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Question 39

The reaction involves electrophilic addition of:

$$\mathrm{HBr/CCl_4}$$

to the alkene:

$$\mathrm{1\text{-}methyl\text{-}1\text{-}vinylcyclohexane}$$

The reaction follows Markovnikov addition and proceeds through a carbocation intermediate.

First, protonation of the double bond occurs.

According to Markovnikov’s rule, the proton adds to the terminal carbon:

$$\mathrm{=CH_2}$$

forming the more stable secondary carbocation.

$$\mathrm{-CH=CH_2 \longrightarrow -C^+H-CH_3}$$

This secondary carbocation lies adjacent to a quaternary carbon of the cyclohexane ring.

To form a more stable carbocation, a:

$$\mathrm{1,2\text{-}methyl\ shift}$$

takes place.

The methyl group migrates from the ring carbon to the carbocation centre.

This produces a highly stable tertiary carbocation on the ring carbon.

Simultaneously, the side chain becomes:

$$\mathrm{-CH(CH_3)_2}$$

(isopropyl group).

Ring expansion is less favoured because converting a six-membered ring into a seven-membered ring increases ring strain.

Finally, the bromide ion attacks the tertiary carbocation.

$$\mathrm{C^+ + Br^- \longrightarrow C-Br}$$

Thus, the major product formed is:

$$\mathrm{1\text{-}bromo\text{-}1\text{-}isopropylcyclohexane}$$

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