An organic compound (A) on treatment with `CHCl_(3) and KOH` gives (Y) and (Z) both of which in turn gives the same compound (T) when distilled with Zn. Oxidation of (T) Yields (S) of formula `C_(7)H_(6)O_(2)`. The sodium salt of (S) with sodalime gives (P) which can also be obtained by distilling (X).
The compound (T) is

Explanation

(i) Since compound (A) on treatment with `CHCl_(3)` and KOH (i.e., Reimer-Tiemann reaction), gives two products B and C, therefore, A must be phenol and B and C must be o-hydroxybenzaldehyde and p-hydroxybenzaldehyde respectively or vive-versa.
(ii) Since both B and C on distillation with Zn dust give the same compound (D), therefore, D must be benzaldehyde.
(iii) since oxidation of D gives E with M.F. `C_(7)H_(6)O_(2)`, therefore, E must be benzoic acid.
(iv) Since sodium salt of E, i.e., benzoic acid upon heating with soda-lime gives compound (F), therefore, The chemical equations for the above reactions are given below.