOHCH 2 OH
18C
(i)approx: 64.5, 58, 60
(ii)propan-1-ol is a liquid, the others are gases
(iii)hydrogen bonding between the alcohol molecules
results in stronger intermolecular attractions than in the
other compounds.
18D
I, IV, VII, VIII are primary; II, III, VI are secondary; V is
tertiary
18E
(i)aldehyde (ii)ketone (iii)aldehyde
(iv)ketone (v)ketone.
18F
(i)‘eth’ is a prefix that refers to two carbon atoms. The
simplest ketone, propanone, contains three.
(ii)
CH 3 CH 2 CH 2 CH 2 CHO CHO CH 3 CH 2 CHCHO
pentanal ||
CH 3 CCH 3 CH 3
CH 3 CHCH 2 CHO | 2-methylbutanal
| CH 3
CH 3 2,2-dimethylpropanal
3-methylbutanal
(iii)
CH 3 CH 2 CH 2 CHO CH 3 CHCH 3
butanal |
CHO
CH 3 COCH 2 CH 3 2-methylpropanal
butanone
18G
(i)butanone CH 3 COCH 2 CH 3
(ii)propanoic acid C 2 H 5 COOH
(iii)octan-2-one C 6 H 13 COCH 3.
18H
(i)CH 3 CH 3 CH 2 COOH
(ii)CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 n–hexaneCH 3 CH 2 CH 2 CHCH 3
CH 32–methylpentaneCH 3 CH 2 CHCH 2 CH 3
CH 33–methylpentaneCH 3 CHCHCH 3
CH 32,3–dimethylbutane
H 3 CCH 3 CCH 3
CH 3CH 3
2,2–dimethylpropaneCH 3CH 3H 3 C CH 3CH 3 CH 3CH 2 CH 3
CH 3CH 3CH 2 CH 3 H 3 C CH 3CH 2 CH 3CH 3CH 3CH 3CH 2 CH 2 CH 3CH 3
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