Exams › JEE Advanced › Chemistry

Propan-1-ol and propan-2-ol cannot be distinguished by which of the following reagents or reaction sequences?

1. PCC followed by Fehling's reagent
2. Treating with NaOH/I2 (iodoform test)
3. Heating with Cu at 300 deg C followed by Fehling's reagent
4. Reaction with semicarbazide

Correct answer: Reaction with semicarbazide

Solution

Semicarbazide reacts only with carbonyl compounds to form semicarbazones, and since neither propan-1-ol nor propan-2-ol has a carbonyl group, no reaction occurs with either — so they cannot be distinguished this way. The other options (PCC + Fehling's, iodoform test, Cu oxidation + Fehling's) differentiate them based on the aldehyde vs. ketone formed upon oxidation.

Related JEE Advanced Chemistry questions

• Which compound listed below cannot be synthesized using the standard Williamson ether synthesis?
• In the reaction between C6H5Br and CH3CH2OH, C6H5Br remains unreactive. However, when C6H5OH reacts with MeCBrMe2, the tert-halide group in CBr undergoes elimination instead of substitution. Therefore, only certain combinations are suitable for ether synthesis.
• When CH3CH2CH2CH2OH is heated with Al2O3 at 350°C, what is the main product formed?
• The compound that does NOT liberate CO₂ on treatment with aqueous sodium bicarbonate solution, is -
• Which type of solution is required to perform the dye test for identifying β-naphthol?
• Among the following alcohols, how many will produce an immediate white turbidity (precipitate) when treated with Lucas reagent (anhydrous ZnCl2 / conc. HCl) at room temperature? The compounds are: (i) 1-butanol, (ii) 2-butanol, (iii) 2-methyl-2-propanol (tert-butanol), (iv) benzyl alcohol.

⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →