Alcohols, Phenols and Ethers · Lesson 6 of 12
Substitution and Dehydration of Alcohols
“Explain leaving-group activation, distinguish substitution from elimination and predict rearranged products without inventing stable primary carbocations.”
• Explain why protonation or reagent activation is needed for OH replacement. • Predict halide products and interpret the Lucas test with its limits. • Distinguish primary concerted dehydration from carbocation pathways. • Track a hydride shift and select the major alkene or halide.
Make a poor leaving group into a better one
An unactivated hydroxyl group would have to depart as OH⁻ in a simple substitution, and hydroxide is a poor leaving group. In acid, oxygen accepts a proton and becomes ROH₂⁺. Breaking its C–O bond now releases neutral water. The leaving group has changed even though the carbon skeleton has not yet changed. A nucleophile can then replace it, or a neighbouring C–H bond can participate in elimination. Reagent, structure and temperature decide between these pathways.
For an unhindered primary alcohol, a halide can attack the saturated carbon while water leaves from the protonated substrate, an SN2 pathway. A tertiary centre is too crowded for ordinary SN2 attack; water loss instead gives a relatively stable carbocation that a halide can capture, an SN1 pathway. Secondary and resonance-stabilised substrates can depend strongly on conditions. If a carbocation forms, rearrangement may change the carbon bearing the halide; never assume all products retain the original OH position.
| Starting alcohol | HBr product in a simple unrearranged case | Reason |
|---|---|---|
| Butan-1-ol | CH₃CH₂CH₂CH₂Br | Primary substitution retains the skeleton |
| 2-Methylbutan-2-ol | (CH₃)₂C(Br)CH₂CH₃ | Tertiary cation capture |
| Propan-2-ol | CH₃CHBrCH₃ | Secondary substitution; condition-dependent pathway |
Other reagents activate oxygen differently. PBr₃ converts many primary and secondary alcohols to bromides through an activated phosphorus-containing leaving group and displacement. SOCl₂ converts alcohols to chlorides; pyridine is often used, and the detailed stereochemical pathway depends on conditions. Do not assign one universal retention or inversion outcome without specifying the reagent environment. Phenolic aryl C–O bonds do not undergo these ordinary alkyl substitutions to give aryl halides.
The Lucas test measures formation of a separate phase
Lucas reagent is concentrated HCl with anhydrous ZnCl₂. ZnCl₂ acts as a Lewis acid that assists C–O activation. Formation of a poorly water-soluble alkyl chloride produces turbidity. For suitable small saturated alcohols at room temperature, tertiary alcohols turn cloudy immediately, secondary alcohols within a few minutes, and ordinary primary alcohols show no prompt turbidity. A slow or negative observation does not mean that a primary alcohol can never form a chloride; heating or other reagents can allow conversion.
This is a kinetic and phase-behaviour comparison under specified conditions. Allylic and benzylic alcohols may react rapidly even when primary because resonance stabilises the developing charge. Poor initial solubility, temperature and mixing can also affect cloudiness. Therefore the test is not a universal substitute for examining structure. “No immediate turbidity” is an observation; “no chemical reaction under all conditions” is an unjustified conclusion.
Problem
Compare butan-1-ol, 2-methylbutan-2-ol and benzyl alcohol in the Lucas test.
- 1.Butan-1-ol is an ordinary primary alcohol and does not give prompt turbidity at room temperature under the standard comparison.
- 2.2-Methylbutan-2-ol is tertiary and rapidly forms a chloride through favourable carbocation formation, giving immediate turbidity.
- 3.Benzyl alcohol is primary by carbon count but has a resonance-stabilised benzylic reaction centre. It can react rapidly and is an exception to using degree alone to infer the Lucas observation.
Dehydration removes water and creates a π bond
In alkene formation, OH is removed from one carbon and H from an adjacent carbon. The latter is called a β hydrogen relative to the OH-bearing carbon. Heat with a suitable protic acid such as H₂SO₄ or H₃PO₄ promotes dehydration; heated alumina is another catalyst. Ethanol gives ethene with concentrated H₂SO₄ at about 443 K. Under milder conditions near 413 K, intermolecular ether formation can compete strongly. Conditions belong with the equation because the same reactant can produce different functional groups.
For secondary and tertiary alcohols, an E1 description is often appropriate: first protonate oxygen; then water leaves to produce a carbocation; finally a weak base such as water removes a β proton and the C–H electron pair forms C=C. The acid catalyst is regenerated. Tertiary alcohols generally dehydrate more readily than comparable secondary alcohols, which react more readily than ordinary primary ones, because the cation-forming barrier is lower for more stabilised centres.
A primary alcohol such as ethanol should not be taught as requiring a freely existing stable ethyl carbocation. After protonation, β-H removal and C–O bond cleavage can occur together in a concerted, E2-like elimination. A base lone pair bonds to β-H, the C–H pair becomes the new C=C bond, and the C–O pair leaves with water. All three electron-pair movements occur in the elimination step. The useful sequence is activation, concerted elimination and acid recovery, rather than an isolated primary-cation intermediate.
Choose the alkene, then check for isomerisation
When more than one β position is available, the more substituted, more stable alkene is often favoured under acid-catalysed conditions, in agreement with the usual Zaitsev trend. In an E1 pathway the carbocation can rearrange before elimination. An initially produced alkene can also undergo acid-catalysed double-bond migration toward a more stable isomer. Therefore distinguish the directly formed dehydration product from the product mixture after continued acid exposure.
Problem
Predict products from 1-methylcyclohexan-1-ol and butan-1-ol under acid-catalysed dehydration conditions.
- 1.The ring alcohol can lose water to give a tertiary cation. Removing a ring β hydrogen gives 1-methylcyclohexene, the more substituted major alkene; an exocyclic methylenecyclohexane may be a competing product.
- 2.For butan-1-ol, direct concerted elimination gives but-1-ene. Under continued hot acid conditions, protonation and deprotonation can move the double bond to but-2-ene, whose more substituted double bond is favoured; trans-but-2-ene is generally more stable than cis.
- 3.Thus a mixture containing substantial or predominant but-2-ene is compatible with acidic conditions, but it should not be justified by assuming a stable free n-butyl cation must first form from the primary alcohol. Product ratios depend on conditions and time.
A hydride shift moves a bond pair, not a bare proton
Rearrangement matters when a more stable carbocation can be obtained by moving a neighbouring group with its bond electron pair. A 1,2-hydride shift moves H with the two electrons of its original C–H bond to an adjacent positively charged carbon. The carbon that loses this bonding pair becomes the new cation. It is not proton loss: a hydride shift changes the charge position while retaining all atoms in the organic intermediate.
Problem
Explain why 3-methylbutan-2-ol can give 2-bromo-2-methylbutane with HBr.
- 1.Write the substrate as CH₃CH(OH)CH(CH₃)CH₃. Protonate OH, then water loss gives CH₃CH⁺CH(CH₃)CH₃, a secondary cation at the original C2.
- 2.The adjacent C3–H bond donates its pair to C2 in a 1,2-hydride shift. The intermediate becomes CH₃CH₂C⁺(CH₃)CH₃, a more stable tertiary cation at the original C3.
- 3.Br⁻ captures that cation to give CH₃CH₂C(Br)(CH₃)CH₃. Numbering the product for lowest substituent locants gives 2-bromo-2-methylbutane. Five carbons are preserved, and Br is on a different original carbon from OH.
- 4.Draw the rearrangement arrow from the neighbouring C–H bond to the cationic carbon. Drawing it from H⁺ would depict a different and incorrect electron movement.
Quick Quiz
Quiz
Why does acid assist replacement of OH by a halide?
Explanation: ROH₂⁺ can lose water, avoiding direct departure of hydroxide.
Which normally gives immediate Lucas turbidity among simple saturated alcohols?
Explanation: Tertiary substitution rapidly produces an immiscible chloride under the usual test conditions.
What is the direct alkene from concerted dehydration of butan-1-ol?
Explanation: β-H removal adjacent to terminal OH initially creates the terminal double bond.
What moves during a 1,2-hydride shift?
Explanation: The pair makes a new C–H bond and relocates the carbocation.
Which condition is associated with ethanol-to-ethene conversion?
Explanation: Hot acid favours dehydration, while a lower temperature can favour ether formation.
Practice Problems
- Compare butan-1-ol and 2-methylbutan-2-ol with HBr and explain whether a free primary cation is needed.
- Write equations for converting butan-1-ol to its bromide with PBr₃ and to its chloride with SOCl₂.
- Describe ethanol dehydration using chemically appropriate electron movements.
- Why may butan-1-ol produce but-2-ene in a hot acidic product mixture?
- Track the hydride shift in 3-methylbutan-2-ol + HBr.
- Why should a rapid Lucas reaction not automatically classify benzyl alcohol as tertiary?
Butan-1-ol gives 1-bromobutane through displacement on an activated/protonated primary centre, ordinarily SN2 without a free primary carbocation. The tertiary alcohol gives 2-bromo-2-methylbutane through favourable water loss and tertiary cation capture. Their same net substitution pattern does not require the same mechanism.
Key Takeaways
• OH is a poor leaving group until protonated or otherwise activated. • Primary and tertiary alcohols often use different halide-substitution pathways. • Lucas turbidity measures formation of a separate chloride phase under specified conditions. • Dehydration removes OH and an adjacent hydrogen to make C=C. • Primary ethanol dehydration need not pass through a free ethyl carbocation. • Secondary and tertiary E1 pathways can include carbocation rearrangements. • Hot acid can isomerise initially formed alkenes. • A hydride shift transfers a C–H bonding pair and relocates positive charge.