$A$ dihaloalkane '$X$',having formula $C_{3}H_{6}Cl_{2}$,on hydrolysis gives a compound that can reduce Tollen's reagent. The compound '$X$' is

Compound $A$ with molecular formula $C_4H_9Br$ is treated with aqueous $KOH$ solution. The rate of this reaction depends upon the concentration of compound $A$ only. When another optically active isomer $B$ of this compound was treated with aqueous $KOH$ solution,the rate of reaction was found to be dependent on the concentration of both the compound and $KOH$.
$(i)$ Write down the structural formula of both compounds $A$ and $B$.
$(ii)$ Out of these two compounds,which one will be converted to the product with inverted configuration?

Which of the following is most reactive towards alcoholic $KOH$?

Given below are two statements:
Statement $I$: $(CH_3)_3C-CH_2-Cl$ will undergo $S_N1$ reaction even though it is a primary halide.
Statement $II$: It will not undergo $S_N2$ reaction very easily even though it is a primary halide.
In the light of the above statements,choose the most appropriate answer from the options given below:

The compounds $A$ and $B$ in the following reaction are,respectively:
$Benzene$ $\xrightarrow{HCHO + HCl}$ $A$ $\xrightarrow{AgCN}$ $B$

Benzene $+ (CH_3CO)_2O$ $\xrightarrow{AlCl_3} X$ $\xrightarrow{Zn-Hg, HCl} Y$ $\xrightarrow{NBS} Z$;
$Z$ is