Ph-CH_2-CH(OH)-CH_3 xrightarrow{K} xrightarro | vedclass

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(B) The reaction proceeds via the Williamson ether synthesis mechanism. $1$. The alcohol $Ph-CH_2-CH(OH)-CH_3$ reacts with potassium $(K)$ to form an

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$Ph-CH_2-CH(OEt)-CH_3$ (retention)

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(B) The reaction proceeds via the Williamson ether synthesis mechanism. $1$. The alcohol $Ph-CH_2-CH(OH)-CH_3$ reacts with potassium $(K)$ to form an alkoxide ion,$Ph-CH_2-CH(O^-)-CH_3$. $2$. This alkoxide ion acts as a nucleophile and attacks the ethyl bromide $(C_2H_5Br)$ via an $S_N2$ mechanism. $3$. In an $S_N2$ reaction,the nucleophile attacks the electrophilic carbon of the alkyl halide. $4$. Since the chiral center at the $CH$ group is not involved in the bond-breaking or bond-forming process,the configuration at the chiral center remains unchanged. $5$. Therefore,the product is $Ph-CH_2-CH(OEt)-CH_3$ with retention of configuration at the chiral center.

$Ph-CH_2-CH(OH)-CH_3$ $\xrightarrow{K}$ $\xrightarrow{C_2H_5Br} (A)$. Product $(A)$ in the above reaction is:

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