Acetaldehyde on treatment with Fehling's solution gives a precipitate of
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(Cu_2O\)
Fehling’s reagent contains copper(II) held in a tartrate complex in alkaline solution. Aldehydes are easily oxidised, so on heating, acetaldehyde reduces copper(II) to copper(I), which separates as copper(I) oxide, and the aldehyde becomes the carboxylate ion:
\( \displaystyle \ce{CH3CHO + 2Cu^2+ + 5OH^- -> CH3COO^- + Cu2O v + 3H2O} \)
Copper goes from +2 to +1, and each \( \ce{Cu^{2+}} \) takes one electron.
Why the other options are incorrect
Option (2) \(Cu\) — Incorrect. Reduction stops at copper(I); metallic copper is not formed. (The silver mirror of Tollens’ test is the case where the free metal separates.)
Option (3) \(CuO\) — Incorrect. CuO contains copper(II), the state the reagent starts in; there would be no reduction. CuO is black, not reddish brown.
Option (4) \(Cu_2O_2\) — Incorrect. There is no such compound of copper as a product of this test.
NCERT (2026-27 edition):
Oxidation – …
