(i) Ⓐ of molecular formula CH₂O is identified as HCHO, formaldehyde.
(ii) Formaldehyde reacts with CH₃MgI followed by hydrolysis to give ethanol, CH₃- CH₂OH B as the product.

(iii) Ethanol Ⓑ, reacts with PCl₅ to give C₂H₅CI, Ethyl chloride © as the product.

(iv) CH₃-CH₂CI © on reaction with alcoholic KOH undergoes dehydrohalogenation to give ethylene CH₂ = CH₂ © as the product.
\(CH_{ 3 }-{ CH }_{ 2 }Cl\overset { alcoholic }{ \underset { KOH }{ \longrightarrow } } \underset { Ethlene }{ { CH }_{ 2 }={ CH }_{ 2 }+HCl } \)

(i) The simple aromatic hydrocarbon is C₆H₆. Benzene (A)
(ii) Benzene reacts with Cl₂ to give chlorobenzene.

(iii) Chlorobenzene reacts with ethyl chloride to form ethyl benzene, © as the product

(iv) Chlorobenzene on reaction Na metal gives Biphenyl compound © as the product.

(i) C₄H₉Br: Two isomers may be there: (A) \({ CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ \underset { \overset { | }{ Br } }{ C } } -{ CH }_{ 3 }\) Tert butyl bromide and \({ CH }_{ 3 }-\underset { \overset { | }{ Br } }{ C } H-{ CH }_{ 2 }-{ CH }_{ 3 }\) α-Bromobutane
(ii)Ⓐ on reaction with alcoholic KOH gives ISO butyleneⒸ as the product.

(iii) 2-bromobutane Ⓑ on reaction with alcoholic KOH follows Saytzeff's rule to give a mixture of olefins in different amounts

(i) An organic compound ® of molecular formula C₂H₂ is CH = CH (Acetylene)

(ii) Acetylene Ⓐ reacts with HCI to give ethylene dichloride Ⓑ as the product.

(i) The aromatic hydrocarbon Ⓐ is Benzene, C₆H₆.
(ii) Benzene reacts. with Cl₂ to give chlorobenzene (C₆H₅CI), as Ⓑ as the product.

(iii) Chlorobenzene Ⓑ acts with sodium hydroxide to give (C₆HsOH) phenol, © as the product.

(iv) Chlorobenzene reacts with ammonia to give Aniline, C₆H₅NH₂Ⓓ as the product.