Bromine Water and Baeyer's Test: How to Detect Double and Triple Bonds
Chemistry · Hydrocarbons · NEET
Both tests find carbon-carbon double or triple bonds (unsaturation). Bromine water is orange and turns colourless when it adds across a C=C or C≡C bond. Baeyer's reagent is cold dilute alkaline KMnO4; it is purple/pink and turns colourless (with a brown MnO2 solid) when it oxidises the double bond to a diol. Memory hook: "Colour dies where the double bond lies."
Bromine water (orange) fades by ADDING across C=C, while Baeyer's reagent (purple KMnO4) fades by OXIDISING C=C to a diol with brown MnO2. Saturated compounds and benzene give no colour change.
Your doubts, answered
Why does bromine water turn colourless with an alkene but not with an alkane?
An alkene has a C=C double bond. The pi electrons attack bromine, so Br2 ADDS across the double bond and forms a colourless dibromide. Since the orange Br2 is used up, the colour disappears. An alkane has only single bonds and no pi electrons, so it cannot add bromine easily (it needs UV light for slow substitution). So the orange colour stays. Rule: colour goes = unsaturation present.
What is Baeyer's reagent and what does it test?
Baeyer's reagent is cold, dilute, alkaline (or neutral) potassium permanganate, KMnO4. It is purple/pink. When it meets a C=C double bond it oxidises it to a vicinal glycol (a diol, two -OH on next-door carbons). The purple colour fades and a brown solid, manganese dioxide (MnO2), forms. This decolourisation is the standard NCERT test for unsaturation.
What is the difference between the bromine water test and Baeyer's test?
Both detect double/triple bonds, but the chemistry differs. Bromine water (orange) does ADDITION: Br2 adds across C=C to give a colourless dibromide, so orange fades to colourless. Baeyer's reagent (purple KMnO4) does OXIDATION: it turns C=C into a diol, so purple fades and brown MnO2 appears. Memory tip: Bromine = ADD, Baeyer = OXIDISE.
Do alkynes also give these tests?
Yes. A triple bond (C≡C) is also unsaturated, so alkynes decolourise both bromine water and Baeyer's reagent, just like alkenes. In fact a very reactive small alkene or alkyne reacts fastest. Both tests only tell you unsaturation is present; they do not tell you if it is a double or triple bond by themselves.
Does benzene decolourise bromine water or Baeyer's reagent?
No. Even though benzene has 'double bonds' in its ring, they are delocalised and very stable (aromatic). Benzene does not do normal addition or easy oxidation, so it does NOT decolourise bromine water or Baeyer's reagent. This is a favourite NEET trap: aromatic rings are unsaturated on paper but do not give the unsaturation test.
Can a compound decolourise bromine water without having a double bond?
Yes, and NEET tests this. Very electron-rich benzene rings, like in phenol (C6H5-OH) and aniline (C6H5-NH2), decolourise bromine WATER by electrophilic SUBSTITUTION (they release HBr), not by addition. So a positive bromine-water result is not proof of a C=C. Plain benzene and saturated compounds like cyclohexane do not react.
⚠️ The NEET trap ✗ Cyclohexane will decolourise bromine water because it is a ring with the formula C6H12. ✓ Cyclohexane does NOT decolourise bromine water. It is fully saturated (all single bonds) with no C=C and no activated ring. Being a ring does not mean unsaturated. CyclohexENE (with a double bond) would decolourise it. 🧠 A ring is not a double bond. Look for C=C or C≡C, or an electron-rich ring (phenol/aniline). No pi bond and no activated ring = colour stays.
Real NEET questions
NEET 2025
Which one of the following compounds does NOT decolourize bromine water?
A · Styrene (vinylbenzene), C6H5-CH=CH2
B · Aniline, C6H5-NH2
C · Cyclohexane, C6H12 ✓
D · Phenol, C6H5-OH
Solution: Bromine water is decolourised by unsaturated or electron-rich compounds. Styrene has a C=C double bond, so it does addition and decolourises it. Aniline and phenol have very electron-rich rings, so they decolourise bromine water by substitution (they release HBr). Cyclohexane is fully saturated, has no C=C and no activated ring, so it cannot react. Answer: (C) Cyclohexane.
NEET 2016 Phase 2
The compound that will react most readily with gaseous bromine has the formula
A · C3H6 ✓
B · C2H2
C · C4H10
D · C2H4
Solution: Bromine adds fastest to a C=C double bond. C4H10 is a saturated alkane (no reaction). C2H2 is an alkyne (triple bond) and C2H4 is an alkene, both react. C3H6 (propene) is an electron-rich alkene, and here it is the most reactive alkene toward addition of bromine, so it reacts most readily. Answer: (A) C3H6.
ReNEET 2026
Statement-I: trans-But-2-ene with Br2 in CCl4 gives meso-2,3-dibromobutane. Statement-II: cis-But-2-ene with alkaline KMnO4 (Baeyer's reagent) gives meso-butane-2,3-diol. Choose the correct answer.
A · Both Statement-I and Statement-II are correct
B · Both Statement-I and Statement-II are incorrect
C · Statement-I is correct but Statement-II is incorrect
D · Statement-I is incorrect but Statement-II is correct ✓
Solution: Bromine addition is ANTI (the two Br add from opposite faces). Anti addition to trans-but-2-ene gives the racemic (d,l) pair, not meso, so Statement-I is wrong. Baeyer's reagent (alkaline KMnO4) does SYN dihydroxylation (both -OH add from the same face). Syn addition to cis-but-2-ene gives meso-butane-2,3-diol, so Statement-II is correct. Answer: (D).
Solved Hydrocarbons NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What colour change confirms a positive bromine water test?
The orange/reddish-brown bromine water turns colourless. That means the double or triple bond used up the bromine by addition.
What colour change confirms a positive Baeyer's test?
The purple/pink KMnO4 fades to colourless and a brown solid, manganese dioxide (MnO2), appears. That confirms unsaturation.
Does Baeyer's test give addition or oxidation?
Oxidation. Cold dilute alkaline KMnO4 oxidises the C=C into a vicinal diol (glycol). It does not add across the bond like bromine does.
Why is Baeyer's reagent called a test for unsaturation?
Because only compounds with C=C or C≡C (unsaturated) decolourise it easily. Saturated alkanes leave it purple, so decolourisation signals a multiple bond.
Can these tests tell a double bond from a triple bond?
No. Both alkenes and alkynes decolourise both reagents. The tests only show that some unsaturation is present, not whether it is a double or triple bond.
Why doesn't benzene decolourise these reagents even though it looks unsaturated?
Benzene's pi electrons are delocalised and very stable (aromatic). It resists addition and easy oxidation, so it does not decolourise bromine water or Baeyer's reagent under normal conditions.
