[" (3) Neo-penty bromide undergoce tedent "],[" dehridrohalogenation Lo sive akere "],[" even through it has no lathydragen "],[" This is due to "]

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

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

Neopentyl bromide undergoes dehydro halogenation 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 :

The formation of oxide ion O^(2-)(g) from oxygen atom requires first an exothermic and then an endothermic step as shown below O(g)+e^(-) rarr O^(-)(g), DeltaH^(-) = - 141 kj mol^(-1) O^(-)(g) +e^(-) rarr O^(2-) (g), DeltaH^(-) =+ 780 kj mol^(-1) Thus, process of formation of O^(2-) in gas phase is unfavourable even through O^(2-) is isoelectronic with neon. It is due to the fact that A) oxygen is more electronegative B) addition of electron in oxygen results in larget size of the ion C) electron repulsion outweights the stability gained by achieving noble gas configuration D) O^(-) ion has comparatively smaller size than oxygen atom