A dibromide $\mathrm{X}\left(\mathrm{C}_4 \mathrm{H}_8 \mathrm{Br}_2\right)$ on dehydrohalogenation gave Y…
A dibromide $\mathrm{X}\left(\mathrm{C}_4 \mathrm{H}_8 \mathrm{Br}_2\right)$ on dehydrohalogenation gave Y which on reduction with Z gave non polar isomer of $\mathrm{C}_4 \mathrm{H}_8$. What are X and Z respectively?
Solution
Dehydrohalogenation: A reaction in which a hydrogen atom and a halogen atom are removed from adjacent atoms in a molecule, forming usually an alkene
