At low temperature, the slow addition of molecular bromine to ${{\rm{H}}_2}{\rm{C}} = {\rm{CH}}-{\rm{C}}{{\rm{H}}_2}-{\rm{C}} \equiv {\rm{CH}}$ gives:

${{\rm{H}}_2}{\rm{C}} = {\rm{CH}}-{\rm{C}}{{\rm{H}}_2}-{\rm{C}}{{\rm{H}}_2}-{\rm{CB}}{{\rm{r}}_3}$

${\rm{C}}{{\rm{H}}_2} = {\rm{CH}}-{\rm{C}}{{\rm{H}}_2}-{\rm{CBr}} = {\rm{CHBr}}$

${\rm{BrC}}{{\rm{H}}_2}-{\rm{CHBr}}-{\rm{C}}{{\rm{H}}_2}-{\rm{C}} \equiv {\rm{CH}}$

${\rm{C}}{{\rm{H}}_3}-{\rm{CB}}{{\rm{r}}_2}-{\rm{C}}{{\rm{H}}_2}-{\rm{C}} \equiv {\rm{CH}}$

JEE CHEMISTRY MCQ: ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES AND TECHNIQUES

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Application +4 | -1

At low temperature, the slow addition of molecular bromine to

{\mathrm{H}_2}\mathrm{C} = \mathrm{CH}-\mathrm{C}{\mathrm{H}_2}-\mathrm{C} \equiv \mathrm{CH}

gives:

View answer and explanation

Correct option: C

{\mathrm{H}_2}\mathrm{C} = \mathrm{CH}-\mathrm{C}{\mathrm{H}_2}-\mathrm{C}{\mathrm{H}_2}-\mathrm{CB}{\mathrm{r}_3}

Option B (

\mathrm{BrC}{\mathrm{H}_2}-\mathrm{CHBr}-\mathrm{C}{\mathrm{H}_2}-\mathrm{C} \equiv \mathrm{CH}

) is correct because at low temperatures, the addition of bromine favors the more reactive double bond over the triple bond.