Preparation of Amines Questions in English

(D) Gabriel phthalimide synthesis is specifically used for the preparation of primary $(1^{\circ})$ aliphatic amines.
It involves the reaction of potassium phthalimide with an alkyl halide,followed by alkaline hydrolysis.
Since aryl halides do not undergo nucleophilic substitution with the phthalimide anion,aromatic amines (like aniline) cannot be prepared by this method.
Secondary and tertiary amines are also not prepared by this method.
Among the given options,ethylamine $(CH_3CH_2NH_2)$ is a primary aliphatic amine,whereas aniline is aromatic,and the others are secondary or tertiary amines.

(B) The reaction of aniline with nitrous acid $(HNO_2)$ at low temperature $(273-278 \ K)$ is known as the diazotization reaction.
In this reaction,aniline reacts with $HNO_2$ (generated in situ from $NaNO_2$ and $HCl$) to form benzene diazonium chloride.
The chemical equation is:
$C_6H_5NH_2 + HNO_2 + HCl \xrightarrow{273-278 \ K} C_6H_5N_2^+Cl^- + 2H_2O$.
Thus,the product $A$ is benzene diazonium salt.

(C) To prepare a $1^{\circ}$-amine from an alkyl halide with the simultaneous addition of one $-CH_2-$ group to the carbon chain,$KCN$ is used as the reagent.
The reaction proceeds as follows:
$1$. Nucleophilic substitution: $R-X + KCN \rightarrow R-CN + KX$
$2$. Reduction: $R-CN + 4[H] \rightarrow R-CH_2-NH_2$
$KCN$ provides the cyanide ion $(CN^-)$,which acts as a nucleophile to extend the carbon chain by one carbon atom,and subsequent reduction of the nitrile group yields the $1^{\circ}$-amine.

(D) In the given reaction,$X$ is $RCONH_2$ and $Y$ is $RCN$.
$LiAlH_4$ is a strong reducing agent that reduces amides $(RCONH_2)$ to primary amines $(RCH_2NH_2)$.
Catalytic hydrogenation of nitriles $(RCN)$ using $H_2/Ni$ also yields primary amines $(RCH_2NH_2)$.

(D) and $B$ are precursors that can be reduced to form the primary amine $R-CH_2-NH_2$.
$1.$ Nitriles $(R-CN)$ undergo catalytic hydrogenation with $H_2/Ni$ to produce primary amines: $R-CN + 2H_2 \xrightarrow{Ni} R-CH_2-NH_2$
$2.$ Amides $(R-CONH_2)$ are reduced by $LiAlH_4$ followed by hydrolysis to produce primary amines: $R-CONH_2 \xrightarrow[ii) H_2O]{i) LiAlH_4} R-CH_2-NH_2$
Therefore,$A = R-CN$ and $B = R-CONH_2$.

(B) In the reaction of $C_6H_5NO_2$ with $Zn$ powder in the presence of alcoholic $KOH$,the major product is azoxybenzene or hydrazobenzene depending on conditions,not aniline.
$C_6H_5OH + NH_3 \xrightarrow[300^{\circ}C]{ZnCl_2}$ gives aniline.
$C_6H_5Cl + NH_3 \xrightarrow[Cu_2O]{200^{\circ}C, \text{high pressure}}$ gives aniline.
$C_6H_5NO_2 + 6[H] \xrightarrow[HCl]{Fe + H_2O}$ is the reduction of nitrobenzene to aniline.
Therefore,the reaction that does not produce aniline as the major product is $C_6H_5NO_2 + Zn \text{ powder } \xrightarrow{\text{alcoholic } KOH}$.

(B) The reaction sequence is as follows:
$1$. The reaction of chlorocyclopentane with ethanolic $KCN$ undergoes a nucleophilic substitution $(S_N2)$ reaction to form cyclopentanecarbonitrile $(C_5H_9CN)$.
$2$. The reduction of the nitrile group $(-CN)$ using sodium amalgam $(Na(Hg))$ in the presence of ethanol $(C_2H_5OH)$ yields a primary amine. This is a standard reduction reaction for nitriles to primary amines,resulting in cyclopentylmethanamine $(C_5H_9CH_2NH_2)$.

