MCQ
Product $(A)$ is
- A
- ✓
- C
- D
✓
Answer
Correct option: B.
b
Hint : Think carefully about the relative stabilization of developing positive charge, when the double bond reacts with an electrophile
Hint : Think carefully about the relative stabilization of developing positive charge, when the double bond reacts with an electrophile
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Similar questions
Match List$-I$ with List$-II :$
→|
List$-I$ (Species) |
List$-II$ (Hybrid Orbitals) |
| $(a)$ $\mathrm{SF}_{4}$ | $(i)$ $\mathrm{sp}^{3} \mathrm{~d}^{2}$ |
| $(b)$ $\mathrm{IF}_{5}$ | $(ii)$ $\mathrm{d}^{2} \mathrm{sp}^{3}$ |
| $(c)$ $\mathrm{NO}_{2}^{+}$ | $(iii)$ $\mathrm{sp}^{3} \mathrm{~d}$ |
| $(d)$ $\mathrm{NH}_{4}^{+}$ | $(iv)$ $\mathrm{sp}^{3}$ |
| $(v)$ $\mathrm{sp}$ |
Choose the correct answer form the options given below :
$\begin{array}{*{20}{c}}
{C{H_2} - C{O_2}K} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_2} - C{O_2}K}
\end{array}\xrightarrow{{electrolysis}}\mathop {(A)}\limits_{\left( {major} \right)} $ (Kolbe electrolysis method)
→{C{H_2} - C{O_2}K} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_2} - C{O_2}K}
\end{array}\xrightarrow{{electrolysis}}\mathop {(A)}\limits_{\left( {major} \right)} $ (Kolbe electrolysis method)
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