Which of the following conversions is a Hofmann elimination reaction?
- A$RCONH_2 \xrightarrow{Br_2, KOH_{(aq)}} RNH_2$
- B$RNH_2 \xrightarrow[\text{excess}]{RX} R_4NX$
- C$R_4NX \xrightarrow[\Delta]{\text{moist } Ag_2O} \text{Alkene} + R_3N$
- D$RCN \xrightarrow{Na / C_2H_5OH} RCH_2NH_2$
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Similar Questions
In the given reaction,$[X]$ is:
$4-\text{nitrostyrene} \xrightarrow{[X]} 4-\text{ethylaniline}$
$4-\text{nitrostyrene} \xrightarrow{[X]} 4-\text{ethylaniline}$
Medium
View SolutionNitrosobenzene can be prepared by oxidizing aniline with which of the following reagents?
Difficult
View SolutionMatch List-$I$ with List-$II$:
Benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$ is reacted with reagents in List-$I$ to form products in List-$II$.
| List-$I$ (Reagent) | List-$II$ (Product) |
| :--- | :--- |
| $A$. Aniline $(C_6H_5NH_2)$ | $I$. Fluorobenzene $(C_6H_5F)$ |
| $B$. $HBF_4, \Delta$ | $II$. Benzonitrile $(C_6H_5CN)$ |
| $C$. $Cu, HCl$ | $III$. $p$-Aminoazobenzene $(C_6H_5-N=N-C_6H_4-NH_2)$ |
| $D$. $CuCN/KCN$ | $IV$. Chlorobenzene $(C_6H_5Cl)$ |
Choose the correct answer from the options given below:
Benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$ is reacted with reagents in List-$I$ to form products in List-$II$.
| List-$I$ (Reagent) | List-$II$ (Product) |
| :--- | :--- |
| $A$. Aniline $(C_6H_5NH_2)$ | $I$. Fluorobenzene $(C_6H_5F)$ |
| $B$. $HBF_4, \Delta$ | $II$. Benzonitrile $(C_6H_5CN)$ |
| $C$. $Cu, HCl$ | $III$. $p$-Aminoazobenzene $(C_6H_5-N=N-C_6H_4-NH_2)$ |
| $D$. $CuCN/KCN$ | $IV$. Chlorobenzene $(C_6H_5Cl)$ |
Choose the correct answer from the options given below:
Assertion : Amines are basic in nature.
Reason : Presence of a lone pair of electrons on the nitrogen atom.
Reason : Presence of a lone pair of electrons on the nitrogen atom.
MediumAIIMS 1999
View SolutionWhich one of the following is the most basic compound?
Medium
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