The optically inactive compound from the following is

What is the principal product of the following reaction?
$CH_3-CH(Br)-CH_2-CH_2-CH(Cl)-CH_3 + NaN_3 \rightarrow \text{Product}$

Consider the following reaction:
$CH_3-CH_2-CH(Br)-CH_3 \xrightarrow{alc. KOH, \Delta} CH_3-CH_2-CH=CH_2 (I) + CH_3-CH=CH-CH_3 (II)$
Which of the following statements are correct?
$a.$ $I$ is the major product of the reaction
$b.$ $II$ is the major product of the reaction
$c.$ Formation of $I$ is in accordance with Saytzeff's rule
$d.$ $II$ is more stable because it is more substituted

Consider the reaction: $1$-methylcyclopentene $+ HBr \xrightarrow{CCl_4} \text{Product}$. Comment upon the optical activity of the product.

In the following reaction $C_{6}H_{5}CH_{2}Br \xrightarrow[(ii)H_{3}O^{+}]{(i)Mg, \text{Ether}} X$
Product $X$ is

$C_6H_6 + CH_3CH_2CH_2Cl + AlCl_3 \to X$ (Major product)
$X$ (Major product) $\xrightarrow{HNO_3 + H_2SO_4} Y$ (Major product) $\xrightarrow{Sn + HCl} Z$ $\xrightarrow{H_2SO_4 + SO_3} W$
$W$ is