Chemistry 9701 · AS & A Level · Hydrocarbons

Hydrocarbons — practice question

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

The mechanism for this reaction involves methylbenzene reacting with a $\text{Br}^+$ ion. This ion is formed when bromine reacts with the catalyst. Complete the equation for the reaction of $\text{Br}_2$ with the catalyst. $\text{Br}_2 + \ldots \rightarrow \text{Br}^+ + \ldots$

(a(ii))[2]

One of the isomers of L is made in much smaller quantities 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 between methylbenzene and the $\text{Br}^+$ ion. Include all relevant curly arrows and charges.

(b(i))[1]

Chlorobutane and chlorobenzene are each added to separate samples of warm aqueous $\text{AgNO}_3$. One chloro-compound reacts slowly, whereas the other shows no reaction. Identify the chloro-compound that reacts and describe the observation.

(b(ii))[2]

Provide two equations to explain any observations in (b)(i).

(b(iii))[2]

Explain why chlorobutane and chlorobenzene differ in 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: $\text{Br}_2 + \text{AlBr}_3 \rightarrow \text{Br}^+ + \text{AlBr}_4^-$

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