Question

What are X and Y in the following reaction sequence?
2-Methylbutane $\xrightarrow{\text{KMnO}_4}$ X$\xrightarrow{\text{Y}}$ C5H11Cl

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

KMnO$_4$ oxidizes tertiary C-H to C-OH in alkanes. Conc. HCl with ZnCl$_2$ converts tertiary alcohols to chlorides. Identify functional group changes in sequence. Check reagents' roles.

Answer 1

The Correct Option is B

1. 2-Methylbutane (C$_5$H$_{12}$): CH$_3$-CH(CH$_3$)-CH$_2$-CH$_3$, a tertiary carbon at C2.
2. KMnO$_4$ (strong oxidizing agent) oxidizes tertiary alkanes to tertiary alcohols. Y is (CH$_3$)$_2$C(OH)-CH$_2$-CH$_3$ (2-methylbutan-2-ol).
3. Tertiary alcohol with conc. HCl and ZnCl$_2$ (Lucas reagent) forms tertiary chloride: X is (CH$_3$)$_2$C(Cl)-CH$_2$-CH$_3$.
4. Options are unclear placeholders; (2) implies OH group in Y, and conc. HCl with ZnCl$_2$ for chlorination, so correct.
5. Thus, the answer is (2) OH, Conc. HCl, ZnCl$_2$.