CH3-CH2-Br - CH3-CH2-Br -Na-dry etherThe product-s formed are

In the above reaction, disproportionation takes place which leads to the formation of an alkene. nbsp;CH3 CH3 nbsp;is also formed due to the abstraction of ...

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Standard XI

Chemistry

Wurtz Halide Reaction

CH3-CH2-Br + ...

Question

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

{CH}_{3} - {CH}_{2} - {CH}_{2} - {CH}_{3}

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H_{2} C = {CH}_{2}

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{CH}_{3} - {CH}_{3}

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All

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Solution

The correct option is D All
In the above reaction, disproportionation takes place which leads to the formation of an alkene. ${CH}_{3} - {CH}_{3}$ is also formed due to the abstraction of a $H^{.}$ formed in the above said disproportionation.

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Similar questions

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

Q. The decreasing order of rate of

S_{N} 2

is :

(I)

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

(II)

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

(III)

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

(IV)

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

Q. Predict the products of the following reactions:
(a)

C H_{3} - C H_{2} - B r + A g C N ⟶ (a q . a l c o h o l)

(b)

C H_{3} - C H_{2} - B r + K N O_{2} ⟶ (a q . a l c o h o l)

(d)

C H_{3} - I + M g d r y e t h e r - ---- \to

Q. Select the member of each pair that shows faster rate of

S_{N} 2

reaction with

K I

in acetone.

(a)

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)

(b)

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)

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)

(d)

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} - C H_{2} - C H_{2} - C H_{2} - B r + C H_{3} O^{-} C H_{3} O H - ---- \to 60^{o} C P r o d u c t (I)

C H_{3} - C H_{2} - C H_{2} - C H_{2} - B r + (C H_{3})_{3} C O^{-} (C H_{3})_{3} C - O H - -------- \to 60^{o} C P r o d u c t (I I)

Select the correct options among the following.

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CH3−CH2−Br + CH3−CH2−Br Na/dry ether−−−−−−−−→The product/s formed are

The correct option is D AllIn the above reaction, disproportionation takes place which leads to the formation of an alkene. CH3−CH3 is also formed due to the abstraction of a H. formed in the above said disproportionation.

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

The correct option is D All
In the above reaction, disproportionation takes place which leads to the formation of an alkene. ${CH}_{3} - {CH}_{3}$ is also formed due to the abstraction of a $H^{.}$ formed in the above said disproportionation.