$B r - C H_{2} - C H_{2} - C H_{3}$ reacts with Na in the presence of ether at $100^{\circ} C$ to produce-

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

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C H_{2} = C = C H_{2}

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C H_{3} - C H - C H_{2} - C H_{2} - C H - C H_{3}

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All of these

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Solution

The correct option is C
$C H_{3} - C H - C H_{2} - C H_{2} - C H - C H_{3}$

Wurtz reaction:
$R X + 2 N a + X R \to R - R + 2 N a X$
So here product is
$C H_{3} - C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H_{3}$
In Wurts coupling reaction coupling of alkyl halide forming a hydrocarbon containing double the number of carbon than alkyl group.

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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} - C H_{3}, C H_{2} = C H_{2}

C H \equiv C H

130^{\circ} C

C H_{3} - C H_{2} - C H_{2} - C H_{2} - C H | C H_{3} - C | C H_{3} - C H_{3} - C H_{2} - C H_{3} | C H - C H_{2} - C H_{2} - C H_{2} - C H_{3}

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

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