📝Class 11–12 Chemistry · Class 12

Alcohols, Phenols and Ethers (Class 12): Reactions and Conversions

Alcohols, Phenols and Ethers (Class 12): Reactions and Conversions
Class 12 · Organic Chemistry

Alcohols, Phenols and Ethers (Class 12): Reactions and Conversions

One idea — how well the negative charge left behind is stabilised — explains the acidity order, the substituent effects and half the reaction outcomes in this chapter.

Class 12 · Organic Chemistry · Boards + JEE + NEET · Published 27 August 2026

In short: Phenol is far more acidic than an alcohol because the phenoxide ion delocalises its negative charge into the ring, while an alkoxide cannot. Electron-withdrawing groups make it more acidic still; electron-donating groups make it less. Almost every acidity question in this chapter is that one sentence applied.

Acidity, explained once and reused

Acidity is about the stability of the conjugate base. Ethoxide holds its charge on a single oxygen. Phenoxide spreads it over the oxygen and three ring carbons by resonance, which is far more comfortable, so phenol gives up its proton much more readily. Add a nitro group and the charge is spread further still; add a methyl group and the ring pushes electron density back, destabilising the anion.

CompoundRelative acidityWhy
EthanolWeakestCharge localised on one oxygen
WaterStronger than ethanolNo electron-donating alkyl group
PhenolStronger stillPhenoxide is resonance stabilised
4-NitrophenolStronger than phenolNitro group withdraws electrons, spreading the charge
4-MethylphenolWeaker than phenolMethyl donates electrons, concentrating the charge

Reactions that reliably appear

Alcohols

Dehydration to alkenes with concentrated H2SO4, following Saytzeff. Oxidation, where the product depends on the class: primary alcohols give aldehydes and then carboxylic acids, secondary give ketones, and tertiary resist oxidation without breaking the carbon skeleton. Reaction with sodium to give alkoxides, which is also the standard test.

Phenols

Electrophilic substitution is unusually easy because –OH activates the ring strongly and directs ortho and para. Bromine water gives 2,4,6-tribromophenol immediately — a favourite one-mark identification. Kolbe’s reaction and Reimer–Tiemann both appear regularly and are worth writing out rather than recognising.

Ethers

Cleavage by HI is the examinable reaction. Which side of the ether keeps the iodine depends on the mechanism: with a primary or secondary alkyl group the attack is SN2 at the less hindered carbon; with a tertiary group the reaction goes SN1 through the more stable carbocation. Reasoning that out is worth more marks than remembering the product.

How to attack a conversion question: work backwards from the target. Ask what single functional group change would produce it, then what would produce that, until you reach the starting material. Forward guessing wastes time; backward chaining almost always lands in two or three steps.

Distinguishing tests worth knowing cold

  • Phenol vs alcohol: neutral FeCl3 gives a violet colour with phenol only.
  • Primary, secondary, tertiary alcohol: Lucas reagent — tertiary turns turbid at once, secondary in about five minutes, primary only on heating.
  • Alcohol containing CH3CH(OH)–: positive iodoform test.

FAQs

Why is phenol more acidic than ethanol?

Because the phenoxide ion delocalises its negative charge into the aromatic ring by resonance, which stabilises it. Ethoxide has no such delocalisation and holds the charge entirely on oxygen, so ethanol is far less willing to lose its proton.

Why does phenol undergo electrophilic substitution more easily than benzene?

The lone pair on the –OH oxygen is donated into the ring, raising electron density at the ortho and para positions. That makes the ring markedly more attractive to electrophiles than plain benzene.

In ether cleavage by HI, which fragment gets the iodine?

It depends on the mechanism. For primary and secondary groups the reaction is SN2 and iodide attacks the less hindered carbon. Where a tertiary carbon is available the reaction goes SN1 through the more stable carbocation, and that carbon takes the iodine.

How should I practise conversions?

By reagent rather than by chapter. Build a sheet of what each reagent does to each functional group, then practise chaining two and three steps backwards from the target. Conversions repeat far more than students expect.

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