A) 2-pentanone
B) ethanol
C) ethanal
D) 3-pentanone
Correct Answer: D
Solution :
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 \[{{C}_{2}}{{H}_{5}}CHO\] give this test CH3CH2CH2COCH3 2-pentanone and 3-pentanone \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}COC{{H}_{3}}\] does not give idoform test. \[\begin{align} & C{{H}_{3}}-COC{{H}_{2}}-C{{H}_{2}}-C{{H}_{3}}+3{{I}_{2}}+4NaOH \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\to \,\,\underset{\begin{smallmatrix} Iodoform\, \\ (Yellow\,ppt.) \end{smallmatrix}}{\mathop{CH{{I}_{3}}}}\,\downarrow \,+\,C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}COONa \\ \end{align}\] \[+3NaI+3{{H}_{2}}O\] \[\begin{align} & C{{H}_{3}}-CHO+3{{I}_{2}}+4NaOH\xrightarrow[{}]{{}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underset{\begin{smallmatrix} Iodoform\, \\ (yellow\,ppt) \end{smallmatrix}}{\mathop{CH{{I}_{3}}}}\,\downarrow \,\,+HCOONa \\ \end{align}\] \[+3NaI+3{{H}_{2}}O\] \[{{C}_{2}}{{H}_{5}}OH+4{{I}_{2}}+6NaOH\xrightarrow[{}]{{}}\underset{\begin{smallmatrix} Iodoform \\ (yellow\,ppt). \end{smallmatrix}}{\mathop{CH{{I}_{3}}}}\,\,\,\downarrow \] \[+HCOONa+5NaI+5{{H}_{2}}O\] In it following reaction is posses \[({{C}_{2}}{{H}_{5}}OH\xrightarrow[{}]{{{I}_{2}}}C{{H}_{3}}-CHO\xrightarrow[{}]{{{I}_{2}}+NaOH}CH{{I}_{3}})\] \[\begin{align} & C{{H}_{3}}-C{{H}_{2}}-CO-C{{H}_{2}}-C{{H}_{3}}+{{I}_{2}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+NaOH\xrightarrow[{}]{{}}No\,reaction \\ \end{align}\]
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