JEE Main · 2021 · 22 Jul 2021 · Shift 2 · General Organic Chemistry

Geometrical isomerism requires two different groups on each doubly bonded carbon, and optical isomerism requires a carbon bearing four different groups. In 3-ethylhex-3-ene, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}({\mathrm{C}}_2{\mathrm{H}}_5)=\mathrm{CHC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$, the third carbon carries an ethyl group from the chain and an ethyl substituent, two identical groups, and no carbon in the molecule has four different groups, so it has no stereoisomers. 3,4-dimethylhex-3-ene has a methyl and an ethyl on each doubly bonded carbon and shows cis-trans isomerism, while 3-methylhex-1-ene and 4-methylhex-1-ene each contain a carbon bonded to four different groups and are optically active. The molecule without stereoisomerism is 3-ethylhex-3-ene.