Organic Chemistry · Alcohols and Phenols

Reactions of Alcohols

9 min read
Constants cross-checked against current references (PubChem, 2026-08): PBr₃ molar mass 270.67 g/mol, density ~2.85 g/mL; pKa water 15.7, pKa hydronium ≈ −1.7, pKa TsOH ≈ −2.8; alcohol O–H pKa ≈ 16–18 (standard organic chemistry references).
Want it in plain words first? Jump to Eli explains — the same idea, no jargon.
On this page 9 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Check yourself
  8. Study tools
  9. Sources & references

In 30 seconds

Every alcohol reaction is a contest between the three reactive features of the –OH carbon: the oxygen lone pairs (base and nucleophile), the O–H bond (weak acid), and the C–O bond (almost impossible to break on its own). Because hydroxide (OH-) is a terrible leaving group, the C–O bond becomes reactive only when the oxygen is first converted into something that can leave — either by protonation (making water the leaving group) or by derivatization (making a tosylate or halide). This single idea — "activate the oxygen before you can break the C–O bond" — organizes the entire topic.

The major reaction families: (1) to alkenes under acid and heat, (2) substitution to alkyl halides with HX, PBr3, or SOCl2, (3) tosylate formation as a gateway to SN2 chemistry, (4) esterification with acid chlorides or anhydrides, and (5) formation with strong bases or alkali metals. Oxidation is the sixth family (next topic).

Why this matters

Alcohols are the most versatile intermediates in organic synthesis: nearly every functional group can be made from an alcohol, and nearly every one can be reduced to one. In industry and medicine, these reactions make ether anesthetics, alkene monomers, alkyl halide building blocks, and ester prodrugs. In the body, the same chemistry governs ethanol metabolism and enzyme activation of hydroxyl groups. For exams, alcohol reactions are a favorite "predict the product" target: the answer depends on three quick decisions — alcohol class (1°/2°/3°), acidic vs basic reagent, and whether the oxygen has been activated.

The college version

Core Concepts

The leaving-group problem: why –OH must be activated

Hydroxide is a strong base (pKa of water ≈ 15.7), so it never leaves on its own. Two fixes exist:

  • Protonation: acid converts –OH into –OH2+, making water (pKa ≈ –1.7) the leaving group — the path for dehydration and HX substitution.
  • Derivatization: tosyl chloride converts –OH into –OTs; tosylate (pKa ≈ –2.8) leaves easily. This preserves stereochemistry at carbon and works where protonation would fail.

Dehydration to alkenes (acid-catalyzed)

Heating an alcohol with a strong acid (H2SO4 or H3PO4) removes water and forms an alkene. The mechanism is E1 for 2° and 3° alcohols: protonation, loss of water to a carbocation, then loss of a β-hydrogen. The alkene obeys — the more substituted alkene predominates. Two exam features:

  • Carbocation rearrangements: hydride or alkyl shifts can change the product skeleton (e.g., 3,3-dimethyl-2-butanol rearranges via a methyl shift before elimination).
  • Reaction order by class: 3° alcohols dehydrate under mild conditions (dilute acid, ~50 °C); 2° alcohols need stronger conditions; 1° alcohols require concentrated acid and ~180 °C and react by an E2-like pathway, since a 1° carbocation is too unstable to form.

Conversion to alkyl halides with HX

Alcohols react with hydrogen halides to give alkyl halides plus water. Reactivity follows HI > HBr > HCl; HCl is so sluggish with 1° and 2° alcohols that it needs ZnCl2 (the ) as a Lewis-acid catalyst.

  • 3° alcohols: rapid S_N1 — protonation, loss of water, carbocation, halide capture. The Lucas test exploits this: 3° alcohols cloud within seconds.
  • 2° alcohols: S_N1 as well, but slow, with possible rearrangement (cloudy in ~5 min).
  • 1° alcohols: S_N2 — bromide or iodide displaces the protonated hydroxyl with inversion. Even so, HCl is too slow; 1° alcohols give no Lucas reaction at room temperature.

Conversion to alkyl halides with PBr3 and SOCl2

Using HBr directly risks carbocation rearrangements for 2° alcohols. Phosphorus tribromide avoids this: an SN2-type displacement at carbon gives inversion and no free carbocation, so the skeleton is preserved. Stoichiometry: 3 mol alcohol per 1 mol PBr3:

3 ROH + PBr3 ⟶ 3 RBr + H3PO3

Similarly, thionyl chloride (SOCl2, usually with pyridine) gives alkyl chlorides with inversion and no rearrangement (SO2 and HCl byproducts).

