Alcohols, Phenols and Ethers · Lesson 3 of 12
Preparing Alcohols
“Choose between alkene addition, carbonyl reduction and carbon–carbon bond formation to obtain a specified alcohol.”
• Predict hydration and hydroboration–oxidation products with their conditions. • Explain electron movement in acid-catalysed alkene hydration. • Select reducing agents for carbonyl compounds, acids and esters. • Design Grignard syntheses and account for carbon atoms and water-sensitive steps.
Choose the method from the carbon skeleton
Preparing an alcohol requires more than finding a reagent that contains oxygen. Ask whether the starting carbon skeleton should remain intact or gain a new carbon group, and identify which carbon must bear OH. Addition of water across C=C preserves the skeleton but may place OH in different positions. Carbonyl reduction preserves the C–C framework while changing C=O to C–OH. Grignard addition constructs a new C–C bond before the oxygen is protonated. These three decisions organise most alcohol preparations in this chapter.
Acid-catalysed hydration: follow the electrons
An alkene reacts with water in dilute aqueous acid to form an alcohol. For propene, the principal product is propan-2-ol. The π bond is electron-rich and hydronium supplies an electrophilic proton. In unsymmetrical alkenes, the usual Markovnikov orientation places H on the carbon already carrying more hydrogens and OH on the more substituted carbon, subject to carbocation stability and possible rearrangement. This is an outcome of the pathway, not a rule about isolated OH⁻ attacking an unactivated double bond.
Step 1: the alkene π-electron pair bonds to H of H₃O⁺, while that hydronium O–H bond returns its electron pair to oxygen. Addition of H to the terminal carbon of propene leaves a secondary carbocation at the middle carbon, rather than a less stable primary one at the end. Formal charge is conserved: neutral alkene plus hydronium gives a carbocation plus water. A curved arrow starts at an electron pair or bond, never at a positive charge.
Step 2: a lone pair on water attacks the carbocation carbon, making a C–O bond. Oxygen now has three bonds and positive formal charge, giving CH₃CH(OH₂⁺)CH₃. Step 3: another water molecule removes an oxygen-bound proton; the O–H bond electrons stay on the alcohol oxygen. Neutral propan-2-ol and H₃O⁺ are produced. Acid is regenerated, and the net added atoms are H and OH from water.
Problem
Explain hydration of ethene and use an alkene to prepare 1-phenylethanol.
- 1.Ethene has equivalent double-bond ends. Proton transfer and water addition, followed by deprotonation, give CH₃CH₂OH regardless of which end is labelled first. A highly unstable ethyl cation drawing is a simplified way to show charge development, not proof of a long-lived freely existing intermediate.
- 2.For 1-phenylethanol, start from styrene, C₆H₅CH=CH₂. Dilute aqueous acid hydration gives C₆H₅CH(OH)CH₃: proton addition leaves a resonance-stabilised benzylic cation at the ring-adjacent carbon.
- 3.The two-carbon side chain is retained. Do not choose benzyl alcohol, C₆H₅CH₂OH, which has only one side-chain carbon.
Hydroboration–oxidation reverses the usual orientation
First treat the alkene with a borane reagent, such as BH₃ in an ether solvent or diborane, B₂H₆. Then add hydrogen peroxide in aqueous NaOH. In the initial addition, B attaches predominantly to the less substituted carbon and H to the other carbon. The two bonds form together on the same face in a concerted syn addition, without a free carbocation. Oxidation replaces C–B by C–O while retaining the configuration at that carbon. The overall outcome is anti-Markovnikov hydration; carbocation rearrangements are avoided.
Writing only “water added” would hide the regiochemical difference: propene plus dilute acid gives propan-2-ol, whereas sequential borane and alkaline peroxide gives propan-1-ol. Reagent numbering is chronological. Peroxide and water are not added at the beginning of a water-sensitive organoborane preparation. For a target with terminal CH₂OH on a suitable alkene skeleton, hydroboration–oxidation is often the natural choice.
Reduction adds hydrogen without a new carbon group
An aldehyde RCHO becomes RCH₂OH; a ketone RC(=O)R′ becomes RCH(OH)R′. Catalytic hydrogenation uses H₂ with finely divided Ni, Pt or Pd under suitable conditions. NaBH₄ is a useful selective reagent for ordinary aldehydes and ketones; LiAlH₄ is stronger and is used in dry ether followed by a separate aqueous work-up. In a hydride-transfer description, hydride attacks the electrophilic carbonyl carbon and the C=O π electrons move to oxygen. Protonation of the resulting alkoxide supplies the O–H bond.
| Substrate | Transformation | Suitable reagent and important limit |
|---|---|---|
| RCHO | RCH₂OH | NaBH₄; LiAlH₄ then work-up; or H₂/metal |
| RC(=O)R′ | RCH(OH)R′ | Same classes of carbonyl reducing agents |
| RCOOH | RCH₂OH | LiAlH₄, dry ether, then water; NaBH₄ is not the ordinary choice |
| RCOOR′ | RCH₂OH + R′OH | LiAlH₄ then hydrolysis; appropriate catalytic hydrogenation is also possible |
A carboxylic acid or ester needs a stronger reduction than an ordinary aldehyde or ketone. For an ester, follow both carbon groups: the acyl carbon becomes the CH₂OH carbon, while the alkoxy group emerges as R′OH. Reducing ethyl ethanoate therefore yields ethanol from each fragment, not a four-carbon alcohol. A catalyst is essential for hydrogenation, and one should not assume identical conditions or selectivity for every reducible functional group.
Problem
A cyclohexanone ring has a –CH₂COOCH₃ side chain next to its carbonyl carbon. Predict the usual NaBH₄ product. Also reduce CH₃CH₂CH(CH₃)CHO.
- 1.Locate the ketone C=O in the ring and the ester C=O in the side chain. Under ordinary NaBH₄ conditions the ring ketone is reduced to a secondary alcohol while the ester remains intact. An unspecified starting stereochemistry does not justify a single asserted stereoisomer.
- 2.The aldehyde CH₃CH₂CH(CH₃)CHO becomes CH₃CH₂CH(CH₃)CH₂OH, 2-methylbutan-1-ol. Number the product from CH₂OH.
- 3.In both cases C–C connectivity is retained. Selectivity follows the reagent and carbonyl class, not a blanket instruction to reduce every oxygen-containing group.
Grignard addition builds a carbon–carbon bond
In RMgX, the C–Mg bond is polarised toward carbon, so the R group behaves as a carbon nucleophile. It attacks the carbonyl carbon; the carbonyl π bond moves to oxygen and gives a magnesium alkoxide. Only after this addition is complete is water or dilute acid added to protonate oxygen. Dry ether is used for the addition because water, alcohols and acids would protonate the carbon reagent instead. For example, CH₃MgBr + H₂O → CH₄ + Mg(OH)Br consumes the reagent without constructing the intended C–C bond.
| Carbonyl partner | After RMgX, then hydrolysis | Alcohol class |
|---|---|---|
| Methanal, HCHO | RCH₂OH | Primary, for a carbon-containing R group |
| Other aldehyde, R′CHO | R′CH(OH)R | Secondary |
| Ketone, R′C(=O)R″ | R′C(OH)(R)R″ | Tertiary |
These formulas show where each carbon comes from. Methanal contributes one carbon, so ethylmagnesium chloride plus methanal produces a three-carbon alcohol, propan-1-ol. Propanone contributes the future OH carbon and its two methyl substituents; addition of methylmagnesium bromide creates the third methyl substituent of 2-methylpropan-2-ol. The alcohol degree is determined by the carbonyl partner and the groups now bonded to the former carbonyl carbon.
Problem
Find the products of butanal hydrogenation, propene hydration in dilute acid, and propanone with CH₃MgBr followed by hydrolysis.
- 1.Butanal CH₃CH₂CH₂CHO + H₂ over a suitable metal catalyst gives CH₃CH₂CH₂CH₂OH, butan-1-ol. Reduction preserves four carbons.
- 2.Propene + water in dilute acid gives CH₃CH(OH)CH₃, propan-2-ol, by the more favourable secondary-cation hydration orientation.
- 3.Propanone + CH₃MgBr in dry ether gives (CH₃)₃COMgBr. Subsequent hydrolysis gives (CH₃)₃COH + Mg(OH)Br, 2-methylpropan-2-ol. Three starting carbonyl carbons plus one reagent carbon give four product carbons.
Problem
Prepare 2-methylpropan-1-ol and cyclohexylmethanol using methanal.
- 1.Write each target as RCH₂OH. The first requires R=(CH₃)₂CH–, so use isopropylmagnesium bromide plus HCHO in dry ether, then aqueous work-up.
- 2.Cyclohexylmethanol is C₆H₁₁CH₂OH, so use cyclohexylmagnesium bromide plus HCHO, then water. Methanal supplies the side-chain CH₂OH carbon; the ring remains unchanged.
- 3.If the aim were cyclohexanol instead, this route would be wrong because it adds an extra carbon. Reverse planning must identify the actual OH-bearing carbon.
An alternative preparation of a suitable primary alcohol uses SN2 substitution of a primary alkyl halide by aqueous hydroxide: C₆H₁₁CH₂Br gives cyclohexylmethanol and CH₃(CH₂)₄Br gives pentan-1-ol. The halogen must be on the carbon that is to receive OH. Aryl halides do not obey this ordinary primary-alkyl substitution rule.
Selecting alkenes for branched and cyclic targets
Problem
Choose alkenes for 1-methylcyclohexan-1-ol, 4-methylheptan-4-ol, pentan-2-ol and 2-cyclohexylbutan-2-ol.
- 1.Use 1-methylcyclohexene for the ring alcohol. Acid hydration favours OH on the methyl-bearing tertiary carbon.
- 2.Use CH₃CH₂CH=C(CH₃)CH₂CH₂CH₃, 4-methylhept-3-ene, for 4-methylheptan-4-ol. Protonation leaves the tertiary centre at C4, which water attacks.
- 3.Use pentan-1-ene for pentan-2-ol. Its usual acid hydration orientation gives the secondary alcohol at C2.
- 4.Use C₆H₁₁C(CH₃)=CHCH₃, 2-cyclohexylbut-2-ene, for 2-cyclohexylbutan-2-ol. The appropriate protonation leaves the tertiary centre bearing the ring and methyl; water addition supplies OH there. These are conventional major-product routes, not claims of guaranteed single products under every condition.
Quick Quiz
Quiz
Which sequence converts propene mainly to propan-1-ol?
Explanation: Hydroboration places B at the terminal carbon; oxidation replaces C–B by C–OH.
What does NaBH₄ ordinarily do to a ketone containing an ester elsewhere?
Explanation: Ordinary NaBH₄ conditions distinguish the more reactive ketone from the ester.
Methanal plus ethylmagnesium chloride, then water, gives which alcohol?
Explanation: Methanal contributes one carbon and the ethyl group contributes two.
Why is water excluded during Grignard addition?
Explanation: Water converts RMgX to RH instead of allowing carbon–carbon bond formation.
What is regenerated in acid-catalysed hydration?
Explanation: Final deprotonation returns the acid catalyst.
Practice Problems
- Prepare propan-1-ol and propan-2-ol from propene and explain the different orientations.
- Choose a carbonyl compound and Grignard reagent for 2-methylbutan-2-ol.
- Reduce ethyl propanoate and state both organic products.
- Describe the electron-pair movements in hydration of propene.
- Prepare cyclohexylmethanol in two distinct ways.
- Predict how to prepare pentan-2-ol and 1-methylcyclohexan-1-ol from alkenes.
- Why does ordinary NaBH₄ reduce butanal but not serve as the standard reagent for butanoic acid?
For propan-1-ol use borane in an ether solvent, then alkaline H₂O₂; the less substituted carbon receives B and then OH. For propan-2-ol use dilute aqueous acid; the orientation producing the more stable secondary carbocation is favoured. Both retain three carbons, but only the first avoids a carbocation pathway.
Key Takeaways
• Acid hydration and hydroboration–oxidation can give different positional alcohols from one alkene. • Acid hydration uses protonation, water attack and deprotonation; acid is regenerated. • Hydroboration is concerted, syn and free of ordinary carbocation rearrangement. • Aldehydes reduce to primary alcohols and ketones to secondary alcohols. • LiAlH₄ reduces acids and esters; ordinary NaBH₄ is selective for aldehydes and ketones. • Grignard addition makes a new C–C bond before a separate protonation step. • Dry conditions prevent proton donors from consuming the Grignard reagent. • Track each carbon and the eventual OH carbon when designing a synthesis.