Question

Devise a detailed mechanism for formation of the major product of the elimination reaction below. I already know the answer, but would you please explain why it is the alpha hydrogen that is being removed by the base?

For the other eliminations, it had always

been the beta hydrogen being removed. Please explian the mechanism :)

4: 04 I H H₂ H LC H Na :0-H H-0-H so Correct GH NO CH3 * Na 5-H H-6-H 2 H-0-- O: Na The strong base, hydroxide, removes an alpha proton giving a resonance-stabilized enolate.

Answer 1

он H LH NOOHH OH ey Loo OH CH3 NaOH, H20 -CH3 PH It cH2 + H₂O H out In the above elimination reaction deprotonation at 2-position as more favourable because d- Hydrogen is more acidic than B-Hydrogen more acidic Hydrogen h Less t c H3 Agdec L o deprotenation OH Hydrogen deprotenation gives stable Carbanton on Lto FL stablecarbonion E CH₂ H SH2

finally the elimination formed corbanron on nrolves reaction to gives diß unsaturated Carbonyl compound - CH3 unstable Stable compound B-Hydride elimination is not possible - Carbonyl compounds But In the case of alkyl halide en presence of strong base B-elimination takes place Example Eror kotosu BuoH Si por essere Hlynotou H above reaction B-Hydrogen elimination reoction takes place