Addition of HBr on; CH ≡ C - CH2 - CH = CH2 and CH ≡ C - CH = CH2 separately gives

Question

Addition of HBr on; CH ≡ C - CH2 - CH = CH2 and CH ≡ C - CH = CH2 separately gives (A)  (B)  (C)  (D) none of these

Tardigrade Admin · Accepted Answer

The first addition will occur on the double bond. Thus, But in CH ≡ C - CH = CH2, he addition will occur at CH ≡ C because the product formed is conjugated alkadiene, which is more stable.

Q. Addition of HBr on;
$C H \equiv C - C H_{2} - C H = C H_{2}$ and $C H \equiv C - C H = C H_{2}$ separately gives

none of these

The first addition will occur on the double bond. Thus,
But in $C H \equiv C - C H = C H_{2},$ he addition will occur at $C H \equiv C$ because the product formed is conjugated alkadiene, which is more stable.

Q. Addition of HBr on; CH≡C−CH2​−CH=CH2​ and CH≡C−CH=CH2​ separately gives

none of these

The first addition will occur on the double bond. Thus, But in CH≡C−CH=CH2​, he addition will occur at CH≡C because the product formed is conjugated alkadiene, which is more stable.

Q. Addition of HBr on;
$C H \equiv C - C H_{2} - C H = C H_{2}$ and $C H \equiv C - C H = C H_{2}$ separately gives

The first addition will occur on the double bond. Thus,
But in $C H \equiv C - C H = C H_{2},$ he addition will occur at $C H \equiv C$ because the product formed is conjugated alkadiene, which is more stable.