Column-I (Reaction)	Column-II (Reagent)
$\text{ l) }RCOOR\to {RCH}_{2}OH+ROH$	$\text{ A) }{CrO}_3+H_2{SO}_4$
$\text{ II) }RC{H}_{2}OH\to RCHO$	$\text{ B) }{H}_2/Ni\left(or\right)Pt$
$\text{ III) }RC{H}_{2}OH\to R-COOH$	$\text{ C) }{B}_2{H}_6/H_{2}O$
$\text{ IV) }RCOOH\to RC{H}_{2}OH$	$\text{ D) }{C}_5{H}_5\stackrel{+}{N}{HCrO}_3{Cl}^{-}$

$\text{ l) }{\mathrm{H}}_2/\mathrm{cat}\text{ and }{\mathrm{LiAlH}}_4$can reduce ester to give two moles of alcohol (one of which is always 1⁰).

II) Controlled oxidation of 1⁰ alcohol gives aldehyde using PCC.

III) 1⁰ alcohol is oxidized to carboxylic acid using Jone’s reagent

IV) ${\mathrm{B}}_2{\mathrm{H}}_6/{\mathrm{H}}_2\mathrm{O}\text{ or }{\mathrm{LiAlH}}_4/{\mathrm{H}}_2\mathrm{O}$ can reduce carboxylic acid to $1^0$ alcohol.