Q.Write the structure of the product in the following reaction: phenol $+\ Zn \xrightarrow{\text{Heat}} ?$

Zinc dust reduces phenol by removing its –OH group, converting it directly to benzene.

The reaction

When phenol vapour is passed over heated zinc dust, the zinc reductively removes the hydroxyl group from the aromatic ring, replacing it with hydrogen and being itself oxidised to zinc oxide:

$C_6H_5OH + Zn \xrightarrow{\Delta} C_6H_6 + ZnO$

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First strip the $-OH$ group off phenol by distilling it with zinc dust to get benzene, then alkylate the benzene with methyl chloride (Friedel-Crafts) to install the methyl group and get toluene.

Step 1 — Reductive removal of $-OH$ (Zn-dust distillation): heating (distilling) phenol with zinc dust reductively cleaves the $C-OH$ bond, replacing $-OH$ with $-H$ and giving benzene:

$C_6H_5OH + Zn \xrightarrow{\Delta} C_6H_6 + ZnO$

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Distilling phenol with zinc dust reduces it, removing the -OH group and giving benzene.

The diagram shows phenol (a benzene ring bearing an -OH group) reacting with zinc dust, drawn over the reaction arrow, to give an unlabelled product.

When phenol vapour is passed over heated zinc dust (i.e. distilled with Zn dust), the -OH group is reductively removed (deoxygenation) and replaced by -H, converting phenol directly into benzene:

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