JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Hydrocarbons

The hydrocarbon $(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{C}-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}(\mathrm{C}{\mathrm{H}}_3{)}_3$ is symmetric, leaving only two kinds of hydrogen: those on the equivalent methyl groups and those on the equivalent methylene carbons. Replacing a methyl hydrogen gives one product, 1-chloro-2,2,5,5-tetramethylhexane. Replacing a methylene hydrogen builds a carbon bearing chlorine, hydrogen, a $tert$-butyl group and a neopentyl group, four different groups, so that carbon is a stereocentre and its product exists as a pair of enantiomers. Counting the single methyl product with the two mirror-image forms of the methylene product gives three isomeric monochloro derivatives.