Concept:Both reactions form a carbocation that rearranges via a 1,2-shift to a tertiary carbocation, giving the same product X.
Explanation:Step 1: For reaction (i), 3-Methylbut-1-ene (
CH2=CH−CH(CH3)2) adds HCl.
Protonation at C-1 gives a secondary carbocation at C-2.
A 1,2-hydride shift from the adjacent tertiary carbon (
−CH(CH3)2) moves the positive charge, forming a tertiary carbocation:
(CH3)2C+−CH2−CH3.
Chloride ion attacks this tertiary carbocation, yielding
(CH3)2CCl−CH2−CH3.
Step 2: For reaction (ii), neopentyl alcohol (
(CH3)3C−CH2OH) reacts with
HCl/anh.
ZnCl2 (Lucas reagent).
ZnCl2 activates the -OH group, and loss of water generates a primary neopentyl carbocation at C-1:
(CH3)3C−CH2+.
A 1,2-methyl shift from the adjacent tertiary carbon converts it to the same tertiary carbocation:
(CH3)2C+−CH2−CH3.
Chloride ion then attacks to give
(CH3)2CCl−CH2−CH3 again.
Thus, X is identical in both cases: 2-chloro-2-methylbutane.
Answer:Option B:
(CH3)2CCl−CH2−CH3 (2-chloro-2-methylbutane).