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Question: If 3-hexanone is reacted with \(NaB{H_4}\) followed by hydrolysis with \({D_2}O\), the product will ...

If 3-hexanone is reacted with NaBH4NaB{H_4} followed by hydrolysis with D2O{D_2}O, the product will be-
A.CH2CH2CH(OH)CH2CH2CH3C{H_2}C{H_2}CH\left( {OH} \right)C{H_2}C{H_2}C{H_3}
B.CH2CH2CD(OH)CH2CH2CH3C{H_2}C{H_2}CD\left( {OH} \right)C{H_2}C{H_2}C{H_3}
C.CH2CH2CH(OD)CH2CH2CH3C{H_2}C{H_2}CH\left( {OD} \right)C{H_2}C{H_2}C{H_3}
D.CH2CH2CD(OD)CH2CH2CH3C{H_2}C{H_2}CD\left( {OD} \right)C{H_2}C{H_2}C{H_3}

Explanation

Solution

NaBH4NaB{H_4} is a mild reducing agent and a good solvent. When it reacts with 33 - hexanone which contains a carbonyl group, the carbonyl group is reduced to hydroxyl group because sodium borohydride reduces aldehyde and ketones to alcohols. When the obtained product is hydrolyzed by heavy water, the hydrogen is replaced with deuterium.

Complete step by step answer:
The formula of 3−hexanone3 - hexanone is given as-CH3CH2CH2COCH2CH3C{H_3}C{H_2}C{H_2}COC{H_2}C{H_3} in which a carbonyl group (ketone) is present. Its common name is ethyl propyl ketone. NaBH4NaB{H_4} reduces aldehydes, acyl chlorides and ketones to alcohols but cannot reduce esters, carboxylic acid, ester or amides as it is a mild reducing agent. When it reacts with NaBH4NaB{H_4} which is a reducing agent, the carbonyl group in the reactants is converted to hydroxyl group and alcohol is formed as a result. The reaction is given as-
CH3CH2CH2COCH2CH3→NaBH4CH3CH2CH(OH)CH2CH3C{H_3}C{H_2}C{H_2}COC{H_2}C{H_3}\xrightarrow{{NaB{H_4}}}C{H_3}C{H_2}CH\left( {OH} \right)C{H_2}C{H_3}
The name of the product formed is 3−pentanol3 - pentanolas it contains hydroxyl group at carbon position at third number from the right. Now when this product is hydrolyzed withD2O{D_2}O, the hydrogen of the hydroxyl group will exchange with deuterium. The reaction is given as-
CH3CH2CH(OH)CH2CH3→D2OCH3CH2CH(OD)CH2CH3C{H_3}C{H_2}CH\left( {OH} \right)C{H_2}C{H_3}\xrightarrow{{{D_2}O}}C{H_3}C{H_2}CH\left( {OD} \right)C{H_2}C{H_3}
This is the final product.

Hence the correct answer is option C.

Note:
The uses of heavy water are-
-It is used as a moderator in nuclear reactors as it slows down the fast moving neutrons due to which nuclear fission can be controlled.
-It is used in nuclear magnetic resonance spectroscopy.
-It is also used as a tracer compound to study the mechanism of various chemical reactions.
-It is used to obtain deuterium by either electrolysis or by its decomposition by sodium metal.