JEE Main · 2025 · 02 Apr 2025 · Shift 1 · General Organic Chemistry

In the carbon chain $\mathrm{HC}\equiv \mathrm{C}-\mathrm{CH}{}_2-\mathrm{CH}(\mathrm{CH}{}_3)\mathrm{CH}{}_3$, removing a hydrogen from the propargylic $\mathrm{CH}{}_2$ at position II gives a radical whose unpaired electron is delocalised onto the neighbouring triple bond, making it the most stabilised of the four. Removing the hydrogen from the terminal $\equiv \mathrm{C}-\mathrm{H}$ at position I leaves the electron on an sp-hybridised carbon; the high s-character binds it tightly and no delocalisation is available, so this radical is the least stable. The most stable radical thus forms at the propargylic carbon and the least stable at the terminal alkyne carbon.