Chlorobenzene does not undergo nucleophilic substitution under ordinary conditions because:

The aryl cation would be very unstable and the back side is shielded by the pi cloud

Chlorine is a poor leaving group in all compounds

The aryl cation would be very unstable and the back side is shielded by the pi cloud

Benzene rings are electron-poor

Chlorobenzene is insoluble in water

To prepare tert-butyl methyl ether by Williamson synthesis, the correct pair is:

Sodium tert-butoxide and methyl iodide

Sodium methoxide and tert-butyl chloride

Sodium tert-butoxide and methyl iodide

Neither pair can give this ether

Haloalkanes, Alcohols, Phenols and Ethers

These four families are grouped together because they interconvert constantly and because they share one governing question: is the reagent going to attack carbon or hydrogen? Attack at carbon is substitution; attack at hydrogen is elimination. Almost every "which product predominates" question in this part of the syllabus is asking you to weigh those two, and the factors that decide it — substrate class, nucleophile strength, base bulk, solvent, temperature — are few enough to hold in mind. Phenols are then included as the instructive exception: they look like alcohols but behave quite differently, and the reason is resonance.

SN1 Versus SN2

The two substitution mechanisms compared
Feature	SN1	SN2
Steps	Two — ionisation then attack	One — concerted
Rate law	First order, only in substrate	Second order, in both
Favoured substrate	3° > 2° > 1° — needs a stable carbocation	1° > 2° > 3° — needs an open back side
Stereochemistry	Racemisation, since the cation is planar	Inversion of configuration
Solvent	Polar protic, which stabilises ions	Polar aprotic, which leaves the nucleophile bare
Nucleophile strength	Barely matters	Matters a great deal

🎯 Why SN2 inverts configuration and SN1 racemises

In SN2 the nucleophile must approach from the side opposite the leaving group, because the leaving group's electron cloud blocks the front. As the new bond forms and the old one breaks, the three remaining groups flip through like an umbrella in the wind — the Walden inversion. Every molecule inverts, so a single enantiomer gives a single inverted product. In SN1 the leaving group departs first, giving a flat sp² carbocation with two equivalent faces; the nucleophile attacks either face with roughly equal probability, so the product is a near-racemic mixture. That is why the stereochemical outcome alone tells you which mechanism operated — and why a question reporting inversion has already told you the substrate must have been primary or secondary.

⚠️ Vinyl and aryl halides do neither

Chlorobenzene and vinyl chloride resist both mechanisms, and it is worth knowing why each fails. SN1 is blocked because ionisation would give an aryl or vinyl cation, which is highly unstable — the empty orbital would be in the plane of the ring, unable to receive any resonance help. SN2 is blocked because the pi cloud shields the back side and the ring simply cannot be approached from behind. On top of that, the C–Cl bond has partial double-bond character from resonance, so it is shorter and stronger than in an alkyl halide. Three independent reasons, all pointing the same way, and that is why aryl halides need harsh conditions such as the Dow process at 623 K and 300 atm.

Elimination competes throughout. E2 is concerted and needs a strong base, favours anti-periplanar geometry, and follows Saytzeff orientation — the more substituted, more stable alkene predominates. E1 shares the carbocation intermediate with SN1. Two conditions push the balance towards elimination: a bulky base such as tert-butoxide, which cannot reach carbon and so takes a proton instead, giving the Hofmann (less substituted) alkene; and high temperature, since elimination increases the number of molecules and so is favoured entropically.

Alcohols, and Why Phenols Are Different

Alcohols are made by hydration of alkenes, by reduction of carbonyls with LiAlH₄ or NaBH₄, and by Grignard addition — RMgX plus formaldehyde gives a primary alcohol, another aldehyde a secondary, a ketone a tertiary. They are oxidised by acidified K₂Cr₂O₇ or KMnO₄: primary to aldehyde then acid, secondary to ketone, tertiary not at all under normal conditions, because a tertiary carbon has no hydrogen left to remove.

