Preparation of Amines Questions in English

(D) The reaction sequence is as follows:
$1$. $CH_3CH_2CH_2-I + NaCN \rightarrow CH_3CH_2CH_2-CN (A) + NaI$ (Nucleophilic substitution reaction).
$2$. $CH_3CH_2CH_2-CN + H_2O \xrightarrow{OH^-} CH_3CH_2CH_2-CONH_2 (B)$ (Partial hydrolysis of nitrile to amide).
$3$. $CH_3CH_2CH_2-CONH_2 + Br_2 + 4NaOH \rightarrow CH_3CH_2CH_2-NH_2 (C) + Na_2CO_3 + 2NaBr + 2H_2O$ (Hoffmann bromamide degradation reaction).
The final product $C$ is $CH_3CH_2CH_2-NH_2$,which is propylamine.

(A) The reaction shown is the first step of the Gabriel phthalimide synthesis.
Phthalimide reacts with $KOH$ to form potassium phthalimide,which acts as a nucleophile.
This nucleophile then undergoes an $S_N2$ reaction with the alkyl halide,$4-bromobenzyl chloride$ $(Br-C_6H_4-CH_2Cl)$.
In $S_N2$ reactions,the nucleophile attacks the carbon atom attached to the leaving group.
Since the $CH_2Cl$ group is more reactive towards $S_N2$ than the aryl bromide ($C-Br$ bond),the nitrogen of the phthalimide attacks the $CH_2$ carbon,displacing the $Cl^-$ ion.
Therefore,the product is $N-(4-bromobenzyl)phthalimide$.

(C) To obtain $3,4,5-$Tribromoaniline,we start with $p-$nitroaniline.
$1.$ The amino group $(-NH_2)$ is a strong activating group,so treatment with excess $Br_2$ in acetic acid leads to bromination at the $2$ and $6$ positions relative to the $-NH_2$ group (which are the $3$ and $5$ positions relative to the $-NO_2$ group),yielding $2,6-$dibromo$-4-$nitroaniline.
$2.$ The amino group is then converted to a diazonium salt using $NaNO_2/HCl$.
$3.$ The diazonium group is replaced by a bromine atom using $CuBr$ (Sandmeyer reaction),resulting in $3,4,5-$tribromonitrobenzene.
$4.$ Finally,the nitro group $(-NO_2)$ is reduced to an amino group $(-NH_2)$ using $Sn/HCl$ to yield $3,4,5-$tribromoaniline.

(A) Step $1$: Anti-Markovnikov addition of $HBr$ to $1$-methylcyclopentene in the presence of benzoyl peroxide gives $1$-bromo-$2$-methylcyclopentane.
Step $2$: Nucleophilic substitution $(S_N2)$ of the bromide with $KCN$ yields $1$-cyano-$2$-methylcyclopentane.
Step $3$: Reduction of the nitrile group $(-CN)$ using $Na(Hg)/C_2H_5OH$ (Mendius reduction) converts it into a primary amine $(-CH_2NH_2)$.
Thus,the final product $P$ is $1$-methyl-$2$-(aminomethyl)cyclopentane.

(B) Step $1$: Reaction of propanoic acid with $SOCl_2$ gives propanoyl chloride $(B)$.
$CH_3CH_2COOH + SOCl_2 \rightarrow CH_3CH_2COCl + SO_2 + HCl$
Step $2$: Reaction of propanoyl chloride with $NH_3$ gives propanamide $(C)$.
$CH_3CH_2COCl + NH_3 \rightarrow CH_3CH_2CONH_2 + HCl$
Step $3$: Reaction of propanamide with $Br_2$ and $KOH$ is the Hofmann bromamide degradation reaction,which yields ethanamine $(D)$.
$CH_3CH_2CONH_2 + Br_2 + 4KOH \rightarrow CH_3CH_2NH_2 + K_2CO_3 + 2KBr + 2H_2O$
Thus,compound $(D)$ is $CH_3CH_2NH_2$.

(D) The reduction of isonitriles $(R-NC)$ using $H_2/Pd$ or $LiAlH_4$ yields secondary amines $(R-NH-CH_3)$.
$R-N \equiv C \xrightarrow{4[H]} R-NH-CH_3$
Therefore,the correct option is $D$.

