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NEET CHEMISTRYHydrocarbonsEasy

Question

Which of the following alkanes cannot be made in good yield by Wurtz reaction?

A

2,3-Dimethylbutane

B

n-Heptane

C

n-Butane

D

n-Hexane

Step-by-Step Solution

  1. Principle of Wurtz Reaction: The Wurtz reaction involves the coupling of two alkyl halide molecules with sodium metal in dry ether to form higher alkanes (2RX+2Na→R−R+2NaX2RX + 2Na \rightarrow R-R + 2NaX). This method is suitable for preparing symmetrical alkanes containing an even number of carbon atoms .
  2. Analysis of Options:
  • n-Butane (C4H10C_4H_{10}): Even number of carbons (4). Symmetrical (C2H5−C2H5C_2H_5-C_2H_5). Can be prepared from bromoethane.
  • n-Hexane (C6H14C_6H_{14}): Even number of carbons (6). Symmetrical (C3H7−C3H7C_3H_7-C_3H_7). Can be prepared from 1-chloropropane.
  • 2,3-Dimethylbutane (C6H14C_6H_{14}): Even number of carbons (6). Symmetrical dimer of isopropyl chloride. Can be prepared effectively.
  • n-Heptane (C7H16C_7H_{16}): Odd number of carbons (7). It is unsymmetrical. To prepare it, one would need two different alkyl halides (e.g., propyl halide and butyl halide). This results in a mixture of three products (propane, octane, and heptane) due to self-coupling and cross-coupling, making separation difficult and the yield of n-heptane poor .
  1. Conclusion: Alkanes with an odd number of carbon atoms, like n-Heptane, cannot be prepared in good yield using the Wurtz reaction.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Hydrocarbons. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYHydrocarbonsfollowingalkanescannotreaction

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