Organic Chemistry · Alcohols and Phenols
Preparation of Alcohols: A Review
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In 30 seconds
You have already met alcohols many times: as products of alkene hydration (Chapters 7–8), of substitution reactions (Chapter 11), and as the functional group whose reactions fill this chapter. This topic is a checkpoint that gathers every method of making an alcohol into one framework: (1) hydration of alkenes — adding H and OH across a double bond; (2) substitution of alkyl halides — replacing a halogen with OH; and (3) carbonyl routes — reduction or Grignard addition to a C=O group, developed in the next two topics. Seeing these as one toolbox turns a list of reagents into a synthesis strategy.
Why this matters
Alcohols are the most versatile intermediates in organic synthesis — the launch point for alkenes, alkyl halides, ethers, esters, and carbonyl compounds. Industry relies on their preparation: ethanol, the solvent and fuel additive, is made on a massive scale by acid-catalyzed hydration of ethylene, and isopropanol (rubbing alcohol) by hydration of propene. In pharmaceutical synthesis, alcohol groups are installed by reduction and Grignard reactions, and the wrong method can mean a rearranged skeleton or an isomer mixture. For exams, problems that ask "how would you prepare this alcohol?" test exactly the decision framework developed here.
The college version
Core Concepts
The three families of alcohol preparations
Every preparation fits one of three patterns. From alkenes, water is added across the C=C bond (hydration). From alkyl halides, the halogen is replaced by OH (substitution). From carbonyl compounds, a hydride or a carbon group is added to the C=O carbon (reduction or Grignard reaction).
Hydration of alkenes: three complementary methods
Acid-catalyzed hydration (H3O+, water, heat) follows Markovnikov's rule: the π electrons of the alkene attack a proton, forming the more substituted carbocation; water then donates a lone pair to that cation; loss of a proton gives the alcohol. Because a free carbocation is involved, rearrangements (1,2-hydride and 1,2-alkyl shifts) can occur, sometimes changing the skeleton.
Oxymercuration–demercuration Hg(OAc)₂/H₂O then NaBH₄; Markovnikov hydration without rearrangements Full entry → (mercuric acetate, Hg(OAc)2, in water/THF, then sodium borohydride) also gives Markovnikov hydration, but through a cyclic Mercurinium ion The cyclic, three-membered intermediate in oxymercuration Full entry → instead of a free carbocation — so no rearrangements occur.
Hydroboration–oxidation BH₃·THF then H₂O₂/NaOH; anti-Markovnikov, syn addition of water Full entry → (borane, BH3 · THF, then alkaline hydrogen peroxide) gives anti-Markovnikov hydration: the OH lands on the less substituted carbon. Addition is syn, no carbocation forms, and terminal alkenes are converted cleanly to primary alcohols.
Substitution of alkyl halides
Hydroxide displaces halide from primary halides by SN2 — one concerted step with inversion of configuration. Tertiary halides react by SN1: ionization to a carbocation, then attack by water (solvolysis). Secondary halides sit in between and give mixtures, especially because hydroxide is a strong base, so E2 elimination competes. This route is most useful when the alkyl halide is already in hand.
Preview: the carbonyl routes
Reduction (next topic) delivers hydride to the carbonyl carbon: aldehydes become primary alcohols, ketones become secondary alcohols. The Grignard reaction (Topic 5) delivers a carbon group instead of hydrogen, building a new C–C bond — the only alcohol preparation that grows the carbon skeleton.
Choosing a method: a decision framework
Ask four questions before committing to a route. (1) What is the substitution of the alcohol carbon? Primary alcohols come from terminal alkenes (hydroboration–oxidation), primary halides (SN2), or aldehydes (reduction); secondary from alkene hydration or ketone reduction; tertiary from a branched alkene or a Grignard reaction with a ketone. (2) Markovnikov or anti-Markovnikov? Only hydroboration–oxidation gives anti-Markovnikov. (3) Can a carbocation rearrange? If the alkene is allylic, benzylic, or could form a secondary cation next to a branch, use oxymercuration–demercuration. (4) Are you building the skeleton? Only the Grignard reaction adds carbon.
How It Works / Step-by-Step Process
- Draw the target alcohol and mark the carbon bearing the OH; count its hydrogens (primary, secondary, or tertiary).
- Decide whether the skeleton is already fixed (hydration, substitution, or reduction) or must be built (Grignard).
- Match the method: terminal alkene plus anti-Markovnikov? → hydroboration–oxidation. Markovnikov with rearrangement risk? → oxymercuration–demercuration. Primary halide? → SN2 with hydroxide. Ketone plus a new C–C bond? → Grignard.
- Before committing, check for Carbocation rearrangement A 1,2-hydride or 1,2-alkyl shift converts a less stable cation to a more stable one Full entry → hazards and stereochemistry expectations.
Common Confusions
| Do Not Confuse | With | Difference |
|---|---|---|
| Markovnikov addition | Anti-Markovnikov addition | Where the OH lands: the more substituted versus the less substituted alkene carbon |
| Oxymercuration–demercuration | Acid-catalyzed hydration | Both are Markovnikov, but only hydration passes through a free carbocation and can rearrange |
| Hydration of an alkene | Hydrogenation of an alkene | Hydration adds H and OH (alkene → alcohol); hydrogenation adds H and H (alkene → alkane) |
| Hydroxide as a nucleophile | Hydroxide as a base | With primary halides SN2 wins; with secondary/tertiary or bulky substrates, E2 elimination competes |
| SN1 pathway | SN2 pathway | SN1: carbocation, racemization, favors tertiary; SN2: one step, inversion, favors primary |
| Grignard addition | Reduction | Both turn a ketone into an alcohol, but Grignard adds a carbon group (new C–C bond) while reduction adds H |
Eli explains
The same idea, in plain words
Explain it like I’m 10
An alcohol is a molecule with an "OH handle". Making one is like attaching a handle to a cup: glue it on where the recipe says (hydration), swap a handle that is already there for an OH handle (substitution), or build a brand-new cup part and glue the handle onto that (Grignard). The trick is picking the method that puts the handle exactly where you want it — and oxymercuration never accidentally breaks the cup while you glue.
Worked example
Example 1: Two alkenes, one alcohol — spotting a rearrangement
Acid-catalyzed hydration of 2-methyl-2-butene, (CH3)2C=CHCH3, protonates directly to the tertiary carbocation (CH3)2C+CH2CH3 (Markovnikov); water adds and deprotonation gives 2-methyl-2-butanol, (CH3)2C(OH)CH2CH3. Now hydrate 3-methyl-1-butene, (CH3)2CHCH=CH2: protonation gives the secondary carbocation (CH3)2CHCH+CH3, which instantly undergoes a 1,2-hydride shift to the same tertiary cation. Both alkenes give the same alcohol — acid-catalyzed hydration erased the difference. Oxymercuration–demercuration of either alkene gives the Markovnikov alcohol without scrambling, which is why it is preferred when rearrangement is possible.
Example 2: Terminal alkene → primary alcohol
1-Hexene, CH3CH2CH2CH2CH=CH2, treated with BH3 · THF followed by H2O2/NaOH, gives 1-hexanol, CH3(CH2)4CH2OH. Boron adds to the less substituted carbon (sterically controlled), and the peroxide step replaces C–B with C–O with retention: net anti-Markovnikov, syn hydration, no carbocation intermediate. Contrast: acid-catalyzed hydration of the same alkene gives 2-hexanol (Markovnikov).
Example 3: Choosing a route to a secondary alcohol
Target: 2-butanol, CH3CH(OH)CH2CH3. Route A: oxymercuration–demercuration of 1-butene, CH2=CHCH2CH3 — Markovnikov OH at C2, no rearrangement, clean. Route B: SN2 of 2-bromobutane with hydroxide — works, but E2 elimination to 2-butene competes because hydroxide is a strong base. Route C: NaBH₄ reduction of 2-butanone, CH3COCH2CH3 — the mildest, provided you have the ketone. All three deliver the same product; choosing between them is the skill this topic teaches.
Key takeaways
- Acid-catalyzed hydration: Markovnikov, goes through a free carbocation, so rearrangements are possible.
- Oxymercuration–demercuration: Markovnikov but no rearrangement — the safe choice for alkene traps.
- Hydroboration–oxidation: anti-Markovnikov, syn addition, no rearrangement; terminal alkenes → primary alcohols.
- SN2 from primary halides (inversion) and SN1 from tertiary halides (carbocation) both install OH; E2 elimination competes when hydroxide is used.
- Reduction and Grignard addition convert carbonyls to alcohols; only Grignard builds a C–C bond.
- Retrosynthetic habit: identify the alcohol carbon, count its hydrogens, then match the family.
Check yourself
6 review questions from the chapter. Try each one, then open the answer.
Which hydration method gives Markovnikov alcohols without the risk of carbocation rearrangements?
Show answer
Oxymercuration–demercuration (Hg(OAc)₂ in water/THF, then NaBH₄) — Markovnikov via a mercurinium ion, no free carbocation, no rearrangement.
Which reagent combination hydrates a terminal alkene to a primary alcohol?
Show answer
Hydroboration–oxidation: BH₃·THF, then H₂O₂/NaOH — anti-Markovnikov, giving the primary alcohol from a terminal alkene.
Why do acid-catalyzed hydration of 3-methyl-1-butene and of 2-methyl-2-butene give the same alcohol?
Show answer
Both funnel through the same tertiary carbocation: 3-methyl-1-butene forms a secondary cation that undergoes a 1,2-hydride shift to (CH3)2C+CH2CH3, the same intermediate formed directly from 2-methyl-2-butene. Both therefore give 2-methyl-2-butanol.
What four questions should you ask when choosing an alcohol preparation?
Show answer
(1) Is the alcohol carbon primary, secondary, or tertiary? (2) Markovnikov or anti-Markovnikov? (3) Can a carbocation rearrange? (4) Must a carbon–carbon bond be built?
Which preparation method builds a new carbon–carbon bond?
Show answer
The Grignard reaction: the R group of RMgX bonds to the carbonyl carbon, so the product has one more carbon attached than the carbonyl started with.
Why does E2 elimination compete when hydroxide is used to convert a secondary alkyl halide to an alcohol?
Show answer
Because hydroxide is a strong base as well as a nucleophile; with secondary halides the elimination pathway (E2) competes with substitution (SN2).
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- Markovnikov addition
- The OH ends up on the more substituted carbon of the alkene
- Anti-Markovnikov addition
- The OH ends up on the less substituted carbon
- Carbocation rearrangement
- A 1,2-hydride or 1,2-alkyl shift converts a less stable cation to a more stable one
- Mercurinium ion
- The cyclic, three-membered intermediate in oxymercuration
- Oxymercuration–demercuration
- Hg(OAc)₂/H₂O then NaBH₄; Markovnikov hydration without rearrangements
- Hydroboration–oxidation
- BH₃·THF then H₂O₂/NaOH; anti-Markovnikov, syn addition of water
- SN2 substitution
- One-step nucleophilic displacement with inversion of configuration
- SN1 solvolysis
- Ionization to a carbocation, then attack by solvent water
Sources & references
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