Question

could someone explain major and minor products please? (in terms of carbocations etc etc)

Answer 1

Do you know that a tertiary carbocation is more stable than a secondary carbocation, which is in turn more stable than a primary carbocation?
If so, consider the reaction: \[CH3-CH=CH2 + HBr \]
The compound loses its double bond(pi) electrons to the HBr atom, forming a carbocation.
\[CH3-CH-CH2+\]
But as i said, a secondary carbocation is more stable. So a hydrogen atom will shift to the site of positive charge. Thus the positive charge gets transferred to the intermediate carbon :-\[CH3-CH+-CH3\]. The nucleophile Br- will attack the carbocation at the site of positive charge. So, two products will be formed. 1-bromo propane and 2-bromopropane.
But, 2-bromopropane will be major because of the stability of the corresponding carbocation. 1-bromopropane will be minor

Answer 2

ok thats great thanks

Answer 3

It will form CH3-CH+-CH3 in the first step itself (Br is highly selective) The reason is that 2 degree carbocation is more stable.(There is no rearrangement as suggested above in this case)
Rearrangements would occur in cases like Now if you add HBr the H+ adds to the terminal carbon since the 2 degree carbon is more stable. Now a H+ would shift to form a 3 degree carbo cation and Br would add to it.