Reactant of Column-$I$ when reacted with acetaldehyde in the presence of $H^{+}$ gives the product given in Column-$II$ and their names are given in Column-$III$. Match the reactants in Column-$I$ with the products in Column-$II$ and their names in Column-$III$.
Column-$I$ (Reagent) Column-$II$ (Product) and Column-$III$ (Name) $A$. Ammonia $(NH_{3})$ $i$. $CH_{3}CH=NCH_{3}$; $m$. Acetaldehyde hydrazone $B$. Amine $(CH_{3}NH_{2})$ $ii$. $CH_{3}CH=NNH_{2}$; $n$. Acetaldehyde oxime $C$. Hydroxyl amine $(NH_{2}OH)$ $iii$. $CH_{3}CH=NH$; $o$. Acetaldehyde imine $D$. Hydrazine $(NH_{2}NH_{2})$ $iv$. $CH_{3}CH=NOH$; $q$. Acetaldehyde methyl imine
| Column-$I$ (Reagent) | Column-$II$ (Product) and Column-$III$ (Name) |
| $A$. Ammonia $(NH_{3})$ | $i$. $CH_{3}CH=NCH_{3}$; $m$. Acetaldehyde hydrazone |
| $B$. Amine $(CH_{3}NH_{2})$ | $ii$. $CH_{3}CH=NNH_{2}$; $n$. Acetaldehyde oxime |
| $C$. Hydroxyl amine $(NH_{2}OH)$ | $iii$. $CH_{3}CH=NH$; $o$. Acetaldehyde imine |
| $D$. Hydrazine $(NH_{2}NH_{2})$ | $iv$. $CH_{3}CH=NOH$; $q$. Acetaldehyde methyl imine |
(A) The reaction of acetaldehyde $(CH_{3}CHO)$ with ammonia derivatives in the presence of $H^{+}$ follows the general reaction: $CH_{3}CHO + H_{2}N-Z \rightarrow CH_{3}CH=N-Z + H_{2}O$.
$A$. Ammonia $(NH_{3})$ reacts to form $CH_{3}CH=NH$ (Acetaldehyde imine) $\rightarrow A-iii-o$.
$B$. Methylamine $(CH_{3}NH_{2})$ reacts to form $CH_{3}CH=NCH_{3}$ (Acetaldehyde methyl imine) $\rightarrow B-i-q$.
$C$. Hydroxylamine $(NH_{2}OH)$ reacts to form $CH_{3}CH=NOH$ (Acetaldehyde oxime) $\rightarrow C-iv-n$.
$D$. Hydrazine $(NH_{2}NH_{2})$ reacts to form $CH_{3}CH=NNH_{2}$ (Acetaldehyde hydrazone) $\rightarrow D-ii-m$.
Correct matching: $A-iii-o, B-i-q, C-iv-n, D-ii-m$.
$A$. Ammonia $(NH_{3})$ reacts to form $CH_{3}CH=NH$ (Acetaldehyde imine) $\rightarrow A-iii-o$.
$B$. Methylamine $(CH_{3}NH_{2})$ reacts to form $CH_{3}CH=NCH_{3}$ (Acetaldehyde methyl imine) $\rightarrow B-i-q$.
$C$. Hydroxylamine $(NH_{2}OH)$ reacts to form $CH_{3}CH=NOH$ (Acetaldehyde oxime) $\rightarrow C-iv-n$.
$D$. Hydrazine $(NH_{2}NH_{2})$ reacts to form $CH_{3}CH=NNH_{2}$ (Acetaldehyde hydrazone) $\rightarrow D-ii-m$.
Correct matching: $A-iii-o, B-i-q, C-iv-n, D-ii-m$.
Explore More
Similar Questions
Predict the products of the following reactions with $CH_3CHO$:
$(i) CH_3CHO + C_6H_5MgBr$
$(ii) CH_3CHO + C_6H_{11}MgBr$
$(iii) CH_3CHO + (CH_3)_2CHMgBr$
Followed by acidic hydrolysis in each case.
$(i) CH_3CHO + C_6H_5MgBr$
$(ii) CH_3CHO + C_6H_{11}MgBr$
$(iii) CH_3CHO + (CH_3)_2CHMgBr$
Followed by acidic hydrolysis in each case.
Medium
View Solution$(EtO)_2CH-CHO + CH_3MgI \xrightarrow{H_3O^{+}} (A)$. The product obtained in the above reaction is:
Medium
View SolutionThe reaction shown in the image is:
Medium
View SolutionGiven below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$:
Assertion $A$: Acetal/Ketal is stable in basic medium.
Reason $R$: The high leaving tendency of alkoxide ion gives the stability to acetal/ketal in basic medium.
In the light of the above statements,choose the correct answer from the options given below:
Assertion $A$: Acetal/Ketal is stable in basic medium.
Reason $R$: The high leaving tendency of alkoxide ion gives the stability to acetal/ketal in basic medium.
In the light of the above statements,choose the correct answer from the options given below:
DifficultJEE MAIN 2023
View SolutionThe reagents $A$,$B$ and $C$ respectively are:
DifficultKCET 2019
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo