Source: Chapter 9 Lecture Notes
Tags: alkoxides, alkyloxoniums, strong bases, strong acids, carbocation rearrangement, alcohols to alkyl halides, SN1, SN2, E1, E2, inorganic esters, PBr3, SOCl2, tosylate, mesylate, leaving groups, organic chemistry
Difficulty: Intermediate Prerequisites: Familiarity with SN1, SN2, E1, and E2 mechanisms from earlier chapters. You should be comfortable with nucleophiles, electrophiles, and leaving group concepts before tackling this material.
Chapter 9 is about transforming the hydroxyl group (–OH) on alcohols into something more useful for reactions. Alcohols are everywhere in organic chemistry, but the –OH group is a poor leaving group on its own, which limits what you can do with it directly. This chapter teaches you the toolkit for activating that –OH: making alkoxides with strong bases, making alkyloxoniums with strong acids, converting alcohols to alkyl halides through several reagent pathways, and understanding when and why carbocations rearrange. If you have not revised SN1/SN2/E1/E2 from earlier chapters, do that first.
Strong bases turn alcohols into alkoxides (strong nucleophiles/bases). Strong acids turn alcohols into alkyloxoniums (good electrophiles with good leaving groups). To convert an alcohol to an alkyl halide, you can use strong acids (HX), phosphorus trihalides (PBr₃/PCl₃), thionyl chloride (SOCl₂), or sulfonyl chlorides (TsCl/MsCl). The choice of reagent controls mechanism, stereochemistry, and whether rearrangement can occur.
Alkoxide
The conjugate base of an alcohol, formed by deprotonation with a strong base (e.g. NaH, LDA). Carries a negative charge on oxygen. Think of it as the "activated" form of an alcohol that can act as a strong nucleophile or a strong base.
Alkyloxonium ion
The conjugate acid of an alcohol, formed by protonation with a strong acid (e.g. HBr, H₂SO₄). The oxygen bears a positive charge and water becomes the leaving group. In simple terms, this is an alcohol that has been turned into a good electrophile.
Leaving group (LG)
An atom or group that departs with the bonding electrons during a reaction. –OH is a bad leaving group; –OH₂⁺ (water), –Br, –Cl, and inorganic esters are good leaving groups.
Inorganic ester
A compound formed when an alcohol reacts with an inorganic acid derivative (PBr₃, SOCl₂, TsCl, etc.). The –OH is replaced by a much better leaving group without going through a carbocation. Think of it as a way to "disguise" the –OH as something easy to kick out.
Phosphorus tribromide (PBr₃)
A reagent that converts alcohols to alkyl bromides via an SN2 mechanism. Produces inversion of configuration. No carbocation is formed, so no rearrangement occurs.
Thionyl chloride (SOCl₂)
A reagent that converts alcohols to alkyl chlorides. Used with pyridine as a base. The mechanism involves formation of an inorganic ester intermediate, then an intramolecular substitution step. Produces inversion of configuration when pyridine is present.
Tosylate (OTs) / Mesylate (OMs)
Sulfonate ester leaving groups installed by reacting an alcohol with TsCl or MsCl. Stereochemistry of the C–O bond is retained during formation. The resulting sulfonate is an excellent leaving group for subsequent SN2 or SN1 reactions.
Carbocation rearrangement
A shift of a hydrogen atom (hydride shift) or an alkyl group (alkyl shift) to an adjacent carbocation centre to form a more stable carbocation. Occurs in any mechanism that passes through a carbocation intermediate (SN1 and E1 pathways).
