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