Does (2-fluoroethyl)benzene undergo elimination via E1cb or E2?

Question

I have seen in two different sources claiming that $\\ce{PhCH2CH2F}$ gives styrene $\\ce{PhCH=CH2}$ when treated with alcoholic $\\ce{KOH}$, but one source suggests an E1cb mechanism, while the other suggests an E2 mechanism. I think that the phenyl group should not be able to promote the formation of a carbanion in an E1cb mechanism. All the examples of E1cb mechanisms I have seen involve deprotonation of compounds which have $\\mathrm{p}K_\\mathrm{a}$ values of 20 or less, and toluene does not fit this.

Answer 1

2- fLuro ethyl benzene undergoes elimination through E1CB mechanism.

Two reaosons:

1. Flourine is a bad leaving group, no E2
2. E1 CB, 2 step process.
3. as the conjugate carbanion formed in the first step is benzylic carbanion, highly stabilized by conjugation with phenyl ring. In the first step, proton elimination by base, as flourine is bad leaving group.

Ph ^-CHCH₂F is benzylic carbanion.

1. Second step :From benzylic carbanion, Flourine leaves forming a double bond. Ph CH=CH2