A) \[C{{H}_{3}}C{{H}_{2}}C\equiv N\xrightarrow{HCl}{{[C{{H}_{3}}C{{H}_{2}}C\equiv NH]}^{+}}C{{l}^{-}}\]
B) \[\xrightarrow{SnC{{l}_{2}}/HCl}{{[C{{H}_{3}}-C{{H}_{2}}CH=N{{H}_{2}}]}^{+}}SnCl_{6}^{2-}\]
C) \[\xrightarrow{{{H}_{2}}O/boil}C{{H}_{3}}C{{H}_{2}}CHO+N{{H}_{4}}Cl\]
D) \[C{{H}_{3}}Br+NaO-\underset{C{{H}_{3}}}{\mathop{\overset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{\overset{|}{\mathop{C}}\,}}\,}}\,}}\,-C{{H}_{3}}\xrightarrow[-NaBr]{}\underset{t-butyl\,methyl\,ether}{\mathop{C{{H}_{3}}-O-\underset{C{{H}_{3}}}{\mathop{\overset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{\overset{|}{\mathop{C}}\,}}\,}}\,}}\,-C{{H}_{3}}}}\,\]
Correct Answer: D
Solution :
(I) \[2\] (II) \[0.25\] (III) \[4\] Multiply Eq. (II) by 2, adding to (I) and then, subtract Eq. (Ill) from the sum \[1\]You need to login to perform this action.
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