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Wurtz Halide Reaction

An alkyl hali...

Question

An alkyl halide X reacts with sodium to form 3,8-dimethyldecane. What is X?

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

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C H_{3} C H_{2} C H_{2} C H_{2} C l

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

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

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Solution

The correct option is D $C H_{3} C H_{2} - C | H C H_{3} - C H_{2} C H_{2} C l$
In Wurtz reaction, two molecules of RCI combine to give R-R. Dimethyldecane can be formed by combining two molecules of $C H_{3} - C H_{2} - C | H C H_{3} - C H_{2} - C H_{2} C l C h l o r o - 3 - m e t h y l p e n t a n e$ .

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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

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

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

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

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_{2} {C H}_{2} {C H}_{2} {C H}_{3}, {C H}_{3} - {C H}_{3} | C H - {C H}_{2} C H_{3}, {C H}_{3} - {C H}_{3} | C_{3} - {C H}_{3}

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