Organic Chemistry · Functional Group Analysis

One test.
Three colours.
Every alcohol confesses.

Victor Meyer's test tells primary, secondary, and tertiary alcohols apart using nothing but colour — blood red, blue, or no colour at all. Below: the full mechanism, the classification it relies on, and a quiz to test what sticks.

PrimaryBlood red
SecondaryBlue
TertiaryColourless
R–OH
THE ALCOHOL FUNCTIONAL GROUP
Introduction

What is an alcohol?

An alcohol is an organic compound in which a hydroxyl group (–OH) is bonded to a saturated carbon atom, written generally as R–OH. That single –OH group governs almost everything about how the molecule behaves: its boiling point, its solubility in water, and — most importantly for this page — how it reacts with a specific sequence of reagents.

Not all alcohols are built the same way. The carbon carrying the –OH group can be attached to one, two, or three other carbon groups, and that difference alone is enough to change the colour a molecule turns during Victor Meyer's test. Before the test makes sense, the classification has to.

Classification

Three degrees of alcohol

Chemists classify alcohols by how many carbon groups sit on the carbon bearing the –OH. That number is called its degree, and it's the entire premise Victor Meyer built his test around.

1° · One R group

Primary alcohol

R–CH₂–OH

The –OH carbon is attached to only one other carbon chain (and two hydrogens). Least hindered, most reactive toward oxidation.

e.g. ethanol, CH₃CH₂OH
2° · Two R groups

Secondary alcohol

R–CH(OH)–R

The –OH carbon sits between two carbon chains and one hydrogen. Intermediate reactivity and steric bulk.

e.g. isopropanol, (CH₃)₂CHOH
3° · Three R groups

Tertiary alcohol

R–C(OH)(R)–R

The –OH carbon is bonded to three carbon chains and no hydrogen. Most hindered — and, as the test shows, unreactive at the final step.

e.g. tert-butanol, (CH₃)₃COH
The Mechanism

Victor Meyer's test, step by step

Each alcohol is pushed through the same four-step sequence. Primary and secondary alcohols make it all the way to a coloured end point; tertiary alcohols stall at the final step. Hover any card for the chemistry behind it.

Primary Alcohol

R–CH₂–OH
R–CH₂–OH
Starting material: a primary alcohol, one R group on the carbinol carbon.
ΔRed P₄ + I₂
R–CH₂–I
Red phosphorus generates PI₃ in situ, converting the alcohol to an alkyl iodide.
Alc.AgNO₂
R–CH₂–NO₂
Silver nitrite displaces iodide via its nitrogen atom, giving the nitroalkane — two α-H atoms remain on this carbon.
0°CHNO₂
Nitrolic acid R–C(=NOH)–NO₂
HNO₂ substitutes one α-H, forming a nitrolic acid — this is the species that turns blood red with base.
ΔAq. KOH
Blood Red Colour

Secondary Alcohol

R–CH(OH)–R
R–CH(OH)–R
Starting material: a secondary alcohol, two R groups on the carbinol carbon.
ΔRed P₄ + I₂
R–CHI–R
Same halogenation, now on a carbon flanked by two R groups.
Alc.AgNO₂
R–CH(NO₂)–R
Iodide is displaced by nitrite through nitrogen, giving a secondary nitroalkane with just one α-H remaining.
0°CHNO₂
Pseudonitrol R₂C(NO₂)(N=O)
With only one α-H available, HNO₂ substitutes it directly — no O–H forms here, unlike the primary case.
ΔAq. KOH
Blue Colour

Tertiary Alcohol

R–C(OH)(R)–R
R–C(OH)(R)–R
Starting material: a tertiary alcohol, three R groups on the carbinol carbon.
ΔRed P₄ + I₂
R–CI(R)–R
Halogenation proceeds as usual on the fully substituted carbon.
Alc.AgNO₂
R–C(NO₂)(R)–R
Nitrite still displaces iodide, but this carbon now has zero α-hydrogens left to react further.
0°CHNO₂
No Reaction
With no α-H available, HNO₂ has nothing to substitute — the molecule is unchanged.
ΔAq. KOH
Colourless
Check yourself

6-question quiz

Pick an answer — you'll see immediately whether it's right.