Preparation of Amines Questions in English

(B) The reaction sequence is as follows:
$1$. The nitration of benzene $(C_6H_6)$ with concentrated $HNO_3$ in the presence of concentrated $H_2SO_4$ yields nitrobenzene $(C_6H_5NO_2)$,which is compound $B$.
$2$. Nitrobenzene $(C_6H_5NO_2)$ is then reduced using $Sn$ and $HCl$ to form aniline $(C_6H_5NH_2)$.
$3$. Therefore,the starting compound $A$ is benzene.

(A) The reaction sequence is as follows:
$1. \, C_2H_5Br \xrightarrow{KCN (X)} C_2H_5CN$ (Ethyl cyanide)
$2. \, C_2H_5CN \xrightarrow{LiAlH_4 (Y)} C_2H_5CH_2NH_2$ (Propylamine,$C_3H_7NH_2$)
Thus,$X = KCN$ and $Y = LiAlH_4$.

(A) When ethyl alcohol is passed over heated alumina $(Al_2O_3)$ in the presence of ammonia,it undergoes an ammonolysis reaction to form ethylamine.
The chemical reaction is:
$C_2H_5OH + NH_3 \xrightarrow{Al_2O_3, \Delta} C_2H_5NH_2 + H_2O$
Thus,the correct product is ethylamine $(C_2H_5NH_2)$.

(B) The reaction proceeds in two main steps:
$1$. The ketone,$3$-methylbutan-$2$-one,reacts with methylamine $(CH_3NH_2)$ to form an imine intermediate $(CH_3-CH(CH_3)-C(CH_3)=NCH_3)$ via nucleophilic addition followed by the elimination of water.
$2$. The imine is then reduced by lithium aluminium hydride $(LiAlH_4)$ to the corresponding secondary amine.
$3$. The final product is $N,2,3$-trimethylbutan-$2$-amine.

(D) $CH_3CH_2CONH_2 \xrightarrow{Br_2/NaOH} CH_3CH_2NH_2$
This is a $Hofmann$ bromamide degradation reaction,which converts an amide into a primary amine containing one carbon atom less than the original amide.

(A) In the Gabriel phthalimide synthesis,phthalimide is first treated with alcoholic $KOH$ to form potassium phthalimide.
This potassium phthalimide then reacts with an alkyl halide (such as $C_2H_5I$) to form an $N$-alkyl phthalimide.
Finally,the $N$-alkyl phthalimide is hydrolyzed with dilute $HCl$ or $NaOH$ to yield a primary amine.
Therefore,the initial step involves the reaction with $KOH$.

(C) The reaction of acetamide $(CH_3CONH_2)$ with $NaOBr$ is known as the Hofmann bromamide degradation reaction.
In this reaction,an amide is converted into a primary amine with one carbon atom less than the original amide.
$CH_3CONH_2 + NaOBr + 4NaOH \rightarrow CH_3NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$.
Thus,the product formed is methylamine $(CH_3NH_2)$.

(C) The reaction is known as the $Hofmann$ bromamide degradation reaction.
In this reaction,an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide to form a primary amine.
The chemical equation is:
$CH_3CONH_2 + Br_2 + 4KOH \xrightarrow{\Delta} CH_3NH_2 + K_2CO_3 + 2KBr + 2H_2O$

(D) The reaction $CH_3CONH_2 \xrightarrow{NaOBr} CH_3NH_2$ is known as the $Hofmann$ bromamide degradation reaction.
In this reaction,an amide is treated with an aqueous or ethanolic solution of $NaOH$ and $Br_2$ to produce a primary amine with one carbon atom less than the original amide.
Therefore,$CH_3CONH_2$ (acetamide) yields $CH_3NH_2$ (methanamine).

(B) The reaction of acetamide with $NaOH + Br_2$ is known as the Hofmann bromamide degradation reaction.
In this reaction,an amide is converted into a primary amine with one carbon atom less than the original amide.
$CH_3CONH_2 + Br_2 + 4NaOH \to CH_3NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$
Thus,acetamide gives methyl amine.