(B) $1$. For the reaction $X \xrightarrow{H_2/Ni} CH_3CH_2CH_2NH_2$,the reduction of a nitrile $(R-CN)$ with $H_2/Ni$ adds four hydrogen atoms to produce a primary amine $(R-CH_2NH_2)$. To obtain $CH_3CH_2CH_2NH_2$ (propan$-1-$amine),$X$ must be $CH_3CH_2CN$ (propanenitrile).
$2$. For the reaction $Y \xrightarrow{Br_2/NaOH} CH_3CH_2CH_2NH_2$,this is the Hofmann bromamide degradation reaction,which converts an amide $(R-CONH_2)$ into a primary amine $(R-NH_2)$ with one less carbon atom. To obtain $CH_3CH_2CH_2NH_2$ (propan$-1-$amine),$Y$ must be $CH_3CH_2CH_2CONH_2$ (butanamide).
$3$. Therefore,$X = CH_3CH_2CN$ and $Y = CH_3CH_2CH_2CONH_2$.

(A) The reaction of benzamide $(C_6 H_5 CONH_2)$ with $LiAlH_4$ followed by hydrolysis is a reduction reaction that converts the amide group $(-CONH_2)$ into an amine group $(-CH_2 NH_2)$. Thus,$P$ is benzylamine $(C_6 H_5 CH_2 NH_2)$.
The reaction of benzamide $(C_6 H_5 CONH_2)$ with $Br_2 / NaOH$ is the Hofmann bromamide degradation reaction,which converts an amide into a primary amine with one carbon atom less. Thus,$Q$ is aniline $(C_6 H_5 NH_2)$.
Therefore,$P = C_6 H_5 CH_2 NH_2$ and $Q = C_6 H_5 NH_2$.

(B) Gabriel-phthalimide synthesis is a standard method for the preparation of pure primary aliphatic amines.
$II$. Hoffmann bromamide degradation reaction is used to convert an amide into a primary amine with one carbon atom less.
$III$. Carbylamine reaction is a test for primary amines,not a preparation method.
$IV$. Sandmeyer reaction is used to convert diazonium salts into aryl halides,not for preparing aliphatic amines.
Therefore,only $I$ and $II$ are used for the preparation of aliphatic primary amines.

(A) The reaction $C_2H_5Br \xrightarrow{\text{excess } NH_3} A$ represents the ammonolysis of an alkyl halide,where $A$ is ethylamine $(CH_3CH_2NH_2)$.
The reaction $B \xrightarrow{Br_2/KOH} A$ is the Hofmann bromamide degradation reaction,which converts an amide into a primary amine with one carbon atom less than the original amide.
Since $A$ is ethylamine $(CH_3CH_2NH_2)$,which has $2$ carbon atoms,the amide $B$ must have $3$ carbon atoms.
Therefore,$B$ is propanamide $(CH_3CH_2CONH_2)$.

(D) The preparation of a secondary amine is best achieved by the reaction of a primary amine with an alkyl halide.
This is a nucleophilic substitution reaction where the primary amine acts as a nucleophile and attacks the alkyl halide to form a secondary amine.
$R-NH_2 + R'-X \rightarrow R-NH-R' + HX$.
Option $A$ (Hoffmann bromamide degradation) produces a primary amine.
Option $B$ (Reduction of a nitrile) produces a primary amine.
Option $C$ (Reaction of an alkyl halide with excess ammonia) primarily produces a primary amine.

(C) The Hoffmann bromamide degradation reaction is a method used to convert a primary amide into a primary amine containing one carbon atom less than the starting amide.
The general reaction is: $R-CONH_2 + Br_2 + 4KOH \rightarrow R-NH_2 + K_2CO_3 + 2KBr + 2H_2O$.
Among the given options,$CH_3CONH_2$ (acetamide) is the only primary amide. The other compounds are a nitrile $(CH_3CN)$,a secondary amide $(CH_3CONHCH_3)$,and an isocyanide $(CH_3NC)$.
Therefore,$CH_3CONH_2$ undergoes the Hoffmann degradation reaction.

