The correct statement regarding a carbonyl compound with a hydrogen atom on its $\alpha-$carbon is:
- A$A$ carbonyl compound with a hydrogen atom on its $\alpha-$carbon rapidly equilibrates with its corresponding enol and this process is known as carbonylation.
- B$A$ carbonyl compound with a hydrogen atom on its $\alpha-$carbon rapidly equilibrates with its corresponding enol and this process is known as keto-enol tautomerism.
- C$A$ carbonyl compound with a hydrogen atom on its $\alpha-$carbon never equilibrates with its corresponding enol.
- D$A$ carbonyl compound with a hydrogen atom on its $\alpha-$carbon rapidly equilibrates with its corresponding enol and this process is known as aldehyde-ketone equilibration.
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Similar Questions
Identify $(A)$ in the following reaction sequence:
$\mathop {(A)}\limits_{\text{gives positive iodoform Test}}$ $\xrightarrow[(ii) H^{+}, H_2O, (iii) \text{conc. } H_2SO_4/\Delta ]{(i) CH_3MgBr} (B)$ $\xrightarrow{O_3/Zn, H_2O}$
$\mathop {(A)}\limits_{\text{gives positive iodoform Test}}$ $\xrightarrow[(ii) H^{+}, H_2O, (iii) \text{conc. } H_2SO_4/\Delta ]{(i) CH_3MgBr} (B)$ $\xrightarrow{O_3/Zn, H_2O}$
How can propan$-2-$one be converted to tertiary butyl alcohol?
Medium
View SolutionThe aldol condensation of $CH_{3}CHO$ results in the formation of
Consider the following reaction:
$H-CHO-(CH_2)_4-OH \xrightarrow{H^+} \text{Cyclic product}$
The above reaction is an example of:
$H-CHO-(CH_2)_4-OH \xrightarrow{H^+} \text{Cyclic product}$
The above reaction is an example of:
Difficult
View SolutionReagent $X$ in the given reaction is:
(Image shows cyclohexanone being converted to cyclohexanol with a deuterium atom attached to the same carbon as the hydroxyl group.)
(Image shows cyclohexanone being converted to cyclohexanol with a deuterium atom attached to the same carbon as the hydroxyl group.)
Easy
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