A) \[{{C}_{2}}{{H}_{5}}OH\]
B) \[C{{H}_{3}}OH\]
C) \[C{{H}_{3}}COC{{H}_{3}}\]
D) \[C{{H}_{3}}-\underset{OH}{\mathop{\underset{|}{\mathop{CH}}\,}}\,-C{{H}_{3}}\]
Correct Answer: B
Solution :
lodoform test is a characteristic test of \[-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-C{{H}_{3}}\]group. Ethyl alcohol and ethyl chloride also give this test as in basic conditions, these are oxidised into acetaldehyde \[(C{{H}_{3}}CHO)\] which give iodoform test. Isopropyl alcohol gives acetone on oxidation, which also give the test. \[R-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-C{{H}_{3}}+3{{I}_{2}}+4NaOH\xrightarrow{{}}\] \[\underset{Iodoform\,(yellow)}{\mathop{CH{{I}_{3}}+RCOONa+3NaI+3{{H}_{2}}O}}\,\] \[C{{H}_{3}}OH,\], on oxidation gives \[HCHO\]. which does not have \[-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-C{{H}_{3}}\]group, hence, it does not give iodoform test.
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