(D) The reduction of nitrobenzene $(C_6H_5NO_2)$ in an acidic medium using $Zn / HCl$ results in the formation of aniline $(C_6H_5NH_2)$ as the main product.
The chemical reaction is:
$C_6H_5NO_2 + 6[H] \xrightarrow{Zn/HCl} C_6H_5NH_2 + 2H_2O$

(B) In the reaction of nitrobenzene $(C_6H_5NO_2)$ with zinc powder in the presence of alcoholic $KOH$,the product formed is azoxybenzene or other reduction products depending on conditions,but not aniline as the major product.
Option $A$ is the ammonolysis of phenol.
Option $C$ is the reaction of chlorobenzene with ammonia (ammonolysis).
Option $D$ is the standard reduction of nitrobenzene to aniline using $Fe/HCl$.

(D) The Hoffmann bromamide degradation reaction is used to prepare primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide.
In this reaction,the carbonyl carbon of the amide is removed as a carbonate,and the alkyl or aryl group migrates to the nitrogen atom.
As a result,the amine formed contains one carbon atom less than the starting amide.
To obtain pentanamine $(C_5H_{11}NH_2)$,which has $5$ carbon atoms,the starting amide must have $6$ carbon atoms,which is hexanamide $(CH_3CH_2CH_2CH_2CH_2CONH_2)$.
The reaction is:
$CH_3CH_2CH_2CH_2CH_2CONH_2 + Br_2 + 4NaOH \rightarrow CH_3CH_2CH_2CH_2CH_2NH_2 + 2NaBr + Na_2CO_3 + 2H_2O$

(C) $1$. $X$ is formed from $C_6H_5CN$ and dissolves in dil. $HCl$. This indicates that $X$ is a basic compound,likely an amine $(C_6H_5CH_2NH_2)$. The reduction of nitrile to primary amine is achieved by $LiAlH_4$ followed by $H_2O$.
$2$. $Y$ is formed from $C_6H_5CN$ and reacts with $2,4-DNP$. This indicates that $Y$ is a carbonyl compound (aldehyde),$C_6H_5CHO$. The partial reduction of nitrile to aldehyde is achieved by $DIBAL-H$ followed by $H_2O$.
$3$. Therefore,$A = LiAlH_4, H_2O$ and $B = DIBAL-H, H_2O$.

(A) Gabriel phthalimide synthesis is used for the preparation of aliphatic primary amines without the loss of carbons. It involves nucleophilic substitution $(S_{N}2)$ of alkyl halides by the anion formed by the phthalimide.
Phthalimide on treatment with ethanolic $KOH$ forms potassium salt of phthalimide which on heating with $RX$ followed by either alkaline hydrolysis or hydrazinolysis with hydrazine produces the corresponding $1^{\circ}$ amine.

(A) The reaction involves the treatment of $2$-bromobenzyl chloride with ammonia $(NH_3)$ in ethanol $(EtOH)$. This is a nucleophilic substitution reaction $(S_N2)$ where the nucleophile $NH_3$ attacks the electrophilic carbon of the $-CH_2Cl$ group,displacing the chloride ion $(Cl^-)$. The bromine atom attached to the benzene ring is relatively inert towards nucleophilic substitution under these conditions. Therefore,the $-CH_2Cl$ group is converted into a $-CH_2NH_2$ group,resulting in the formation of $2$-bromobenzylamine.

(C) Nitriles $(R-C\equiv N)$ are reduced to primary amines $(R-CH_2-NH_2)$ using strong reducing agents.
$1$. $\text{LiAlH}_4$ (Lithium Aluminium Hydride) is a strong reducing agent that reduces nitriles to primary amines.
$2$. $\text{H}_2/\text{Ni}$ (Catalytic hydrogenation) also reduces nitriles to primary amines.
$3$. $\text{Na(Hg)}/\text{C}_2\text{H}_5\text{OH}$ is known as the Mendius reduction,which reduces nitriles to primary amines.
$4$. $\text{Sn} + \text{HCl}$ is primarily used for the reduction of nitro compounds to amines.
$5$. $\text{Br}_2/\text{aq. NaOH}$ is used for the Hofmann Bromamide degradation of amides to primary amines with one carbon atom less.
Therefore,the correct reagents are $A, C$,and $D$.