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
`R_(2)CH-underset(X)underset(|)C-R_(2) '+:B^(-) "(base)" to R_(2)C=CR_(2)`
`+H-B+X^(-)`
this reaction is an example of:

Options

A. `E_(1)` reaction
B. `E_(2)`reaction
C. `E_(1)` cb reaction
D. first order reaction

Answer

B