A) 2-pentanone
B) ethanol
C) ethanol
D) 3-pentanone
Correct Answer: D
Solution :
| [d] Iodoform test is given by compounds which have |
| \[\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,H \\ & C{{H}_{3}}-CO-group\,\,or\,C{{H}_{3}}-\overset{|}{\mathop{\underset{|}{\mathop{C}}\,}}\,-group. \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,OH \\ \end{align}\] |
| Hence, 2-pentanone, \[{{C}_{2}}{{H}_{5}}CHO\] and \[{{C}_{2}}{{H}_{5}}OH\] give this test \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}COC{{H}_{3}}\] |
| 2-pentanone and 3-pentanone \[C{{H}_{3}}C{{H}_{2}}COC{{H}_{2}}C{{H}_{3}}\] does not give idoform test. |
| \[C{{H}_{3}}-COC{{H}_{2}}-C{{H}_{2}}-C{{H}_{3}}+3{{I}_{2}}+4NaOH\]\[\to \,\,\underset{\begin{smallmatrix} Iodoform\, \\ (Yellow\,ppt.) \end{smallmatrix}}{\mathop{CH{{I}_{3}}}}\,\downarrow \,+\,C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}COONa\]\[+3NaI+3{{H}_{2}}O\] |
| \[C{{H}_{3}}-CHO+3{{I}_{2}}+4NaOH\xrightarrow[{}]{{}}\]\[\underset{\begin{smallmatrix} Iodoform\, \\ (yellow\,ppt) \end{smallmatrix}}{\mathop{CH{{I}_{3}}}}\,\downarrow \,\,+HCOONa\]\[+3NaI+3{{H}_{2}}O\] |
| \[{{C}_{2}}{{H}_{5}}OH+4{{I}_{2}}+6NaOH\xrightarrow[{}]{{}}\underset{\begin{smallmatrix} Iodoform \\ (yellow\,ppt). \end{smallmatrix}}{\mathop{CH{{I}_{3}}}}\,\,\,\downarrow \] \[+HCOONa+5NaI+5{{H}_{2}}O\] |
| In it following reaction is posses |
| \[({{C}_{2}}{{H}_{5}}OH\xrightarrow[{}]{{{I}_{2}}}C{{H}_{3}}-CHO\xrightarrow[{}]{{{I}_{2}}+NaOH}CH{{I}_{3}})\] |
| \[C{{H}_{3}}-C{{H}_{2}}-CO-C{{H}_{2}}-C{{H}_{3}}+{{I}_{2}}\]\[+NaOH\xrightarrow[{}]{{}}No\,reaction\] |
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