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Grade 12th passAIPMT

i want the know the solution of all three parts.plz help as soon as possible.

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Profile image of manisha
7 Years agoGrade 12th pass
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1 Answer

Profile image of Vallakatla Aditya
7 Years ago
For the 1st question  we will get 2 Br’s on middle carbon and 2 h’s on right end carbon CH3-CBr2-CH3 
For the 2nd question  we will get hydroxyl group on benzyl carbon then after on rearrangement we get ketone gp
as Ph-CO-CH2CH3  because HgSO4 &H2SO4 gives markonikov product.
For the 3rd question  we will get 2 Br’s on two double bonded carbons and it is Anti addition  we get as
CH3-CH3-CHBr-CHBr-CH(CH3)2
 

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