(A) The reduction of nitroalkanes with $Sn/HCl$ (or $Fe/HCl$) is a standard method for the preparation of primary amines.
The reaction for nitroethane is:
$CH_3CH_2NO_2 + 6[H] \xrightarrow{Sn/HCl} CH_3CH_2NH_2 + 2H_2O$.
Thus,the product obtained is ethanamine.

(C) Gabriel phthalimide synthesis involves the following steps:
$1$. Phthalimide reacts with $alc. KOH$ to form the potassium salt of phthalimide.
$2$. The potassium salt of phthalimide reacts with an alkyl halide $(R-X)$ to form $N$-alkyl phthalimide.
$3$. $N$-alkyl phthalimide undergoes alkaline hydrolysis (using $NaOH_{(aq)}$) to yield a primary amine $(R-NH_2)$ and the sodium salt of phthalic acid (sodium phthalate).
Phthalic acid itself is not formed as a product; instead,its sodium salt is obtained. Therefore,phthalic acid is not obtained at any stage.

(B) The reaction shown is the conversion of aniline to benzene diazonium chloride.
This process is known as diazotisation.
Aniline reacts with nitrous acid $(HNO_2)$,which is generated in situ by the reaction of $NaNO_2$ and $HCl$ at a low temperature $(273-278 \ K)$,to form benzene diazonium chloride.
Therefore,the correct reagent '$A$' is $NaNO_2+HCl$ at low temperature.

(B) Step $1$: $CH_3Br$ reacts with $KCN$ to form methyl cyanide $(CH_3CN)$ as product $A$.
$CH_3Br + KCN \rightarrow CH_3CN + KBr$
Step $2$: Methyl cyanide $(CH_3CN)$ undergoes reduction with $Na / C_2H_5OH$ (Mendius reduction) to form ethyl amine $(CH_3CH_2NH_2)$ as product $B$.
$CH_3CN + 4[H] \xrightarrow{Na / C_2H_5OH} CH_3CH_2NH_2$
Therefore,the product $B$ is ethyl amine.

(D) When aniline $(C_6H_5NH_2)$ reacts with nitrous acid $(HNO_2)$,which is generated in situ by the reaction of $NaNO_2$ and $HCl$ at a low temperature of $273-278 \ K$,it undergoes a diazotization reaction.
This reaction converts the amino group $(-NH_2)$ into a diazonium group $(-N_2^+Cl^-)$,resulting in the formation of benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$.

(B) $LiAlH_4$ is a strong reducing agent that reduces amides to corresponding amines.
The reaction is: $CH_3CONH_2 \xrightarrow[\text{ether}]{LiAlH_4} CH_3CH_2NH_2$ (Ethanamine).
Thus,the reactant '$A$' is acetamide $(CH_3CONH_2)$.

(D) Step $1$: Reaction of $CH_3Br$ with $KCN$ gives methyl cyanide $(CH_3CN)$ as $A$.
$CH_3Br + KCN \rightarrow CH_3CN + KBr$
Step $2$: Reduction of methyl cyanide $(CH_3CN)$ with $Na / C_2H_5OH$ (Mendius reduction) yields ethylamine $(CH_3CH_2NH_2)$ as $B$.
$CH_3CN + 4[H] \xrightarrow{Na / C_2H_5OH} CH_3CH_2NH_2$

(C) The reagent $Sn / HCl$ is a common reducing agent used to reduce nitro compounds $(R-NO_2)$ to primary amines $(R-NH_2)$.
The balanced chemical equation for the reduction of a nitroalkane is:
$R-NO_2 + 6[H] \xrightarrow{Sn / HCl} R-NH_2 + 2H_2O$.

(D) The Gabriel phthalimide synthesis is a method used for the preparation of primary aliphatic amines. The reaction involves the following steps:
$1$. Phthalimide reacts with ethanolic $KOH$ to form potassium phthalimide.
$2$. Potassium phthalimide reacts with an alkyl halide $(R-X)$ to form $N$-alkylphthalimide.
$3$. $N$-alkylphthalimide undergoes alkaline hydrolysis (using $NaOH_{\text{(aq.)}}$) to yield the corresponding primary amine $(R-NH_2)$ and sodium phthalate.

(C) Gabriel phthalimide synthesis is used for the preparation of primary aliphatic amines.
It cannot be used for the preparation of aromatic amines because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide due to the partial double bond character of the $C-X$ bond and the instability of the phenyl cation.

