NEET 2020 Chemistry Question 134 Solution

Alkyl halides on treatment with sodium metal in dry ethereal solution give higher alkanes containing twice the number of carbon atoms: $$2\text{R}-\text{X} + 2\text{Na} \xrightarrow{\text{dry ether}} \text{R}-\text{R} + 2\text{NaX}$$ Because two identical alkyl groups are coupled, the product is always a symmetrical alkane with an even number of carbon atoms.

- $$n$$-Hexane ($$6\text{ C}$$): Formed by coupling two 1-propyl groups ($$2\text{ C}_3\text{H}_7\text{Br} + 2\text{Na} \to \text{C}_6\text{H}_{14}$$) in good yield. - 2,3-Dimethylbutane ($$6\text{ C}$$): Symmetrical alkane formed by coupling two isopropyl groups ($$2 (\text{CH}_3)_2\text{CH-Br} + 2\text{Na} \to (\text{CH}_3)_2\text{CH}-\text{CH}(\text{CH}_3)_2$$) in good yield. - $$n$$-Butane ($$4\text{ C}$$): Formed by coupling two ethyl groups ($$2\text{ C}_2\text{H}_5\text{Br} + 2\text{Na} \to \text{C}_4\text{H}_{10}$$) in good yield.

$$n$$-Heptane contains 7 carbon atoms (an odd number). To synthesize it, two different alkyl halides must be used: 1-halopropane ($$3\text{ C}$$) and 1-halobutane ($$4\text{ C}$$). Three different radical coupling reactions occur simultaneously: 1. $$\text{C}_3\text{H}_7\cdot + \text{C}_3\text{H}_7\cdot \to \text{Hexane (6 C)}$$ 2. $$\text{C}_3\text{H}_7\cdot + \text{C}_4\text{H}_9\cdot \to \text{Heptane (7 C)}$$ 3. $$\text{C}_4\text{H}_9\cdot + \text{C}_4\text{H}_9\cdot \to \text{Octane (8 C)}$$ This generates a complex mixture of three alkanes whose boiling points are very close, making separation extremely difficult and the yield of heptane very poor.

NCERT explicitly states that the Wurtz reaction cannot be used for the preparation of unsymmetrical alkanes having an odd number of carbon atoms. Thus, $$n$$-heptane is the correct answer.