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.

  • $\begingroup$ According to the Evans $\mathrm pK_\mathrm a$ table, toluene is more acidic than propene, so the stabilisation of the phenyl group is larger than that of the allyl group. But I will acknowledge that that is a tough call for an $\mathrm{E1cb}$ mechanism, especially if the base is something weak like $\ce{KOH}$. $\endgroup$
    – Jan
    Sep 30, 2017 at 14:45
  • $\begingroup$ There's no reason why either mechanism is wrong. It might depend on conditions. The only way to know with greater certainty is to run the mechanistic experiments. $\endgroup$
    – Zhe
    Feb 13, 2018 at 18:29
  • 1
    $\begingroup$ Run Zhe's experiment in ROMetal/ROD. Check if deuterium is incorporated in unreacted starting material at the benzylic position. Incorporation of deuterium means that at least some of the elimination occurs stepwise; no incorporation means a concerted elimination. $\endgroup$
    – user55119
    Feb 13, 2018 at 21:07
  • $\begingroup$ Please mention the "sources" along with page number/ topic. That's really going to be helpful for those answering your questions. Thanks. $\endgroup$
    – Archer
    Aug 7, 2018 at 18:08

1 Answer 1


I think that it is easier for the molecule $\ce{Ph-CH2-CH2-F}$ to undergo E1cb mechanism than E2 mechanism. It can be also possible that the molecule undergoes elimination through a mixture of E1cb and E2, but it can be said that majority portion of the molecules will prefer the E1cb rather than any other pathway.

First of all, an E2 mechanism goes via simultaneous cleavage of the $\ce{C_{\beta}-H}$ and $\ce{C_{\alpha}-LG}$ bonds (LG being the leaving group). So, the rate of the reaction depends on the ease of breaking these two bonds. In this case the leaving group is fluoride ($\ce{F-}$), which has a very strong bond with the $\alpha$-carbon, making it a bad leaving group. So, due to this reason, the rate of E2 should lower for this molecule.
enter image description here

Now, if we check the plausibility of E1cb reaction pathway, we will find that it is most suitable in this case.

Firstly, the proton abstraction by the base gives a highly resonance stabilised benzylic carbanion intermediate, and this carbanion formation is the slow step. As this intermediate's energy will decrease, the activation energy will decrease resulting in an increase in the rate constant as, $k = A\exp(-E_\mathrm a/RT)$, helping to increase the rate for E1cb.

Secondly, the breaking of $\ce{C-F}$ bond doesn't happen in the rate determining step, which also doesn't alter the rate of the reaction in this case.

enter image description here

So, observing the above mentioned cases, it can be definitely concluded that E1cb is the more preferable pathway to undergo $\beta$-elimination to form Styrene.

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    $\begingroup$ Subscripts should be as such: $\mathrm{S_N1}$, $\mathrm{S_N2}$, $\mathrm{E1}$, $\mathrm{E2}$, $\mathrm{E1cb}$. For the elimination ones, you don't need MathJaX anyway as there are no subscripts. On topic, calling the benzylic anion "highly" resonance stabilised is a bit disingenuous. Toluene has a $\mathrm pK_\mathrm a$ of roughly 40. That's not a very stable anion. $\endgroup$ Oct 1, 2018 at 17:59
  • $\begingroup$ Okay but relaive to the other pathway, E1cb provides better feasibility by this stabilisation. And we are mainly concerned about relative rates. $\endgroup$
    – Soumik Das
    Oct 1, 2018 at 18:13

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