Tosylate esters: the gateway to SN2

Tosyl chloride in pyridine converts –OH to –OTs with retention of configuration (the C–O bond never breaks). The tosylate is then displaced by any good nucleophile (CN-, N3-, RO-, I-, RS-) in a clean SN2 step with inversion:

R–OH TsCl, pyridine⟶ R–OTs Nu-⟶ R–Nu + TsO-

This "activate then displace" sequence converts an alcohol into almost any other functional group with controlled stereochemistry — one clean inversion, never a scrambled product.

Esterification with acid chlorides and anhydrides

Alcohols react with acid chlorides or anhydrides (usually with pyridine to absorb the HCl) to give esters:

ROH + CH3COCl ⟶ ROCOCH3 + HCl

This is nucleophilic acyl substitution: the alcohol's oxygen attacks the carbonyl carbon, then chloride leaves. Phenols react the same way (topic 10). Acetylation of hydroxyl groups is also how the body tags molecules for transport or excretion, and how aspirin is made from salicylic acid.

Alkoxide formation

Alcohols are weak acids (pKa ≈ 16–18), so strong bases deprotonate them. With sodium or potassium metal:

2 ROH + 2 Na ⟶ 2 RO-Na+ + H2

With sodium hydride, the byproduct is hydrogen gas as well. Alkoxides are the nucleophiles/bases in Williamson ether synthesis (Chapter 18) and in many eliminations. Practical warning: an alcohol's O–H proton is acidic, so alcohols quench Grignard and organolithium reagents — protect an –OH (topic 8) before using an organometallic reagent on the same molecule.

Common Confusions

Do not confuseWithDifference
Dehydration vs "dehydration of a hydrate"Water loss from an alcohol (C–O cleavage)Alcohol dehydration forms a C=C; the hydrate phrase refers to inorganic water of crystallization
PBr3 vs HBrBoth make alkyl bromidesPBr3: S_N2, inversion, no rearrangement. HBr on 2°/3° alcohols can rearrange via a carbocation
Retention vs inversionTosylation (retention) vs S_N2 displacement (inversion)TsCl only changes O–H; the subsequent nucleophilic attack flips the stereocenter
E1 vs E2 dehydrationMechanism for 1° vs 3° alcohols3° alcohols eliminate through a carbocation (E1); 1° alcohols use an E2-like pathway — no carbocation
Lucas positive vs Lucas negative3°/2° vs 1° alcoholsCloudiness timing: 3° immediate, 2° ~5 min, 1° no reaction at room temperature
Alcohol + Na vs alcohol + NaHBoth give alkoxidesNa: 2 ROH → 2 RO⁻Na⁺ + H₂. NaH: ROH → RO⁻Na⁺ + H₂ (1:1, no H₂ from the alcohol's H)
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Imagine –OH is a guest that refuses to leave a party (the carbon). The only way to get the guest out is to make it uncomfortable: squirt acid on it so it becomes water (then it happily leaves), or dress it up as a tosylate (then a strong "nudger" can push it out). Once the –OH is gone, whatever molecule was waiting outside takes its seat — sometimes with a twist, like a chair flip (inversion).

Worked example

Example 1: Predict the major dehydration product

Problem. Dehydrate 2-methyl-2-butanol with sulfuric acid and heat. Predict the major alkene.

Approach. The alcohol is 3°, so the mechanism is E1: protonate –OH, lose water to give the 3° carbocation (CH3)2C+CH2CH3, then remove a β-hydrogen. Zaitsev's rule says the more substituted alkene wins. Two β-positions exist: a methyl group (2-methyl-1-butene, disubstituted) and the CH2 of the ethyl group (2-methyl-2-butene, trisubstituted).

Answer. The major product is 2-methyl-2-butene (trisubstituted, Zaitsev), with 2-methyl-1-butene as the minor product:

(CH3)2C(OH)CH2CH3 H2SO4, Δ⟶ (CH3)2C=CHCH3 + H2O

Example 2: Convert 1-butanol to 1-bromobutane without rearrangement

Problem. You need 1-bromobutane from 1-butanol. Why is PBr3 better than HBr, and how much PBr3 is required for 0.250 mol of alcohol?

Step 1 — reagent choice. For 1-butanol, HBr would give a clean S_N2 product too, but for 2° alcohols HBr can scramble stereochemistry through a carbocation; PBr3 always gives inversion without rearrangement, so it is the general-purpose choice for a stereochemically defined bromide.

