$CH_2=CH_2 \xrightarrow{Br_2/H_2O} A$. In the above reaction,the compound $A$ is:

The predominant product formed when $3$-methyl-$2$-pentene reacts with $HOCl$ is:

$A$ hydrocarbon of formula $C_6H_{10}$ absorbs only one molecule of $H_2$ upon catalytic hydrogenation. Upon ozonolysis,the hydrocarbon yields $OHC-CH_2-CH_2-CH_2-CH_2-CHO$. The hydrocarbon is:

The reaction of propene with $HOCl$ proceeds via the addition of which of the following in the first step?

Identify the correct statements with respect to $cis/trans$-but$-2$-ene from the following:
$I$. $cis$-but$-2$-ene is more polar than $trans$-but$-2$-ene
$II$. Melting point of $cis$-but$-2$-ene is greater than that of $trans$-but$-2$-ene
$III$. Boiling point of $cis$-but$-2$-ene is greater than that of $trans$-but$-2$-ene
Correct answer is

The main product of the following reaction $(1 \ mol)$ will be:
$C_6H_5-CH_2-CH=CH_2 + HBr \rightarrow \text{Products}$