Ethylbenzene, C₆H₅CH₂CH₃, is treated in three different ways…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
Ethylbenzene, , is treated in three different ways.
- Chlorine, in ultraviolet light, replaces one hydrogen atom.
- Chlorine, with iron(III) chloride, in the dark, replaces one hydrogen atom.
- Hot alkaline potassium permanganate is added, and the mixture is then acidified.
Name the main product of each reaction. For (a), say which carbon of the side chain is attacked and why. Then count the carbon atoms of the product of (c), and the carbon atoms that were lost.
Need a reminder? The key ideaShow
A group on the ring directs the next substituent. , and alkyl groups activate and direct ortho and para. Halogens deactivate and direct ortho and para. , , , and deactivate and direct meta. Light and heat attack the side chain. A Lewis acid in the dark attacks the ring.
The worked solution
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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. (a) A free-radical chain. The hydrogen atom on the carbon next to the ring is removed, because the radical formed is stabilised by the ring. The product is 1-chloro-1-phenylethane, .
Step 2. (b) The Lewis acid makes the ring react. The ethyl group directs ortho and para. The products are o-chloroethylbenzene and p-chloroethylbenzene, with the para isomer in the larger amount.
Step 3. (c) The side chain is oxidised to a carboxyl group, whatever its length. The product is benzoic acid, .
Step 4. Carbon atoms. Ethylbenzene has carbon atoms. Benzoic acid has . One carbon atom is lost, as carbon dioxide.
Answer: (a) 1-chloro-1-phenylethane, (b) o- and p-chloroethylbenzene, (c) benzoic acid. The acid has carbon atoms, and was lost.
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