The principal organic product formed in the reaction is:
$C H_{2} = C H (C H_{2})_{8} C O O H + H B r P e r o x i d e - ---- \to$
$C H_{3} C H (B r) (C H_{2})_{8} C O O H$
$C H 2 = C H (C H_{2})_{8} C O B r$
$B r C H_{2} (C H_{2})_{9} C O O H$
$C H_{2} = C H (C H_{2})_{7} C H (B r) C O O H$

Question

The principal organic product formed in the reaction is-CH-2 - CH-CH-2-8 COOH - HBr xrightarrow - -Peroxide- CH-3CH-Br-CH-2-8 COOHCH2 - CH-CH-2-8 COBrBrCH-2 -CH-2-9 COOHCH-2 - CH-CH-2 -7 CH-Br-COOH

Toppr Admin · Accepted Answer

CH2-CH-CH2-8COOH-HBrPeroxide-x2212-x2212-x2212-x2212-x2212-x2192- BrCH2-CH2-9COOH-xA0-A molecule of HBr is added to C-C double bond- Br is added to less substituted terminal carbon atom-xA0- -The anti markonikov-apos-s addition of HBr to unsymmetrical alkenes in presence of peroxides-

The principal organic product formed in the reaction is:CH2=CH(CH2)8COOH+HBrPeroxide−−−−−→ CH3CH(Br)(CH2)8COOHCH2=CH(CH2)8COBrBrCH2(CH2)9COOHCH2=CH(CH2)7CH(Br)COOH

CH2=CH(CH2)8COOH+HBrPeroxide−−−−−→ BrCH2(CH2)9COOH. A molecule of HBr is added to C=C double bond. Br is added to less substituted terminal carbon atom. (The anti markonikov's addition of HBr to unsymmetrical alkenes in presence of peroxides)

The principal organic product formed in the reaction is:
$C H_{2} = C H (C H_{2})_{8} C O O H + H B r P e r o x i d e - ---- \to$
$C H_{3} C H (B r) (C H_{2})_{8} C O O H$
$C H 2 = C H (C H_{2})_{8} C O B r$
$B r C H_{2} (C H_{2})_{9} C O O H$
$C H_{2} = C H (C H_{2})_{7} C H (B r) C O O H$