Question

Match the column $I$ with column $II$ and mark the appropriate choice.

Column $I$		Column $II$
(A)	Clemmensen reduction	(i)	Conc. $KOH$
$(B)$	Rosenmund reduction	(ii)	$Zn/Hg+$ conc. $HCl$
$(C)$	Iodoform reaction	(iii)	$H_2/Pd-BaS{O}_4$
$(D)$	Cannizzaro reaction	(iv)	$NaOH+I_2$

• A. $\left(A\right)\to \left(i\right),\left(B\right)\to \left(iii\right),\left(C\right)\to \left(ii\right),\left(D\right)\to \left(iv\right)$
• B. $\left(A\right)\to \left(iii\right),\left(B\right)\to \left(iv\right),\left(C\right)\to \left(i\right),\left(D\right)\to \left(ii\right)$
• C. $\left(A\right)\to \left(ii\right),\left(B\right)\to \left(iii\right),\left(C\right)\to \left(iv\right),\left(D\right)\to \left(i\right)$
• D. $\left(A\right)\to \left(iv\right),\left(B\right)\to \left(i\right),\left(C\right)\to \left(ii\right),\left(D\right)\to \left(iii\right)$

Answer 1

Answer: (C)

Clemmensen reduction: $Zn\mid Hg+$ conc.HCl
Clemmensen reduction reduces Aldehydes and ketones to corresponding hydrocarbons.
Rosenmund reduction: $H_2/Pd-BaS{O}_4$
The Rosenmund reduction is used to prepare aldehyde from acyl halide.
Iodoform reaction: $I_2+NaOH$
Idoform reaction used for the identification of methyl ketone.
Cannizzaro:Conc.$KOH$
Cannizzaro's reaction is a disproportionation reaction.
It is shown by the aldehydes or
ketones which do not have $\alpha$-hydrogen atoms.