A-LevelChemistryCarbonyl compounds9701/22 · May/June 2018 · Q416 Marks

W is CH3COCH2CH3.

(a)The reaction between W and alkaline aqueous iodine produces a yellow precipitate.
(a)(i)Give the name of the compound formed as a yellow precipitate in this reaction.[1]
(a)(ii)Give the name of W.[1]
(b)There are two structural isomers of W that are also carbonyl compounds. Draw the structures of these two isomers of W.[2]
(c)(i)Identify the type of reaction occurring in reaction 1.[1]
(c)(ii)Identify the reagent for reaction 1.[1]
(d)Reaction 2 is carried out by adding a mixture of HCN and NaCN to W. The product, X, is formed as a mixture of two isomers.
(d)(i)Complete the mechanism for this reaction. Include the structure of the intermediate formed and all necessary charges, dipoles, lone pairs and curly arrows.[4]
(d)(ii)State the name of the type of isomerism shown by X.[1]
(d)(iii)Explain fully why X shows this type of isomerism.[2]
(e)If X is treated with ammonia and the product hydrolysed, a compound, Y, is obtained that contains 51.3% C, 9.40% H, 12.0% N and 27.3% O by mass.
(e)(i)Show that the empirical formula of Y is C5H11NO2.[2]
(e)(ii)The empirical formula of Y is C5H11NO2 and the M₁ of Y is 117. Deduce the molecular formula of Y. You must explain your reasoning.[1]

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 2018). The mark or mark type is shown in brackets.

[1] iodoform / tri-iodomethane
[1] butanone
[1] CH3CH2CH2CHO
[1] (CH3)2CHCHO
[1] reduction
[1] NaBH4 / sodium borohydride OR LiAlH4 / lithium aluminium hydride
[1] lone pair on C of CN⁻ and curly arrow to C of C=O
[1] correct dipole on C=O and curly arrow from = to O
[1] correct intermediate anion [diagram showing O- and CN attached to C]
[1] curly arrow from lone pair on O to H+
[1] optical
[1] (X has a) chiral centre / asymmetric carbon atom OR (X has a) C atom attached to four different groups / atoms / chains
[1] non-super(im)posable mirror images
[2] M1 is for the process of taking the % of each element and dividing by its relative atomic mass. C 51.3/12 = 4.28, H 9.40/1 = 9.40, N 12.0/14 = 0.857, O 27.3/16 = 1.71
M2 is for dividing the smallest %/A, into each of the remaining values to produce the correct ratio. 4.28/0.857 = 4.9942 ≈ 5, 9.40/0.857 = 10.9685 ≈ 11, 0.857/0.857 = 1 ≈ 1, 1.71/0.857 = 1.9953 ≈ 2
[1] C5H11NO2 AND because the EFM = RFM

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