What is the ratio of (x/y) in the above react | vedclass

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Question

(D) In Reaction-$1$,the substrate has two $-OH$ groups (one alcoholic,one thiol) and two $-COCl$ groups.Each $-OH$ group reacts with $1$ mole of $CH_3

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) In Reaction-$1$,the substrate has two $-OH$ groups (one alcoholic,one thiol) and two $-COCl$ groups.Each $-OH$ group reacts with $1$ mole of $CH_3MgBr$ as an acid-base reaction ($2$ moles total).Each $-COCl$ group reacts with $2$ moles of $CH_3MgBr$ as a nucleophilic addition-elimination reaction ($4$ moles total).Thus,$x = 2 + 4 = 6$.In Reaction-$2$,the substrate has one $-COOEt$ group and one $-CH_2Cl$ group.The $-COOEt$ group reacts with $2$ moles of $CH_3MgBr$ to form a tertiary alcohol.The $-CH_2Cl$ group is an alkyl halide and does not react with $CH_3MgBr$ under these conditions.Thus,$y = 2$ (Note: The provided solution had $y=3$,but based on the product shown,only $2$ moles are consumed for the ester).However,if we re-evaluate the product,the ester group consumes $2$ moles. The ratio $(x/y) = 6/2 = 3$.

What is the ratio of $(x/y)$ in the above reactions?

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