Alcohols, Phenols and Ethers · Lesson 8 of 12
Electrophilic Substitution in Phenols
“Explain ring activation and use reagent conditions and competing directing groups to predict nitration and bromination products.”
• Explain why OH activates ortho and para ring positions. • Predict phenol nitration and bromination under different conditions. • Use both directing and steric effects in substituted phenols. • Check names against structures when acylated phenols are involved.
The ring is a second reaction site
Phenol can lose its OH proton, but it can also react at the aromatic ring while retaining OH. In electrophilic aromatic substitution, an electrophile replaces a ring hydrogen. The ring π system attacks the electrophile, briefly loses aromaticity in a sigma complex, then loses H⁺ to restore the aromatic system. The net result is substitution, not permanent addition across one drawn benzene double bond.
An OH oxygen lone pair can overlap with the ring π system. Resonance contributors place increased electron density at the two ortho positions and the para position. Attack there allows additional stabilisation of the intermediate through oxygen donation; meta attack lacks that same contributor. Thus OH is strongly activating and ortho/para directing. The oxygen also withdraws inductively because of its electronegativity, but for the ring substitutions discussed here, resonance donation dominates the directing and activating outcome.
Do not confuse two uses of resonance. Phenoxide stabilisation helps explain why phenol can lose H⁺. Ring activation explains where a carbon electrophile or other electrophile attacks. The actual delocalised structure supports both effects, but one is acid–base chemistry at O–H and the other is substitution of C–H. Predict the bond being changed before invoking a memorised rule.
Nitration: dilution and extent matter
Because phenol is activated, dilute nitric acid suffices for mononitration. Both ortho and para positions are available. The ortho product can form an intramolecular hydrogen bond, while the para product associates more through intermolecular hydrogen bonds; this accounts for their steam-separation behaviour. The product orientation and physical separation are therefore connected by structure, but they answer different questions.
An alternative route first sulphonates phenol with concentrated sulphuric acid to give phenol-2,4-disulphonic acid, then uses nitric acid to introduce nitro groups at the activated positions, replacing the sulphonic groups as well. The sulphonic substituents are temporary groups, not additional carbon skeleton. Each installed nitro group withdraws electron density and makes the remaining phenolic proton more acidic; 2,4,6-trinitrophenol is much more acidic than phenol.
Phenol-2,4-disulphonic acid has three ring hydrogens, one OH hydrogen and two SO₃H hydrogens, giving molecular formula C₆H₆O₇S₂. In the final trinitrophenol the ring retains two hydrogens and OH supplies one, giving C₆H₃N₃O₇. Counting hydrogens in acidic groups as well as on the ring is essential for balancing both stages.
Bromination: the medium changes how far substitution proceeds
Bromine in a low-polarity solvent such as CS₂ or CHCl₃ at low temperature gives mainly monobrominated phenols, with para generally favoured and ortho also formed. A Lewis acid such as FeBr₃ is not required because the OH-activated ring is already reactive enough to polarise and react with bromine. Lower temperature and controlled reagent exposure help limit further substitution; they are part of the reaction identity.
The white solid is an organic substituted phenol, not an inorganic bromide precipitate. Its poor water solubility explains the phase change. Three ring hydrogens are replaced by three bromines, so three HBr molecules are formed. Bromine colour loss alone is not a uniquely identifying observation because alkenes and other reactive substances also consume bromine; use the full context, including the precipitate and starting structure.
When two groups direct, draw the ring positions first
In 3-methylphenol, number OH as C1 and methyl as C3. Both groups direct an incoming electrophile toward C2, C4 and C6. Position C2 lies between OH and methyl and is crowded. C4 is para to OH and ortho to methyl; C6 is ortho to OH and para to methyl. Mononitration therefore favours the less crowded jointly directed C4/C6 possibilities rather than treating every free ring site as equivalent. Exact ratios require conditions and data; directing rules predict a set of favoured sites, not a universal numerical mixture.
Problem
Identify the principal positions indicated for nitration of 3-methylphenol.
- 1.Label the original ring before placing nitro: OH at C1 and CH₃ at C3. The less crowded common directing sites are C4 and C6.
- 2.The two favoured mononitro connectivities are 3-methyl-4-nitrophenol and, after preferred renumbering, 5-methyl-2-nitrophenol for substitution at the original C6.
- 3.Putting nitro at both original C4 and C6 gives 5-methyl-2,4-dinitrophenol after preferred numbering. A first nitro group is meta directing and deactivating; the remaining C4/C6 site is also meta to that nitro, reinforcing the illustrated pattern.
- 4.The original carbon labels are useful for reaction tracking, while the final product name may number in the opposite direction. Relabelling a ring must not move an actual substituent.
An ester linkage must be read in the correct direction
Phenyl methanoate is C₆H₅–O–C(=O)H. Phenyl ethanoate is C₆H₅–O–C(=O)CH₃. They differ by one carbon. Both attach the ring through oxygen, so the –OC(=O)R group remains ortho/para directing, although its oxygen donation is reduced by competition with the adjacent carbonyl. This differs from a ring directly bonded to a carbonyl group, Ar–C(=O)R, which is generally meta directing.
Problem
Predict the orientation of mononitration of phenyl methanoate and distinguish it from phenyl ethanoate.
- 1.For phenyl methanoate, retain H on the acyl carbon: H–C(=O)–O–C₆H₅. Nitration gives chiefly the para orientation in the usual steric comparison, with an ortho product also possible.
- 2.The para product is 4-nitrophenyl methanoate, HCOO–C₆H₄–NO₂, with NO₂ para to the O-formyl substituent. Do not replace HCO with CH₃CO; that would create a different ester.
- 3.Phenyl ethanoate gives the corresponding nitrophenyl ethanoate products. Similar directing effects do not make their names or formulas interchangeable.
Quick Quiz
Quiz
Why does OH direct ring substitution to ortho and para?
Explanation: Resonance donation increases reactivity and supports the relevant intermediates.
What is observed with phenol and bromine water?
Explanation: Strong activation allows three substitutions and formation of a poorly soluble organic solid.
Which condition favours monobromination over the aqueous tribromo outcome?
Explanation: Reaction medium and controlled conditions affect the extent of bromination.
Which structure is phenyl methanoate?
Explanation: The methanoate acyl carbon carries H, whereas ethanoate carries CH₃.
Which position in 3-methylphenol is crowded between OH and methyl?
Explanation: OH at C1 and CH₃ at C3 flank C2, reducing its accessibility despite favourable directing.
Practice Problems
- Explain why phenol brominates without the Lewis acid needed for ordinary benzene bromination.
- Write balanced equations and conditions for mono- and tribromination of phenol.
- Explain why dilute and concentrated nitric acid do not represent the same phenol transformation.
- Predict favoured sites for mononitration and dinitration of 3-methylphenol.
- A drawing labelled nitrophenyl methanoate contains –OC(=O)CH₃. Identify and repair the error.
- Distinguish ortho/para direction by –OC(=O)CH₃ from meta direction by –C(=O)CH₃.
The OH oxygen donates electron density into the ring and stabilises the substitution intermediate. The activated ring can react with polarised Br₂ without added FeBr₃. This is greater ring reactivity, not loss of OH or cleavage of the ring.
Key Takeaways
• Phenol undergoes substitution at ring C–H while retaining OH. • OH resonance donation activates the ring and favours ortho/para attack. • Dilute nitric acid gives mononitrophenols; extensive nitration gives the trinitro derivative. • Cold low-polarity bromination and bromine water give different extents of substitution. • Bromine water produces a white organic tribromophenol precipitate. • Combined directing groups must be considered together with steric crowding. • Phenyl methanoate and phenyl ethanoate have different acyl groups. • Read the atom attached to the ring before assigning a directing effect.