A) ethyl alcohol and 2-propanol
B) ehtane and hemi acetal
C) 2-propanol and acetal
D) propane and methyl acetate
Correct Answer: C
Solution :
(i) \[C{{H}_{3}}=\underset{\begin{smallmatrix} \text{ }\!\!|\!\!\text{ } \\ \text{H} \end{smallmatrix}}{\overset{\begin{smallmatrix} \text{O} \\ \text{ }\!\!|\!\!\text{ }\!\!|\!\!\text{ } \end{smallmatrix}}{\mathop{\text{C}}}}\,\xrightarrow[(ii){{H}_{3}}{{O}^{+}}]{(i)C{{H}_{3}}MgBr}\underset{2-\text{Propanol}}{\mathop{C{{H}_{3}}\underset{\begin{smallmatrix} \text{ }\!\!|\!\!\text{ } \\ \text{C}{{\text{H}}_{\text{3}}} \end{smallmatrix}}{\overset{\begin{smallmatrix} \text{H} \\ \text{ }\!\!|\!\!\text{ } \end{smallmatrix}}{\mathop{\text{C}}}}\,-OH}}\,\] (ii) \[C{{H}_{3}}=\underset{{}}{\overset{\begin{smallmatrix} \text{H} \\ \text{ }\!\!|\!\!\text{ } \end{smallmatrix}}{\mathop{\text{C}}}}\,=O+{{C}_{2}}{{H}_{5}}OH\xrightarrow[{}]{HCl}\] \[\underset{Hemiacetal}{\mathop{C{{H}_{3}}-\underset{\begin{smallmatrix} \text{ }\!\!|\!\!\text{ } \\ \text{OH} \end{smallmatrix}}{\overset{\begin{smallmatrix} \text{H} \\ \text{ }\!\!|\!\!\text{ } \end{smallmatrix}}{\mathop{\text{C}}}}\,-O{{C}_{2}}{{H}_{5}}}}\,\xrightarrow[{{C}_{2}}{{H}_{5}}OH]{HCl}C{{H}_{3}}-\underset{{}}{\overset{\begin{smallmatrix} \text{H} \\ \text{ }\!\!|\!\!\text{ } \end{smallmatrix}}{\mathop{\text{C}}}}\,\underset{-Acetal-}{\mathop{{{(O{{C}_{2}}{{H}_{5}})}_{2}}}}\,\]
You need to login to perform this action.
You will be redirected in
3 sec