Strong bases generate alkoxides
Reagents: NaH, LDA, NaNH₂, and similar strong bases
The base deprotonates the alcohol's O–H bond
Product: an alkoxide ion (RO⁻) plus H₂ gas (when using NaH)
NaOH is not strong enough to fully deprotonate most alcohols, so it does not cleanly generate alkoxides
Result: alkoxides are strong nucleophiles and strong bases, which feeds into SN2 and E2 reactivity
Strong acids create alkyloxoniums
Reagents: HX (HBr, HCl, HI), H₂SO₄, HNO₃, H₃PO₄
The acid protonates the –OH group, converting it to –OH₂⁺
Water is now the leaving group (a good leaving group)
The resulting alkyloxonium is a good electrophile
Opens up SN1, SN2, and E1 pathways, plus a new "reactivity pattern" involving inorganic esters
Primary (1°) alcohols + HX → SN2-like
Cl⁻ from HCl is a weak nucleophile and a stable, weak base
The reaction is concerted (no carbocation intermediate), so inversion of stereochemistry occurs
SN2-like because the halide attacks the backside of the carbon as water departs
Secondary (2°) and tertiary (3°) alcohols + HX → SN1 / E1
Protonation first gives the alkyloxonium
Loss of water gives a carbocation (2° or 3° leaving group)
Conditions are acidic, which favours carbocation formation
All the usual rules for SN1 vs E1 apply: carbocation stability, possibility of rearrangement
Rearrangements are possible wherever a carbocation forms
Alcohols treated with H₂SO₄ at elevated temperature (e.g. 165 °C) undergo elimination (dehydration) to form alkenes
H₂SO₄ protonates the –OH, then water leaves to give a carbocation
The bisulfate ion (HSO₄⁻) is a non-nucleophilic, weak, large, stable base
It abstracts a proton to give the alkene product (E1 pathway)
Orbital basis for rearrangement
In a carbocation, the empty p orbital on the cationic carbon sits next to a C–H or C–C bond on the adjacent carbon
If that adjacent bond is aligned (anti-periplanar) with the empty p orbital, hyperconjugation weakens it, making a shift favourable
1,2-Hydride shift
A hydrogen migrates with its bonding electrons from an adjacent carbon to the carbocation centre
The transition state involves partial bonding of the hydrogen to both carbons
Rehybridisation occurs: the original carbocation centre becomes sp³, and the carbon that lost the hydrogen becomes sp² (the new carbocation)
Drives the system towards a more stable carbocation (e.g. 2° → 3°)
1,2-Alkyl (methyl) shift
An alkyl group migrates instead of a hydrogen
Concerted with departure of the leaving group in some cases
Look for situations where a shift would produce a more stable carbocation, and expect rearrangement to happen immediately if it can
The central problem: –OH is a bad leaving group; –Cl and –Br are good leaving groups. The whole section is about how to make the swap.
Method 1: Strong acids (HX)
Harsh reaction conditions
Anti-periplanar (ABC) geometry required for E2 side reactions
Rearrangements are possible (because carbocations form with 2°/3° substrates)
Best suited for simple cases where stereochemistry and rearrangement are not concerns
Method 2: Inorganic esters → SN2 pathway
General advantages of the inorganic ester route:
Mild conditions
No carbocation forms, so no rearrangement
Inversion of configuration (SN2)
Inorganic esters are good leaving groups
A. PBr₃ or PCl₃ (phosphorus trihalides)
Alcohol reacts with PBr₃ to form an inorganic ester (alkyl dibromophosphite)
Bromide ion (Br⁻) is released as a by-product and acts as the nucleophile
SN2 attack on the carbon gives the alkyl bromide with inversion
PBr₃ is a Lewis acid
The HO–P–Br by-product can react two more times (PBr₃ can convert up to three equivalents of alcohol)
B. SOCl₂ (thionyl chloride)
Alcohol reacts with SOCl₂ to form a chlorosulfite ester (inorganic ester)
Pyridine is used as the base to scavenge HCl
Mechanism proceeds through addition of the alcohol to SOCl₂, then intramolecular elimination releases SO₂ gas and Cl⁻
Cl⁻ attacks via SN2 (with pyridine present), giving inversion
Gaseous SO₂ escapes, driving the reaction forward
Produces: alkyl chloride + SO₂ (g) + Cl⁻ (captured by pyridine)
C. Sulfonyl chlorides (TsCl, MsCl) – non-SN2 at the carbon
TsCl (tosyl chloride) or MsCl (mesyl chloride) react with the –OH to form a sulfonate ester
The C–O bond is not broken during this step, so stereochemistry at carbon is retained
The sulfonate (OTs or OMs) is an excellent leaving group for a subsequent SN2 or SN1 reaction
Pyridine is used as the base
Common sulfonyl groups: tosylate (OTs), triflate (OTf), mesylate (OMs)
When you need to convert an alcohol to a different functional group with specific stereochemistry, the choice of reagent determines the outcome.
Example: converting an alcohol to a nitrile with inversion
Option 1 (HBr, then NaCN): HBr can cause rearrangement and loss of stereochemistry. Not ideal.