(B) The reaction of an amide with bromine and caustic potash $(KOH)$ is known as the $Hofmann$ bromamide degradation reaction.
This reaction results in the formation of a primary amine with one carbon atom less than the parent amide.
To prepare ethylamine $(CH_3CH_2NH_2)$,which has $2$ carbon atoms,the starting amide must be propionamide $(CH_3CH_2CONH_2)$,which has $3$ carbon atoms.
The chemical equation is:
$CH_3CH_2CONH_2 + Br_2 + 4KOH \to CH_3CH_2NH_2 + K_2CO_3 + 2KBr + 2H_2O$

(C) The reduction of nitroalkanes using reducing agents like $Sn/HCl$ or $Fe/HCl$ results in the formation of primary amines.
$CH_3CH_2NO_2 + 6[H] \xrightarrow{Sn/HCl} CH_3CH_2NH_2 + 2H_2O$
Thus,the reduction of nitroalkanes yields amines.

(A) $Acetamide$ $(CH_3CONH_2)$ changes into $methylamine$ $(CH_3NH_2)$ by the $Hofmann$ bromamide degradation reaction.
In this reaction,an amide is treated with bromine $(Br_2)$ in an aqueous or ethanolic solution of sodium hydroxide $(NaOH)$,which leads to the degradation of the amide and the formation of a primary amine with one carbon atom less than the original amide.

(C) The correct answer is $(C)$.
Methyl isocyanide $(CH_3NC)$ on reduction with hydrogen in the presence of a catalyst gives $N$-methylmethanamine,which is a secondary amine ($2^\circ$ amine).
The reaction is:
$CH_3-NC + 4[H] \xrightarrow{\text{Reduction}} CH_3-NH-CH_3$ ($N$-methylmethanamine).

(A) When ethanol $(C_2H_5OH)$ is reacted with ammonia $(NH_3)$ in the presence of alumina $(Al_2O_3)$ at high temperature,it undergoes an ammonolysis reaction to form ethylamine $(C_2H_5NH_2)$.
The chemical equation is: $C_2H_5OH + NH_3 \xrightarrow{Al_2O_3} C_2H_5NH_2 + H_2O$.

(B) The reaction proceeds as follows:
$1$. Acetamide $(CH_3CONH_2)$ on heating with $P_2O_5$ (a dehydrating agent) undergoes dehydration to form methyl cyanide $(CH_3CN)$ as product $A$.
$2$. Methyl cyanide $(CH_3CN)$ on reduction with $4H$ (using reagents like $LiAlH_4$ or $Na/C_2H_5OH$) forms ethylamine $(CH_3CH_2NH_2)$ as product $B$.
Reaction sequence: $CH_3CONH_2$ $\xrightarrow{P_2O_5, \Delta} CH_3CN$ $\xrightarrow{4H} CH_3CH_2NH_2$.

(A) Nitroalkanes,such as $CH_3-CH_2-NO_2$,undergo reduction in the presence of a catalyst like $Sn/HCl$ or $H_2/Ni$ to form primary amines.
The reaction is: $CH_3-CH_2-NO_2 + 6[H] \to CH_3-CH_2-NH_2 + 2H_2O$.

(C) The reaction sequence is as follows:
$1$. $CH_3COOH + PCl_5 \rightarrow CH_3COCl + POCl_3 + HCl$. Product $(A)$ is Acetyl chloride $(CH_3COCl)$.
$2$. $CH_3COCl + NH_3 \rightarrow CH_3CONH_2 + HCl$. Product $(B)$ is Acetamide $(CH_3CONH_2)$.
$3$. $CH_3CONH_2 + NaBrO \rightarrow CH_3NH_2 + Na_2CO_3 + NaBr$. This is the Hofmann bromamide degradation reaction.
Product $(C)$ is Methylamine $(CH_3NH_2)$.

(A) The reaction of an amide $(CH_3CONH_2)$ with sodium $(Na)$ and alcohol $(ROH)$ is a reduction reaction.
This is a specific method for the reduction of amides to primary amines.
The reaction proceeds as follows: $CH_3CONH_2 + 4[H] \xrightarrow{Na + ROH} CH_3CH_2NH_2 + H_2O$.
Therefore,$Z$ is $CH_3CH_2NH_2$.

(C) The catalytic hydrogenation of nitrobenzene in the presence of platinum $(Pt)$ acts as a reducing agent to convert the nitro group $(-NO_2)$ into an amino group $(-NH_2)$.
The chemical equation is: $C_6H_5NO_2 + 6H \xrightarrow{Pt} C_6H_5NH_2 + 2H_2O$.
Thus,the product formed is aniline.

