Put butan-2-ol in a flask with sodium bromide and nothing happens. Change the reagent and the same alcohol gives 2-bromobutane, 2-chlorobutane, but-2-ene or a nitrile. The reagent also decides what happens at the carbon that carried the OH. Its configuration can be kept, inverted or lost, and with some reagents the halogen ends up on a different carbon altogether. This page is about making that choice on purpose.
Alcohols are worth the effort because they sit at the center of synthesis. They can be made from alkenes, carbonyl compounds and alkyl halides, and they can be turned into halides, alkenes, ethers, esters, aldehydes, ketones and acids. Many multistep routes pass through one.
Why the OH has to be changed first
Start with the flask that does nothing. For bromide to bond to C2 of butan-2-ol, the C–O bond has to break and the OH has to leave as hydroxide. Hydroxide is the conjugate base of water, pKa 15.7, so it is a strong base, and strong bases do not leave. Leaving groups works through the same failure with butan-1-ol.
The fix is called activating the alcohol: changing the OH so that what leaves is a weak base. There are three ways to do it. The first is to protonate the oxygen, giving R–OH₂⁺. When the C–O bond breaks, what leaves is neutral water. Water's conjugate acid, H₃O⁺, has pKa −1.7, so water is a weak base. (The protonated alcohol itself is a strong acid, pKa about −2.)
The second way is to turn the alcohol into a sulfonate ester such as a tosylate, R–OTs. What leaves then is the tosylate ion, pKaH −2.8, with its charge shared over three oxygens. The third is to use SOCl₂ or PBr₃, which activate the OH and replace it with a halogen in the same flask.
Sulfonate esters: TsCl, MsCl and triflates
Stir butan-1-ol with TsCl (p-toluenesulfonyl chloride) and pyridine, a mild base, and you get butyl tosylate. The alcohol's oxygen bonds to sulfur, chloride leaves, and pyridine takes the H off the oxygen, so the HCl formed ends up as pyridinium chloride. Leaving groups draws each arrow of that step. The point to carry forward is that the C–O bond is never touched.
Two relatives do the same job. MsCl (methanesulfonyl chloride) gives the mesylate, R–OMs, with a CH₃ on sulfur in place of the tolyl ring. It is smaller and cheaper, and questions treat tosylates and mesylates as interchangeable. Triflic anhydride, a reagent that carries two CF₃SO₂ groups, gives the triflate, R–OTf, with a CF₃ on sulfur. The three fluorines pull electron density away from the sulfonate by induction, so the triflate ion is an even weaker base and leaves far faster than the other two. It is the one to use when a tosylate will not react.
All three leave as sulfonate ions with the charge spread over three oxygens, so they sit with iodide at the top of the leaving-group ranking. At carbon, a sulfonate ester behaves like an alkyl halide. Butyl tosylate and sodium cyanide give pentanenitrile by SN2, exactly as 1-bromobutane would. Protonation cannot do this: in strong acid, a basic nucleophile such as cyanide would be protonated itself. Tosylation installs the leaving group in its own step, and you choose the nucleophile afterward.
Alcohol to alkyl halide
Shake 2-methylpropan-2-ol (tert-butyl alcohol) with concentrated HCl at room temperature, and 2-chloro-2-methylpropane separates within minutes. The acid protonates the OH, water leaves, and chloride captures the tertiary carbocation. That is an SN1 reaction, and it is fast because a tertiary cation forms easily.
Butan-1-ol reacts with HBr only on heating, and by a different route, as in Leaving groups. A primary cation is too unstable to form, so bromide attacks the carbon from the side opposite the oxygen at the same moment water leaves. That is an SN2 reaction.
So with HX (HCl, HBr or HI), a tertiary alcohol reacts by SN1 and a primary one by SN2. A secondary alcohol sits in between and can react either way. When it reacts through the cation, the cation can rearrange before the halide arrives, and that is the main drawback of the HX route.
PBr₃ turns an alcohol into an alkyl bromide, and SOCl₂ (thionyl chloride), usually with pyridine, turns it into an alkyl chloride. Butan-2-ol gives 2-bromobutane with PBr₃ and 2-chlorobutane with SOCl₂. Neither reagent is a strong acid, and neither makes a carbocation, so neither rearranges. When a secondary alcohol has to become a halide with its skeleton intact, these are the reagents to use.
Both work in two steps. In step 1 the alcohol's oxygen bonds to phosphorus or sulfur and a halide ion leaves, as drawn in Leaving groups. The oxygen now carries a group that leaves easily. With PBr₃ that group departs as the neutral molecule HO–PBr₂. With SOCl₂, pyridine then removes the H from the oxygen, giving an alkyl chlorosulfite, R–O–S(=O)Cl. Its –O–S(=O)Cl group falls apart into SO₂ gas and a chloride ion as it leaves. In step 2 the halide released in step 1 attacks the carbon from the side opposite the oxygen. That is an SN2 step, so the carbon's configuration inverts.
One exception is worth knowing, because exam questions turn on it. Without pyridine, SOCl₂ can deliver the chlorine from the chlorosulfite itself to the same face the oxygen leaves from. That pathway is called SNi (substitution, nucleophilic, internal), and it gives retention of configuration. Questions almost always say "SOCl₂, pyridine" and expect inversion, so read the conditions before you commit to an answer.
Question. What does 3,3-dimethylbutan-2-ol give with (a) HBr, and (b) PBr₃?
Step 1 — what does each reagent do to the OH? HBr protonates it, so water leaves and C2 becomes a secondary cation. PBr₃ bonds the oxygen to phosphorus, and bromide then attacks C2 directly. One route makes a cation and the other never does.
Step 2 — follow the cation. Look at C2's neighbor. C3 carries three methyls and no hydrogen, so no hydride can move. A whole methyl group can, taking its bonding pair with it: a 1,2-methyl shift. The shift moves the positive charge to C3, which is now tertiary. Secondary to tertiary is downhill, and the shift is faster than the bromide's arrival. Carbocations draws every arrow of this route, protonation included.
Step 3 — capture. Bromide bonds to the tertiary cation. The main HBr product is 2-bromo-2,3-dimethylbutane, with the bromine on a carbon that never carried the OH.
Step 4 — the PBr₃ route. PBr₃ avoids a free cation, which is why the skeleton is kept. Bromide attacks C2 from the back, and the product is 3-bromo-2,2-dimethylbutane, with that carbon inverted. (The name numbers the chain from the other end, so the bromine carbon is C3.)
What each reagent does to a stereocenter
The problem. You need 2-bromobutane with a single, known configuration. HBr will not do it. Butan-2-ol is secondary, so HBr protonates the oxygen, water leaves, and much of the reaction passes through a flat secondary cation that bromide can attack from either face. The product is largely racemic, so you cannot give it a single descriptor.
Step 1 — make the tosylate. TsCl and pyridine convert the alcohol into butan-2-yl tosylate. The oxygen bonds to sulfur and nothing happens at C2, so its four groups stay where they were. The descriptor is still (S): OTs outranks ethyl just as OH did, and ethyl still outranks methyl.
Step 2 — substitute. Add NaBr. Bromide attacks C2 from the side opposite the tosylate in one SN2 step, and the other three groups flip over like an umbrella in the wind. That is the one inversion in the sequence.
Step 3 — name the product. At C2 of 2-bromobutane the priority order is Br > ethyl > methyl > H. Bromine takes oxygen's place at the top, so the ranking is unchanged. The groups have flipped, so the descriptor flips too: the product is (R)-2-bromobutane.
The rule behind the example comes from where each reagent does its chemistry. Count inversions, not steps.
| Reagent | Where the bonds change | At the stereocenter |
|---|---|---|
| TsCl or MsCl, pyridine | O–H and S–Cl; the C–O bond is untouched | retention; the SN2 you run next inverts once |
| PBr₃; SOCl₂ with pyridine | O bonds to P or S, then the halide attacks carbon from the back | one inversion |
| SOCl₂, no base | the chlorine is delivered to the same face (SNi) | retention |
| HX (secondary or tertiary alcohol) | water leaves and a flat carbocation forms | largely racemized; the skeleton may rearrange |
| hot H₂SO₄ | water leaves, then a β-H is lost | stereocenter destroyed (C=C forms); may rearrange first |
Dehydration: elimination of water
Heat butan-2-ol with concentrated sulfuric acid, and but-2-ene distills out, with a little but-1-ene. The acid protonates the OH, and water leaves to give the secondary cation. A water molecule then removes a β-hydrogen, one on a carbon next to the positive carbon, as in E1. The pair from that C–H bond becomes the new π bond.
This is the E1 mechanism from E1, with water as the leaving group once the acid has protonated the OH. Step 2 is slow and sets the rate. Step 3 is fast, and it decides which alkene forms. The base in step 3 can be as weak as water, because the hydrogen it removes sits next to a full positive charge. Concentrated H₃PO₄ works the same way as H₂SO₄.
Where there is a choice of β-hydrogen, the reaction follows Zaitsev's rule: the major product is the more substituted alkene, the one with more carbons attached to the C=C. Butan-2-ol has β-hydrogens on C1 and on C3. Losing one from C3 gives but-2-ene, with two carbons on the double bond, so but-2-ene is the major product.
Tertiary alcohols dehydrate most easily, then secondary, then primary, which is the order of carbocation stability. A primary alcohol reacts only under forcing conditions, and it avoids the primary cation by losing water and the β-hydrogen together, in a concerted E2-like step.
Dehydration is the reverse of the acid-catalyzed hydration in Addition reactions, through the same intermediates in the opposite order. The conditions decide which way it runs. Dilute acid and plenty of water push the equilibrium toward the alcohol. Concentrated acid, heat, and distilling the alkene off as it forms push it toward the alkene. Both are Le Châtelier's principle at work.
For a molecule that cannot survive strong acid, POCl₃ in pyridine dehydrates the alcohol under mild, basic conditions. The alcohol's oxygen bonds to phosphorus and a chloride leaves, giving a dichlorophosphate ester, R–O–P(=O)Cl₂. That group is a good leaving group, and no acid was needed to install it.
Pyridine then removes a β-hydrogen in one concerted E2 step, and the E2 geometry rule from E2 applies: the hydrogen and the leaving group must be anti-periplanar, in one plane and pointing in opposite directions, 180° apart. Pyridine does two jobs. It is the base for the elimination, and it takes up the HCl released while the ester forms.
Because no cation forms, POCl₃ and pyridine do not rearrange the skeleton. Take 3,3-dimethylbutan-2-ol again. C3 has no hydrogen, so the only β-hydrogens are on C1, and the product is 3,3-dimethylbut-1-ene. The acid route gave 2,3-dimethylbut-2-ene from the same alcohol.
Choosing the reagent
Start with butan-2-ol, a secondary alcohol, and let the reagent decide.
PBr₃ gives 2-bromobutane. The oxygen bonds to phosphorus first, then bromide attacks from the back. C2 is inverted, and with no cation the skeleton cannot move.
SOCl₂ and pyridine give 2-chlorobutane by the same logic: the chlorosulfite forms, chloride attacks from the back, and C2 is inverted with no rearrangement.
Concentrated H₂SO₄ and heat give but-2-ene. The OH is protonated, water leaves to give the secondary cation, and a β-hydrogen is lost from C3. Zaitsev's rule favors but-2-ene over but-1-ene because it is disubstituted rather than monosubstituted.
TsCl and pyridine, then NaCN give 2-methylbutanenitrile. Tosylation happens at sulfur, so C2 keeps its configuration. Cyanide then does one SN2 on the tosylate, and that step alone inverts C2.
Four reagents give four products, but only two answers to the question "what happens to the configuration at C2?" Three routes invert it once. Dehydration removes the stereocenter altogether, because C2 becomes part of the flat C=C.
A preview: oxidation
This paragraph is a preview; Oxidation & Reduction teaches it. Oxidizing an alcohol at the carbon that carries the OH (the carbinol carbon) replaces a C–H bond on that carbon with a bond to oxygen. So you can count how far an alcohol can go. A primary alcohol has two such C–H bonds and can be oxidized twice, to an aldehyde and then to a carboxylic acid. A secondary alcohol has one and stops at the ketone. A tertiary alcohol has none and does not react. Some reagents stop a primary alcohol at the aldehyde: PCC (a chromium reagent used without water) and the Swern oxidation (no metal). Chromium(VI) in water carries it on to the acid. Why water makes the difference is explained in that chapter.
Alcohols as acids
Alcohols are weak acids, with pKa around 16. A strong base such as NaH removes the O–H proton to give an alkoxide, RO⁻, with H₂ bubbling off, as shown in pKa. Sodium metal also gives the alkoxide, and here too H₂ bubbles off. An alkoxide is a much stronger base and a much better nucleophile than the alcohol it came from. It is the nucleophile of the Williamson ether synthesis in Ether chemistry.
"Around 16" hides a trend that surprises most students: methanol 15.5, water 15.7, ethanol 16, tert-butanol 18. Methanol is more acidic than water, and acidity falls as the carbon bearing the OH gains alkyl groups. The main reason is solvation, as Factors affecting acidity explains. Water molecules crowd closely around small methoxide and share its charge, but bulky tert-butoxide keeps them at a distance, so it is stabilized less. "Alcohols are weaker acids than water" is therefore a fair rule of thumb from ethanol upward, and false for methanol.
Alcohols also react with carboxylic acids and their derivatives to give esters. That chemistry belongs to Carboxylic Acids & Derivatives, which comes later.
Where alcohols come from
You have already made alcohols three ways. Acid-catalyzed hydration puts the OH on the more substituted carbon of an alkene (Addition reactions). Hydroboration–oxidation puts it on the less substituted carbon (Markovnikov / anti-Markovnikov). And hydroxide can do an SN2 on a primary alkyl halide (Substitution & Elimination). The most useful routes, adding a carbon nucleophile to a carbonyl group or reducing one, come in Carbonyl Chemistry and later chapters.
What carries forward
Activating an alcohol is a move you will make in many multistep syntheses: nothing happens at that carbon until the OH becomes a leaving group. The alkoxides made here are the nucleophiles of Ether chemistry. And oxidation (taught in Oxidation & Reduction) ties alcohols to Carbonyl Chemistry and Carboxylic Acids & Derivatives, because most aldehydes and ketones you meet there can be traced back to an alcohol.