✓
Answer
d
(d) $C{H_3} - C{H_2} - \mathop {\mathop {{\text{ }}C}\limits^| H - }\limits^{C{H_3}\,\,\,\,\,\,\,\,\,} \mathop {\mathop {{\text{ }}C}\limits^| H - }\limits^{C{H_3}\,\,\,\,\,\,\,\,\,} \mathop {\mathop {{\text{ }}C}\limits^| H - }\limits^{C{H_3}\,\,\,\,\,\,\,\,\,} C{H_3}$, it is a chiral compound.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.