Chemistry 9701 · AS & A Level · Hydrocarbons

Hydrocarbons — practice question

Under the stated conditions, 2-methylbut-1-ene reacts with acidified manganate(VII) ions to give two organic products, X and Y. X then reacts straight away with the acidified manganate(VII) ions, producing carbon dioxide and water. Y has structural formula $\text{CH}_3\text{CH}_2\text{COCH}_3$.
(a)[1]

Draw the skeletal formula for 2-methylbut-1-ene.

(b(i))[1]

State the precise conditions required for acidified manganate(VII) ions to react with 2-methylbut-1-ene in this way.

(b(ii))[1]

Name the reaction type occurring to the functional group in 2-methylbut-1-ene in the reaction in (b)(i).

(c)[1]

Draw the structural formula of X.

(d)[2]

Describe a chemical test and the expected observation(s) that would confirm the presence of the carbonyl functional group in Y.

(e)[2]

The infra-red spectrum of 2-methylbut-1-ene is shown. Predict two major differences that would appear between the spectra of $Y$, $\text{CH}_3\text{CH}_2\text{COCH}_3$, and 2-methylbut-1-ene. Give reasons for your predictions. Your response should refer only to the part of each spectrum above $1500\,\text{cm}^{-1}$.

(f(i))[1]

Propanoic acid, $\text{CH}_3\text{CH}_2\text{CO}_2\text{H}$, is reduced by $\text{LiAlH}_4$. Write an equation to show this change. Use $[\text{H}]$ to represent one hydrogen atom from the reducing agent.

(f(ii))[1]

Name the organic product formed in this reaction.

(g(i))[1]

Organic compound $W$ is an ester and a structural isomer of propanoic acid. State the molecular formula of $W$.

(g(ii))[1]

Draw one possible structure of $W$.

Worked solution & mark scheme

This 12-mark question has a full step-by-step worked solution and mark scheme. One marking point: The correct displayed alkene structure is shown.

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