🎯 Why phenol is a million times more acidic than ethanol

Phenol has pK_a about 10 and ethanol about 16 — a factor of a million in acid strength, and both differences trace to what happens after the proton leaves. The phenoxide ion delocalises its negative charge into the aromatic ring, spreading it over the oxygen and three ring carbons, so it is far more stable than an alkoxide where the charge sits entirely on one oxygen. Since a more stable conjugate base means a stronger acid, phenol wins comfortably. The practical consequence is the standard distinguishing test: phenol dissolves in NaOH but not in NaHCO₃, because it is stronger than water but weaker than carbonic acid, whereas an alcohol dissolves in neither and a carboxylic acid dissolves in both.

Named reactions of phenol
Reaction	Reagent	Product
Reimer–Tiemann	CHCl₃ + NaOH	Salicylaldehyde — the ortho aldehyde
Kolbe–Schmitt	NaOH then CO₂, then acid	Salicylic acid — the aspirin precursor
Bromination in water	Bromine water	2,4,6-tribromophenol — all three at once
Coupling	Diazonium salt in alkali	An azo dye, coupling at para
Colour test	Neutral FeCl₃	Violet colouration

⚠️ Bromine water gives the tribromide, but bromine in CS₂ gives monobromide

Phenol is so strongly activated that in water it takes up three bromines at once, giving 2,4,6-tribromophenol as a white precipitate. In a non-polar solvent such as CS₂ at low temperature, however, the reaction stops at monosubstitution and gives mainly p-bromophenol. The difference is that water assists ionisation of the phenol to the even more reactive phenoxide, whereas CS₂ does not. So a question that specifies the solvent is specifying the answer, and treating "bromination of phenol" as a single reaction with a single product is how candidates lose the mark.

Ethers are prepared by Williamson synthesis, an SN2 reaction of an alkoxide with an alkyl halide — which means the halide must be primary, since a tertiary halide would eliminate instead. That is why tert-butyl methyl ether is made from sodium tert-butoxide plus methyl iodide, and not from methoxide plus tert-butyl chloride. Ethers are otherwise unreactive, which is exactly why they are used as solvents, but they do cleave with hot concentrated HI, and the attack follows whichever side gives the better SN1 or SN2 pathway.

Key takeaways

SN1 favours 3° and racemises; SN2 favours 1° and inverts. The stereochemistry alone identifies the mechanism.
Polar protic solvents favour SN1; polar aprotic solvents favour SN2 by leaving the nucleophile unencumbered.
Aryl and vinyl halides resist both mechanisms — unstable cation, blocked back side, and a shortened C–X bond.
A bulky base gives the Hofmann alkene; ordinary bases give the Saytzeff, more substituted one.
Primary alcohols oxidise to acids, secondary to ketones, tertiary not at all — no hydrogen to remove.
Phenol is far more acidic than an alcohol because phenoxide delocalises its charge into the ring.
Phenol dissolves in NaOH but not NaHCO₃; an acid dissolves in both and an alcohol in neither.
Williamson synthesis is SN2, so the halide must be primary — choose which partner supplies the alkoxide.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Hydrolysis of an optically active 2-bromobutane gives a product with inverted configuration. The mechanism is:
1. SN1
2. SN2
3. E1
4. E2
Show answer
Answer: B — SN2
Clean inversion is the signature of SN2, because the nucleophile must attack from the side opposite the leaving group and the other three groups flip through in a Walden inversion. SN1 would proceed through a planar carbocation attacked equally from both faces and so would give a largely racemic product. Both eliminations would give an alkene rather than a substituted product at all.
Chlorobenzene does not undergo nucleophilic substitution under ordinary conditions because:
1. Chlorine is a poor leaving group in all compounds
2. The aryl cation would be very unstable and the back side is shielded by the pi cloud
3. Benzene rings are electron-poor
4. Chlorobenzene is insoluble in water
Show answer
Answer: B — The aryl cation would be very unstable and the back side is shielded by the pi cloud
Both pathways are closed off. SN1 would require an aryl cation whose empty orbital lies in the ring plane and gets no resonance stabilisation, and SN2 requires back-side attack that the pi cloud physically prevents. Resonance also gives the C–Cl bond partial double-bond character, shortening and strengthening it. Chlorine is a perfectly good leaving group in alkyl halides, so the first option misplaces the cause.
Which reagent dissolves phenol but not ethanol?
1. NaHCO₃ solution
2. NaOH solution
3. Water
4. Dilute HCl
Show answer
Answer: B — NaOH solution
Phenol is acidic enough to be deprotonated by NaOH, forming soluble sodium phenoxide, while ethanol is far too weak an acid to react. NaHCO₃ is a weaker base and does not deprotonate phenol, which is exactly what distinguishes phenol from a carboxylic acid — the acid effervesces with bicarbonate, the phenol does not. This three-way test separates alcohol, phenol and acid with two reagents.
Reaction of 2-bromo-2-methylpropane with sodium tert-butoxide gives mainly:
1. An ether by SN2
2. An alkene by elimination
3. An alcohol by SN1
4. No reaction
Show answer
Answer: B — An alkene by elimination
The substrate is tertiary, so SN2 is sterically impossible, and tert-butoxide is both a strong base and very bulky, so it cannot reach the crowded carbon and abstracts a proton instead. Elimination therefore dominates and gives 2-methylpropene. This combination — tertiary substrate plus bulky base — is the standard setup for pushing a reaction entirely towards elimination.
Phenol treated with bromine water gives:
1. p-bromophenol only
2. 2,4,6-tribromophenol
3. o-bromophenol only
4. No reaction without a catalyst
Show answer
Answer: B — 2,4,6-tribromophenol
The hydroxyl group activates the ring so strongly that in aqueous medium all three available ortho and para positions are brominated at once, precipitating 2,4,6-tribromophenol. Monosubstitution to p-bromophenol requires a non-polar solvent such as CS₂ at low temperature. No Lewis acid catalyst is needed at all, which is itself a measure of how activated phenol is compared with benzene.
Which alcohol cannot be oxidised by acidified K₂Cr₂O₇ under normal conditions?
1. Ethanol
2. Propan-2-ol
3. 2-methylpropan-2-ol
4. Butan-1-ol
Show answer
Answer: C — 2-methylpropan-2-ol
Oxidation of an alcohol requires removal of a hydrogen from the carbon bearing the OH group, and in a tertiary alcohol that carbon carries three alkyl groups and no hydrogen at all. Primary alcohols therefore go to aldehyde and then acid, secondary alcohols stop at the ketone, and tertiary alcohols resist oxidation until conditions harsh enough to break carbon–carbon bonds are used.
The Reimer–Tiemann reaction converts phenol into:
1. Salicylic acid
2. Salicylaldehyde
3. Picric acid
4. Anisole
Show answer
Answer: B — Salicylaldehyde
Chloroform with alkali generates dichlorocarbene, which attacks the ring ortho to the hydroxyl group and after hydrolysis leaves an aldehyde — salicylaldehyde. Salicylic acid comes from the Kolbe–Schmitt reaction with carbon dioxide instead, and these two are the pair most often confused because both give ortho products from phenol.
To prepare tert-butyl methyl ether by Williamson synthesis, the correct pair is:
1. Sodium methoxide and tert-butyl chloride
2. Sodium tert-butoxide and methyl iodide
3. Either pair works equally well
4. Neither pair can give this ether
Show answer
Answer: B — Sodium tert-butoxide and methyl iodide
Williamson synthesis is an SN2 reaction, so the halide must be primary or methyl. Pairing methoxide with tert-butyl chloride puts the leaving group on a tertiary carbon, where the alkoxide acts as a base and eliminates to give 2-methylpropene instead of the ether. Reversing the roles — bulky group as the alkoxide, methyl as the halide — gives clean substitution, which is the general rule for planning any Williamson step.