Chemistry 9701 · AS & A Level · Hydrocarbons

Hydrocarbons — practice question

In the dark, bromine reacts with methylbenzene in the presence of a suitable catalyst to produce HBr and compound L, $\text{C}_7\text{H}_7\text{Br}$. L is one of the three isomers that may be produced in this reaction.
(a(i))[1]

In the mechanism, methylbenzene reacts with a $\text{Br}^+$ ion. Bromine forms this ion when it reacts with the catalyst. Complete the equation for the reaction of $\text{Br}_2$ with the catalyst. $\text{Br}_2 + \ldots \longrightarrow \text{Br}^+ + \ldots$

(a(ii))[2]

One of the isomers of L is produced in much smaller amounts than the other two isomers. Draw the structure of this isomer and explain why it is formed in the smallest amount.

(a(iii))[3]

Complete the mechanism in Fig. 8.1 for the reaction of methylbenzene with the $\text{Br}^+$ ion. Include every relevant curly arrow and all charges.

(b(i))[1]

Chlorobutane and chlorobenzene are each added separately to warm aqueous $\text{AgNO}_3$ samples. One of the chloro-compounds reacts slowly, while the other remains unreactive. Identify the chloro-compound that reacts and describe the observation.

(b(ii))[2]

Write two equations that explain the observations in (b)(i).

(b(iii))[2]

Explain why chlorobutane and chlorobenzene differ in their reactivity with warm aqueous $\text{AgNO}_3$.

Worked solution & mark scheme

This 11-mark question has a full step-by-step worked solution and mark scheme. One marking point: $\mathrm{Br_2 + AlBr_3 \rightarrow Br^+ + AlBr_4^-}$

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