Question

Consider the given reaction:

\(\rm CH_3 CHO \xrightarrow[(ii) \ NaBH_4]{(i) \ dil \ NaOH}\)

1. \(\rm H_3C - CH_2 - CH_2 - COOH\)
2. \(\begin{array}{*{20}{c}} {\rm {H_3}C - C - C{H_2} - C{H_2} - COOH}\\ {|| \ \ \ \ \ \ \ \ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\ {\rm O \ \ \ \ \ \ \ \ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \end{array}\)
3. \(\begin{array}{*{20}{c}} \rm{{H_3}C - CH - C{H_2} - C{H_2} - OH}\\ {| \ \ \ \ \ \ \ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\ {\,\,\,\,\,\,\rm OH \ \ \ \ \ \ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \end{array}\)
4. \(\rm H_3C - CH_2 - CH_2 - CHO\)

Answer 1

Correct Answer - Option 3 : \(\begin{array}{*{20}{c}} \rm{{H_3}C - CH - C{H_2} - C{H_2} - OH}\\ {| \ \ \ \ \ \ \ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\ {\,\,\,\,\,\,\rm OH \ \ \ \ \ \ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \end{array}\)

Concept:

Aldol reaction:

• The aldehydes and ketones undergo a number of reactions due to the acidic nature of α
   -hydrogen.
• The acidity of α-hydrogen atoms of carbonyl compounds is due to the strong electron-withdrawing effect of the carbonyl group and resonance stabilization of the conjugate base.
• Aldehydes and ketones having at least one α -hydrogen undergo a reaction in the presence of dilute alkali as a catalyst to form β-hydroxy aldehydes (aldol) or β-hydroxy ketones (ketol), respectively. This is known as the Aldol reaction.

Reduction of aldehydes and ketones:

• Aldehydes and ketones are reduced to the corresponding alcohols by the addition of hydrogen in the presence of catalysts (catalytic hydrogenation).
• The usual catalyst is a finely divided metal such as platinum, palladium, or nickel.
• It is also prepared by treating aldehydes and ketones with sodium borohydride (NaBH₄) or lithium aluminum hydride (LiAlH₄).
• Aldehydes yield primary alcohols whereas ketones give secondary alcohols.

\(RCHO\xrightarrow{{NaB{H_4}}}R - C{H_2}OH\)

Explanation:

The given reaction is:-

\(\rm CH_3 CHO \xrightarrow[(ii) \ NaBH_4]{(i) \ dil \ NaOH}\)

Step 1: Acetaldehyde with dil.NaOH

• The reaction of acetaldehyde with acetaldehyde in the presence of dilute NaOH is the kind of Aldol reaction by which obtained 3-hydroxybutanal as the product.

\(\mathop {C{H_3}CHO}\limits_{{\text{Acetaldehyde}}} {\mkern 1mu} + C{H_3}CHO{\mkern 1mu} \overset {Dilute\,NaOH} \leftrightarrows \mathop {C{H_3} - \mathop {\mathop {CH}\limits_| }\limits_{OH} - C{H_2} - CHO}\limits_{{\kern 1pt} {\text{3 - hydroxybutanal}}{\kern 1pt} {\kern 1pt} {\text{(aldol)}}} \)

Step 2: Reduction of aldehyde by NaBH₄

• Aldehydes on reduction with NaBH4 gives primary alcohol.
• 3-hydroxybutanal formed in the first reaction gets reduced by NaBH₄ to yield 1,3-Butanediol or Butylene glycol.

\(\mathop {C{H_3} - \mathop {\mathop {CH}\limits_| }\limits_{OH} - C{H_2} - CHO}\limits_{{\text{3 - hydroxybutanal}}} \xrightarrow{{NaB{H_4}}}\mathop {C{H_3} - \mathop {\mathop {CH}\limits_| }\limits_{OH} - C{H_2} - C{H_2} - OH}\limits_{{\text{1,3 - Butanediol}}} \)

The overall reaction is as follows:

\(\mathop {{\text{2}}\,{\text{C}}{{\text{H}}_{\text{3}}}{\text{CHO}}}\limits_{{\text{Acetaldehyde}}} \xrightarrow[{{\text{(ii)}}\;{\text{NaB}}{{\text{H}}_{\text{4}}}}]{{{\text{(i)}}\;{\text{dil}}\;{\text{NaOH}}}}\mathop {{{\text{H}}_{\text{3}}}{\text{C - }}\mathop {\mathop {{\text{CH}}}\limits_{\text{|}} }\limits_{{\text{OH}}} {\text{ - C}}{{\text{H}}_{\text{2}}}{\text{ - C}}{{\text{H}}_{\text{2}}}{\text{ - OH}}}\limits_{{\text{1,3 - Butanediol}}} \)

Hence, the final product of the given reaction is (1,3-Butanediol)

 \(\begin{array}{*{20}{c}} \rm{{H_3}C - CH - C{H_2} - C{H_2} - OH}\\ {| \ \ \ \ \ \ \ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\ {\,\,\,\,\,\,\rm OH \ \ \ \ \ \ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \end{array}\)