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NCERT Class XII Chemistry
Chapter - Haloalkanes and Haloarenes
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Question : 21 of 31
Marks: +1, -0
Primary alkyl halide C4H9Br\mathrm{C_4H_9Br} (A) reacted with alcoholic KOH to give compound (B). Compound (B) is reacted with HBr to give (C) which is an isomer of (A). When (A) is reacted with sodium metal it gives compound (D).
C8H18\mathrm{C_8H_18} which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (A) and write the equations for all the reactions.
Solution:  
The given reaction sequence is :
When (A) is treated with Na metal, Wurtz reaction takes place. It is given that (A) is a primary alkyl halide.
∴ The possible structures of C4H9Br\mathrm{C_4H_9Br} (A) are
(I) H3C−C∣CH3H−CH2Br\mathrm{(I)}\ \mathrm{H_3C-}\underset{\mathrm{CH}_3}{\underset{\vert}{\mathrm{C}}}\mathrm{H-}\mathrm{CH_2Br}
(II) CH3−CH2−CH2−CH2−Br\mathrm{(II)}\ \mathrm{CH_3-CH_2-CH_2-CH_2-Br}
However, it is given that (A) on reaction with Na metal gives alkane C8H18\mathrm{C_8H_18} which is not the same as the one obtained from the Wurtz reaction of n-butylbromide. Therefore (A) ≠ II\mathrm{II}.
Thus,
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