Answer
Verified
37.2k+ views
Hint: Haloform reaction is oxidising in nature. It requires a methyl group near a carbonyl carbon to give a positive test for this reaction.
Complete step-by-step answer:
The Haloform reaction is actually used to distinguish aldehydes and ketones which have a methyl group attached to the carbonyl carbon from the rest of the mixture. This reaction is oxidising in nature and therefore its products are a carboxylic acid which has one carbon atom less and a methyl halide. This is the same methyl group that was attached to the alpha carbon atom (the carbon atom bearing the carbonyl oxygen). The reaction is shown as below:
The “R” group here can be a hydrogen atom, making it an aldehyde or can be another alkyl group making it a ketone. This reaction does not affect other double bonds. So in a way the aromatic compounds can also undergo this process.
This reaction can also happen with alcohols, because they oxidise to carbonyl compounds. An example with ethyl alcohol is as below:
As you can see, ethyl alcohol loses a methyl group and then oxidises into formaldehyde. The by product is same as any Haloform reaction, which is a methyl halide.
Let’s look into the options that are given one-by-one:
The carbonyl group here is in the middle of the compound and it has no methyl groups attached to it. Methyl groups are only possible at any end of a molecular chain or at the end of a branched chain.
The carbonyl group has a phenyl group at its side and no other methyl group.
The methyl group is attached to the carbon bearing the hydroxyl group. This will respond to the oxidation reaction because it will convert into its corresponding carbonyl compound, which will be as shown below:
This will also respond to the given Haloform reaction as the methyl group is directly attached to the carbonyl carbon. The product of the reaction is:
Note:
The products of this reaction cannot be converted back into their original forms. This is because the compound that undergoes this reaction loses a methyl group, hence becoming shorter. Therefore this reaction is not used to oxidise carbonyl compounds into carboxylic acids or alcohols into aldehydes or ketones. This is only a reaction that can detect the presence of certain specific compounds. The by-products of this reaction are also a nuisance to separate.
Complete step-by-step answer:
The Haloform reaction is actually used to distinguish aldehydes and ketones which have a methyl group attached to the carbonyl carbon from the rest of the mixture. This reaction is oxidising in nature and therefore its products are a carboxylic acid which has one carbon atom less and a methyl halide. This is the same methyl group that was attached to the alpha carbon atom (the carbon atom bearing the carbonyl oxygen). The reaction is shown as below:
The “R” group here can be a hydrogen atom, making it an aldehyde or can be another alkyl group making it a ketone. This reaction does not affect other double bonds. So in a way the aromatic compounds can also undergo this process.
This reaction can also happen with alcohols, because they oxidise to carbonyl compounds. An example with ethyl alcohol is as below:
As you can see, ethyl alcohol loses a methyl group and then oxidises into formaldehyde. The by product is same as any Haloform reaction, which is a methyl halide.
Let’s look into the options that are given one-by-one:
The carbonyl group here is in the middle of the compound and it has no methyl groups attached to it. Methyl groups are only possible at any end of a molecular chain or at the end of a branched chain.
The carbonyl group has a phenyl group at its side and no other methyl group.
The methyl group is attached to the carbon bearing the hydroxyl group. This will respond to the oxidation reaction because it will convert into its corresponding carbonyl compound, which will be as shown below:
This will also respond to the given Haloform reaction as the methyl group is directly attached to the carbonyl carbon. The product of the reaction is:
Note:
The products of this reaction cannot be converted back into their original forms. This is because the compound that undergoes this reaction loses a methyl group, hence becoming shorter. Therefore this reaction is not used to oxidise carbonyl compounds into carboxylic acids or alcohols into aldehydes or ketones. This is only a reaction that can detect the presence of certain specific compounds. The by-products of this reaction are also a nuisance to separate.
Recently Updated Pages
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
The path difference between two waves for constructive class 11 physics JEE_MAIN
What is the difference between solvation and hydra class 11 chemistry JEE_Main
IfFxdfrac1x2intlimits4xleft 4t22Ft rightdt then F4-class-12-maths-JEE_Main
Sodium chloride is purified by passing hydrogen chloride class 11 chemistry JEE_Main
Consider the following oxyanions PO43P2O62SO42MnO4CrO4S2O52S2O72 class 11 chemistry JEE_Main
Other Pages
Formula for number of images formed by two plane mirrors class 12 physics JEE_Main
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
Explain the construction and working of a GeigerMuller class 12 physics JEE_Main
A currentcarrying coil is placed in a magnetic field class 12 physics JEE_Main
In the given circuit the current through the 5mH inductor class 12 physics JEE_Main
In a family each daughter has the same number of brothers class 10 maths JEE_Main