Explanation

(b)
Molecular formula of $A = C_{2}Cl_{3}OH$
As (A) reduces Fehling’s solution and on oxidation gives a monocarboxylic acid (B). It means (A) must be an aldehyde.
$CCl_{3}CHO$
(A)
This is further confirmed by the reaction
$C_{2}H_{5}OH + Cl_{2}\underset{oxidation}{\overset{\ \ \ \ \ \ \ \ \lbrack O\rbrack\ \ \ \ \ \ \ \ \ \ \ \ \ \ }{\rightarrow}}CH_{3}CHO\underset{chlorination}{\overset{{\ \ \ \ \ \ \ \ \ \ \ \ \ \ Cl}_{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ }{\rightarrow}}CCl_{3}CHO$
A=Chloral [$CCl_{3}CHO$]