In the given conversion,the best yield will b | vedclass

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(B) The starting material is a polycyclic ketone. To introduce an acetyl group at the specific position,we first need to reduce the carbonyl group to

Vedclass · Accepted Answer

$A = Zn(Hg), HCl$; $B = CH_3 - CO - Cl, AlCl_3$

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(B) The starting material is a polycyclic ketone. To introduce an acetyl group at the specific position,we first need to reduce the carbonyl group to a methylene group using Clemmensen reduction $(Zn(Hg), HCl)$.After the reduction,the aromatic system is activated,allowing for a Friedel-Crafts acylation using $CH_3 - CO - Cl$ in the presence of $AlCl_3$ to introduce the acetyl group.Therefore,the correct sequence is $A = Zn(Hg), HCl$ followed by $B = CH_3 - CO - Cl, AlCl_3$.

In the given conversion,the best yield will be obtained with:

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