(A) $1$. The reaction of acetanilide with a nitrating mixture $(conc. HNO_3 + conc. H_2SO_4)$ leads to electrophilic aromatic substitution,primarily yielding $p$-nitroacetanilide as product $A$ due to the steric hindrance at the ortho position.
$2$. The subsequent step involves the hydrolysis of the amide group $(-NHCOCH_3)$ in $p$-nitroacetanilide using acid $(H^+)$ or base $(OH^-)$.
$3$. This hydrolysis converts the $-NHCOCH_3$ group back into an amino group $(-NH_2)$,resulting in the formation of $p$-nitroaniline as product $B$.

(D) The reaction sequence is: $CH_{3}CH_{2}NO_{2}$ $\xrightarrow[\Delta]{HNO_{2}} A$ $\xrightarrow{LiAlH_{4}} B$.
Nitroalkanes like nitroethane $(CH_{3}CH_{2}NO_{2})$ react with nitrous acid $(HNO_{2})$ to form nitrolic acids or pseudonitroles. However,in the context of standard organic synthesis problems,the reduction of a nitro group $(NO_{2})$ using $LiAlH_{4}$ yields a primary amine.
$A$ is the nitro compound $CH_{3}CH_{2}NO_{2}$.
Reduction of $CH_{3}CH_{2}NO_{2}$ with $LiAlH_{4}$ gives $CH_{3}CH_{2}NH_{2}$.
Therefore,$B$ is $CH_{3}CH_{2}NH_{2}$ (Ethanamine).

(C) Gabriel phthalimide synthesis is used for the preparation of aliphatic primary amines.
It involves the nucleophilic substitution of an alkyl halide with the phthalimide anion.
Aromatic primary amines,such as $Aniline$,cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the phthalimide anion under these conditions.

(A) The reduction of ketoximes $(R_2C=NOH)$ using sodium $(Na)$ in ethanol $(C_2H_5OH)$ is a standard method for the preparation of primary amines ($1^{\circ}$ amines).
For example,acetoxime $(CH_3-C(=NOH)-CH_3)$ upon reduction with $Na/C_2H_5OH$ yields isopropylamine $(CH_3-CH(NH_2)-CH_3)$,which is a $1^{\circ}$ amine.

(D) The Mendius reduction involves the reduction of nitriles (alkyl cyanides) to primary amines using sodium $(Na)$ in ethanol $(C_2H_5OH)$.
The reaction is represented as:
$R-C \equiv N + 4[H] \xrightarrow{Na / C_2H_5OH} R-CH_2-NH_2$ (primary amine).

(B) The reaction of phthalimide with an alkyl halide in the presence of a base followed by alkaline hydrolysis is known as the Gabriel phthalimide synthesis.
This method is specifically used for the preparation of primary aliphatic amines $(R-NH_2)$.
It cannot be used for the preparation of aromatic amines $(Ar-NH_2)$ because aryl halides do not undergo nucleophilic substitution with the phthalimide anion.

(C) Amides react with bromine and caustic soda to give their corresponding primary amines. This reaction is known as Hofmann's bromamide degradation reaction.
The chemical equation is:
$CH_{3}CONH_{2} + Br_{2} + 4KOH \xrightarrow{343K} CH_{3}NH_{2} + 2KBr + K_{2}CO_{3} + 2H_{2}O$
Thus,acetamide $(CH_{3}CONH_{2})$ gives methanamine $(CH_{3}NH_{2})$.

(A) The reaction shown is the alkaline hydrolysis of $N$-alkyl phthalimide,which is a step in the Gabriel phthalimide synthesis.
The reaction is: $N$-alkyl phthalimide + $2NaOH_{(aq)} \rightarrow \text{Sodium phthalate} + R-NH_2$ (Primary amine).
Therefore,the product '$A$' is a primary amine.

(B) The reduction of acetamide $(CH_3CONH_2)$ to ethylamine $(CH_3CH_2NH_2)$ using a strong reducing agent like $LiAlH_4$ or $Na/C_2H_5OH$ is represented by the following chemical equation:
$CH_3CONH_2 + 4[H] \xrightarrow{LiAlH_4/ether} CH_3CH_2NH_2 + H_2O$
From the stoichiometry of the reaction,it is clear that $4$ moles of nascent hydrogen $([H])$ are required to reduce $1$ mole of acetamide to $1$ mole of ethylamine.
Therefore,the correct option is $B$.

