MHTCET 16 Apr 2026 Shift 1 Paper

Concept:In the Wurtz reaction, two alkyl halides react with sodium metal in dry ether to form a higher alkane by coupling of alkyl groups.Explanation:The general Wurtz reaction is:$$2 R-X + 2\mathrm{Na} \rightarrow R-R + 2\mathrm{Na}X$$The given mixture contains ethyl iodide, $\mathrm{C_2H_5I}$, and $n$-propyl iodide, $\mathrm{C_3H_7I}$.Coupling can occur between any two alkyl groups present in the mixture.When two ethyl groups couple, we get $n$-butane:$$\mathrm{C_2H_5-C_2H_5} \rightarrow n\text{-butane}$$When an ethyl group couples with a $n$-propyl group, we get $n$-pentane:$$\mathrm{C_2H_5-C_3H_7} \rightarrow n\text{-pentane}$$When two $n$-propyl groups couple, we get $n$-hexane:$$\mathrm{C_3H_7-C_3H_7} \rightarrow n\text{-hexane}$$Propane cannot be formed because coupling always joins two alkyl fragments and does not produce a smaller alkane.Answer:The alkane NOT obtained is Option B: Propane.