The product$(s)$ formed during the following reaction is/are:
$Ph-CH_2-C^*H_2-OTs \xrightarrow{AcONa, AcOH} \text{Product(s)} \ [C^* = \text{isotopic carbon}]$

Which one of the following compounds will be most reactive for $S_N1$ reactions?

Given below are two statements:
Statement $I$: $C-Cl$ bond is stronger in $CH_2=CH-Cl$ than $CH_3-CH_2-Cl$.
Statement $II$: The given optically active molecule on hydrolysis gives a solution that can rotate the plane polarized light.
In the light of the above statements, choose the correct answer from the options given below.

Consider the following statements:
$(1)$ Bridgehead halides are inert towards both $S_{N^1}$ and $S_{N^2}$ reactions (until one of the ring size is an eight-membered ring).
$(2)$ The first step in both $S_{N^1}$ and $E_1$ reactions is the same.
$(3)$ $S_{N^2}$ reactions proceed with total retention of configuration.
$(4)$ $E_2$ eliminations are favored by the use of a solvent of low polarity and a high concentration of a strong base.
Which of the above statements are correct?

Arrange the following reactants in increasing order of reactivity towards $S_{N^1}$ solvolysis reaction with methanol.

In the following reaction sequence,
$Br-CH_2-CH(Br)-Ph$ $\xrightarrow[2. NaNH_2]{1. Alc. KOH} X$ $\xrightarrow[4. Conc. HNO_3/H_2SO_4]{3. HgSO_4/dil. H_2SO_4, Heat} Y$
$X$ and $Y$ respectively are,