CAIE A-Level Chemistry 30.1 Arenes

Practise selecting arene reagents, drawing electrophilic-substitution mechanisms and predicting substitution sites.

Exam points

select reagents and conditions for halogenation, nitration, Friedel–Crafts reactions and hydrogenation
draw electrophile formation, the arenium intermediate and restoration of the delocalised ring
predict ring or side-chain products from conditions and explain substituent effects on reactivity and position

Covered syllabus points

30.1.1—Chemistry of arenes 30.1.2—Mechanism: electrophilic substitution in arenes 30.1.3—Whether halogenation will occur 30.1.4—In the electrophilic substitution of arenes,

[Maximum number: 5]

Cumene is an aromatic hydrocarbon used in the synthesis of other useful chemicals.

cumene

(a)

Cumene can be synthesised via a Friedel-Crafts alkylation reaction, as shown in Fig. 5.2.

Fig. 5.2

[ 5 ]

(i)

The first step of the reaction forms the $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}^{+}$ cation.

Identify a suitable reagent for the formation of this cation from 2-bromopropane, $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHBr}$ .

[ 1 ]

$\mathrm{AlBr}_{3}$

(ii)

Complete Fig. 5.3 to show the mechanism for the reaction of benzene with the $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}^{+}$ cation.

Include all relevant curly arrows and charges.

Fig. 5.3

[ 3 ]

Figure for Answer to Question (ii) — CAIE A-Level Chemistry A2

M1: curly arrow from inside the hexagon ring

AND towards positively charged carbon atom (in the electrophile $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}^{+}$ )

M2: structure of the intermediate

M3: curly arrow from C-H bond into the ring AND formation / loss of $\mathrm{H}^{+}$

(iii)

The Friedel-Crafts alkylation of benzene by 1-bromopropane, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}$ , also produces cumene as the major product.

The $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2}{ }^{+}$ cation formed in the first step quickly rearranges to form the $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}^{+}$ cation.

Suggest why this is the case. Explain your answer.

[ 1 ]

$2^{\circ}$ carbocation / $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}^{+}$

AND is more stable due to greater positive inductive effect (of alkyl groups)

[Maximum number: 6]

Benzene can be used to make benzoic acid in the two-step process shown in Fig. 7.1.

Fig. 7.1

(a)

Give the reagents and conditions for step 1 and step 2.
step 1
step 2

[ 2 ]

$\mathrm{CH}_{3} \mathrm{Cl}+\mathrm{AlCl}_{3}[1]$

hot alkaline) $\mathrm{MnO}_{4}^{-}$ [1]

(b)

When treated with $\mathrm{Cl}_{2}$ under suitable conditions, methylbenzene forms compound J .

When treated with $\mathrm{Cl}_{2}$ under different conditions with different reagents, methylbenzene forms compound K.

Suggest and draw structures of compounds J and K in the boxes. The molecular formula of each compound is given.

Figure for Question (b) — CAIE A-Level Chemistry A2

State the reagents and conditions required to form each product.
to form compound J
to form compound K

[ 4 ]

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Covered syllabus points

30.1 Arenes question 1

Question (a)

Question (i)

Question (ii)

Question (iii)

30.1 Arenes question 2

Question (a)

Question (b)