(D) The correct answer is $D$.
Mendius reduction involves the reduction of nitriles $(R-CN)$ to primary amines $(R-CH_2NH_2)$ using reducing agents like sodium in ethanol or methanol ($Na / C_2H_5OH$ or $Na / CH_3OH$) or lithium aluminum hydride $(LiAlH_4)$.
Explanation of other options:
- $A$: This is Hofmann bromamide degradation,where an amide $(R-CONH_2)$ reacts with bromine and alkali to form a primary amine.
- $B$: This is the $N$-alkylation of amines,where an amine reacts with excess alkyl halide to form a quaternary ammonium salt.
- $C$: This is Hofmann elimination,where a quaternary ammonium hydroxide undergoes thermal decomposition to form an alkene and a tertiary amine.

(D) Mendius reduction is a chemical reaction that involves the reduction of alkyl cyanides (nitriles) to primary amines using sodium in ethanol $(Na/C_2H_5OH)$.
The reaction is represented as: $R-CN + 4[H] \xrightarrow{Na/C_2H_5OH} R-CH_2NH_2$.
This reaction does not involve the loss of a carbon atom as $CO_2$.
Therefore,the statement that one carbon atom is lost as $CO_2$ is incorrect.

(C)
Primary amines can be obtained by the reduction of alkyl cyanide with sodium and ethanol. The chemical reaction is:
$R-C \equiv N + 4[H] \xrightarrow{Na/ethanol} R-CH_2-NH_2$
This reaction is known as Mendius reduction.

(D) The molecular formula $C_4H_{11}N$ corresponds to a saturated amine. Primary amines have the general structure $R-NH_2$.
For a $C_4$ alkyl group,the possible primary amines are:
$1$. $CH_3-CH_2-CH_2-CH_2-NH_2$ (butan-$1$-amine)
$2$. $CH_3-CH_2-CH(CH_3)-NH_2$ (butan-$2$-amine)
$3$. $(CH_3)_2CH-CH_2-NH_2$ ($2$-methylpropan-$1$-amine)
$4$. $(CH_3)_3C-NH_2$ ($2$-methylpropan-$2$-amine)
Thus,there are $4$ possible primary amines.

(B) primary amine has the general structure $R-NH_2$. For the molecular formula $C_4H_{11}N$,the alkyl group $R$ must be $C_4H_9$.
The possible isomers for the butyl group $(C_4H_9-)$ are:
$1$. $n$-butylamine: $CH_3CH_2CH_2CH_2NH_2$
$2$. Isobutylamine: $(CH_3)_2CHCH_2NH_2$
$3$. sec-butylamine: $CH_3CH_2CH(NH_2)CH_3$
$4$. tert-butylamine: $(CH_3)_3CNH_2$
Thus,there are $4$ possible primary amines.

(D) is methyl cyanide $(CH_3-CN)$ formed by the nucleophilic substitution of $CH_3-I$ with $KCN$.
The reduction of $CH_3-CN$ using $Na / C_2H_5OH$ is known as the Mendius reduction,which produces ethylamine $(CH_3-CH_2-NH_2)$ as product $B$.
$CH_3-I + KCN \rightarrow CH_3-CN (A) + KI$
$CH_3-CN + 4[H] \xrightarrow{Na / C_2H_5OH} CH_3-CH_2-NH_2 (B)$

(D) Step $1$: The reaction of methyl iodide $(CH_3I)$ with alcoholic $KCN$ is a nucleophilic substitution reaction $(S_N2)$ which yields methyl cyanide $(CH_3CN)$ as product $A$.
$CH_3-I + KCN \rightarrow CH_3-CN + KI$
Step $2$: The reduction of methyl cyanide $(CH_3CN)$ with sodium $(Na)$ in the presence of ethanol $(C_2H_5OH)$ is known as Mendius reduction,which produces ethylamine $(CH_3CH_2NH_2)$ as product $B$.
$CH_3-CN + 4[H] \xrightarrow[C_2H_5OH]{Na} CH_3-CH_2-NH_2$

(A) Step $1$: Reaction of $CH_3Br$ with $KCN$ is a nucleophilic substitution reaction.
$CH_3Br + KCN \rightarrow CH_3CN + KBr$
Here,$X$ is $CH_3CN$ (methyl cyanide or acetonitrile).
Step $2$: Reduction of $CH_3CN$ with $Na / C_2H_5OH$ (Mendius reduction) yields a primary amine.
$CH_3CN + 4[H] \xrightarrow{Na / C_2H_5OH} CH_3-CH_2-NH_2$
Thus,$Y$ is $CH_3-CH_2-NH_2$ (ethylamine).

