The principle organic product formed in the reaction given below:
$C H_{2} = C H (C H_{2})_{8} C O O H + H B r p e r o x i d e \to$

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

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

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

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

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Solution

The correct option is C $C H_{2} = B r C H_{2} (C H_{2})_{8} C O O H$
The product forms in aledane with anti markonikov rule due to presence of peuride halogen attaches to less substituted carbon atom.

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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

A . C H_{3} - C H = C H_{2} + H C l ⟶ C H_{3} - C | C l H - C H_{3}

B . C H_{3} - C H = C H_{2} + H B r ⟶ C H_{3} - C H_{2} - C H_{2} B r

C . C H_{3} - C H = C H_{2} + H B r p e r o x i d e - ---- \to C H_{3} - C H_{2} - C H_{2} - B r

D . C H_{3} - C H = C H_{2} + H I p e r o x i d e - ---- \to C H_{3} - C | I H - C H_{3}

{CH}_{3} - {CH}_{2} - Br + {CH}_{3} - {CH}_{2} - Br Na / dry ether - ------- \to The product / s formed are

C H_{3} - C H = C H_{2} + H B r p e r o x i d e \to ?

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

C H_{3} - C H_{2} - C H_{2} - C H_{2} - C H_{3} | C | C H_{3} - C H_{2} - C H_{2} - O H H^{+} - - \to Δ (A) (m a j o r)

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