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Q8.15(vii):
How will you bring about the following conversion in not more than two steps? (vii) Benzaldehyde to 3-Phenylpropan-1-ol

Solution :

Step 1: Aldol Condensation with Acetone (Claisen-Schmidt Reaction)

To convert Benzaldehyde ($C_6H_5CHO$) to a carbon chain suitable for forming $3$-phenylpropan-$1$-ol, we must first increase the carbon count. Benzaldehyde reacts with acetaldehyde (ethanal) via a crossed-aldol condensation in the presence of dilute $NaOH$ to form cinnamaldehyde ($3$-phenylprop-$2$-enal).

Reaction:
$C_6H_5CHO + CH_3CHO \xrightarrow{dil. NaOH, \Delta} C_6H_5CH=CH-CHO + H_2O$

[Justification: Per the Claisen-Schmidt condensation mechanism, the alpha-hydrogen of the aldehyde is abstracted by the base to form an enolate, which attacks the carbonyl carbon of the benzaldehyde followed by dehydration.]

Step 2: Catalytic Hydrogenation

The resulting $3$-phenylprop-$2$-enal contains both a carbon-carbon double bond and a carbonyl group. To achieve $3$-phenylpropan-$1$-ol, we perform a complete catalytic hydrogenation using hydrogen gas in the presence of a metal catalyst like $Ni$, $Pd$, or $Pt$. This reduces both the alkene and the aldehyde functional groups simultaneously.

Reaction:
$C_6H_5CH=CH-CHO + 2H_2 \xrightarrow{Ni, \Delta} C_6H_5CH_2CH_2CH_2OH$

[Justification: Hydrogenation of unsaturated aldehydes over $Ni$ or $Pd$ catalysts results in the saturation of the $\pi$-bonds. The alkene reduces to an alkane, and the aldehyde reduces to a primary alcohol.]

Benzaldehyde CH3CHO, OH- Cinnamaldehyde 2H2, Ni 3-Phenylpropan-1-ol

Summary of Transformation

Step Reagent Intermediate/Product
1 $CH_3CHO / dil. NaOH$ Cinnamaldehyde
2 $H_2 / Ni$ $3$-Phenylpropan-$1$-ol

Final Solution: The conversion is achieved by first reacting benzaldehyde with acetaldehyde in the presence of dilute $NaOH$ to obtain $3$-phenylprop-$2$-enal, followed by complete catalytic hydrogenation using $H_2/Ni$ to yield $3$-phenylpropan-$1$-ol.


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