Question

Ph Reaction Ais (Select] B) = + + HF = H-F X Reaction Bis [Select] + H-1 Reaction Cis Select] orga D) PnH.Br – yang Reaction Dis [Select] E) O + H-Br + H-Br Reaction Eis Select]

1) Correct rxn-no change

2) product is wrong constitutional isomer

3) product is wrong stereoisomer

4) no product will form

using 1-4, which describes each reaction?

Answer 1

1 describes each reaction as all reactions are correct. All reactions are HX type acid addition to the alkene in which most stable carbocation is formed in the first step of it's mechanism.When carbocation is formed then an unhybridised p orbital is generated on the carbon bearing positive charge and $X^-$ can attack from both side to the the two lobes of p orbital with equal probability in the second step.

B) & C) tertiary carbocation is more stable than secondary or primary, hence the above mentioned product is formed.

A) since the $X^-$ can attack from both sides of p orbital hence both the enantiomers are formed.

E)The most stable tertiary carbocation is formed,and halide approaches from both sides of p orbital on carbon bearing positive charge hence the above mentioned product.

D) Here the most stable carbocation is in which carbon adjacent to the phenyl ring bears positive charge (benzylic carbocation) as it is resonace stabilized.So at first secondary carbocation is formed,then hydride from very next carbon(which bears methyl group in above the plane)is shifted to it and a tertiary carbocation is formed.After that another hydride shift from the carbon adjacent to the ph ring occurs and most stable carbocation is formed, during this process hydride from carbon adjacent to the ph ring can attack from both sides of the p orbital generated on the tertiary carbon and hence specific stereochemistry of that carbon is finally halide attacks from both sides of p orbital on bezylic carbon and forms the above product.