Option 2 (PBr₃ in DCM, then NaCN in THF): two sequential SN2 steps. PBr₃ gives inversion (alcohol → bromide), NaCN gives a second inversion (bromide → nitrile). Two inversions = retention of original configuration. This is the controlled approach.
Option 3 (TsCl, then NaCN): TsCl retains configuration at carbon (just swaps the leaving group). Then NaCN gives one inversion (tosylate → nitrile). Net result: one inversion overall.
The point: mechanism matters. Counting inversions lets you predict (and control) the stereochemical outcome.
Students often assume NaOH is strong enough to generate alkoxides cleanly. It is not; you need a stronger base such as NaH or LDA.
Students frequently forget that rearrangement is possible any time a carbocation forms. If you are using HX with a 2° or 3° alcohol, always check for rearrangement.
A common mistake is thinking PBr₃ and SOCl₂ proceed through carbocations. They do not. Both work via inorganic ester intermediates and SN2 displacement, so no rearrangement occurs.
Students mix up retention and inversion when sulfonyl chlorides are involved. TsCl retains configuration at carbon (it only replaces –OH with –OTs). The inversion happens in the next step, when a nucleophile displaces the tosylate.
⚠️ Know which reagents produce inversion, which produce retention, and which can cause rearrangement. This is a classic exam question format: "which reagent gives the correct stereochemical outcome?"
⚠️ PBr₃ and SOCl₂ both avoid carbocations. If the exam asks for conditions that prevent rearrangement, these are your answers.
⚠️ Be able to trace a multi-step synthesis and count inversions to predict the final stereochemistry (as in the alcohol → nitrile example).
⚠️ Dehydration with H₂SO₄ is E1. The bisulfate ion is too bulky and weak to be a good nucleophile, so substitution does not compete.
True or False: NaOH is strong enough to fully deprotonate an alcohol to generate an alkoxide.
Fill in the blank: PBr₃ converts an alcohol to an alkyl bromide with _______ of configuration.
True or False: SOCl₂ reactions proceed through a carbocation intermediate.
Fill in the blank: When TsCl reacts with an alcohol, the stereochemistry at carbon is _______.
True or False: Carbocation rearrangement is possible when treating a 3° alcohol with HBr.
Answers: 1. False. 2. Inversion. 3. False. 4. Retained. 5. True.
Q: A secondary alcohol is treated with HBr. What mechanism(s) operate, and is rearrangement possible?
A: SN1 and E1 mechanisms operate. HBr protonates the –OH to form an alkyloxonium, water leaves to give a 2° carbocation, and then the bromide ion attacks (SN1) or a proton is lost (E1). Rearrangement is possible because a carbocation intermediate forms.
Q: Why is PBr₃ preferred over HBr when you need to preserve stereochemistry?
A: PBr₃ converts the alcohol to an alkyl bromide via an SN2 mechanism (through an inorganic ester intermediate). There is no carbocation, so no rearrangement occurs, and the reaction gives clean inversion of configuration.
Q: You want to convert (R)-2-butanol to (S)-2-butanenitrile. Outline a two-step sequence and explain the stereochemical outcome.
A: Step 1: treat with TsCl and pyridine. This replaces –OH with –OTs while retaining configuration at carbon (still R). Step 2: treat with NaCN (SN2). The cyanide ion attacks the backside, giving inversion (R → S). Net result: one inversion, so the product is (S)-2-butanenitrile.
Q: What is the driving force behind a 1,2-hydride shift in a carbocation?
A: The shift moves the positive charge to a more substituted (more stable) carbon. The C–H bond on the adjacent carbon is aligned with the empty p orbital of the cationic carbon, and migration of the hydride with its electrons produces a more stable carbocation.
Q: In a dehydration reaction with H₂SO₄, why does elimination dominate over substitution?
A: The conjugate base of H₂SO₄ (bisulfate, HSO₄⁻) is large, stable, and non-nucleophilic. It is too weak a nucleophile to attack the carbocation, so it acts only as a base, abstracting a proton to form the alkene.
This material builds directly on the SN1/SN2/E1/E2 framework from earlier chapters. Every decision about which reagent to use depends on those mechanisms. It also sets up Chapter 9's later sections on ether synthesis (Williamson and acidic), since alkoxides are the key nucleophile in Williamson ether synthesis. The concept of protecting groups (covered in Part 2) relies on the ability to selectively activate and deactivate –OH groups using the techniques from this section.
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