Neopentyl bromide undergoes dehydro halogenation to give alkene even though it has no `beta-` hydrogen. This is due to `:`

Neopentyl bromide undergoes dehydrohalogenation to give alkene even though it has no beta -hydrogen. This is due to :

The removal of two atoms or groups one generally hydrogen (H^(+)) and the other a leaving group (L^(-)) resulting in the formation of unsaturated compound is known as elimination reaction. In E_(1) (elimination) reactions the C-L bond is broken heterolytically (in step 1) to form a carbocation (as in S_(N^(1)) reaction) in which (L^(-)) is lost (rate determining step). The carbocation (in step 2) loses a proton from the beta- carbon atom by a base (nucleophile) to form an alkene. E_(1) reaction is favoured in compounds in which the leaving group is at secondar (2^(@)) or tertiary (3^(@)) Position. In E_(2) (elimination) reactions two sigma bonds are broken and a pi- bond is formed simultaneously. E_(2) reactions occur in one step through a transition state. E_(2) reactions are most common in haloalkanes (particulary 1^(@) ) and better the leaving group higher is the E_(2) reaction. In E_(2) reactions, both the leaving groups should be antiplaner. E_(1) cb (Elimination unimolecular conjugate base) reaction involves the removal of proton by a conjugate base (step 1) to produce carbanion which loses a leaving group to form an alkene (step 2) and is a slow step Neopentyl bromide undergoes dehydrohalogenation to give alkene even though it has no beta- hydrogen.This is due to :

Neopentyl bormide, undergoes dehydrohalogenation to give alkenes even though it has no beta-H . This is due to

Neopentyl bromide undergoes nucleophilic substitution reactions very slowly. Why?

The dehydro halogenation of neopentyle bromide with alco.KOH mainly given: