Chemistry - Hydrocarbon Question with Solution | TestHub

This conversion occurs as follows to give iodoform

${\text{CH}}_{\text{3}}-\text{CH}\text{=}{\text{CH}}_{\text{2}}\stackrel{\text{HBr}}{\to }\underset{\left(\text{X}\right)}{{\text{CH}}_{\text{3}}-\underset{\text{Br}}{\underset{\text{| }}{\text{CH}}}-{\text{CH}}_{\text{3}}}\stackrel{\text{Aq}\text{.NaOH}}{\to }\underset{\left(\text{Y}\right)}{{\text{CH}}_{\text{3}}-\underset{\text{OH}}{\underset{\text{| }}{\text{CH}}}-{\text{CH}}_{\text{3}}}\stackrel{\text{NaOH}\text{+}{\text{I}}_{\text{2}}}{\to }\underset{\left(\text{Z}\right)}{{\text{CHI}}_{\text{3}}}\text{+}{\text{CH}}_{\text{3}}\text{COONa}$