Question

A secondary haloalkane ‘M’ when heated with alcoholic KOH produces another compound ‘N’ which on ozonolysis gives a mixture of ethanal and methanal. Identify ‘M’ and ‘N’

Answer 1

• Let the secondary haloalkane M is $R-CH(Cl)-CH_3$.
• It on reaction with alcoholic KOH releases HCl by taking one hydrogen atom from one R group and form an alkene.
• Thus the reaction is $R-CH(Cl)-CH_3+KOH\rightarrow R-CH=CH_2$
• This alkene on ozonolysis gives a mixture of ethanal and methanal.
• $R-CH=CH_2+O_3\rightarrow R-CHO+HCHO$
• Thus R must be methyl group as the chemical formula of the product ethanal is $CH_3CHO$.
• So, M will be propanaldhyde $H_3C-CH(Cl)-CH_3$ and N will be $CH_3-CH=CH_2$.