(D) The reaction of chlorobenzene with ammonia in the presence of $Cu_2O$ at $570 \ K$ is a nucleophilic substitution reaction used for the preparation of aniline.
The chemical equation is:
$2C_6H_5Cl + 2NH_3 \xrightarrow{Cu_2O, \ 570 \ K} 2C_6H_5NH_2 + Cu_2Cl_2 + H_2O$
Thus,the product obtained is aniline.

(C) The formation of diazonium salts,known as the diazotization reaction,occurs when a $1^{\circ}$ aromatic amine reacts with nitrous acid $(HNO_2)$ in the presence of an excess of mineral acid (like $HCl$) at a low temperature of $273-278 \ K$.
$Ar-NH_2 + HNO_2 + 2HX \xrightarrow{273-278 \ K} Ar-N_2^+X^- + 2H_2O$
Therefore,the correct reactant is a primary aromatic amine.

(B) The reduction of nitrobenzene to aniline in an acidic medium is typically carried out using a metal and a mineral acid,such as $Sn/HCl$ or $Fe/HCl$.
The reaction is:
$C_6H_5NO_2 + 6[H] \xrightarrow{Sn/HCl} C_6H_5-NH_2 + 2H_2O$
Therefore,the correct reducing agent is $Sn/HCl$.

(A) In aryl halides $(ArX)$,the carbon-halogen bond acquires partial double bond character due to resonance,making it resistant to nucleophilic substitution by $CN^-$. Thus,$ArX$ does not react with $KCN$ under normal conditions to form $ArCN$.
$(B)$ $ArN_2^+ + CuCN \to ArCN + N_2 + Cu^+$ (Sandmeyer reaction).
$(C)$ $ArCONH_2 \xrightarrow{P_2O_5} ArCN + H_2O$ (Dehydration of amide).
$(D)$ $ArCONH_2 + SOCl_2 \to ArCN + SO_2 + 2HCl$ (Dehydration of amide).

(C) Gabriel's phthalimide synthesis involves the reaction of potassium phthalimide with an alkyl halide followed by alkaline hydrolysis.
This reaction specifically yields $1^{\circ}$ (primary) aliphatic amines.
It cannot be used for the preparation of aromatic amines because aryl halides do not undergo nucleophilic substitution with the phthalimide anion under these conditions.

(A) The preparation of $p-$nitroiodobenzene from $p-$nitroaniline involves two main steps:
$1$. Diazotization: $p-$nitroaniline reacts with $NaNO_2$ and $HCl$ at $0-5^{\circ}C$ to form $p-$nitrobenzenediazonium chloride.
$2$. Iodination: The diazonium salt is then treated with $KI$ (potassium iodide) to replace the diazonium group with an iodine atom,yielding $p-$nitroiodobenzene.
This is a standard method for introducing iodine into an aromatic ring via a diazonium intermediate.

(D) The Friedel-Crafts reaction is used for the alkylation or acylation of aromatic rings,such as the preparation of alkylbenzene or acetophenone.
It is not a method used for the preparation of primary amines.

(C) . The Hofmann bromamide degradation reaction is used to convert an amide into a primary amine containing one carbon atom less than the original amide.
$CH_3CONH_2 + Br_2 + 4KOH \to CH_3NH_2 + K_2CO_3 + 3KBr + 2H_2O$.
Since the number of carbon atoms in the product is less than in the reactant,it is known as a method for stepping down a series.

(A) $LiAlH_4$ is a strong reducing agent capable of reducing amides to primary amines.
The reaction is: $RCONH_2 \xrightarrow{LiAlH_4} RCH_2NH_2$.
$NaBH_4$ is generally too weak to reduce amides to amines.

(B) The reduction of carbylamines (isocyanides) yields secondary amines.
The reaction is: $R-NC + 4[H] \xrightarrow{Ni/H_2} R-NH-CH_3$.
Thus,the correct option is $B$.

(C) The reaction of acetamide with $Br_2$ and caustic soda $(NaOH)$ is known as the Hofmann bromamide degradation reaction.
In this reaction,an amide is converted into a primary amine with one carbon atom less than the original amide.
$CH_3CONH_2 + Br_2 + 4NaOH \to CH_3NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$
Thus,acetamide $(CH_3CONH_2)$ yields methyl amine $(CH_3NH_2)$.