Step 2 — stoichiometry. 1 mol PBr3 per 3 mol ROH:

3 CH3CH2CH2CH2OH + PBr3 ⟶ 3 CH3CH2CH2CH2Br + H3PO3

Step 3 — dimensional analysis. Convert moles of alcohol to moles of PBr3, then to grams (molar mass of PBr3 = 30.97 + 3 × 79.90 = 270.67 g/mol):

0.250 mol ROH × 1 mol PBr33 mol ROH = 0.0833 mol PBr3

0.0833 mol PBr3 × 270.67 gmol = 22.6 g PBr3

Answer. 22.6 g of PBr3 (about 8.9 mL at density 2.85 g/mL) converts 0.250 mol of 1-butanol.

Example 3: Tosylate strategy to install a nitrile with inversion

Problem. Convert (R)-2-butanol into (S)-2-butanenitrile (CH3CH2CH(CN)CH3).

Step 1 — tosylation. Treat (R)-2-butanol with TsCl/pyridine. The C–O bond is untouched, so the product is (R)-2-butyl tosylate (retention).

Step 2 — displacement. Cyanide is a good nucleophile; SN2 at the tosylate carbon inverts configuration: (R) becomes (S).

Step 3 — net result. One inversion overall, with no carbocation intermediate and no rearranged byproducts:

(R)-CH3CH2CH(OH)CH3 1. TsCl/pyridine 2. NaCN⟶ (S)-CH3CH2CH(CN)CH3

Key takeaways

  • Hydroxide never leaves on its own; activate –OH by protonation (→ H2O) or tosylation (→ OTs-) before breaking the C–O bond.
  • Dehydration (acid + heat) gives alkenes via E1 for 2°/3° alcohols; products obey Zaitsev's rule; watch for hydride/methyl shifts.
  • HX reactivity: HI > HBr > HCl. 3° → S_N1 (fast, Lucas positive), 2° → S_N1 (slower), 1° → S_N2 (inversion).
  • PBr3 and SOCl2 give alkyl halides without rearrangement (inversion at carbon); 3 mol ROH per 1 mol PBr3.
  • TsCl/pyridine converts –OH to –OTs with retention; the tosylate then undergoes SN2 with inversion.
  • Acid chlorides/anhydrides + alcohol → ester (nucleophilic acyl substitution).
  • Na or NaH converts alcohols to alkoxides; alcohols destroy Grignard reagents, so protect –OH first.
  • Lucas test: 3° cloudy immediately, 2° in ~5 min, 1° no reaction.

Check yourself

5 review questions from the chapter. Try each one, then open the answer.

  1. Why can't hydroxide act as a leaving group, and what two strategies fix that?

    Show answer

    Hydroxide is a strong base (pKa of water ≈ 15.7), so it is a poor leaving group. Strategy 1: protonate –OH to –OH2+ so that water leaves. Strategy 2: derivatize –OH to a tosylate (–OTs), which is an excellent leaving group.

  2. Predict the major product of dehydrating 2-butanol with H2SO4 and heat. Which rule decides the answer?

    Show answer

    2-Butene (more substituted, Zaitsev's rule). Dehydration of 2-butanol gives 2-butene as the major product and 1-butene as the minor product.

  3. An unknown alcohol turns cloudy instantly with Lucas reagent. What does that tell you, and what reaction occurs?

    Show answer

    It is a 3° alcohol. Lucas reagent (ZnCl₂ in concentrated HCl) protonates the –OH; the carbocation is captured by chloride, and the insoluble alkyl chloride clouds the solution. 3° react instantly; 2° take minutes; 1° do not react at room temperature.

  4. How many grams of PBr3 are needed to convert 0.500 mol of cyclohexanol to bromocyclohexane?

    Show answer

    0.500 mol ROH × (1 mol PBr₃ / 3 mol ROH) = 0.1667 mol PBr₃ × 270.67 g/mol = 45.1 g.

  5. (R)-2-Pentanol is converted to its tosylate, then treated with sodium azide. What is the stereochemical outcome of the final product?

    Show answer

    The final product is (S)-2-pentyl azide. Tosylation retains configuration ((R) stays (R)); the S_N2 displacement by azide inverts it to (S).

Keep learning

Ready to build on this? Continue to the next lesson.

Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Dehydration
Loss of water from an alcohol to form an alkene
Zaitsev's rule
The more substituted alkene forms preferentially
Lucas reagent
ZnCl2 in concentrated HCl
Tosylate (OTs)
p-Toluenesulfonate ester of an alcohol
PBr₃
Phosphorus tribromide; converts ROH to RBr (3:1 stoichiometry)
S_N1 / S_N2
Unimolecular/bimolecular nucleophilic substitution
Alkoxide
RO-; conjugate base of an alcohol
Carbocation rearrangement
Hydride or alkyl shift to a more stable cation

Sources & references

  1. openstax.org — Organic Chemistry

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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