[Harshit Dhiman]

the question says that what will be the major product if the compound

react with 1 equivalent of HBr
I predicted the answer as
but the given answer is

Also the intermediate in both of the compounds are resonance structures of the same compounds
i am unable to predict which on is more stable and why
please help me out of this

[Alwin Kristen]

Bromine attacks to the less hindered end to give prenyl bromide. Methyl groups stabilizes carbocation, but protons of methyl group makes the surrounding around it crowded. The other resonance structure,Allylic cation, is also very stable, because p-orbitals of double bond interacts with empty p-orbital of carbon at the end. Indeed the electrophilic addition by proton from HBr happens to the double bond at the left, because positive charge delocalizes between carbocation and allylic cation. If addition happens to the other double bond methyl group cannot contribute to stability of cation. After electrophilic addition, nucleophile attacks to the less hindered end, because there is simply more space to go in. The reaction is similar to Sn1 substitution to allylic compound. Bromine attacks to the end. You can think it this way. Carbocation takes its electrons where it gets it most easy, from double bond nearby. Then Br attacks to the end of allylic cation, because Br is the only at the surrounding to have free electrons to react.

J. Clayden, Organic Chemistry, 2nd edition, 435.

[orgopete]

The predicted product is half-right. This is how I reason this problem. If this reaction were carried out at low temperature, then the tertiary bromide can probably be isolated as the major product. I reason that because of stability issues, a greater charge is found on the tertiary carbon and trapping it gives the kinetic product.

However, we can easily understand this bromide can ionize to give a tertiary allylic carbocation. If bromide were to react at the less stable resonance form, the resulting product would be more stable as a primary bromide and as a trisubstituted double bond. Thus, the resultant product forms as the thermodynamic product.

This is an example of forming different kinetic and thermodynamic products.