
Tautomerism is exhibited by
A.${{\text{R}}_3}{\text{CN}}{{\text{O}}_{\text{2}}}$
B.${\text{RC}}{{\text{H}}_{\text{2}}}{\text{N}}{{\text{O}}_{\text{2}}}$
C.${\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{3}}}{\text{CNO}}$
D.${\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{NH}}$
Answer
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Hint: To solve this question, you must recall the principles of tautomerism. Tautomerism involves hydrogen changing places with a double bond.
Complete step by step answer:
We know that the basic requirements for a compound to show tautomerism are the presence of a double bond and the presence of hydrogen atoms on the carbon atom adjacent to the double bond, that is the $\alpha {\text{ - C}}$.
In option A, ${{\text{R}}_3}{\text{CN}}{{\text{O}}_{\text{2}}}$ is the given compound.
Double bond is present between nitrogen and oxygen in the nitro group, but the $\alpha {\text{ - C}}$ is attached to three alkyl groups. Thus, this compound does not exhibit tautomerism.
In option B, ${\text{RC}}{{\text{H}}_{\text{2}}}{\text{N}}{{\text{O}}_{\text{2}}}$ is the given compound.
In this compound, a double bond is present between the nitrogen and oxygen atom in the nitro group.
Also, the $\alpha {\text{ - C}}$ atom has 2 hydrogen atoms attached to it. As a result, we can conclude that this compound exhibits tautomerism.
In the option C, ${\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{3}}}{\text{CNO}}$ is the given compound.
Double bond is present between the nitrogen and oxygen atom, but the adjacent carbon atom is bonded to three methyl groups and has no hydrogen atoms to be able to exhibit tautomerism.
In the option D, ${\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{NH}}$ is the given compound.
This compound is a secondary amine and has two $\alpha {\text{ - C}}$ atoms and six hydrogen atoms available. But there are no double bonds in the compound. Thus, it will not exhibit tautomerism.
We can clearly say that the compound given in the option exhibits tautomerism.
The correct option is B.
Note:
The most common form of tautomerism exhibited is keto-enol tautomerism. It involves the shift of a proton to the carbonyl oxygen forming an enol. Generally, the keto form is more stable.
Tautomers are structural isomers and it is important to make sure that you don’t confuse them as contributing structures in resonance.
Complete step by step answer:
We know that the basic requirements for a compound to show tautomerism are the presence of a double bond and the presence of hydrogen atoms on the carbon atom adjacent to the double bond, that is the $\alpha {\text{ - C}}$.
In option A, ${{\text{R}}_3}{\text{CN}}{{\text{O}}_{\text{2}}}$ is the given compound.
Double bond is present between nitrogen and oxygen in the nitro group, but the $\alpha {\text{ - C}}$ is attached to three alkyl groups. Thus, this compound does not exhibit tautomerism.
In option B, ${\text{RC}}{{\text{H}}_{\text{2}}}{\text{N}}{{\text{O}}_{\text{2}}}$ is the given compound.
In this compound, a double bond is present between the nitrogen and oxygen atom in the nitro group.
Also, the $\alpha {\text{ - C}}$ atom has 2 hydrogen atoms attached to it. As a result, we can conclude that this compound exhibits tautomerism.
In the option C, ${\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{3}}}{\text{CNO}}$ is the given compound.
Double bond is present between the nitrogen and oxygen atom, but the adjacent carbon atom is bonded to three methyl groups and has no hydrogen atoms to be able to exhibit tautomerism.
In the option D, ${\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{NH}}$ is the given compound.
This compound is a secondary amine and has two $\alpha {\text{ - C}}$ atoms and six hydrogen atoms available. But there are no double bonds in the compound. Thus, it will not exhibit tautomerism.
We can clearly say that the compound given in the option exhibits tautomerism.
The correct option is B.
Note:
The most common form of tautomerism exhibited is keto-enol tautomerism. It involves the shift of a proton to the carbonyl oxygen forming an enol. Generally, the keto form is more stable.
Tautomers are structural isomers and it is important to make sure that you don’t confuse them as contributing structures in resonance.
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