CH3-CH2-Br - CH3-CH2-Br - Na - dry etherThe product-s formed areA- CH3-CH2-CH2-CH3B- H 2 C - CH 2C- CH3-CH3D- All

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 XII

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

An alkane with molecular formula $C_{6} H_{14}$ gives two monochloro alkanes on chlorination. What is the structure of alkane?

Q. Which amongst the following is a positional isomer of

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

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)

Q. Identify the type of the following reaction of carbon compounds.
a. CH₃ -CH₂ -CH₂-OH → CH₃ -CH₂ -COOH
b. CH₃ -CH₂ -CH₃ → 3CO₂ + 4H₂O
c. CH₃ -CH = CH -CH₃ + Br₂ → CH₃ -CHBr - CHBr -CH₃
d. CH₃ -CH₃ + Cl₂→ CH₃ -CH₂ -Cl + HCl
e. CH₃-CH₂ -CH₂ -CH₂ -OH → CH₃ -CH₂ -CH=CH₂ + H₂O
f. CH₃-CH₂ -COOH + NaOH → CH₃ -CH₂ -COO - Na⁺ + H₂O
g. CH₃ -COOH + CH₃ -OH → CH₃ -COO- CH₃ + H₂O

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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.