Question

Product obtained in following reactions $(1),(2)\And (3)$ is:



(a) $A=B,C$ is different
(b) $A=C,B$ is different
(c) $B=C,A$ is different
(d) $A=B=C$ is same

Answer 1

Alcoholic potassium hydroxide $KOH$ acts as a strong base; it is used in the elimination reactions where unsaturated products are obtained from saturated compounds.

Complete step by step solution: Three reactions $(1),(2)\And (3)$ are given that involves $1,2,3,4,5,6-$ chloro cyclohexane, which is treated with three moles of alcoholic potassium hydroxide. The reaction is a type of $\beta -$ elimination reaction that results in double bonds in the product.
The first reaction is accompanied by the removal of hydrogen from$\beta -$ position which is at the anti position, so this will be the type of ${{E}_{2}}$ anti-elimination reaction. This reaction is carried in the way that alternate double bond formation takes place when $\beta -$ hydrogens from $1,3,5$ positions get removed along with chlorine from $2,4,6$ positions as $3$ moles of $KOH$ are used. Hence the product formed will be $1,3,5$ trichlorobenzene.

This product will be obtained in the first case. While in the second and third case, the product obtained will be the same as in reaction $(2)$ four $Cl$ atoms are on the same side while two are on opposite sides which will retain the symmetry and $1,3,5$ trichlorobenzene. Same will happen in reaction $(3)$ as four $Cl$ atoms are on the same side and two on the opposite. The geometry in all the three reactions will be such that the elimination will occur of $Cl$ and hydrogen to retain $Cl$ at $1,3,5$ positions along with double bonds.

Therefore, option (d) is correct as products $A=B=C$ are the same.

Note: Anti-elimination takes place when the $\beta -$ hydrogen is removed from the opposite side to that of the leaving group (here chlorine). Syn elimination takes place when $\beta -$ hydrogen is removed from the same side to that of the leaving group.