- A+1
- B-1
- C+2
- D-2
Explanation:
Cl is in -1 oxidation state since it has lower atomic number than iodine. Let the oxidation state of iodine be x.
x + (−1) × 2 = −1
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$C{H_3}CHBrC{H_2}C{H_3}\xrightarrow{{alc.\,KOH}}$
$(i)$ $C{H_3}CH = CHC{H_3}$ (major product)
$(ii)$ $C{H_2} = CHC{H_2}C{H_3}$ (minor product)

$\mathop {{{(C{H_3})}_3}C\mathop C\limits^ + HC{H_2}Cl}\limits_1 $ $\mathop {{{(C{H_3})}_3}C\mathop C\limits^ + HC{H_3}}\limits_2 $ $\mathop {\begin{array}{*{20}{c}}
{{{(C{H_3})}_2}C\mathop C\limits^ + {{(C{H_3})}_2}} \\
{|\,\,\,\,} \\
{Cl\,\,\,}
\end{array}}\limits_3 $ $\mathop {{{(C{H_3})}_2}\mathop C\limits^ + CH{{(C{H_3})}_2}}\limits_4 $