
- A$NH_2 -NH_2/ KOH$
- B$Zn -Hg/ HCl$
- C$Red\, P + HI$
- ✓All

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$\mathrm{Li}, \mathrm{Be}, \mathrm{B}, \mathrm{C}, \mathrm{N}$
Choose the correct answer from the options given below:

$A$ $+$ $\begin{array}{*{20}{c}}
{C{H_3} - C - C{H_3}} \\
{||\,} \\
{O\,\,\,}
\end{array}$ $→$ $B$ $'B'$ is
$\mathop C\limits_6 {H_3} - \mathop C\limits_5 H = \mathop C\limits_4 H - \mathop C\limits_3 {H_2} - \mathop C\limits_2 \equiv \mathop C\limits_1 H$
The state of hybridization of carbons $1, 3$ and $5$ are in the following sequence
$CH_3-CH = CH -CH_2 -CH_2 -CH(CH_3)_2$
$(A)$ $(B)\,\,\,(C)$ $(D)$ $(E)\,\,\,(F)$
arrange them in decreasing order of reactivity towards free radical substitution