What are reactant X and product Y in the foll | vedclass

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(A) The reaction sequence involves the conversion of an alcohol to a tosylate followed by an $S_N2$ reaction with sodium azide $(NaN_3)$.$1$. The star

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(A) The reaction sequence involves the conversion of an alcohol to a tosylate followed by an $S_N2$ reaction with sodium azide $(NaN_3)$.$1$. The starting material $X$ is cis$-3-$methylcyclopentanol,where the $-OH$ group and the $-CH_3$ group are on the same side of the cyclopentane ring.$2$. Reaction with $p$-toluenesulfonyl chloride $(TsCl)$ in pyridine converts the $-OH$ group into a good leaving group (tosylate,$-OTs$) without changing the stereochemistry at the chiral center.$3$. The subsequent reaction with $NaN_3$ proceeds via an $S_N2$ mechanism,which involves an inversion of configuration at the carbon atom bearing the leaving group.$4$. Since the starting alcohol is cis$-3-$methylcyclopentanol,the inversion of configuration results in the formation of trans$-3-$azidocyclopentane as product $Y$.

$(I)$ Compound	$(II)$ $C-X$ Bond Length $(pm)$	$(III)$ Dipole Moment $(D)$
$(A)$ $H_3C-F$	$(i)$ $214$	$(1)$ $1.636$
$(B)$ $H_3C-Cl$	$(ii)$ $193$	$(2)$ $1.830$
$(C)$ $H_3C-Br$	$(iii)$ $178$	$(3)$ $1.847$
$(D)$ $H_3C-I$	$(iv)$ $139$	$(4)$ $1.860$

What are reactant $X$ and product $Y$ in the following sequence of reactions?
Reactant $X$ $\xrightarrow{\text{Pyridine, } CH_3C_6H_4SO_2Cl}$ Intermediate $\xrightarrow{NaN_3, \text{Ethanol-water}}$ Product $Y$

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What are reactant $X$ and product $Y$ in the following sequence of reactions?Reactant $X$ $\xrightarrow{\text{Pyridine, } CH_3C_6H_4SO_2Cl}$ Intermediate $\xrightarrow{NaN_3, \text{Ethanol-water}}$ Product $Y$