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Hint: Both the above compounds are mixed ketones, one is acetophenone and another one is cyclohex-1-enyl phenyl ketone. In one compound, there is presence of acetyl group whereas in another, benzoyl group is present. Identify the reaction which needs a compound having acetyl group as one of its reactants.
Complete step by step solution:
-Let’s have a look at the compounds given in the question.
The first compound is Acetophenone. It is a carbonyl compound categorized as a ketone having one phenyl group and one methyl group as substituents.
The second compound is cyclohex-1-enyl phenyl ketone or 1-Benzoylcyclohex-1-ene. It is also a carbonyl compound having one Phenyl group and one cyclohex-1-enyl group as substituents.
-One of the reactions which carbonyl compounds give is the Haloform reaction, also popularly known as Iodoform test.
-Haloform reaction is a reaction in which a ketone having a methyl group as one of its substituents reacts with sodium hypohalite and gets oxidized to give sodium salt of carboxylic acid (having one less carbon atom) and haloform.
-In simpler terms, iodoform test is the reaction in which acetyl group from a carbonyl compound reacts with sodium hydroxide and iodine in the presence of heat to form iodoform (yellow precipitate) and sodium salt of carboxylic acid.
-In this reaction, sodium hypohalite is generated in-situ with the help of sodium hydroxide and iodine on warming.
-One of the important characteristics of this reaction is that, if any $C=C$ double bond is present in the molecule; it isn’t attacked by hypohalite.
Therefore, iodoform test is the simple chemical test which can be used to distinguish between the two given compounds.
Note: Remember iodoform test is only given by compounds containing acetyl groups in them. Acetaldehyde is the only aldehyde which gives iodoform test. Some alcohols like ethanol and isopropyl alcohol also give iodoform test because sodium hypohalite first reacts with alcohols and oxidizes them to form acetaldehyde and acetone respectively.
Complete step by step solution:
-Let’s have a look at the compounds given in the question.
The first compound is Acetophenone. It is a carbonyl compound categorized as a ketone having one phenyl group and one methyl group as substituents.
The second compound is cyclohex-1-enyl phenyl ketone or 1-Benzoylcyclohex-1-ene. It is also a carbonyl compound having one Phenyl group and one cyclohex-1-enyl group as substituents.
-One of the reactions which carbonyl compounds give is the Haloform reaction, also popularly known as Iodoform test.
-Haloform reaction is a reaction in which a ketone having a methyl group as one of its substituents reacts with sodium hypohalite and gets oxidized to give sodium salt of carboxylic acid (having one less carbon atom) and haloform.
-In simpler terms, iodoform test is the reaction in which acetyl group from a carbonyl compound reacts with sodium hydroxide and iodine in the presence of heat to form iodoform (yellow precipitate) and sodium salt of carboxylic acid.
-In this reaction, sodium hypohalite is generated in-situ with the help of sodium hydroxide and iodine on warming.
-One of the important characteristics of this reaction is that, if any $C=C$ double bond is present in the molecule; it isn’t attacked by hypohalite.
Therefore, iodoform test is the simple chemical test which can be used to distinguish between the two given compounds.
Note: Remember iodoform test is only given by compounds containing acetyl groups in them. Acetaldehyde is the only aldehyde which gives iodoform test. Some alcohols like ethanol and isopropyl alcohol also give iodoform test because sodium hypohalite first reacts with alcohols and oxidizes them to form acetaldehyde and acetone respectively.
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