CH_3-CH(OAc)-Et (d= dextro rotatory) xrightar | vedclass

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Question

(A) The reaction is the alkaline hydrolysis (saponification) of an ester. In this reaction,the $HO^{-}$ nucleophile attacks the carbonyl carbon of the

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$CH_3-CH(OH)-Et$ $(d)$

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(A) The reaction is the alkaline hydrolysis (saponification) of an ester. In this reaction,the $HO^{-}$ nucleophile attacks the carbonyl carbon of the acetate group. The bond between the oxygen atom of the alcohol part and the carbonyl carbon of the ester is broken. Since the chiral center (the carbon atom attached to the $OAc$ group) is not involved in the bond-breaking or bond-forming process,the configuration at the chiral center remains unchanged. Therefore,the product $CH_3-CH(OH)-Et$ retains the same optical activity as the reactant,which is $d$ (dextro rotatory).

$CH_3-CH(OAc)-Et$ ($d=$ dextro rotatory) $\xrightarrow{HO^{-}}$ Product of the reaction is

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