H214C - CH -CH3 -Product of the above reaction is -

The correct option is D both (a) and (b) H_2 14C=CH CH_3 #10;Low cone Br_2High #10;temperature⟶ H_2 #10;14C=CH CH_2 Br+Br 14CH_2 CH=CH_2 ...

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Organic Compounds

H214C = CH ...

Question

$H_{2} 14 C = C H - C H_{3} l o w c o n c . o f B r_{2} - --------- \to o r h i g h t e m p (?)$
Product of the above reaction is :

H_{2} 14 C = C H - C H_{2} - B r

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H_{2} C = C H - 14 C H_{2} - B r

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14 C | B r H_{2} - C | B r H - C H_{3}

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both (a) and (b)

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Similar questions

S_{N} 2

K I

C H_{3} - C H_{2} - C H_{2} - C H_{2} - C l a n d C H_{3} - C H ∣ C H_{3} - C H_{2} - C l

(I) (I I)

C H_{3} - C H_{2} - C H_{2} - C l a n d C H_{3} - C H_{2} - C H_{2} - B r

(I) (I I)

C H_{3} - C H ∣ C H_{3} - C H_{2} - C H_{2} - C l a n d C H_{3} - C H_{3} ∣ C ∣ C H_{3} - C H_{2} C l

(I) (I I)

C H_{3} - C H_{2} - C H_{2} - B r ∣ C H - C H_{3} a n d C H_{3} - C H_{3} ∣ C H - C H_{2} - B r ∣ C H - C H_{3}

(I) (I I)

{C H}_{3} - B r | C | H - C H_{2} - C H_{3} + a l c . K O H \to X

{C H}_{3} - C H | C H_{3} - C H = C H_{2} O_{3} - \to Y + Z

(A) C H_{3} - C H_{2} - C | B r H - + C H - C H_{3}

(B) C H_{3} - C H_{2} - C | I H - + C H - C H_{3}

S_{N} 2

B r - C H_{2} - C H_{2} - B r

N = C - C H_{2} - B r

C H_{3} - C | B r H - C H_{3}

C H_{3} - C H_{2} - B r

C H_{3} - C H_{2} - \oplus C H_{2}

C H_{3} - C H | F - \oplus C H_{2}

C H_{3} - C H | B r - \oplus C H_{2}

C H_{3} - C H | C l - \oplus C H_{2}

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