(B) The reaction of $CH_3Br$ with $AgNO_2$ is a nucleophilic substitution reaction that yields nitromethane $(A = CH_3NO_2)$.
$CH_3Br + AgNO_2 \rightarrow CH_3NO_2 + AgBr$
Reduction of nitromethane with $Sn / HCl$ (a reducing agent) converts the nitro group $(-NO_2)$ into an amine group $(-NH_2)$,yielding methanamine $(B = CH_3NH_2)$.
$CH_3NO_2 + 6[H] \xrightarrow{Sn / HCl} CH_3NH_2 + 2H_2O$

(B) The reaction proceeds as follows:
$1$. $CH_3Br$ reacts with $AgNO_2$ to form nitromethane $(A)$ as the major product: $CH_3Br + AgNO_2 \longrightarrow CH_3NO_2 + AgBr$.
$2$. Nitromethane $(CH_3NO_2)$ is then reduced by $Sn / HCl$ (a reducing agent) to form methylamine $(B)$: $CH_3NO_2 + 6[H] \xrightarrow{Sn / HCl} CH_3NH_2 + 2H_2O$.
Therefore,product $B$ is $CH_3NH_2$.

(A) When an alkyl halide $(R-X)$ is reacted with an excess of alcoholic ammonia $(NH_3)$,the reaction is an ammonolysis reaction.
Since ammonia is in large excess,the probability of the alkyl halide molecule colliding with an ammonia molecule is much higher than colliding with the formed amine.
Therefore,the reaction stops at the formation of the primary amine $(R-NH_2)$:
$R-X + NH_3 (\text{alc.}) \xrightarrow{\Delta} R-NH_2 + HX$
Thus,the correct option is $A$.

(A) The reaction of sodium $\alpha$-chloroacetate with aqueous sodium nitrite involves the substitution of the chlorine atom by the nitro group,followed by decarboxylation upon boiling with water.
The reaction proceeds as follows:
$Cl-CH_2-COONa + NaNO_2 \rightarrow O_2N-CH_2-COONa + NaCl$
Then,the intermediate $O_2N-CH_2-COONa$ undergoes decarboxylation in the presence of water:
$O_2N-CH_2-COONa + H_2O \rightarrow CH_3NO_2 + NaHCO_3$
Thus,the final product is nitromethane $(CH_3NO_2)$.

(C) When bromoethane $(CH_3CH_2Br)$ is heated with an excess of alcoholic ammonia,it undergoes a nucleophilic substitution reaction (ammonolysis) to form ethanamine $(CH_3CH_2NH_2)$ as the primary product.
$CH_3CH_2Br + NH_3 (\text{excess}) \rightarrow CH_3CH_2NH_2 + HBr$
Since ammonia is in excess,the reaction stops at the primary amine stage.

(C) Let us analyze each reaction:
$I.$ Ammonolysis of alkyl halides $(R-X + NH_3 \rightarrow R-NH_2)$ yields amines.
$II.$ Reduction of nitriles $(R-C \equiv N + 4[H] \rightarrow R-CH_2NH_2)$ yields primary amines.
$III.$ Acidic hydrolysis of nitriles $(R-C \equiv N + 2H_2O + H^+ \rightarrow R-COOH + NH_4^+)$ yields carboxylic acids,not amines.
$IV.$ Reduction of amides $(R-CONH_2 + 4[H] \rightarrow R-CH_2NH_2 + H_2O)$ yields primary amines.
Therefore,only reaction $III$ does not yield an amine.

(B) The given reaction sequence is:
$CH_3CN$ $\xrightarrow{\text{Reduction}} CH_3CH_2NH_2$ $\xrightarrow{HNO_2} CH_3CH_2OH$
Here,$A$ is $CH_3CN$ (ethane nitrile),$B$ is $CH_3CH_2NH_2$ (ethylamine),and the final product is ethanol.
Therefore,the compound $A$ is ethane nitrile.

