The product (D) in the following reaction seq | vedclass

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(A) $1.$ $PCC$ oxidizes the primary alcohol to an aldehyde: $HOCH_2CH_2CH_2COOCH_2CH_3 \xrightarrow{PCC} O=CHCH_2CH_2COOCH_2CH_3$ $(A)$.$2.$ One molar

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$H_2C=CHCH(OCOCH_3)CH_2CH_2COOH$

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(A) $1.$ $PCC$ oxidizes the primary alcohol to an aldehyde: $HOCH_2CH_2CH_2COOCH_2CH_3 \xrightarrow{PCC} O=CHCH_2CH_2COOCH_2CH_3$ $(A)$.$2.$ One molar equivalent of vinyl magnesium bromide reacts with the more reactive aldehyde group: $O=CHCH_2CH_2COOCH_2CH_3$ $\xrightarrow{H_2C=CHMgBr}$ $\xrightarrow{NH_4Cl/H_2O} H_2C=CHCH(OH)CH_2CH_2COOCH_2CH_3$ $(C)$.$3.$ Saponification of the ester with $KOH/H_2O$ followed by acidification with $H_3O^{+}$ gives the carboxylic acid: $H_2C=CHCH(OH)CH_2CH_2COOH$.$4.$ Acetylation of the secondary alcohol with acetic anhydride in pyridine yields the final product $(D)$: $H_2C=CHCH(OCOCH_3)CH_2CH_2COOH$.

The product $(D)$ in the following reaction sequence is: $HOCH_2CH_2CH_2COOCH_2CH_3$ $\xrightarrow{PCC} (A)$ $\xrightarrow[{(1 \ molar \ equivalent)}]{H_2C=CHMgBr} (B)$ $\xrightarrow{NH_4Cl/H_2O} (C)$ $\xrightarrow[H_2O]{KOH}$ $\xrightarrow{H_3O^{+}}$ $\xrightarrow[pyridine]{(CH_3CO)_2O} (D)$

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