Comment(0) Step 3 of 6 Treatment of 1-bromo-2-methylpropane with potassium terti
ID: 944915 • Letter: C
Question
Comment(0) Step 3 of 6 Treatment of 1-bromo-2-methylpropane with potassium tertiary butoxide in the presence of THF gives the less substituted alkene as major product as the base is sterically hindered. Thus, the reaction is as followe Br (CH3)3COK THF Comment(C) Step 4 of6 The mechanism of this reaction is as follows: The secondary carbocation is formed by expelling out the leaving group. Next the secondary carbocation is converted to tertiary carbocation by the 1,2-hydride shift. Now, the abstracts the proton and forms the required alkene. 1 2-hydride shiftExplanation / Answer
(CH3)3CO-(aq) is a bulky nucleophile , hence when it tries to undergo nucleophilci substitution the molecule experiences repulsions , hence (CH3)3CO- (aq) acts as bad nulecophile , instead it acts as good base removing H+ from less substituted carbon since its easier to remove there with less repulsions form other part of molecule.
Hence we get eliminated product not SN2 product