A) \[C{{H}_{3}}C{{H}_{2}}CH=NNHCON{{H}_{2}}\]
B) \[C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,=NNHCON{{H}_{2}}\]
C) \[C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,=NCONHN{{H}_{2}}\]
D) \[C{{H}_{3}}C{{H}_{2}}CH=NCONHN{{H}_{2}}\]
Correct Answer: A
Solution :
[a] \[\underset{A}{\mathop{{{C}_{3}}{{H}_{8}}O\xrightarrow{{{K}_{2}}C{{r}_{2}}{{O}_{7}}/{{H}^{+}}}{{C}_{3}}{{H}_{6}}O}}\,\] \[\underset{B\,(-CHO)}{\mathop{\xrightarrow{{{H}_{2}}NCONHN{{H}_{2}}}C}}\,\] Since B reduces Tollen's reagent, it indicates that it has an \[-CHO\] group, so it must be\[C{{H}_{3}}C{{H}_{2}}CHO\]. Hence \[\underset{[A]}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}OH}}\,\to \underset{[B]}{\mathop{C{{H}_{3}}C{{H}_{2}}CHO}}\,\xrightarrow{{{H}_{2}}NHCON{{H}_{2}}}\]\[\underset{[C]}{\mathop{C{{H}_{3}}C{{H}_{2}}CH}}\,=NNHCON{{H}_{2}}\]You need to login to perform this action.
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