A ketone A (C₄H₈O) which undergoes Iodoform reaction gives compound B on reduction. B on heating…

A ketone A ($\text{C}_4\text{H}_8\text{O}$) which undergoes Iodoform reaction gives compound B on reduction. B on heating with $\text{conc. H}_2\text{SO}_4$ at $443\text{ K}$ gives a compound C which forms ozonide D. D on hydrolysis with Zn dust gives only E. Identify the compounds A to E. Write the Iodoform reaction with compound A.

Answer

Equation: $\mathrm{CH_{3}COCH_{2}CH_{3}} + 3\mathrm{I_{2}} + 4\mathrm{NaOH} \xrightarrow{\text{iodoform}\text{ }\text{reaction}} \mathrm{CHI_{3}} + \mathrm{CH_{3}CH_{2}COONa} + 3\mathrm{NaI} + 3\mathrm{H_{2}O}$

A - butan-2-one: CH3COCH2CH3

B - butan-2-ol: CH3CH(OH)CH2CH3

C - but-2-ene: CH3CH=CHCH3

D - 3,5-dimethyl-1,2,4-trioxolane: C4H8O3

E - ethanal: CH3CHO

A = butan-2-one ($\mathrm{CH_3COCH_2CH_3}$) B = butan-2-ol ($\mathrm{CH_3CH(OH)CH_2CH_3}$) C = but-2-ene ($\mathrm{CH_3CH=CHCH_3}$) D = 3,5-dimethyl-1,2,4-trioxolane (but-2-ene ozonide) E = ethanal ($\mathrm{CH_3CHO}$) Iodoform reaction of A: butan-2-one gives yellow iodoform and sodium propanoate.