Question

The reaction of the given compound with HBr gives:

A.

B.

C.

D.

Answer 1

Hint: The addition of hydrogen bromide to an alkene is an electrophilic addition reaction. This reaction takes place according to Markownikoff’s rule. It states that when HBr is added to a carbon-carbon double bond of an alkene, the negative part of the reagent is added to the more substituted carbon atom.

Complete Step by Step Answer:
The given compound has the following structure.
Its IUPAC name is 4-(1-Prop-1-enyl)phenol.
It contains an aliphatic portion comprising three carbon atoms and an aromatic portion comprising a benzene ring.
The aliphatic part has a three-membered carbon chain including a carbon-carbon double bond.
This reaction happens in three steps.
Proton from HBr is contributed to a double-bonded alkene group forming a secondary carbocation.
This secondary carbonation undergoes rearrangement to form a more stable benzyl carbocation through a 1,2-hydride shift.
Due to the formation of this carbocation, the positive charge on the carbon atom gets delocalized due to resonance.
The mechanism is as follows:

Image: Mechanism of formation of 4-(1-Bromopropyl)phenol.
This stable benzyl carbocation then takes a bromide ion to form 4-(1-Bromopropyl)phenol.
4-(1-Prop-1-enyl)phenol on reaction with HBr forms 4-(1-Bromopropyl)phenol
as a major product.
The structure of 4-(1-Bromopropyl)phenol is represented by option C.
So, option C is correct.

Note: Benzyl and allyl carbocations are the most stable carbocation followed by tertiary, secondary, primary, and methyl. 1,2 Hydride Shift is a change in configuration reaction in which hydrogen shifts from one carbon atom to another carbon in a chemical compound. This tendency of movement concerns two adjacent atoms. This occurs to produce a more stable carbocation.