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: $A-iii-o, B-i-q, C-iv-n, D-ii-m$.
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$CH_3-CH(OH)-CH_2-CHO$
$3HCHO + A \xrightarrow[40^o C]{Na_2CO_3} (B) (82\%)$
Product $(B)$ of the above reaction is
$3HCHO + A \xrightarrow[40^o C]{Na_2CO_3} (B) (82\%)$
Product $(B)$ of the above reaction is
Medium
View SolutionMatch the following reactions in Column-$1$ with their products in Column-$2$.
Column-$1$ (Reaction) Column-$2$ (Product) $A$. Benzaldehyde + $Zn-Hg/HCl$ $I$. Cyclohexanol $B$. Benzoyl chloride + $H_2/Pd-BaSO_4$ $II$. Benzyl alcohol $C$. Cyclohexanone + $NH_2NH_2/KOH$ $III$. Toluene $IV$. Benzaldehyde $V$. Cyclohexane
The correct match is:
| Column-$1$ (Reaction) | Column-$2$ (Product) |
| $A$. Benzaldehyde + $Zn-Hg/HCl$ | $I$. Cyclohexanol |
| $B$. Benzoyl chloride + $H_2/Pd-BaSO_4$ | $II$. Benzyl alcohol |
| $C$. Cyclohexanone + $NH_2NH_2/KOH$ | $III$. Toluene |
| $IV$. Benzaldehyde | |
| $V$. Cyclohexane |
The correct match is:
DifficultTS EAMCET 2021
View SolutionMatch the reactions in Column-$I$ with their products in Column-$II$.
| Column-$I$ (Reagents) | Column-$II$ (Product) |
| :--- | :--- |
| $(A) \ CH_3CHO + 2C_2H_5OH \xrightarrow{HCl_{(g)}} $ | $(i) \ (CH_3)_2C(OCH_2)_2$ |
| $(B) \ CH_3COCH_3 + CH_2OH-CH_2OH \xrightarrow{HCl_{(g)}} $ | $(ii) \ CH_3CH(OH)CN$ |
| $(C) \ CH_3CHO + HCN$ $\xrightarrow{OH^-}$ $\xrightarrow{H_3O^+}$ | $(iii) \ CH_3CH(OC_2H_5)_2$ |
| $(D) \ CH_3CHO + HCN \xrightarrow{OH^-} $ | $(iv) \ CH_3CH(OH)COOH$ |
| Column-$I$ (Reagents) | Column-$II$ (Product) |
| :--- | :--- |
| $(A) \ CH_3CHO + 2C_2H_5OH \xrightarrow{HCl_{(g)}} $ | $(i) \ (CH_3)_2C(OCH_2)_2$ |
| $(B) \ CH_3COCH_3 + CH_2OH-CH_2OH \xrightarrow{HCl_{(g)}} $ | $(ii) \ CH_3CH(OH)CN$ |
| $(C) \ CH_3CHO + HCN$ $\xrightarrow{OH^-}$ $\xrightarrow{H_3O^+}$ | $(iii) \ CH_3CH(OC_2H_5)_2$ |
| $(D) \ CH_3CHO + HCN \xrightarrow{OH^-} $ | $(iv) \ CH_3CH(OH)COOH$ |
Medium
View SolutionWhich statement is correct for product '$X$'?
$Cl-CO-CH_2-C(C \equiv CH)(CH_2-CN)-CH_2-CH_2-CO-CH_3 \xrightarrow{H_2/Pd/BaSO_4} X$
$Cl-CO-CH_2-C(C \equiv CH)(CH_2-CN)-CH_2-CH_2-CO-CH_3 \xrightarrow{H_2/Pd/BaSO_4} X$
Difficult
View SolutionIf the general formula of oxime and semicarbazone is $C=N-Z$,what is '$Z$' in oxime $(I)$ and semicarbazone $(II)$?
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