Start with toluene, C6H5–CH3. Its CH3 carbon is not part of the ring, but it is bonded straight to a ring carbon. A carbon in that spot is called benzylic: an sp³ carbon attached directly to a benzene ring. The CH2 of ethylbenzene and the CH of isopropylbenzene are benzylic too.
A benzylic carbon behaves unlike any other sp³ carbon in the molecule, and the reason is resonance. When a cation, a radical or an anion forms on the benzylic carbon, its empty orbital, odd electron or lone pair spreads into the ring. That one idea explains every reaction on this page.
The same boundary separates two kinds of OH. An OH on the benzylic carbon makes a benzylic alcohol. It reacts like an ordinary alcohol, but once protonated it loses water much more easily, because the cation left behind is benzylic. An OH on a ring carbon makes a phenol, which behaves like no ordinary alcohol and is the subject of the next section.
Why this one position gets special treatment
Take the cation first. When bromide leaves benzyl bromide, C6H5CH2Br, the CH2 carbon is left with a positive charge right next to the ring’s double bonds. Resonance can then move the charge off that carbon and into the ring.
The first curved arrow below moves the electrons of the nearest ring double bond toward the CH2 carbon. They become a C=CH2 double bond, and the plus moves onto an ortho carbon, the ring carbon next to the point of attachment. Two more arrows move it to the para carbon, across the ring, and then to the other ortho carbon.
So four carbons share the charge: the benzylic carbon, both ortho carbons and the para carbon. The two meta carbons never carry it. The phenoxide ion and the Meisenheimer anion from the SNAr section follow the same pattern: ortho and para to the atom where the charge starts, never meta.
The radical and the anion use the same four carbons. Pull a hydrogen atom off toluene’s CH3 and you get the benzyl radical, whose odd electron spreads into the ring. Pull off a proton instead and you get the benzyl anion, whose lone pair spreads the same way.
Each intermediate gains a great deal from the ring. The table compares each one with the alkyl version from earlier chapters.
| Intermediate | Benzylic version is | Roughly as stable as |
|---|---|---|
| Carbocation | Delocalized over the ring | About a tertiary cation |
| Radical | Delocalized over the ring | A tertiary radical, or better |
| Carbanion | Delocalized over the ring | Far better than an alkyl one |
Substitution at a benzylic position
Benzyl chloride, C6H5CH2Cl, is a primary alkyl halide. An ordinary primary halide such as 1-chloropropane reacts only by SN2. Benzyl chloride reacts well by both SN1 and SN2.
SN1 is fast because the cation is benzylic, and so stabilized by the resonance above. A primary benzylic cation is about as stable as a tertiary alkyl cation, so a primary benzylic halide ionizes where an ordinary primary halide never would. Benzyl chloride even undergoes solvolysis in water or ethanol, where the solvent itself is the nucleophile.
SN2 is fast too, for a different reason. Halfway through an SN2 reaction, the carbon under attack is partly bonded to the nucleophile and partly bonded to the leaving group. At that moment its other three groups lie flat around it, as around an sp² carbon, and a p orbital runs along the line from nucleophile to leaving group. On a benzylic carbon that p orbital lies parallel to the ring’s p orbitals and overlaps with them. The overlap spreads out the charge that builds up in the transition state, and that lowers the barrier.
A benzylic CH2 is also unhindered. Besides the ring it carries only two hydrogens, so nothing blocks the nucleophile’s path to the back of the carbon. The SN2 section of Substitution & Elimination puts the gain at roughly ten to a hundred times the rate of a comparable ordinary substrate.
The conditions also decide the stereochemistry. (R)-(1-Bromoethyl)benzene, C6H5CH(Br)CH3, has a stereocenter at its benzylic carbon. (NBS makes this compound, as a racemate, from ethylbenzene in the next section.) In water, a weak nucleophile, it reacts by SN1. The bromide leaves first, and water can attack the flat cation from either face, so the 1-phenylethanol that forms is racemic. As in the SN1 section, the ion pair can leave a slight excess of the inverted product.
With a good nucleophile in acetone, the same bromide reacts by SN2 instead. The nucleophile attacks from the side opposite the bromine, and the product is inverted. One substrate gives two stereochemical outcomes, chosen by the conditions.
That double ability is rare. Most substrates are good at one mechanism and poor at the other, because what favors one disfavors the other. A benzylic halide is good at both. Its fast SN2 is why benzyl bromide turns up so often as an alkylating agent, a reagent that attaches an alkyl group to a nucleophile. The malonic ester and enamine alkylations in Enolate Chemistry are two places it is used.
Radical bromination, and the reagent that makes it selective
Ethylbenzene has two kinds of C–H on its side chain: two on the benzylic CH2 and three on the CH3. Treat it with NBS and light, and bromine replaces a benzylic hydrogen almost exclusively. The product is (1-bromoethyl)benzene.
Radical halogenation usually gives mixtures, because a radical will take a hydrogen from almost anywhere in a molecule. A benzylic position changes that. Bond dissociation energies, the energy needed to break a bond into two radicals, show why. The radical halogenation section gave about 101 kcal/mol for a primary C–H and 96 for a tertiary one. A benzylic C–H takes only about 90 in toluene, and about 87 in ethylbenzene, where the benzylic carbon is secondary.
The gap, at least 6 kcal/mol even against a tertiary C–H, comes from the stability of the benzylic radical. Bromination is the selective halogenation: its transition state comes late, so nearly all of a difference in bond strength shows up in the barrier. With a gap this large, bromine takes the benzylic hydrogen almost every time.
The reagent is NBS (N-bromosuccinimide) with light or a radical initiator. You met it for allylic bromination in How Reactions Happen. It releases Br2 slowly, which keeps the concentration too low for Br2 to add across a C=C bond by the ionic route.
The ring is not brominated either. Putting a bromine on a ring carbon needs Br2 with a Lewis acid such as FeBr3, which is electrophilic aromatic substitution from Aromatic Chemistry.
One more benzylic bond that breaks easily
Hydrogen over a palladium catalyst normally reduces C=C and C≡C bonds and leaves single bonds alone. A benzylic C–O bond is the exception. Take a benzyl ether, R–O–CH2–C6H5. With H2 and Pd/C, the bond between the oxygen and the benzylic carbon breaks, and a hydrogen caps each end. The products are the free alcohol, R–OH, and toluene. Breaking a single bond with H2 this way is called hydrogenolysis.
The reason is the one running through this page. A benzylic bond is weaker than an ordinary one, because the ring stabilizes the benzylic fragment that forms as the bond breaks.
That makes a benzyl ether a useful protecting group for an alcohol, a group put on to hide the OH during other steps and taken off afterward. Benzyl bromide and a base put it on: the base turns the alcohol into an alkoxide, which attacks the benzylic carbon by SN2 (a Williamson ether synthesis). It survives most acids and bases, including the aqueous acid that removes an acetal, and H2 with Pd/C takes it off under neutral conditions.
The cost is that H2/Pd reduces anything else it can reach in the same step. A C=C bond elsewhere in the molecule is hydrogenated while the benzyl group comes off, so the protecting group and the rest of the molecule have to be compatible.
Oxidation: the whole side chain becomes a carboxyl
Heat propylbenzene with KMnO4 in aqueous base, then add acid. The product is benzoic acid, C6H5COOH: the three-carbon chain has been cut back to one carbon, and that carbon is now a carboxyl. The rule is general. Any alkyl side chain with at least one benzylic hydrogen is cut back to a single carbon and oxidized to –COOH.
tert-Butylbenzene is the exception, and it shows how the reaction starts. Its benzylic carbon carries three methyl groups and no hydrogen. The oxidation begins by attacking a benzylic C–H, so with none to attack, tert-butylbenzene does not react.
Write the workup as its own step. Under the basic reaction conditions the product is the carboxylate ion, C6H5COO−, and the carboxylic acid appears only after H3O+ is added. Hot Na2Cr2O7 in H2SO4 does the same oxidation and gives the acid directly, because the mixture is already acidic.
The ring itself survives. Hot permanganate cleaves an ordinary C=C bond, as in Alkenes & Alkynes, yet it leaves the benzene ring alone. Aromatic stabilization makes the ring inert to conditions that oxidize almost anything else.
Step 1. Work backward from the carboxyl. Electrophilic aromatic substitution cannot put a COOH group on a plain benzene ring, but oxidizing an alkyl side chain can. So the last step is hot KMnO4, then H3O+.
Step 2. When does the nitro group go on? Not last: by then the ring carries a carboxyl, a deactivating meta director. The nitro group has to go on while the ring still carries the methyl, which is an ortho/para director.
Step 3. When does the methyl go on? First. Friedel–Crafts alkylation fails on a ring that carries a nitro group, so the methyl cannot follow the nitration.
The sequence: First, CH3Cl with AlCl3. Benzene is used in excess, so the electrophile meets benzene far more often than toluene, which limits polyalkylation. Second, HNO3 with H2SO4. This gives both the ortho and the para isomer, with more ortho than para, and the para isomer is separated from it. Third, hot KMnO4 and acidic workup. Any other order fails: nitrating first blocks the Friedel–Crafts step.
To make the meta isomer, 3-nitrobenzoic acid, swap the last two steps. Oxidize toluene to benzoic acid first, and the carboxyl then directs the nitro group meta.
What carries forward
Benzylic is the position next to the ring, and everything about it follows from resonance into the ring. Cations, radicals and anions are all stabilized there. Substitution runs by either mechanism, NBS brominates there selectively, and hot permanganate cuts any side chain with a benzylic hydrogen back to a carboxylic acid.
When you plan a synthesis by working backward from the target, a benzoic acid should make you ask which alkylbenzene it could come from. Alkylating and then oxidizing is a common way to get a carboxyl onto a ring.