The correct options are
A Both the compounds from same product on treatment with alcoholic KOH.
C Both the compounds from same product on reduction.
(A) Ethylene chloride and ethylidene chloride on treatment with alc. KOH show elimination reaction and ethyne as the product.
CH3−CHCl2Ethylidene chlorideAlc.KOH(excess)−−−−−−−−−−−→CH≡CH
(B) Both these compounds from different products on treatment with aq. NaOH.
Cl−CH2−CH2ClAq.NaOH−−−−−−→HO−CH2−CH2−OHEthylene glycol
CH3−CHCl2Aq.NaOH−−−−−−→CH3−CH(OH)2−H2O−−−−→CH3CHOethanal
(c) Both these compounds form same products on reduction.
Cl−CH2−CH2−ClReduction−−−−−−→H3C−CH3ethane+2HCl
CH3−CHCl2→H3C−CH3ethane+2HCl
(d) Both these compounds are optically inactive.