The major product of the given reaction is3-m | vedclass

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(A) The reaction of $3-methoxyphenol$ with $OHCCH_2COCl$ involves the esterification of the phenolic $-OH$ group with the acid chloride group $(-COCl)

Vedclass · Accepted Answer

$7-$methoxy-2H-chromen$-2-$one

Vedclass · Answer

(A) The reaction of $3-methoxyphenol$ with $OHCCH_2COCl$ involves the esterification of the phenolic $-OH$ group with the acid chloride group $(-COCl)$,as acid chlorides are more reactive than aldehydes.This forms an intermediate ester: $3-methoxyphenyl \ 2-formylacetate$.In the presence of concentrated $H_2SO_4$ and heat,this intermediate undergoes an intramolecular Pechmann-type condensation (cyclisation) between the aldehyde group and the ortho-position of the benzene ring.Since the starting material is $3-methoxyphenol$,the cyclisation can occur at the $2$ or $6$ positions. The $6$-position is less sterically hindered than the $2$-position (which is between the $-OMe$ and the ester group),leading to $7-methoxy-2H-chromen-2-one$ as the major product.

The major product of the given reaction is
$3-methoxyphenol$ $\xrightarrow[{(ii) \ H_2SO_4, \ heat}]{{(i) \ OHCCH_2COCl}}$

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The major product of the given reaction is$3-methoxyphenol$ $\xrightarrow[{(ii) \ H_2SO_4, \ heat}]{{(i) \ OHCCH_2COCl}}$