A compound A has a molecular formula C₂Cl₃OH. It reduces F...
Explanation:
The compound A has a molecular formula CClOH, which means it contains carbon (C), chlorine (Cl), oxygen (O), and hydrogen (H) atoms.
Reduction of Fehling Solution:
Compound A is capable of reducing Fehling solution. Fehling solution is a mixture of copper sulfate (CuSO4) and alkaline tartrate, which is used to test for the presence of reducing sugars. When a reducing sugar is present, such as in compound A, it reacts with Fehling solution to form a red precipitate of copper(I) oxide (Cu2O). This indicates the presence of an aldehyde or a reducing sugar.
Oxidation to Monocarboxylic Acid:
Compound A can be oxidized to form a monocarboxylic acid, which means it can be converted into a compound containing a carboxyl group (COOH). This indicates the presence of an aldehyde group (CHO) in compound A.
Action of Chlorine on Ethyl Alcohol:
Compound A is obtained by the action of chlorine on ethyl alcohol. Chlorine (Cl2) reacts with ethyl alcohol (C2H5OH) to form a compound with the formula CClOH. This compound is called chloral (trichloroacetaldehyde).
Conclusion:
Based on the information given, compound A with the molecular formula CClOH is chloral. Chloral is formed by the reaction of chlorine with ethyl alcohol and it possesses an aldehyde group that allows it to reduce Fehling solution. On oxidation, chloral can be converted into a monocarboxylic acid. Therefore, the correct answer is option C, chloral.
A compound A has a molecular formula C₂Cl₃OH. It reduces F...
When ethyl alcohol reacts with Cl2 it oxidises it to ethanal and then each of the alpha hydrogen is replaced by cl and thus chloral is formed i. e. CCL3CHO which is the required compound and being an aldehyde it gives fehling test.
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