CH_3-(CH_2)_3-OH xrightarrow[Et_3N]{CH_3-SO_2 | vedclass

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(B) Step $1$: The alcohol $CH_3-(CH_2)_3-OH$ reacts with methanesulfonyl chloride $(CH_3SO_2Cl)$ in the presence of triethylamine $(Et_3N)$ to form a

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$CH_3-(CH_2)_3-^{14}CO_2H$

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(B) Step $1$: The alcohol $CH_3-(CH_2)_3-OH$ reacts with methanesulfonyl chloride $(CH_3SO_2Cl)$ in the presence of triethylamine $(Et_3N)$ to form a good leaving group,the mesylate $(A)$,which is $CH_3-(CH_2)_3-OSO_2CH_3$.Step $2$: The mesylate $(A)$ undergoes an $S_N2$ reaction with $K^{14}CN$. The $^{14}CN^-$ nucleophile attacks the primary carbon,displacing the mesylate group to form the nitrile $(B)$,which is $CH_3-(CH_2)_3-^{14}CN$.Step $3$: The nitrile $(B)$ undergoes acid-catalyzed hydrolysis $(H_3O^{\oplus})$ to form the corresponding carboxylic acid $(C)$,which is $CH_3-(CH_2)_3-^{14}COOH$.

$CH_3-(CH_2)_3-OH$ $\xrightarrow[Et_3N]{CH_3-SO_2-Cl} (A)$ $\xrightarrow{K^{14}CN} (B)$ $\xrightarrow{H_3O^{\oplus}} (C)$
Product $(C)$ is

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$CH_3-(CH_2)_3-OH$ $\xrightarrow[Et_3N]{CH_3-SO_2-Cl} (A)$ $\xrightarrow{K^{14}CN} (B)$ $\xrightarrow{H_3O^{\oplus}} (C)$Product $(C)$ is