Q. 385.0( 1 Vote )
Ethylene dichlori A. Both the compounds form same product on treatment with alcoholic KOH.
B. Both the compounds form same product on treatment with aq. NaOH.
C. Both the compounds form same product on reduction.
D. Both the compounds are optically active.
Ethylidene dichloride CH3—CHCl2 and Ethylene chloride ClCH2—CH2Cl are isomers. Both molecules are not optically active as the carbon atoms are not surrounded by different groups.
When haloalkanes are treated with alcoholic KOH, they undergo elimination reaction where there is an elimination of hydrogen atom from β-carbon atom and a halogen atom from the α-carbon atom. Both these compounds lose hydrogen and chlorine atoms to give ethyne molecule.
In case of treatment with aqueous NaOH, the molecules undergo nucleophilic substitution, where the –Cl groups are replaced with –OH molecules. Due to the positions of the halides on different carbon atoms, the products will also be different.
Reduction of the two compounds is carried out with Zn dust in alcohol, and the end product is of these is the corresponding alkene, hence both compounds on reduction give alkene.
Options (i) and (iii) are correct.
Rate this question :