
Fehling’s solution can make a distinction between:
$
A.C{H_3}CHO\,and\,{C_6}{H_3}CHO \\
B.C{H_3}CHO\,and\,{H_3}C - (CO) - C{H_2} - OH \\
C.{H_3}C - CH(OH) - (CO) - C{H_3}\,and\,HCHO \\
D.C{H_3}CHO \,and \,HCHO \\
$
Answer
577.5k+ views
Hint: Aldehydes differ from ketones in their oxidation reactions. Aldehydes are easily oxidised to carboxylic acids on treatment with common oxidizing agents like nitric oxide, potassium permanganate etc. even mild oxidizing agents can oxidize aldehydes like tollen’s reagent and fehling’s reagent.
Complete step by step solution:
Let us understand what is a fehling’s test, fehling reagent comprises of two solutions fehling solution A and fehling solution B. fehling solution A is aqueous copper sulphate and fehling solution B is alkaline sodium, potassium tartrate (rochelle's salt). These two solutions are mixed in equal amounts before the test. On heating an aldehyde with fehling's reagent, a reddish brown ppt. is obtained. Aldehydes are oxidized to corresponding carboxylate anions. During the oxidation of aldehydes to acids, the cupric ions are reduced to cuprous ions which are precipitated as red cuprous oxide. aromatic aldehydes do not respond to this test.
In the above given options we observe that in option A, a compound benzaldehyde is given and as discussed above it does not reduce fehling’s solution as it is an aromatic aldehyde. Therefore aliphatic and aromatic aldehydes can be separated by this method.
In the above options are a combination of ketones and aldehydes. Aldehydes are characterized and in particular are differentiated from ketones through their ease of oxidation: aldehydes give a positive test with tollen’s reagent; ketones do not. Methyl ketones are characterized through iodoform tests.
Therefore the correct option is A.
Note:
Ketones are generally oxidized under vigorous conditions i.e. strong oxidizing agents and at elevated temperatures. Their oxidation involves C-C bond cleavage to afford a mixture of carboxylic acids having lesser number of carbon atoms than the parent ketone.
Complete step by step solution:
Let us understand what is a fehling’s test, fehling reagent comprises of two solutions fehling solution A and fehling solution B. fehling solution A is aqueous copper sulphate and fehling solution B is alkaline sodium, potassium tartrate (rochelle's salt). These two solutions are mixed in equal amounts before the test. On heating an aldehyde with fehling's reagent, a reddish brown ppt. is obtained. Aldehydes are oxidized to corresponding carboxylate anions. During the oxidation of aldehydes to acids, the cupric ions are reduced to cuprous ions which are precipitated as red cuprous oxide. aromatic aldehydes do not respond to this test.
In the above given options we observe that in option A, a compound benzaldehyde is given and as discussed above it does not reduce fehling’s solution as it is an aromatic aldehyde. Therefore aliphatic and aromatic aldehydes can be separated by this method.
In the above options are a combination of ketones and aldehydes. Aldehydes are characterized and in particular are differentiated from ketones through their ease of oxidation: aldehydes give a positive test with tollen’s reagent; ketones do not. Methyl ketones are characterized through iodoform tests.
Therefore the correct option is A.
Note:
Ketones are generally oxidized under vigorous conditions i.e. strong oxidizing agents and at elevated temperatures. Their oxidation involves C-C bond cleavage to afford a mixture of carboxylic acids having lesser number of carbon atoms than the parent ketone.
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