Eduqas 2018 unit 2 paper

Question 1

Why can but-2-ene form EZ isomers but but - 1- ene cannot?

Answer 1

E-Z isomerism occurs due to restricted rotation about the
double bond
but-2-ene has two different groups attached to the carbons in the
double bond while but-1-ene has two hydrogens attached to one
carbon in the double bond

Question 2

name the alkene monomer that can be polymerised to give the following polymer
H CH3
C-C
(H CH2CH3 )

Answer 2

propene

Question 5

Explain why sulphur compounds should be removed from fuels before they are used

Answer 5

sulphur combines with oxygen to form sulphur dioxide
reacts with rainwater to form acid rain

Question 6

Enthalpy change for reaction = -1690

Answer 6

bonds broken
(8 × C—H) + (2 × 348) + (5 × 496)
bonds formed
(6 × 743) + (8 × 463) = 8162
‒1690 = (8 × C—H) + 3176 – 8162
C—H = 412

Question 7

Mechanism for reaction to form chloromethane

Answer 7

Cl2 → 2Cl.
Cl + CH4 → CH3 + HCl (1)
CH3 + Cl2 → CH3Cl + Cl.
e.g. CH3. + Cl. → CH3Cl

Question 8

boiling temp for butane
ethanoic acid
propan 1 ol ?

Answer 8

butane ‒1°C, propan-1-ol 97°C, ethanoic acid 118°C
butane shows no hydrogen bonding between molecules
ethanoic acid has more/stronger hydrogen bonds between molecules
than propan-1-ol

Question 9

Explaining your reasoning
describe the features that would be seen in the 13 C NMR spectrum and the mass spectrum of the remaining compound

Answer 9

A is ethanoic acid, B is propan-1-ol
13C NMR for butane has 2 peaks between 5 and 40 ppm
13C NMR ethanoic acid has 2 peaks but propan-1-ol has 3 peaks
both have peaks at m/z 60 due to molecular ion (butane does not)
both have peaks at m/z 45 due to COOH+
and CH2CH2OH+
respectively (butane does not)
mass spectrum for butane has peaks at m/z 15, 29, 43 and 58
 propan-1-ol has a peak at m/z 29 due to CH3CH2
+
and m/z 31 due
to CH2OH+
(ethanoic acid does not)
ethanoic acid has a peak at m/z 43 due to CH3CO+
(propan-1-ol does not)

Question 10

How would you expect the infrared spectrum of ethanoic acid to differ from that of propan - 1ol

Answer 10

only ethanoic acid would have an absorption at 1650-1750 cm‒1
due to
presence of the C=O bon
ethanoic acid would have an absorption at 2500-3200 cm‒1
due to the
presence of the O—H bond, while the absorption due to O—H in
propan-1-ol would be at 3200-3550 cm‒1

Question 11

Identify compounds of A - E?

Answer 11

A is but-1-ene
B is 2-bromobutane
C is 1-bromobutane
D is butan-1-ol
E is butanoic acid

Question 12

Name type of reaction taking place when compound B is converted back to compound A

Answer 12

elimination

Question 13

functional groups present?

Answer 13

C=C and alcohol

Question 14

molecular formula of compound?

Answer 14

C10H18O

Question 17

Enthalpy change of combustion of Pentan 2 ol

Answer 17

ΔH = ‒mcΔT
n
ΔT = 6.8°C, m = 200 g, n = 2.043 × 10‒3

ΔH = ‒2796000
ΔH = ‒2796

Question 18

Each reading on the thermometer is accurate = 0.1 degrees
calculate the percentage error

Answer 18

0.2 × 100 = 2.9
6.8

Question 19

experimental improvement that would reduce error due to heat loss

Answer 19

use less water (e.g. 50 cm
burn for longer

Question 20

Standard enthalpy of combustion?

Answer 20

ΔfH
θ
reactants = ‒380 and ΔfH
θ
products = ‒3686 (1)
ΔcH
θ
= ‒3686 – (‒380) = ‒3306 (1)

Question 21

why 02 has a value of 0 for its standard enthalpy change of formation?

Answer 21

it is an element in its standard state

Question 22

describe a chemical test that the student could use to clearly distinguish between the alcohols

Answer 22

reflux both with acidified potassium dichromate
colour changes from orange to green with pentan-2-ol
no change with 2-methylbutan-2-ol

Question 23

complete the diagram by drawing a thermometer and clearly label flow of water

Answer 23

thermometer bulb adjacent to outlet leading to condenser
water in through lower tube and out through upper tube