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Conversion of tertiary haloalkane

can someone teach me how do you convert this reactant to get the product?
I consider this as free radical substituition reaction and in the presence of UV , then the H atom in the 2 CH3 will also be substituted.... i want ony the H atom on the carbon bonded to C to be substituted only.
Original post by wilson dang
can someone teach me how do you convert this reactant to get the product?
I consider this as free radical substituition reaction and in the presence of UV , then the H atom in the 2 CH3 will also be substituted.... i want ony the H atom on the carbon bonded to C to be substituted only.


I'll give you a clue to start with, (pretty sure I have an effective synthesis in mind)

The tertiary proton is fairly acidic as the conjugate base is heavily stabilised by the phenyl ring.... start by forming the anion (with a suitable base.... this is important) and then it's one reagent from the product.
Reply 2
Original post by JMaydom
I'll give you a clue to start with, (pretty sure I have an effective synthesis in mind)

The tertiary proton is fairly acidic as the conjugate base is heavily stabilised by the phenyl ring.... start by forming the anion (with a suitable base.... this is important) and then it's one reagent from the product.

i cant understand what do you mean. can you explain further ? thanks...
Deprotonate the starting material at the most acidic position using a strong and non-nucleophillic base
Reply 4
This reaction can be brought about by the free radical substitution using N - bromosuccinamide
Original post by GDN
This reaction can be brought about by the free radical substitution using N - bromosuccinamide


I was going to suggest using NBS as an electrophillic bromine source for this transformation!

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