The reaction of $CH_3-CH=CH-C_6H_4-OH$ with $HBr$ gives:

Identify the compound that undergoes $S_{N}1$ mechanism most rapidly.

Which of the following will give $S_N2$ reaction at the fastest rate?

The correct decreasing order of $SN^2$ reactivity of the following alkyl halides is:
$(I)$ $CH_2=CH-Br$
$(II)$ $CH_3-Br$
$(III)$ $CH_3-CH(Br)-CH_3$
$(IV)$ $CH_3-CH_2-Br$

In the above reaction,the maximum Saytzeff product will be obtained when $X$ is:

An alkene $X$ $(C_4H_8)$ on reaction with $HBr$ gave $Y$ $(C_4H_9Br)$. Reaction of $Y$ with benzene in the presence of anhydrous $AlCl_3$ gave $Z$,which is resistant to oxidation with $KMnO_4-KOH$. What are $X, Y, Z$ respectively?