A) 15.4V
B) 2.35V
C) 3.85V
D) 1.5V
Correct Answer: D
Solution :
\[A{{g}_{2}}C{{O}_{3}}\] and \[=\frac{2.76\times 216}{276}=2.16g\] \[\begin{align} & C{{H}_{3}}-CH-C{{H}_{2}}C{{H}_{2}}OH\xleftarrow{{{H}_{2}}O} \\ & | \\ & C{{H}_{3}} \\ & \text{3-meethyl butanol} \\ \end{align}\] \[\begin{align} & C{{H}_{3}}-CH-C{{H}_{2}}C{{H}_{2}}OMgBr \\ & | \\ & C{{H}_{3}} \\ \end{align}\] \[Ph-C\equiv C-C{{H}_{3}}+{{H}_{2}}O\xrightarrow{H{{g}^{2+}}/{{H}^{+}}}\] \[\begin{align} & OH \\ & | \\ & Ph-C=CH-C{{H}_{3}} \\ \end{align}\] \[\begin{align} & OOH \\ & ||| \\ & Ph-C-C{{H}_{2}}C{{H}_{3}}\xleftarrow[-{{H}_{2}}O]{}Ph-C-C{{H}_{2}}C{{H}_{3}} \\ & | \\ & OH \\ \end{align}\] \[\xleftarrow[{{H}_{2}}O]{H{{g}^{2+}}/{{H}^{+}}}\] = 1.5 VYou need to login to perform this action.
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