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(A) The reaction of $1,2$-dihydronaphthalene with $Br_2$ proceeds via the formation of a cyclic bromonium ion intermediate. In the presence of a nucle

Vedclass · Accepted Answer

$1-$ethoxy$-2-$bromo$-1,2,3,4-$tetrahydronaphthalene

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(A) The reaction of $1,2$-dihydronaphthalene with $Br_2$ proceeds via the formation of a cyclic bromonium ion intermediate. In the presence of a nucleophilic solvent like ethanol $(EtOH)$,the solvent attacks the more substituted carbon of the bromonium ion to form the major product. The benzylic position is more stable for the development of partial positive charge,leading to the formation of $1$-ethoxy-$2$-bromo-$1,2,3,4$-tetrahydronaphthalene as the major product.

The major product of the following reaction is:

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