The principal organic product formed in the reaction is:
$C H_{2} = C H (C H_{2})_{8} C O O H + H B r P e r o x i d e - ---- \to$

C H_{3} C H (B r) (C H_{2})_{8} C O O H

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C H 2 = C H (C H_{2})_{8} C O B r

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B r C H_{2} (C H_{2})_{9} C O O H

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C H_{2} = C H (C H_{2})_{7} C H (B r) C O O H

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $B r C H_{2} (C H_{2})_{9} C O O H$
$C H_{2} = C H (C H_{2})_{8} C O O H + H B r P e r o x i d e - ---- \to$ $B r C H_{2} (C H_{2})_{9} C O O H$ .
A molecule of $H B r$ is added to $C = C$ double bond. $B r$ is added to less substituted terminal carbon atom. (The anti markonikov's addition of HBr to unsymmetrical alkenes in presence of peroxides)

Suggest Corrections

Similar questions

C H_{2} = C H (C H_{2})_{8} C O O H + H B r p e r o x i d e \to

C_{6} H_{5} - C H_{2} - C H = C H_{2} + H B r \to

C_{6} H_{5} - C H = C H_{2} + H B r P e r o x i d e - ---- \to

C H_{3} - C H_{3} | C | O H - H | C = C H_{2} + H B r \to

{C H}_{3} - C H = {C H}_{2} + H B r ⟶ ?

End product in this reaction is $C H C H + H B r \to X K O H \to$

Join BYJU'S Learning Program