Question

Study the reaction steps carefully and identify the products X (major) and Y respectively. $\underset{\mathrm{Propane}}{{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_3}+\underset{\mathrm{Bromine}}{{\mathrm{Br}}_2}\to \mathrm{X}\to \mathrm{Y}$

Answer 1

Bromination:

• Halogenation of alkanes is a radical process that is often indiscriminate (Dihalogenated compounds are possible because radical mechanisms can be extremely unselective)
• Bromination is a chemical process in which a compound reacts with bromine, resulting in the addition of bromine to the molecule. After bromination, the result will have different properties from the starting reactant.
• Because we are replacing $\mathrm{H}$ an atom with $\mathrm{Br}$ an atom, the reaction is termed a substitution reaction.
• Alkanes are transformed into alkyl bromides when they are treated with bromine (${\mathrm{Br}}_2$) and light. When there are no double bonds, bromination of tertiary carbons is selective.

The reaction of Propane with Bromine:

• Here propane reacts with bromine to give $1-$Bromo propane and $2-$Bromopropane. The main product of a reaction is the one that is the most stable and so most likely to form since secondary carbocations are more stable than the primary carbocations $2-$Bromopropane is a major product.$\underset{\mathrm{Propane}}{{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_3}+\underset{\mathrm{Bromine}}{{\mathrm{Br}}_2}\stackrel{\mathrm{light}}{\to }\underset{1-\mathrm{Bromopropane}}{{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{Br}}+\underset{2-\mathrm{Bromopropane}\left(\mathrm{Major}\right)\left(\mathrm{X}\right)}{{\mathrm{CH}}_3{\mathrm{CHBrCH}}_3}$
• Propene is produced by an elimination reaction when either bromopropane isomer is treated with an alcohol solution of $\mathrm{KOH}$ ($\mathrm{H}$and $\mathrm{Br}$ are lost as $\mathrm{KOH}$ eliminates hydrogen halide). $\underset{2-\mathrm{Bromopropane}}{{\mathrm{CH}}_3{\mathrm{CHBrCH}}_3}\stackrel{\mathrm{KOH}}{\to }\underset{\mathrm{Propene}\left(\mathrm{Y}\right)}{{\mathrm{CH}}_3{\mathrm{CHCH}}_2}$

Therefore from the reaction “X” is $2-$Bromopropane and Y is Propene