Aromatic HydrocarbonshardNUMERICAL

Consider the reaction of benzene with 1-chloro-2,2-dimethylpropane (neopentyl chloride) in the preseAromatic Hydrocarbons Chemistry Question

Question

Consider the reaction of benzene with 1-chloro-2,2-dimethylpropane (neopentyl chloride) in the presence of anhydrous $AlCl_3$. The reaction involves a carbocation rearrangement. How many $sp^3$ hybridized carbon atoms are present in the major alkylbenzene product formed?

Answer: 5

💡 Solution & Explanation

Neopentyl chloride ($(CH_3)_3C-CH_2-Cl$) forms a primary carbocation-like complex, which rapidly undergoes a 1,2-methyl shift to yield the highly stable tert-pentyl carbocation: $CH_3-C^+(CH_3)-CH_2-CH_3$. Attack on benzene yields tert-pentylbenzene. The tert-pentyl group has exactly 5 carbon atoms (one central, two methyls, one methylene, one terminal methyl), all of which are $sp^3$ hybridized.

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