Elimination Reactions of Alcohols

by James

in Alcohols, Organic Chemistry 2

Here’s a summary of what we talk about in today’s post.


Hydrohalic acids (HX)  plus alcohols give substitution products…

We just saw that treating an alcohol with a strong hydrohalic acid – think HCl, HBr, or HI – resulted in the formation of alkyl halides. With a tertiary alcohol like the one drawn below, this proceeds through an SN1 mechanism. [Protonation of alcohol, then loss of H2O to form a carbocation, then attack of nucleophile on carbocation].

2-hcl alcohols sn1

…But H2SO4, H3PO4, and TsOH Give Elimination Products!

You’d be forgiven for thinking that if we treated an alcohol with H2SO4 (sulfuric acid) the same type of thing would occur, and the carbocation would be attacked by the OSO3H anion to make the product below. Very reasonable to propose. Of course, in practice, it doesn’t work that way!

3-h2so4 alcohols

So what happened here?

First, look at what bonds formed and broke. We formed C-C (π) and broke C-OH and C-H. (We also formed H-O , in that molecule of water that forms as a byproduct). This is the pattern of an elimination reaction.

Now let’s ask: How could this have formed? If you look closely, note that we’ve broken a C-H bond on the carbon adjacent to the carbocation and formed a new C-C π bond at that spot. It’s reasonable to propose that instead of attacking the carbocation to form a new substitution product, a  base removed a proton adjacent to the carbocation and formed the alkene. [That carbon adjacent to the carbocation is often referred to as the “β (beta) carbon”. The carbocation itself is the “α (alpha) carbon”].

We’ve seen this type of process before actually! This is an E1 process [elimination (E) , unimolecular (1) rate determining step]. You might also remember that elimination reactions tend to follow “Zaitsev’s rule” – we always form the most substituted alkene [or to put it another way, we remove a proton from the carbon with the fewest attached hydrogens] because alkene stability increases as we increase the number of attached carbons.

4-elimination mech

[By the way, you might ask – why “heat” ? Heat generally tends to favour elimination reactions.]

By no means is H2SO4 the only acid that does this. Phosphoric acid (H3PO4) as well as “tosic acid” (p-toluenesulfonic acid) also tend to form elimination products.

5-tsoh or phosphoric acid

Why Does H2SO4 (Or H3PO4 or TsOH) Give Elimination Products But HCl, HBr, HI give Substitution Products?

So why do we get elimination reactions with H2SO4 as acid (or H3PO4, or TsOH) whereas we get substitution reactions with HCl, HBr, and HI?  The answer is that the HSO4–  anion is a very poor nucleophile , being quite stabilized by resonance. In the diagram below, note how that negative charge is delocalized over three different oxygens [the same is true for the TsOand H2PO4 anions].  Compare that to halide anions, where the negative charge cannot be spread over more than one atom. The upshot is that delocalization of charge results in a slower reaction of HSO4– as a nucleophile compared to deprotonation of C-H by a base, and the alkene product dominates.

6-resonance stabilized

So the bottom line here is that heating tertiary alcohols with these acids will result in loss of water [“dehydration”] and formation of an alkene [elimination].

What About Secondary Alcohols?

Heating a secondary alcohol with sulfuric acid or phosphoric acid? Same deal as with tertiary alcohols: expect an alkene to form. As with all elimination reactions, there are two things to watch out for: first, the “most substituted” alkene (Zaitsev) will be the dominant product, and also, don’t forget that trans alkenes will be favoured (more stable) than cis alkenes due to less steric strain.


There is one last thing to watch out for with secondary alcohols, though… like a bad nightmare, they keep coming back.

Carbocation Rearrangements… Again?

As we saw with the reactions of HCl, HBr, and HI with secondary alcohols, we have to watch out for carbocation rearrangement reactions. If a more stable carbocation can be formed through migration of an adjacent hydride (H- )  or an alkyl group, then that migration will occur.

For example, treatment of the alcohol below with H2SO4 leads to formation of a secondary carbocation, followed by a hydride shift to give a tertiary carbocation, followed by deprotonation at whichever β carbon leads to the most substituted alkene.

8-hydride shift

Ring Expansion Followed By Elimination

It’s also possible for alkyl shifts to occur to give a more stable carbocation. A classic example of this are expansions of strained rings (like cyclobutanes) to give less strained rings (like cyclopentanes).

For example in the case below the key step is where the C3-C4 bond breaks to form the C2-C4 bond, resulting in a new (tertiary) carbocation on C-3 as well as a less strained ring. Since there isn’t a good nucleophile around, elimination occurs in such a way that the most substituted alkene is formed.

9-ring expansion

What About Primary Alcohols?

