A) Methanol
B) Ethyl carbinol
C) Pentanol\[-3\]
D) Methyl carbinol
Correct Answer: D
Solution :
Key Idea: Iodoform test is given by alcohols which produce carbonyl compounds having \[C{{H}_{3}}CO\] group. (i)\[C{{H}_{3}}OH\xrightarrow{[O]}HCHO\], no\[C{{H}_{3}}CO\]group, no iodoform test. (ii)\[\underset{\begin{smallmatrix} methyl \\ carbinol \end{smallmatrix}}{\mathop{{{C}_{2}}{{H}_{5}}OH}}\,\xrightarrow{[O]}C{{H}_{3}}CHO\], has \[C{{H}_{3}}CO\] group, gives iodoform test (iii)\[\underset{\begin{smallmatrix} ethyl \\ carbinol \end{smallmatrix}}{\mathop{{{C}_{2}}{{H}_{5}}C{{H}_{2}}OH}}\,\xrightarrow{[O]}{{C}_{2}}{{H}_{5}}CHO\], no\[C{{H}_{3}}CO\] group no iodoform test. (iv)\[C{{H}_{3}}-C{{H}_{2}}-CH(OH)-C{{H}_{2}}-C{{H}_{3}}\xrightarrow{[O]}\]\[C{{H}_{3}}-C{{H}_{2}}-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-C{{H}_{2}}-C{{H}_{3}}\] No \[C{{H}_{3}}CO\] group, no iodoform test. \[\therefore \]Only methyl carbinol gives iodoform test.
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