A carbonyl compound with molecular weight 86, does not reduce Fehling's solution but forms crystalline bisulphite derivative and gives iodoform test. The possible compound can be
Answer & explanation
Correct answer: option 2
The correct answer is option 2. 2-pentanone and 3-methyl-2-butanone.
To determine the carbonyl compound based on the given tests and properties, let us analyze each option:
Fehling's Solution Test:
Fehling's solution is used to test for the presence of reducing sugars or aldehydes. It consists of two separate solutions: Fehling's A (copper sulfate in aqueous solution) and Fehling's B (potassium sodium tartrate in aqueous sodium hydroxide). Aldehydes can reduce Fehling's solution, forming a red precipitate of copper(I) oxide.
Crystalline Bisulphite Derivative:
Carbonyl compounds, specifically aldehydes and ketones, can react with sodium bisulphite (\(\text{NaHSO}_3 \)) to form crystalline derivatives. This reaction is used to differentiate between aldehydes and ketones.
Iodoform Test:
The iodoform test is used to detect the presence of a methyl ketone (a compound with the structure \(\text{CH}_3\text{CO}^-\)) or a compound that can be oxidized to a methyl ketone. It involves the reaction of a methyl ketone with iodine in the presence of a base to produce yellow precipitates of iodoform (\(\text{CHI}_3\)).
Now, let us evaluate the options provided:
1. 2-pentanone and 3-pentanone:
2-Pentanone (\( \text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3\)) is a methyl ketone and gives positive results for iodoform test. It forms a crystalline bisulphite derivative.
3-Pentanone (\( \text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3\)) is also a methyl ketone and gives positive results for iodoform test. It forms a crystalline bisulphite derivative.
2. 2-pentanone and 3-methyl-2-butanone:
2-Pentanone (\( \text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3\)) is a methyl ketone and gives positive results for iodoform test. It forms a crystalline bisulphite derivative.
3-Methyl-2-butanone (\( \text{CH}_3\text{COCH}_2\text{CH}( \text{CH}_3)\text{CH}_3 \)) is not a methyl ketone and does not give a positive iodoform test.
3. 2-pentanone and pentanal:
Pentanal (\( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CHO}\)) is an aldehyde and reduces Fehling's solution, but it does not form a crystalline bisulphite derivative.
4. 3-pentanone and 3-methyl-2-butanone:
3-Pentanone (\( \text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3\)) is also a methyl ketone and gives positive results for iodoform test. It forms a crystalline bisulphite derivative.
3-Methyl-2-butanone is not a methyl ketone and does not give a positive iodoform test.
Conclusion: Based on the tests and properties described (non-reduction of Fehling's solution, formation of crystalline bisulfite derivative, and positive iodoform test), the possible compounds are 2-pentanone and 3-pentanone.
Therefore, the correct answer is: 2-pentanone and 3-pentanone.