0
$\begingroup$

Let’s say you take ethanol: $\ce{CH3-CH2-OH}$ and you add an acid and heat. You can get water, the acid, and $\ce{CH2=CH2}$ (a dehydration reaction). However, you can also get $\ce{CH3-CH2-O-CH2-CH3}$, water, and an acid (symmetric ether synthesis). How would you make one reaction preferable over another?

$\endgroup$
2
$\begingroup$

By modifying the conditions, in this case: the concentration.

The first and rapid step of the mechanism is the protonation of the alcohol:

$$\ce{HA + CH3CH2-OH <=> A- + CH3CH2-OH2+}$$

From then on, it depends what happens:

  • If the molecule is left alone, i.e. its concentration is low, something that has a just enough basicity might at the correct time be close enough to a methyl hydrogen to allow the elimination to happen. $\mathrm{E1}$ is not an option since the primary carbocation is extremely unstable and Wagner-Meerwein rearrangements for enhanced stability are not possible, thus the elimination must preceed by the $\mathrm{E2}$ mechanism. If $\ce{A-}$ is the only basic compound present, it must accept the proton. $$\ce{A- + H-CH2-CH2-OH2+ -> A-H + H2C=CH2 + H2O}$$

  • If the molecule happens to find one if its kind before the former can take place, the two of them could engage in a substitution reaction by the $\mathrm{S_N2}$ mechanism. If the molecule’s concentration is high, this gets more likely. And if it happens, the second molecule’s hydroxy group is nucleophilic enough (even though alcohols are generally bad nucleophiles) to displace the extremely good leaving group $\ce{-OH2+}$ and liberate water. The extraneous proton can later be removed as the ether sees fit. $$\ce{CH3CH2-OH2+ + HO-CH2CH3 -> CH3CH2-(OH+)-CH2CH3 + H2O}$$ $$\ce{CH3CH2-(OH+)-CH2CH3 + A- <=> CH3CH2-O-CH2CH3 + HA}$$

Thus, the first reaction will always happen but be rather slow. Its rate is close enough to being independent of the molecule’s concentration. The second reaction’s rate is directly dependent on the molecule’s concentration and will get faster the higher the concentration is. High concentrations (and low acid concentrations) can thus favour the substitution mechanism over the elimination one.

This will hold true more or less for all cases where two reactions compete against one another. It is always possible to modify the conditions to prefer one or the other. However, it is often not as simple as in this simple case.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.