The final class of alcohols to be concerned about is primary alcohols. You might ask: if we treat a primary alcohol (say, 1-butanol) with a strong acid like H2SO4, will also get elimination to an alkene?

Yes, alkenes can be formed this way (along with some formation of symmetrical ethers [see this previous post]). Here’s an example.


There is a catch however:  the E1 pathway (formation of a primary carbocation) is not the most likely pathway here. Primary carbocations tend to be extremely unstable, and it’s more likely that the reaction passes through an E2 mechanism where the transition state will be lower in energy. Notice what happens here: first we protonate the alcohol to give the good leaving group OH2+ , and then a weak base (which I’m leaving vague, but could be H2O, OSO3H, or another molecule of the alcohol) could then break C-H, leading to formation of the alkene.

11-e2 primary

The Bottom Line

If you see a tertiary or secondary alcohol with H2SO4, TsOH, or H3PO4 (and especially if you see “heat”) think: carbocation formation followed by elimination reaction (E1).

And if you see that a more stable carbocation could be formed through migration of an adjacent H or alkyl group, expect that to happen.

If you see a primary alcohol with H2SO4, TsOH, or H3PO4, expect symmetrical ether formation accompanied by elimination to form the alkene.

Next Time…

So far we’ve learned two ways to convert alcohols to alkenes:

1) convert them to a good leaving group, and then add base (2 steps)

2) add strong acid with heat (one step)

Ideally, we’d like to just use one step. But strong acid can lead to complications (carbocation rearrangements, cough cough) and we might ask: “isn’t there an easier way”?
There is! That’s what we’ll cover in the next post.

Next Post: Elimination Of Alcohols To Alkenes With POCl3



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{ 16 comments… read them below or add one }


I posted a message a few days ago, but somehow it was erased. There it goes again:
– “… we remove a proton from the carbon with the most attached hydrogens”; it’s the carbon with the FEWEST attached hydrogens!
– please check the formulas of acids and their corresponding anions in the text; some appear like this: “H2SO4 as acid (or H3PO4” (they are written correctly in the images).



Don’t know why that comment didn’t post. Thank you for your keen eye, as always!



thank you so much for these information but i have a small question …is there a difference between Elimination and dehydration ??



Dehydration specifically refers to loss of water. Elimination in the sense of this post refers to formation of a double bond. There is overlap between the two when dehydration leads to formation of a double bond.


uday raj singh chauhan

sorry I put my e mail wrong, posting my question again.
why not a SN2 reaction after protonation of primary alcohols??? H2O is a good leaving group and primary carbon is not hindered, a perfect recipe for SN2. HSO4- can attack through SN2, why not? why elimination? given that HSO4- is a week base too.



HSO4- is an extremely poor nucleophile for the SN2.


zuko kai

Plus there is heat involved in the reaction..which is favourable for elimination reactions…thank u n feel free to correct if wrong


Organic Chemistry student

I would assume that secondary alcohols can undergo both E1 and E2 reactions. In your post, you are suggesting that secondary alcohols favor an E1 mechanism. Is that true only if a secondary carbocation can rearrange to give a tertiary? Is it safe to say that otherwise, secondary alcohols can undergo both E1 and E2?



The identity of the acid is important. In the case of H2SO4 or H3PO4, there simply is no sufficiently strong base present to cause an E2 reaction to occur. Loss of H2O to form a carbocation followed by elimination will be the favoured pathway.
Note that secondary alkyl halides can undergo E2 reactions just fine. The issue with alcohols here is that we are using strong acid to turn the OH into a good leaving group. If we add a strong base here (to perform an E2) it will just end up neutralizing this species.
As far as rearrangement is concerned, it will generally only be favoured in a situation where a more stable carbocation will form. That is true for the conversion of secondary carbocations to tertiary carbocations. It *can* be true that rearrangements of tertiary carbocations occur, but generally only in situations where they would be more stabilized (e.g. tertiary carbocation to a resonance-stabilized tertiary carbocation )



When ethanol is heated at 140*C in the presence of conc. Sulphuric acid. Is this a beta elimination reaction??



Likely formation of diethyl ether.



What happens if you use two cis or trans OH in the educt?



It’s somewhat possible that you might get some epoxide formation, or even formation of a ketone/aldehyde. Depends on the structure of the substrate.



Why we use H2SO4 in case of alcohols reacting with HBr and that of we use H3PO4 in case of alcohols reacting with HI . why



They should be equivalent.



just want to thankyou for this clear explanation. i was really confused why H2SO4 was only explained as forming E1 E2 products but not SN1 SN2. couldnt find the answer anywhere until i stumbled on this page.


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