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Among the following, reaction(s) which give(s) tert-butyl benzene as the major product is (are):
A.
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B.
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C.
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D.
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Answer
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Hint:We need a tertiary alkane which has a less hydrogen atom at the center of it can act as Lewis base i.e., the tert-butyl cation This can lead to formation of the tert-butylbenzene.

Complete step by step solution:
In order to get the major product as tert-butylbenzene, we have to first produce the tert-butyl which will react with benzene which can give us tert-butylbenzene.
(a) Here we have sodium ethoxide and tert butyl bromide. They will react to the elimination reaction and remove the bromide group from the compound. Then we will get
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The compound formed will not react with benzene, so we cannot get tert-butylbenzene.

(b) Here we have $AlC{l_3}$which is a Lewis acid and we have chlorine which is Lewis base. The chlorine will attack $AlC{l_3}$, which will form a cation.
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This will go hydride shift to form tert-butyl cation.
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Which can react with a benzene ring, where one hydrogen is removed and the cation is attached to it to form tert-butylbenzene.
Hence option (b) is correct.

(c) We have an alkene and sulphuric acid. The ${H_2}S{O_4}$ will react with alkene and ${H^ + }$ gets attached to the alkene.
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After the hydride shift we will get tert-butyl cation. Which will react with a benzene ring to give tert-butylbenzene.
Hence option (c) is correct.

(d) We have given $B{F_3}$ in presence of ether. So is generally a gas so to trap it we use ether. So $B{F_3}$ acts as Lewis acid and oxygen act as Lewis base. This oxygen acts the $B{F_3}$ and we will get
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After the hydride shift we will get tert-butyl cation. Which will react with a benzene ring to give tert-butylbenzene.
Hence option (d) is correct.

Note:

The hydride shift happens because the tertiary cations are stronger as it involves maximum +I effect and it has maximum hyperconjugation (9 alpha hydrogens). Thus, we can say that tertiary carbocations are more stable than benzyl carbocation.