Ace Your Next Organic Chemistry Exam.

With these Downloadable PDF Study Guides

Our Study Guides

Alcohols, Epoxides and Ethers

By James Ashenhurst

Making Alkyl Halides From Alcohols

Last updated: April 15th, 2019

Making Alkyl Halides From Alcohols

In today’s post we show that treating alcohols with HCl, HBr, or HI (which all fall under the catch-all term “HX” where X is a halide) results in the formation of alkyl halides.


We’ve said many times in this series of posts that alcohols are poor substrates for SN1 and SN2 reactions.  That’s because the hydroxyl ion (HO-) is a poor leaving group, and therefore not likely to either 1) depart of its own accord, leaving behind a carbocation (SN1 pathway) or 2) to be displaced by an incoming nucleophile (which would be an SN2 reaction).

However we’ve also seen that treating an alcohol with acid leads to an interesting “personality adjustment“: the alcohol (R-OH) is converted to its conjugate acid, (R-OH2+) which now possesses a decent leaving group (the weak base we know as water, H2O).  We saw, in a previous post,  how this allows for formation of (symmetrical) ethers from alcohols, either via SN2 pathway (with primary alcohols) or an SN1 pathway (tertiary alcohols).

We might ask: can this be extended to form other functional groups besides ethers?

Sure thing! Treating alcohols with HCl, HBr, or HI (which all fall under the catch-all term “HX” where X is a halide) results in the formation of alkyl halides.

This occurs in a two step process: first, the alcohol is protonated to give its conjugate acid. Secondly, a substitution occurs.


Notice how that second step (substitution) was left vague in the diagram above.  That’s because, as we’ve seen,  the type of substitution pathway depends on the substrate.

Methyl And Primary Alcohols Form Alkyl Halides Via SN2 Reaction

Knowing how sensitive the SN2 reaction is to steric hindrance, we should expect that for methyl alcohol and for primary alcohols, the SN2 pathway dominates. And it does!



Note how in each case we begin by protonating the alcohol, creating a good leaving group which is then displaced by the conjugate base of the acid. Alkyl chlorides, bromides, and iodides can each be made this way.

For Tertiary Alcohols, The SN1 Pathway Dominates

Likewise, understanding the trends of carbocation stability, we should expect that conversion of tertiary alcohols to alkyl halides proceeds through an SN1 pathway. And it does.


Note in the last example that beginning with a chiral starting material will lead to a mixture of inversion and retention (often called, “racemization“) because it goes through the (flat) intermediate carbocation.

A Good Rule of Thumb For Secondary Alcohols With HX: Assume SN1

Methyl, primary, and tertiary alcohols all represent pretty straightforward cases.

“So what about secondary alcohols?” you might ask. Ah yes. This is where things get interesting – and is, therefore, the stuff of which exam questions are made.

In the lab, treatment of secondary alcohols with HX leads to a mixture of products from SN1 and SN2 pathways. For practical purposes it is generally not a useful process, especially if you care about preserving stereochemistry.

However, your introductory textbook and course notes are not “the lab”. The purpose of a course is to introduce you to important concepts in organic chemistry. And from an instructor’s standpoint, it so happens that the conversion of secondary alcohols to secondary alkyl halides by HX is an excellent opportunity to bring up the subject of carbocation rearrangements. This falls under the purview of the SN1 pathway. So a good rule of thumb is to assume – for the purposes of your course – that secondary alcohols treated with HX will proceed through an SN1 mechanism.

Rearrangements, Rearrangements, and More Rearrangements

We’ve covered rearrangements  (hydride and alkyl shifts)  before in the context of SN1 reactions. But it’s worth touching on again.

The basic premise is this. Carbocations are unstable, electron-poor species. Their stability generally increases with the number of attached carbons, which serve to donate electron density. Hence, the stability of carbocations increases in the direction methyl < primary < secondary < tertiary.  We also saw that carbocations are stabilized by resonance.

