1-methylcyclohexene માં DBr ના ઇલેક્ટ્રોફિલિક | vedclass

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Question

(D) The electrophilic addition of $DBr$ to $1$-methylcyclohexene proceeds via the formation of a tertiary carbocation intermediate at the $C1$ positio

Vedclass · Accepted Answer

$(a)$ અને $(b)$ બંને

Vedclass · Answer

(D) The electrophilic addition of $DBr$ to $1$-methylcyclohexene proceeds via the formation of a tertiary carbocation intermediate at the $C1$ position.
Following the formation of the planar carbocation,the bromide ion $(Br^-)$ can attack from either the top or bottom face of the carbocation.
Since the deuterium atom $(D)$ adds to the $C2$ position,the resulting product can have the $D$ and $Br$ atoms in either a $cis$ or $trans$ configuration relative to each other.
Therefore,both $cis$ and $trans$ isomers are formed as products.

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