Alkenes

Reaction of HBr with propene in the presence of peroxide gives n-propyl bromide. This addition reaction is an example of anti-Markownikoffs
addition reaction. (i.e. it is completed in form of free radical addition) Mechanism of this reaction is represented as follows:

Step II Benzoate free radical forms bromine free radical with HBr.
free radical from benzoic acid and give final product of reaction.




Step V Benzoate free radicals are changed into benzoyl peroxide for the termination of free radical chain.

Read more from the chapter: Haloalkanes and Haloarenes

Reactivity of Propene towards Bromine

Propene is more reactive towards (gaseous) than (ethylene), (acetylene), and butane. This is due to the greater electron density in propene.

Explanation:

In the case of alkene reactions with bromine, the reactivity depends on the availability of electron density for the electrophilic addition reaction. In the case of propene:

• Propene: The double bond in propene has more electron density due to the alkyl group (–CH₃) attached to the carbon of the double bond, making it more nucleophilic and more reactive towards electrophilic reagents like .
• Ethylene (CH₂=CH₂): Ethylene has less electron density than propene as there are no alkyl groups donating electron density to the double bond.
• Acetylene (HC≡CH): Acetylene has a triple bond, which is electron-deficient compared to the double bond in propene, making it less reactive towards bromine.
• Butane: Butane is a saturated hydrocarbon, meaning there is no double bond available for electrophilic attack, and thus it doesn't react with bromine in the same way as the alkenes do.

Conclusion:

Propene is the most reactive towards gaseous bromine because the electron density around the double bond is higher compared to the other molecules, facilitating a faster electrophilic addition reaction.

To determine compound X, we analyze the details of its reaction with ozone(O₃) followed by zinc and water reduction(Zn/H₂O). This is an ozonolysis reaction, which breaks down alkenes into carbonyl compounds.

1. Products given are formaldehyde(HCHO) and 2-methyl propanal(CH₃CH(CH₃)CHO). Formaldehyde indicates one terminal carbon in a double bond, as it is derived from an alkene ending with a -CH=O group.

2. For 2-methyl propanal to form, the remaining carbon chain must include a propyl group attached to a carbon with a methyl group, which was part of a double bond before ozonolysis.

Let's examine the given options:

3. From the table, 3-Methylbut-1-ene yields the required products of formaldehyde and 2-methyl propanal after ozonolysis. Thus, the correct compound X is 3-Methylbut-1-ene.