As we saw in a previous series of posts – consult this if you need more hand holding! –  carbocations can undergo 1,2 shifts of C-H and C-C bonds, resulting in new carbocations. Such rearrangements are most likely to occur if they can result in a more stable carbocation. For example, the rearrangement of a secondary to a tertiary carbocation, is a favoured (energetically “downhill”) process, whereas a rearrangement from a tertiary to a secondary carbocation (energetically “uphill”) is unlikely.

Anytime a reaction proceeds through a carbocation intermediate, we need to be on the lookout to see if it is adjacent to a carbon which can generate a more stable carbocation through a shift of a C-H or C-C bond.

There are three cases in particular to watch out for.

Case 1 – the hydride shift

Look for a secondary alcohol adjacent to a tertiary carbon. Note this common example, where protonation leads to loss of water, followed by a hydride shift and then trapping of the carbocation by the halide ion.


Another example of a favourable rearrangement is when a secondary carbocation is adjacent to an “allylic” or “benzylic” hydrogen. Rearrangement results in a secondary carbon which is stabilized by resonance.


Case 2 a) – the alkyl shift 

Look for a secondary alcohol adjacent to a quaternary carbon (i.e. a carbon attached to 4 other carbons). Note how this is essentially the exact same process as the hydride shift above, except that CH3 is migrating, not H.

Case 2 b)- a special case of alkyl shifts – ring expansions or contractions

Look for a secondary alcohol that is adjacent to a strained ring (cyclobutane in the classic case).  Once the secondary carbocation is generated, a bond in the strained ring migrates, leading to expansion of the ring by one. This is particularly favourable in the case of cyclobutane to cyclopentane since cyclobutane is highly strained (ca 26kcal/mol) whereas cyclopentane has only mild ring strain.

7-ring strain

This type of alkyl shift commonly gives students a hard time, which of course makes it a favourite exam problem of instructors. Although the curved arrow drawn is no different than that for the previous two cases, I think the main difficulty is in mapping the product from the starting material. In this respect I recommend two things:

  • Number the carbons (not necessarily IUPAC – just number to keep track of them). For instance here the arrow in Step 3 (the alkyl shift) shows us breaking C2-C3 and forming C1-C3. Applying the rules of curved arrows implies that C1 will then be neutral and C2 will become a carbocation. That’s all that’s happening.  No other bonds are formed or broken in this step. It takes practice to get this right.
  • Draw the ugly version first. THEN redraw to make it look good.

Ring contractions are also possible, although are not as favourable as the opening of strained rings. The same principles apply.

What Doesn’t Work? 

It’s always helpful to know what doesn’t work.

First of all, since we’re dealing with substitution reactions here, some familiar rules apply.  Only alkyl alcohols (alcohols on sp3 hybridized carbons) will undergo SN1 and SN2 reactions. Both the SN1 and SN2 pathways involve buildup of positive charge on carbon, and sp2 and sp hybridized carbocations are extremely unstable. This attempted SN1 of phenol, for example, will fail miserably:


You might also wonder if we can use reagents like HCN, HOAc, or HN3 to convert alcohols to nitriles, esters, and azides respectively. Generally, no. The problem is that each of these are fairly weak acids (pKa 4 and above)  so these will only give a low concentration of the protonated alcohol. Since the reaction rate is proportional to concentration, formation of these products will be slow. [With azides, there are also potential complications with a different type of rearrangement, but as a curtiusy we’re not going to deal with that schmidt right now : – ) ]

9-hx that doesn't work

For our purposes, conversion of alcohols to other substitution products using strong acid is limited to HCl, HBr, HI, and the special case of H+/ROH which gives symmetrical ethers. A good rule of thumb is that the conjugate acid of the nucleophile should have a pKa of 0 or less in order for the reaction to occur.

So what’s the point?

So alcohols can be converted to alkyl halides. You might ask, “why should we care?”. The answer is that, as we said, converting an alcohol (which has a poor leaving group) into an alkyl halide (which has a great leaving group) now allows us to do all kinds of functional group interconversions that were not previously possible. The SN2 is a very useful and powerful reaction, for example. Once a primary alcohol has been converted to a primary alkyl halide, we can then treat it with all varieties of nucleophiles to make a multitude of functional groups.

