Question

\(\xrightarrow{\mathrm{C_2H_5ONa}} A\), \(A\) is formed by Claisen condensation. Which is/are true about \(A\)?

• A. \(A\) forms oxime
• B. \(A\) shows tautomerism
• C. \(A\) shows iodoform test
• D. All of the above are true

Hint:

Ethyl acetoacetate exhibits keto-enol tautomerism and gives characteristic carbonyl reactions.

Answer 1

The Correct Option is D

The given reaction involves Claisen condensation, which typically occurs between esters in the presence of a strong base such as sodium ethoxide (\(\mathrm{C_2H_5ONa}\)). In this reaction, two molecules of the ester undergo a condensation reaction to form a β-keto ester.

Considering the ester involved is ethyl acetate (\(\mathrm{CH_3COOC_2H_5}\)), the product \(A\) will be ethyl acetoacetate (\(\mathrm{CH_3COCH_2COOC_2H_5}\)). We can now examine the given options with respect to compound \(A\):

1. Formation of Oxime: Ketones, like the one in \(\mathrm{CH_3COCH_2COOC_2H_5}\), can form oximes by reacting with hydroxylamine. Thus, \(A\) forms an oxime.
2. Tautomerism: \(\mathrm{CH_3COCH_2COOC_2H_5}\) shows keto-enol tautomerism. The enol form is stabilized by intramolecular hydrogen bonding.
3. Iodoform Test: Compound \(A\) contains the \(\mathrm{-COCH_3}\) group, making it positive for the iodoform test. The methyl ketone group is essential for this reaction, producing a yellow precipitate of iodoform (\(\mathrm{CHI_3}\)).
   

Based on the above points, all the statements about \(A\) are true.

Conclusion: All of the above are true.