Q.What is the action of hydrazine on cyclopentanone in presence of KOH in ethylene glycol?

Clemmensen's reduction uses Zn-Hg amalgam + conc. HCl to reduce C=O to CH$_2$.

Clemmensen's reduction reduces the carbonyl group of an aldehyde or ketone directly to a $-CH_2-$ group, using zinc amalgam (Zn/Hg, which suppresses side reactions of zinc) in the presence of concentrated hydrochloric acid, usually under reflux:

$R_2C=O \xrightarrow[conc.\ HCl]{Zn(Hg)} R_2CH_2$

$H_2$/Ni reduces the ketone C=O to give butan-2-ol, $CH_3CH(OH)CH_2CH_3$.

$C_2H_5-CO-CH_3$ is butan-2-one (ethyl methyl ketone). On catalytic hydrogenation ($H_2$/Ni, $\Delta$), the carbonyl group $C=O$ is reduced to $CH-OH$:

$CH_3CH_2-CO-CH_3 \xrightarrow[\Delta]{H_2/Ni} CH_3CH_2-CH(OH)-CH_3$

The product is butan-2-ol, written in option (b) as $CH_3-CH(OH)-CH_2-CH_3$ — the same molecule.

$LiAlH_4$ selectively reduces the $-CHO$ group to $-CH_2OH$ without touching the isolated C=C double bond.

$LiAlH_4$ is a powerful nucleophilic hydride reducing agent that reduces carbonyl groups (aldehydes, ketones, carboxylic acids, esters) to alcohols, but it does not reduce isolated carbon-carbon double bonds. In $CH_3-CH=CH-CH_2-CHO$, only the aldehyde $-CHO$ is reduced (to $-CH_2OH$, after aqueous work-up with $H_3O^+$), while the $C=C$ double bond is left intact:

$CH_3-CH=CH-CH_2-CHO \xrightarrow[\text{ii) }H_3O^+]{\text{i) }LiAlH_4} CH_3-CH=CH-CH_2-CH_2-OH$