However nice it is to be able to do this, though,  it’s far from ideal.  We have to use strong acid, which can often cause complications if we have acid-sensitive functional groups on our molecule. Furthermore, all those pesky rearrangements on secondary carbons are a hassle. They can screw with our stereochemistry and lead to undesired products.  You might ask, “isn’t there some way to get around that?”.

Yes! We’ll talk about a very nice way around this dilemma in the next post!

Next Post – Tosylates And Mesylates

Related Posts:


Comment section

31 thoughts on “Making Alkyl Halides From Alcohols

  1. Hey James! When is the next article coming out?
    I was wondering how to make a anti-Markovnikov halide from an anti-Markovnikov alcohol. Thanks for your help!

  2. PKa value of HCl is less than for H3O+. So, Cl- remains a weaker base than water. You mentioned in one of the older articles on Sn2 that a weaker base cant displace a stronger base. How can primary alcohols then react with HCl via an Sn2??

    1. Good question.
      The “weaker base can’t displace a stronger base” applies, as a rule of thumb, when the difference in pKa is about 8 or greater. When the difference is less, there will be some equilibrium between the two acids, however minor.

      As it turns out, chloride ion is actually superior nucleophile to water, reacting with alkyl halides at faster rates. Practically, there will be water as a byproduct of a substitution reaction which might lead to some minor reversibility. This can be removed by adding an excess equivalent of HCl, which will protonate H2O to give H30+ (which of course is a much poorer nucleophile).

  3. For making alkyl halide from alcohol, HX is used. This HX is dry and made by reacting NaX with H2SO4 or in case of HI with H3PO4. So why NaX is not used directly to convert alcohol to alkyl halide?

  4. Possible to do ring expansion on cyclobutane to cyclohexane? Is there a point where a methyl shift is preferred over ring expansion, such as when rings become very large?

    1. Not possible to go from cyclobutane to cyclohexane directly, since the carbocation would have to be adjacent to the 4 membered ring and it always involves a 1,2 shift.
      When ring strain is not an issue (carbocations on 5, 6, 7 membered rings) then you can have mixtures of alkyl shift and ring expanded/contracted products.

  5. I think I’ve spotted a typo. In case 2a), the last structure should have a blue Methyl group rather than a blue H.
    Thanks for a great page!!

  6. Is it necessary to use ZnCl2 in the reaction of HCl and Alcohols for preparing Alkyl Halides or can we simply react Alcohols with HCl for preparing Alkyl Halides? Why?

  7. Hi, I believe you have an error on the alkyl shift photo. Going from the second to last step to the final you lose a methyl group and in its place is a hi-lighted blue.

  8. When observing the reaction of butan-2-ol with HCl to produce 2-chlorobutane and water, what can we expect to see? Or will there be no observable reaction? And is it true that when tertiary alcohols react with HCl, the substance becomes more turbid yet with primary alcohols, the substance remains colourless/less turbid than secondary or tertiary reactions? Or does that only pertain to the Lucas Reagent?

    1. The advantage is that substitution is more predictable. You know that the carbon-nucleophile bond will form on the carbon bearing the leaving group. In contrast, when performing electrophilic addition on an alkene, regioselectivity can be a problem. The major product will be determined by Markovnikov’s rule, and even then the selectivity can be bad (4:1). What if you want the anti-Markovnikov product? Or what if you have an alkene where both ends are equally substituted? These problems don’t arise with substitution.

  9. Thank you so much for putting in a lot of efforts and bringing masterpieces to us. Your notes have always helped me and makes Organic Chemistry super fun and interesting as well.

  10. Hi,
    How would reaction change if we have 2-bromo-1-propanol react with HBr ? instead of just propanol ?

    1. Not the best example, since I’m not sure that’s a very stable compound (it easily snaps shut to form an epoxide) but eventually the alcohol would still be replaced with Br, to obtain 1,2-dibromopropane.

Leave a Reply to Ameya

Your email address will not be published.

This site uses Akismet to reduce spam. Learn how your comment data is processed.