A-LevelChemistryHydrocarbons9701/22 · May/June 2017 · Q423 Marks

Draw the skeletal formula of A.

(a)(i)Draw the skeletal formula of A.[1]
(a)(ii)The hydrocarbon A, C₄H₁₀, has a branched isomer. Suggest why unbranched A has a higher boiling point than its branched isomer.[2]
(a)(iii)Give the structural formula of B.[1]
(a)(iv)Explain why B shows geometrical isomerism.[2]
(a)(v)Draw the mechanism of the reaction of B with bromine, Br₂. Include all necessary charges, dipoles, lone pairs and curly arrows.[4]
(a)(vi)Explain the origin of the dipole on Br₂ in this mechanism.[1]
(b)(i)Identify C, D and E.[3]
(b)(ii)Write the equation for the reaction between E and aqueous sodium carbonate.[1]
(c)(i)Name the yellow precipitate produced by the reaction between F and alkaline aqueous iodine.[1]
(c)(ii)Give the structural formula of F and of G.[2]
(c)(iii)Explain the meaning of the term chiral centre.[1]
(d)(i)Identify the bonds responsible for the principal peaks above 1500cm⁻¹ in each spectrum. spectrum of H ................................................................................................ spectrum of I .................................................................................................[2]
(d)(ii)Name H and I. H ................................................................................................ I .................................................................................................[2]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.

[1] (A = ) [Skeletal formula of butane]
[1] (A / straight chain) has strong(er) (temporary dipole-) induced dipole (attractions) ora
[1] (because A / straight chain has) bigger (surface) area / more (points of) contact (in unbranched isomer) ora OR (so) more energy required to break the intermolecular forces ora
[1] CH3CHCHCH3 OR CH3CH=CHCH3
[1] No rotation / restricted / limited rotation of C=C / (carbon) double bond
[1] One (of the two) methyl groups / one (of the two) H (atoms) is on each C (of C=C)
[1] arrow from the C=C double bond drawn to the bromine
[1] dipole on Br2 in correct orientation AND arrow from the Br-Br bond to the Br⁻
[1] correct carbocation / bromonium ion from the structure with C=C drawn
[1] Br⁻ with lone pair, negative charge AND arrow from lone pair to the carbon atom of intermediate OR using both arrows shown (in alternative diagram)
[1] electrons in pi bond induce it (the dipole) OR (high) electron density in pi bond / double bond / C=C repels electrons (away from nearest Br) OR polarised by (high) electron density in pi bond / double bond / C=C
[1] C = (2-)methylpropan-2-ol / (CH3)3COH / any unambiguous structure
[1] D = (2-)methylpropan-1-ol / (CH3)2CHCH2OH / any unambiguous structure
[1] E = (2-)methylpropanoic acid /(CH3)2CHCO2H / any unambiguous structure
[1] 2C4H8O2 + Na2CO3 → 2C4H7O2Na + H2O + CO2
[1] triiodomethane
[1] F = CH3CH2CH2COCH3
[1] G = C2H5CH(CH3)CHO
[1] a (tetrahedral) atom with four different groups / atoms / substituents attached OR a carbon (atom) with four different groups / atoms / substituents attached
[1] H C=O (group / bond) AND O–H (group / bond)
[1] I C=O (group / bond) AND C-H (group / bond)
[1] H = ethanoic acid
[1] I = methyl methanoate