JEE Main 2019ChemistryBiomoleculesMediumMCQ

JEE Main 2019 — Biomolecules Question with Solution

JEE Main 2019 (10 Jan Shift 2)

Question

The correct match between item I and item II is
 
Item I (Compound) Item II (Reagent)
a. Lysine p. 1-naphthol
b. Furfural q. Ninhydrin
c. Benzyl alcohol r. KMnO4
d. Styrene s. Ceric ammonium nitrate

Choose an option

Show full solutionCorrect option: B
Correct answer
Ba→q, b→p, c→s, d→r

Step-by-step explanation

Lysine→Ninhydrin (this is given out by all the amino acids and Lysine is an amino acid.)

Furfural→1-naphthol.

Rapid furfural test is used to distinguish between glucose and fructose. In this test, the sugar solution is added to 1-naphthol and concentrated HCl.

Benzyl alcohol → ceric ammonium nitrate.

Ceric ammonium nitrate test is used for testing of alcohols.

Styrene undergoes oxidation when treated with KMnO4.

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About this question

This is a previous-year question from JEE Main 2019, covering the Biomolecules chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.