The major product (P) formed in the below reaction is H - C ≡ C - CH 2- CH = CH 2 -[Br2 (1 mol)][CCl4,253 K] P

Question

The major product (P) formed in the below reaction is H - C ≡ C - CH 2- CH = CH 2 ->[Br2 (1 mol)][CCl4,253 K] P (A) H - C ≡ C - CH 2- CH ( Br )- CH 2( Br ) (B) CH ( Br )2- C ( Br )2- CH 2- CH = CH 2 (C) CH 2( Br )- CH ( Br )- CH 2- CH = CH 2 (D) CH 2= CH ( Br )- CH ( Br )- CH = CH 2

Tardigrade Admin · Accepted Answer

Treatment of alkene with Br 2 in CCl 4 solvent gives vicinal dibromides. Bromine added to opposite faces of the double bond (anti addition). CH ≡ C - CH 2- CH = CH 2

Q. The major product $(P)$ formed in the below reaction is
$H - C \equiv C - C H_{2} - C H = C H_{2} B r_{2} (1mol) CC l_{4}, 253 K P$

$H - C \equiv C - C H_{2} - C H (B r) - C H_{2} (B r)$

$C H (B r)_{2} - C (B r)_{2} - C H_{2} - C H = C H_{2}$

$C H_{2} (B r) - C H (B r) - C H_{2} - C H = C H_{2}$

$C H_{2} = C H (B r) - C H (B r) - C H = C H_{2}$

Treatment of alkene with $B r_{2}$ in $CC l_{4}$ solvent gives vicinal dibromides. Bromine added to opposite faces of the double bond (anti addition).

$C H \equiv C - C H_{2} - C H = C H_{2}$

Q. The major product (P) formed in the below reaction is H−C≡C−CH2​−CH=CH2​Br2​ (1mol)CCl4​,253 K​P

H−C≡C−CH2​−CH(Br)−CH2​(Br)

CH(Br)2​−C(Br)2​−CH2​−CH=CH2​

CH2​(Br)−CH(Br)−CH2​−CH=CH2​

CH2​=CH(Br)−CH(Br)−CH=CH2​