CH_3-CO-CH_2-CH_2-CH_2-CO_2H xrightarrow[(ii) | vedclass

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(A) The starting material is $5$-oxohexanoic acid $(CH_3-CO-CH_2-CH_2-CH_2-COOH)$.Step $(i)$: $NaBH_4$ is a selective reducing agent that reduces the

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$5-$hydroxyhexanoic acid lactone

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(A) The starting material is $5$-oxohexanoic acid $(CH_3-CO-CH_2-CH_2-CH_2-COOH)$.Step $(i)$: $NaBH_4$ is a selective reducing agent that reduces the ketone group to a secondary alcohol,while leaving the carboxylic acid group unaffected.This yields $5$-hydroxyhexanoic acid $(CH_3-CH(OH)-CH_2-CH_2-CH_2-COOH)$.Step $(ii)$: In the presence of $H^+$,the $5$-hydroxyhexanoic acid undergoes intramolecular esterification (cyclization) to form a cyclic ester,known as a lactone.The hydroxyl group at the $5$-position attacks the carboxylic acid group to form a six-membered ring lactone,specifically $\delta$-lactone ($5$-hydroxyhexanoic acid lactone).

$CH_3-CO-CH_2-CH_2-CH_2-CO_2H \xrightarrow[(ii) H_2O, H^+]{(i) NaBH_4} ?$
What will be the final product of this transformation?

Similar Questions

The major product expected from the following reaction is:
$2-(\text{carbamoyl})-3-(\text{hydroxymethyl})-6-(\text{hydroxy})\text{benzoic acid} \xrightarrow{HCl_{(g)} / CCl_4} ?$

Compound $(B)$ is

The conversion of $CH_{3}CH_{2}COOH$ to the product shown below can be accomplished by:
$CH_{3}CH_{2}CH(OCH_{2}CH_{2}O)$

Identify the correct option based on the following reaction sequence:
Acetic anhydride $\xrightarrow[(ii) EtOH/H^+/\Delta]{(i) H_3O^+} (A)$ $\xrightarrow[(ii) H_3O^+/\Delta]{(i) NaOEt} (B)$

The product $(A)$ obtained is:

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$CH_3-CO-CH_2-CH_2-CH_2-CO_2H \xrightarrow[(ii) H_2O, H^+]{(i) NaBH_4} ?$What will be the final product of this transformation?