(C) The Hofmann bromamide degradation reaction is a characteristic reaction of primary amides.
In this reaction,a primary amide $(R-CONH_2)$ reacts with bromine $(Br_2)$ in the presence of a strong base like potassium hydroxide $(KOH)$ to form a primary amine $(R-NH_2)$ with one carbon atom less than the original amide.
The balanced chemical equation is:
$R-CONH_2 + Br_2 + 4KOH \xrightarrow{} R-NH_2 + 2KBr + K_2CO_3 + 2H_2O$
Therefore,the compound that undergoes this reaction is $R-CONH_2$.

(B) $1$. The reaction of benzene with a mixture of concentrated $HNO_3$ and concentrated $H_2SO_4$ (nitrating mixture) undergoes electrophilic aromatic substitution to form nitrobenzene as the intermediate.
$2$. Nitrobenzene is then reduced using $Sn/HCl$ (a reducing agent) under heating conditions to form aniline $(C_6H_5NH_2)$.
$3$. The overall reaction is: $C_6H_6$ $\xrightarrow{HNO_3/H_2SO_4} C_6H_5NO_2$ $\xrightarrow{Sn/HCl, \Delta} C_6H_5NH_2$.

(A) The given reaction is the reduction of nitrobenzene to aniline using a metal-acid reducing agent $(Sn + HCl)$.
Nitrobenzene $(C_6H_5NO_2)$ reacts with $6[H]$ provided by the $Sn + HCl$ mixture to form aniline $(C_6H_5NH_2)$ and water $(2H_2O)$.
The reaction is: $C_6H_5NO_2 + 6[H] \xrightarrow{Sn/HCl} C_6H_5NH_2 + 2H_2O$.
Comparing this with the given reaction,'$X$' corresponds to the $-NH_2$ group.
Therefore,the correct option is $(A)$.

(A) The Gabriel phthalimide synthesis is used for the preparation of primary aliphatic amines.
$Aryl$ halides do not undergo nucleophilic substitution reactions with the anion formed by phthalimide easily due to the partial double bond character of the $C-X$ bond and the instability of the phenyl cation.
Therefore,aromatic primary amines like $Aniline$ $(C_6H_5NH_2)$ cannot be prepared by this method.

(B) The reduction of an oxime $(R_2C=NOH)$ using sodium $(Na)$ and ethanol $(EtOH)$ is a standard method for the preparation of primary amines.
In this reaction,the oxime $(CH_3)_2C=NOH$ (acetone oxime) is reduced to isopropylamine $(CH_3)_2CHNH_2$.
Isopropylamine is a primary ($p$-) amine because the nitrogen atom is attached to only one carbon atom.
Therefore,the correct option is $B$.

(C) The $Hofmann$ $bromamide$ $degradation$ $reaction$ is specifically used for the preparation of primary $(1^o)$ amines from amides. In this reaction,an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide,resulting in the formation of a primary amine with one carbon atom less than the original amide. The reaction is represented as: $R-CONH_2 + Br_2 + 4NaOH \rightarrow R-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$.

(C) The reaction $R - CN + 2H_2O \xrightarrow{H^+} R - COOH + NH_3$ represents the acid-catalyzed hydrolysis of a nitrile (cyanide) to a carboxylic acid and ammonia,not an amine.
In contrast,the other reactions are standard methods for preparing amines:
$1$. $R - X + NH_3 \rightarrow R - NH_2$ (Ammonolysis of alkyl halides).
$2$. $R - CH = NOH + 4[H] \rightarrow R - CH_2 - NH_2 + H_2O$ (Reduction of oximes).
$3$. $R - CONH_2 + 4[H] \rightarrow R - CH_2 - NH_2 + H_2O$ (Reduction of amides).

(B) The reagent $Fe/HCl$ is a standard reducing agent used to reduce nitro compounds to primary amines.
In this reaction,the nitro group $(-NO_2)$ attached to the benzene ring $(\phi)$ is reduced to an amino group $(-NH_2)$.
The reaction is: $\phi - NO_2 + 6[H] \xrightarrow{Fe/HCl} \phi - NH_2 + 2H_2O$.
Therefore,the correct conversion is $\phi - NO_2 \rightarrow \phi - NH_2$.

(A) The reaction of an amide $(CH_3CONH_2)$ with sodium and alcohol $(Na/ROH)$ is a reduction reaction.
$CH_3CONH_2 + 4[H] \xrightarrow{Na/ROH} CH_3CH_2NH_2 + H_2O$.
Here,the amide is reduced to the corresponding primary amine,which is $CH_3CH_2NH_2$ (ethanamine).

