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Question 38

The product formed in the following reaction.

image

is:

The reaction shown is an acid-catalysed alkylation reaction used industrially for preparation of highly branched alkanes.

The acid catalyst $$\mathrm{(H^+)}$$ protonates the double bond of $$\mathrm{2\text{-}Methylpropene}$$.

According to Markovnikov’s rule, protonation generates a highly stable tertiary carbocation $$\mathrm{(CH_3)_3C^+}$$.

This tert-butyl carbocation attacks another molecule of isobutylene, forming a larger $$\mathrm{C_8}$$ tertiary carbocation intermediate.

A hydride transfer then occurs from $$\mathrm{Isobutane}$$ to the carbocation.

This neutralises the intermediate and regenerates the tert-butyl carbocation.

The final major product formed is $$\mathrm{2,2,4\text{-}Trimethylpentane}$$, commonly called $$\mathrm{Isooctane}$$.

Its condensed structural formula is $$\mathrm{(CH_3)_3C-CH_2-CH(CH_3)_2}$$.

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