A) 1, 2 and 5
B) 1, 3 and 4
C) 1, 2 and 3
D) 2, 4 and 5
E) 1, 2 and 4
Correct Answer: E
Solution :
Carbonyl compounds having\[C{{H}_{3}}CO\]group and alcohols having\[C{{H}_{3}}CH(OH)\]group undergo iodoform reaction. \[\underset{ethanol}{\mathop{C{{H}_{3}}C{{H}_{2}}OH}}\,+4{{I}_{2}}+6NaOH\xrightarrow[{}]{\Delta }\] \[\underset{iodoform}{\mathop{CH{{I}_{3}}}}\,+HCOONa+5NaI+5{{H}_{2}}O\] \[\underset{2-bu\tan ol}{\mathop{C{{H}_{3}}C{{H}_{2}}\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{3}}}}\,+4{{I}_{2}}+6NaOH\xrightarrow[{}]{\Delta }\] \[\underset{iodoform}{\mathop{CH{{I}_{3}}}}\,+C{{H}_{3}}C{{H}_{2}}COONa+5NaI+5{{H}_{2}}O\] \[\underset{ethanal}{\mathop{C{{H}_{3}}CHO}}\,+3{{I}_{2}}+NaOH\xrightarrow{\Delta }\underset{iodoform}{\mathop{CH{{I}_{3}}}}\,\] \[+HCOONa+3HI\]
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