(A) The Hofmann bromamide degradation reaction is specific to primary amides $(R-CONH_2)$.
When a primary amide reacts with $Br_2$ and $KOH$,it undergoes rearrangement to form a primary amine $(R-NH_2)$ with one carbon atom less than the original amide.
The reaction is: $R-CONH_2 + Br_2 + 4KOH \rightarrow R-NH_2 + K_2CO_3 + 2KBr + 2H_2O$.
Thus,the correct compound is $R-CONH_2$.

(B) The Hofmann hypobromide degradation reaction is specific to primary amides $(RCONH_2)$.
$RCONH_2 + Br_2 + 4NaOH \rightarrow RNH_2 + Na_2CO_3 + 2NaBr + 2H_2O$.
Carbamide,also known as urea $(NH_2CONH_2)$,is a diamide. When subjected to the Hofmann hypobromide reaction,it does not yield a simple primary amine $(RNH_2)$ as the final product because the reaction conditions lead to further degradation or decomposition of the intermediate species.
Isobutanamide ($CH_3)_2CHCONH_2$,Ethanamide $(CH_3CONH_2)$,and Benzamide $(C_6H_5CONH_2)$ are all primary amides and will successfully undergo the Hofmann hypobromide reaction to form their respective primary amines.

(B) Acetone oxime is $(CH_3)_2C=NOH$.
Upon catalytic hydrogenation,the $C=N$ bond is reduced to a $C-NH_2$ bond.
The reaction is: $(CH_3)_2C=NOH + 2H_2 \xrightarrow{Ni/Pt} (CH_3)_2CHNH_2 + H_2O$.
The product $(CH_3)_2CHNH_2$ is propan$-2-$amine,also known as isopropylamine.

(C) The conversion of an amide to a primary amine with one carbon atom less is known as the Hofmann bromamide degradation reaction.
Acetamide $(CH_3CONH_2)$ reacts with bromine $(Br_2)$ in the presence of a strong base like sodium hydroxide $(NaOH)$ to form methylamine $(CH_3NH_2)$.
The chemical equation is: $CH_3CONH_2 + Br_2 + 4NaOH \rightarrow CH_3NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$.
Therefore,the correct reagent is $Br_2 + NaOH$.

(A) The conversion of an amide $(CH_3CONH_2)$ to a nitrile $(CH_3CN)$ is a dehydration reaction.
This is typically achieved using a strong dehydrating agent like phosphorus pentoxide $(P_2O_5)$ or phosphorus oxychloride $(POCl_3)$.
Among the given options,$P_2O_5$ is the standard dehydrating agent used for this transformation.

(C) The reaction used to prepare an amine from an amide is the Hofmann bromamide degradation reaction.
In this reaction,an amide $(R-CONH_2)$ reacts with bromine $(Br_2)$ and an alkali $(KOH)$ to form a primary amine $(R-NH_2)$ with one carbon atom less than the parent amide.
The chemical equation is:
$R-CONH_2 + Br_2 + 4KOH \to R-NH_2 + K_2CO_3 + 2KBr + 2H_2O$

(C) The reaction sequence is as follows:
$1$. $CH_3CONH_2 \xrightarrow{PCl_5} CH_3CN$ (Compound $A$ is acetonitrile).
$2$. $CH_3CN \xrightarrow{Na/EtOH} CH_3CH_2NH_2$ (Compound $B$ is ethylamine).
Reaction $II$ involves the reduction of a nitrile $(R-CN)$ to a primary amine $(R-CH_2NH_2)$ using sodium and ethanol. This specific reduction is known as the Mendius reduction.

(A) The Mendius reaction involves the reduction of nitriles (cyanoalkanes) to primary amines using sodium and ethanol.
The reaction is represented as:
$R-CN + 4[H] \xrightarrow{Na/C_2H_5OH} R-CH_2NH_2$

(B) The reaction of an amide with $Br_2$ and $KOH$ is known as the Hofmann bromamide degradation reaction,which produces a primary amine with one carbon atom less than the parent amide.
To obtain ethylamine $(CH_3CH_2NH_2)$,which has $2$ carbon atoms,the starting amide must be propionamide $(CH_3CH_2CONH_2)$,which has $3$ carbon atoms.
The reaction is: $CH_3CH_2CONH_2 + Br_2 + 4KOH \to CH_3CH_2NH_2 + K_2CO_3 + 2KBr + 2H_2O$.