Addition of $HI$ on the double bond of propene yields isopropyl iodide and not $n$-propyl iodide as the major product. This is because the addition proceeds through

The chemical reaction is shown below:
$C_6H_5-CH_2-CH=CH_2 + HCl \rightarrow X$,$X$ is:

An alkene $X$ $(C_6H_{12})$ on ozonolysis gives acetaldehyde and ethyl methyl ketone. What is the product formed when $X$ reacts with $HBr$?

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 $C$.

For the reaction: $CH_2=CH-CH=CH_2 + HBr (1 \text{ eq}) \xrightarrow{80^{\circ}C} P_1 + P_2$ (where $P_1$ is the major product),identify $P_1$.

The major product of the following reaction is: