Question
An unknown alcohol undergoes the following: it gives a positive iodoform test, can be dehydrated to an alkene, and forms an ester with acetic acid. Identify the type of alcohol and write the reactions for dehydration, oxidation, and esterification.
(JEE Main / NEET pattern)
Solution — Step by Step
A positive iodoform test means the alcohol has the structure (secondary alcohol with a methyl group adjacent to the OH-bearing carbon) or (ethanol). Since it is also easily dehydrated, we are likely dealing with ethanol or a secondary alcohol like 2-propanol ().
Let us take ethanol () as our working example.
flowchart TD
A["R-OH\n(Alcohol)"] -->|"conc. H₂SO₄, 170°C"| B["Alkene\n(Dehydration)"]
A -->|"K₂Cr₂O₇/H⁺ or KMnO₄"| C["Aldehyde → Acid\n(Oxidation)"]
A -->|"CH₃COOH, conc. H₂SO₄"| D["Ester\n(Esterification)"]
A -->|"HBr, HCl, or PBr₃"| E["R-X\n(Halogenation)"]
B --> F["CH₂=CH₂"]
C --> G["CH₃CHO → CH₃COOH"]
D --> H["CH₃COOC₂H₅"]
E --> I["C₂H₅Br"]Dehydration:
Oxidation (mild — to aldehyde):
Oxidation (strong — to acid):
Esterification (Fischer):
Halogenation:
- Dehydration: conc. H₂SO₄ at 170°C gives alkene; at 140°C gives ether (Williamson-like)
- Oxidation depends on the reagent: PCC/PDC stops at aldehyde; KMnO₄/K₂Cr₂O₇ goes all the way to acid
- Esterification is reversible — conc. H₂SO₄ acts as both catalyst and dehydrating agent
- For halogenation of primary alcohols, Lucas test is negative at room temperature (no turbidity)
Why This Works
Alcohols are versatile because the O-H bond and the C-O bond can both be broken depending on conditions. Dehydration breaks C-O (and adjacent C-H) to form a double bond. Oxidation breaks O-H and C-H to increase the oxidation state of carbon. Esterification is nucleophilic acyl substitution. Halogenation replaces OH with a halide via an SN2 or SN1 mechanism depending on the alcohol class.
The reaction pathway depends entirely on the reagent and conditions — the same alcohol gives completely different products. This is why organic chemistry is all about “reagent control.”
Alternative Method — Using Lucas Test to Classify First
If the question gives an unknown alcohol, classify it first using the Lucas test (ZnCl₂ + conc. HCl): 3° gives turbidity immediately, 2° in 5-10 minutes, 1° shows no reaction at room temperature. Then apply the appropriate reaction map.
For NEET, the most frequently tested alcohol reaction is dehydration — especially Zaitsev’s rule for which alkene is the major product. For JEE Main, oxidation selectivity (PCC vs KMnO₄) and Lucas test appear almost every year. Master these three and you cover 80% of alcohol questions.
Common Mistake
Students confuse the temperature conditions for dehydration: 170°C gives alkene (elimination), 140°C gives ether (dehydration of two alcohol molecules). Writing the wrong temperature means wrong product — and zero marks. Another common slip: saying tertiary alcohols can be oxidised to ketones easily. Tertiary alcohols resist mild oxidation because there is no hydrogen on the carbon bearing OH. Strong oxidants break the C-C bond instead, giving a mixture.