What is $A$ in the following reaction?
$C_6H_5-CH_2-CH=CH_2 + HCl \rightarrow A$

Alkenes convert into alcohols by:

Consider compounds $A$, $B$ and $C$ with following structural formulae: $A = CH_{3} - CH_{2} - CH_{2} - CH_{2} - CH_{2} - OH$, $B = CH_{2} = CH - CH_{2} - CH_{2} - CH_{3}$, $C = HO - CH_{2} - CH_{2} - CH(OH) - CH_{3}$. For the conversion of $B$ from $A$, reagent $(D)$ required is . . . . . . and structural formula of product $(E)$ obtained when $C$ undergoes same reaction using excess reagent $(D)$ is . . . . . . .

An alkene $A$ (molecular formula $C_{5}H_{10}$) on ozonolysis gives a mixture of two compounds $B$ and $C$. Compound $B$ gives a positive Fehling's test and also forms iodoform on treatment with $I_{2}$ and $NaOH$. Compound $C$ does not give Fehling's test but forms iodoform. Identify the compounds $A, B$ and $C$. Write the reaction for ozonolysis and the formation of iodoform from $B$ and $C$.

Which of the following alkenes does not undergo anti-Markovnikov addition of $HBr$?

In the following reaction,$A$ and $B$ respectively are
$A \xrightarrow{HBr} C_2H_5Br \xrightarrow{B} A$