(i) The organic compound (A) which is sodium salt of phenolic acid is sodium salicylate.
(ii) (A) on heating with soda lime gives Compound (B) phenol. Phenol gives violet colouration with neutral perchloride.

(iii) Phenol on treatment with C₆H₅COCl in the presence of NaOH gives on ester 'C':
C₆H₅OH + C₆H₅COCl \(\overset { NaOH }{ \longrightarrow } \underset { (C) }{ { C }_{ 6 }{ H }_{ 5 }OCO{ C }_{ 6 }{ H }_{ 5 }+HCl } \)

Compound	Compound Name	Formula
A	Sodium benzoate'
B	Phenol	C6H5OH
C	Phenyl benzoate	C6H5OCO C6H5

(i) An organic compound (A) C₆H₆O gives violet colour with neutral FeCl₃ solution.
(ii) With NH₃ in the presence of anhydrous ZnCI₂,(A) gives (B) (C₆H₇N).

(i) Compound (A) with molecular formula C₃H₈O₃ is glycerol.
(ii) Glycerol on heating with P₂O₅ gives an unsaturated compound (B).

(iii) Glycerol with cooled mixture of cone. H₂SO₄ and fuming HNO₃ form compound (C).

(i) Compound (A) is carbonyl compound, it is acetone
(ii) (A) on reduction with LiAlH₄ gives (B) it gives blue colour in Victor Meyer'stest.
\({ CH }_{ 3 }-\underset { \overset { || }{ \underset { (A) }{ O } } }{ C } -{ CH }_{ 3 }\overset { { LiAIH }_{ 4 } }{ \underset { \left[ H \right] }{ \longrightarrow } } { CH }_{ 3 }-{ CH }_{ 3 }-\underset { \overset { | }{ \underset { (B) }{ OH } } }{ CH } -{ CH }_{ 3 }\) 
(iii) (B) reacts with SOCl₂to give (C).

(iv) (C) on treatment with alcoholic KOH, forms (D) by elimination reaction.
\({ CH }_{ 3 }-\underset { \overset { | }{ \underset { (C) }{ Cl } } }{ C } H-{ CH }_{ 3 }\overset { alc.KOH }{ \longrightarrow } \underset { (D) }{ { CH }_{ 3 }CH={ CH }_{ 2 }+HCl } \) 

(i) \( \underset { (A) }{2 \mathrm{C}_{2} \mathrm{H}_{5}} \mathrm{OH}+2 \mathrm{Na} \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{ONa}+\mathrm{H}_{2} \uparrow \) 
(ii) Ethyl alcohol on mild oxidation gives acetaldehyde (B) It answers idoform test.
\({ CH }_{ 3 }-\underset{(A)}{{ CH }_{ 2 }OH}\overset { Mild\left( O \right) }{ \longrightarrow } \underset { (B) }{ CH_{ 3 }CHO } +{ H }_{ 2 }O\) 
\( \underset {(B)}{\mathrm{CH}_{3} \mathrm{CHO}+}3 \mathrm{I}_{2}+\mathrm{NaOH} \overset { B }{ \underset { Iodoform }{ \longrightarrow } } \mathrm{CHI}_{3}+\mathrm{HCOONa}+3 \mathrm{HI}\)
(iii) Acetaldehyde when treated with Cone H₂SO₄ undergoes polymerisation and the product obtained is paraldehyde (C).

Compound	Compound Name	Formula
A	Ethyl alcohol	CH3CH2OH
B	Acetaldehyde	CH3CHO
C	Paraldehyde