[helper]

Hello,
my question is attached.

can  I say that the alkyl halid is disturbed and therefor the nucleophile can not reach the back side of the molecule. moreover, can we say the the Br is electronegative so it withdraw the electrons so the H is more acidity. and for the following reasons the elimination is favored over the substitution?

Thank you.

[Archer]

Probably hydration of the olefin with strong acid and water, Markovikov, followed by oxidation to the ketone.

[orgopete]

Re: elimination rather than substitution with dibromide.

I think this is a good question, and one that I don't have an answer for. A rule of thumb, is a secondary halide give the elimination product if the (conjugate) base has a pKa of greater than 12. There is a limit to the effectiveness of substitution reactions with strong bases. Ammonia is a very weak acid, therefore amide anion will be a very strong base. Presumably, the vicinal regiochemistry is sufficient that elimination becomes the major pathway.

[helper]

Thank you. I agree with you maybe it is a spacial reaction in which we can not just use the rules we have already known but it is more complicated- and in practice this what happens maybe due to kinetics' factors as you said:)

[magician4]

I would strongly support Archer's approach: this is what most probably your teacher will want to hear


regards

Ingo

[helper]

yes this is true but what I were asking about the stage before the reaction of water and acid. And I think that orgopete's answer is correct.

[magician4]

yes this is true but what I were asking about the stage before the reaction of water and acid.

there is no such state in our alternative pathway:

CH₃CH₂CH₂CH₂Br --(KOtBu)-->  CH₃CH₂CH=CH₂ --(H⁺,H₂O)--> CH₃CH₂CH(OH)CH₃ --(Ox.)--> CH₃CH₂-CO-CH₃


regards

Ingo