11.3 Alkenes, Set B Answer Key To print: Ctrl + P, paper A4, margins Default.
Learners' AlleyDesigned by Khow Willard
11.3

Alkenes

Set BAnswer key
A

Multiple choice

3 marks
1

B(CH3)2C=CHCH2CH2C(CH3)=CHCH2CH2OH[1]

Note: 1° alcohols can be oxidised to carboxylic acids using hot acidified potassium manganate(VII). Products of oxidation:

ACH3CO2H, CH3COCH2CH2COCH3 and HO2CCH2CO2H BCH3COCH3, HO2CCH2CH2COCH3 and HO2CCH2CO2H CCH3COCH3, HO2CCH2CH2CO2H and CH3CH2CO2H DCH3COCH3, CH3COCH2COCH3 and HO2CCH2CO2H
2

CIt gives a diol with cold, alkaline potassium dichromate(VI).[1]

Incorrect: only potassium manganate(VII), not potassium dichromate(VI), can oxidise alkenes.

ACorrect. Free radical substitution can occur with any H (not bonded to C=C) on the alkyl chain. BCorrect. Each carbon atom in the C=C has two different groups attached to it. DCorrect. Both benzene and Muscalure are non-polar molecules that can form instantaneous dipole–induced dipole interactions with one another. Muscalure can thus dissolve in benzene and be removed from fly paper.
3

D1,2-dibromopropane[1]

B

Structured questions

21 marks
4
(a)
KOH in ethanol, heat[1]
(b)

D: CH3CH=C(CH3)CH=CH2 or CH3CH=CHC(CH3)=CH2[1]

B: CH3CH2CH(CH3)CH2CH3 or CH3CH(CH3)CH2CH2CH3[1]

(c)

Any 2 of: CH3CHBrCH(CH3)CHBrCH3,   CH3CH2CBr(CH3)CH2CH2Br,   CH3CHBrCH(CH3)CH2CH2Br,   CH3CH2CBr(CH3)CHBrCH3[2]

(d)

Condition for cis-trans isomerism to exist: the carbon atoms that are part of the C=C must each have two different groups or atoms bonded to them. Number of stereoisomers = 21 = 2[1]

  [2]

In cyclohexene, only the cis isomer exists (with the alkyl groups on the same side of C=C). There is no trans isomer due to ring strain. The trans isomer could only be formed by breaking the carbon–carbon single bond in cyclohexene.[1]

5
(a)

The sp2 hybrid orbitals of each carbon atom of the C=C bond lie in a trigonal plane, while the sp3 hybrid orbitals of the adjacent carbon atom are arranged tetrahedrally. In the "trans" form, the sp3 hybrid orbital of one of the adjacent carbon atoms would not be able to overlap head-on effectively with the sp2 hybrid orbital of the doubly bonded carbon atom to form a C–C σ bond. or The trans isomer of cyclohexene, if formed, would experience considerable strain in the ring and would be (energetically) unstable.[1]

As such, the trans isomer of cyclohexene does not exist, and cyclohexene cannot exhibit geometrical isomerism.[1]

(b)

Regardless of whether cis-hex-3-ene or trans-hex-3-ene reacts with hydrogen bromide, the same carbocation intermediate, CH3CH2CH2C+HCH2CH3, is formed.[1]

As the positively charged carbon atom is sp2 hybridised, it is equally probable for the bromide ion to attack from above or below the trigonal plane containing the positively charged carbon atom. This leads to the formation of a pair of enantiomers in a 1 : 1 ratio.[1]

(c)
CH3CH2COOH[1] and HOOCCH2COOH[1]
6
(a)

CH3CH=CH2 → CH3CH(OSO3H)CH3   (cold, concentrated H2SO4)[1]

CH3CH(OSO3H)CH3 → CH3CH(OH)CH3 + H2SO4   (H2O, warm)[1]

(b)

CH3CHClCH3 → CH3CH=CH2 + NaCl + H2O   (NaOH, ethanol, heat)[1]

CH3CH=CH2 → CH3CHBrCH2Br   (Br2/CCl4, dark)[1]

(c)

CH3CH2CH2CH2OH → CH3CH2CH=CH2   (excess concentrated H2SO4, 170 °C)[1]

CH3CH2CH=CH2 → CH3CH2CH(OSO3H)CH3   (concentrated H2SO4, cold), then CH3CH2CH(OSO3H)CH3 → CH3CH2CH(OH)CH3 + H2SO4   (H2O, warm)[1]