A) ethanol
B) ethanal
C) methanol
D) propanone
Correct Answer: C
Solution :
Key Idea: lodoform test is given by compounds having \[\text{C}{{\text{H}}_{\text{3}}}\text{CO}\]or \[\text{C}{{\text{H}}_{1}}\text{OH}\]group. \[C{{H}_{3}}-\underset{has\,C{{H}_{2}}OH\,group}{\mathop{\underset{(a)}{\mathop{\underset{H}{\overset{H}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}}\,}}\,}}\,-OH\] \[\underset{\text{has}\,\text{C}{{\text{H}}_{\text{3}}}\text{COgroup}}{\mathop{\underset{\,\,\,\,\,\,\,\,(b)}{\mathop{C{{H}_{3}}-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-H}}\,}}\,\] \[\underset{\text{has}\,\text{C}{{\text{H}}_{\text{3}}}\text{CO}\,\text{group}}{\mathop{\underset{\,\,\,\,\,\,\,\,(d)}{\mathop{C{{H}_{3}}-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-C{{H}_{3}}}}\,}}\,\] \[\therefore \]All of them give iodoform test. \[\text{C}{{\text{H}}_{\text{3}}}\text{OH}\](choice c) does not have \[\text{C}{{\text{H}}_{\text{3}}}\text{CO}\]or \[\text{C}{{\text{H}}_{\text{2}}}\text{OH}\]group and so it does not give iodoform test.
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