(A) The reaction described is the Hofmann bromamide degradation reaction.
In this reaction,an amide is treated with bromine $(Br_{2})$ and an aqueous solution of a strong base like potassium hydroxide $(KOH)$ to produce a primary amine.
The reaction is as follows:
$R-CONH_{2} + Br_{2} + 4KOH \rightarrow R-NH_{2} + K_{2}CO_{3} + 2KBr + 2H_{2}O$
This reaction is used to decrease the carbon chain length by one carbon atom,which is removed as a carbonate ion $(CO_{3}^{2-})$.

(C) Hofmann's bromamide reaction is a degradation reaction used to convert a primary amide into a primary amine with one carbon atom less than the original amide.
The general chemical equation is:
$RCONH_{2} + Br_{2} + 4KOH \longrightarrow RNH_{2} + K_{2}CO_{3} + 2KBr + 2H_{2}O$
Thus,the reaction converts $\text{Amide} \longrightarrow \text{Amine}$.

(D) Aryl halides,such as chlorobenzene,do not undergo nucleophilic substitution reactions with potassium phthalimide under ordinary conditions because the $C-Cl$ bond in chlorobenzene has partial double bond character and the electron-rich benzene ring repels the nucleophile.
Therefore,the Gabriel phthalimide synthesis cannot be used for the preparation of aniline.
Thus,method $(d)$ is incorrect for the preparation of aniline.

(C) Gabriel phthalimide synthesis is a method used to prepare pure primary amines.
$1$. Phthalimide is reacted with alcoholic $KOH$ to form potassium phthalimide.
$2$. Potassium phthalimide is then reacted with an alkyl halide to form $N$-alkyl phthalimide.
$3$. Finally,$N$-alkyl phthalimide undergoes alkaline hydrolysis using aqueous $NaOH$ to yield the primary amine and phthalic acid (as a salt).
$HCl$ is not used in this process as the final step requires basic hydrolysis to release the amine.

(D) The first step is nitration because the nitro group $(-NO_2)$ is a meta-directing group,which is required to place the bromine atom at the $m$-position relative to the amino group.
Next,bromination of nitrobenzene is performed to introduce the bromine atom at the $m$-position.
Finally,the reduction of the nitro group to an amino group $(-NH_2)$ yields $m$-bromoaniline.
The reaction sequence is:
Benzene $\xrightarrow{\text{nitration}}$ Nitrobenzene $\xrightarrow{\text{bromination}}$ $m$-Bromonitrobenzene $\xrightarrow{\text{reduction}}$ $m$-Bromoaniline

(D) Benzyl amine is $C_6H_5CH_2NH_2$.
Option $A$ gives $N$-methylaniline.
Option $B$ gives benzyl isocyanide followed by reduction to $C_6H_5CH_2NHCH_3$.
Option $C$ is the Hofmann bromamide degradation,which gives aniline $(C_6H_5NH_2)$.
Option $D$ is the reduction of benzamide $(C_6H_5CONH_2)$ using $LiAlH_4$,which yields benzyl amine $(C_6H_5CH_2NH_2)$.

(B) The given reaction involves two different chemical transformations of benzamide $(C_6H_5CONH_2)$:
$1$. Reaction with $LiAlH_4$ followed by $H_2O$: This is a reduction reaction. $LiAlH_4$ reduces the amide group $(-CONH_2)$ to an amine group $(-CH_2NH_2)$. Thus,$Y$ is benzylamine $(C_6H_5CH_2NH_2)$.
$2$. Reaction with $Br_2$ and $NaOH$: This is the Hoffmann bromamide degradation reaction. It converts an amide $(-CONH_2)$ into a primary amine with one carbon atom less $(-NH_2)$. Thus,$X$ is aniline $(C_6H_5NH_2)$.
Therefore,$X$ is $C_6H_5NH_2$ and $Y$ is $C_6H_5CH_2NH_2$.

(C) Gabriel phthalimide synthesis is used for the preparation of primary aliphatic amines.
It cannot be used for the preparation of aromatic primary amines (like Phenylamine or Aniline) because the aryl halides do not undergo nucleophilic substitution reaction with the potassium phthalimide salt easily due to the partial double bond character of the $C-X$ bond in aryl halides.