Two organic compounds K and L were treated with a few drops of Fehling's solutions respectively . K formed a brick-red precipitate while L, remains unaffect...

Assessment: JAMB UTME - Chemistry - 2011 Subject: Chemistry

Question 1 Report

Two organic compounds K and L were treated with a few drops of Fehling's solutions respectively . K formed a brick-red precipitate while L, remains unaffected. The compound K is an
Answer Details
Fehling's solution is a chemical test used to distinguish between reducing sugars (which have a free aldehyde or ketone group) and non-reducing sugars (which do not have a free aldehyde or ketone group). When a reducing sugar is added to Fehling's solution and heated, it reduces the copper(II) ions in the solution to form a brick-red precipitate of copper(I) oxide. In this case, compound K formed a brick-red precipitate when treated with Fehling's solution, indicating that it is a reducing sugar or a compound that contains a free aldehyde or ketone group. On the other hand, compound L remained unaffected, indicating that it is not a reducing sugar and does not contain a free aldehyde or ketone group. Among the given options, the only compound that contains a free aldehyde or ketone group is an alkanal. Therefore, the compound K is an alkanal.

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