Preparation of Amines Questions in English

(B) The reaction of a nitrile $(R-CN)$ with excess $LiAlH_4$ followed by acidic workup $(H_3O^+)$ results in the reduction of the nitrile group to a primary amine $(R-CH_2NH_2)$.
In this case,the reactant is butanenitrile $(CH_3CH_2CH_2CN)$.
The reduction process is:
$CH_3CH_2CH_2CN \xrightarrow[2. H_3O^+]{1. LiAlH_4 (excess)} CH_3CH_2CH_2CH_2NH_2$
The product formed is butan$-1-$amine.

(A) The reaction sequence is as follows:
$1$. The final product is $CH_3CH_2CH_2NH_2$ (propan$-1-$amine),which is formed by the Hofmann bromamide degradation reaction.
$2$. The Hofmann bromamide degradation reaction converts an amide $(RCONH_2)$ into a primary amine $(RNH_2)$ with one carbon atom less.
$3$. Therefore,the intermediate amide must be butanamide $(CH_3CH_2CH_2CONH_2)$.
$4$. The amide is formed by the reaction of an acid chloride $(RCOCl)$ with ammonia $(NH_3)$.
$5$. Thus,the compound $(X)$ is butanoyl chloride $(CH_3CH_2CH_2COCl)$.

(D) The reaction sequence is as follows:
$1$. The starting material is $p$-nitrobenzonitrile $(O_2N-C_6H_4-CN)$.
$2$. Treatment with $1 \ mole$ of $HO^-$ (alkaline hydrolysis) converts the nitrile group $(-CN)$ into an amide group $(-CONH_2)$.
$3$. The resulting intermediate is $p$-nitrobenzamide $(O_2N-C_6H_4-CONH_2)$.
$4$. Subsequent treatment with $HO^-$ and $Br_2$ constitutes the Hofmann bromamide degradation reaction.
$5$. This reaction converts the amide group $(-CONH_2)$ into a primary amine group $(-NH_2)$,resulting in $p$-nitroaniline $(O_2N-C_6H_4-NH_2)$.
$6$. Thus,the final product $(A)$ is $p$-nitroaniline.

(B) $1.$ Reaction of $PhCOOH$ with $PCl_5$ followed by $NH_3$ yields benzamide $(PhCONH_2)$,which is product $A$.
$2.$ Dehydration of benzamide $(PhCONH_2)$ with $P_4O_{10}$ under heating gives benzonitrile $(PhCN)$.
$3.$ Catalytic hydrogenation of benzonitrile $(PhCN)$ with $H_2/Ni$ reduces the nitrile group to a primary amine,resulting in benzylamine $(PhCH_2NH_2)$,which is product $B$.

(A) The reaction of ethyl bromide with $AgCN$ produces ethyl isocyanide as the major product because $AgCN$ is a covalent compound.
$C_2H_5Br + AgCN \rightarrow C_2H_5NC (X) + AgBr$
Reduction of isocyanides with reducing agents like $LiAlH_4$ or catalytic hydrogenation yields secondary amines.
$C_2H_5NC + 4[H] \rightarrow C_2H_5NHCH_3 (Y)$
Thus,$Y$ is ethyl methyl amine.

(A) $1$. Compound $A$ is $C_7H_6O_2$. When treated with aqueous ammonia and heated,it forms compound $B$. This suggests $A$ is benzoic acid $(C_6H_5COOH)$,which reacts with $NH_3$ to form ammonium benzoate,and upon heating,it dehydrates to form benzamide $(C_6H_5CONH_2)$ as compound $B$.
$2$. Compound $B$ (benzamide,$C_6H_5CONH_2$) reacts with $Br_2$ and $KOH$ (Hofmann bromamide degradation) to form aniline $(C_6H_5NH_2)$ as compound $C$.
$3$. The molecular formula of aniline is $C_6H_7N$,which matches the given formula for $C$.
$4$. Therefore,compound $A$ is benzoic acid.

(B) The Gabriel phthalimide synthesis is used for the preparation of primary aliphatic amines.
In this reaction,potassium phthalimide reacts with an alkyl halide $(R-X)$ to form an $N-$alkylphthalimide,which on hydrolysis gives a primary amine $(R-NH_2)$.
This reaction involves an $S_N2$ mechanism. Therefore,it works best with primary alkyl halides.
Secondary and tertiary alkyl halides undergo elimination reactions instead of substitution,and aryl halides do not undergo this reaction due to the partial double bond character of the $C-X$ bond.
Among the given options,$n-$butylamine is a primary amine derived from a primary alkyl halide ($n-$butyl bromide),which readily undergoes $S_N2$ substitution.
$t-$butylamine (tertiary) and triethylamine (tertiary) cannot be prepared by this method.
Thus,the correct answer is $n-$butylamine.

(D) The conversion of $N$-ethylphthalimide to ethylamine is the final step of the Gabriel phthalimide synthesis.
This reaction involves the hydrolysis of $N$-ethylphthalimide using an aqueous base (like $NaOH$) followed by acidification,or more commonly,treatment with hydrazine $(NH_2NH_2)$ to cleave the phthalimide ring and release the primary amine.
In the context of the provided options,$NH_2NH_2$ is the standard reagent used to liberate the primary amine from the $N$-substituted phthalimide.

(A) Cyanobenzene $(C_6H_5CN)$ can be reduced to benzylamine $(C_6H_5CH_2NH_2)$ using strong reducing agents like $LiAlH_4$ or catalytic hydrogenation $(H_2/Ni)$.
Option $A$ involves $HCl/H_2O$ (hydrolysis) which converts the nitrile to a carboxylic acid,and $NaBH_4$ is not strong enough to reduce nitriles to amines.
Option $C$ involves the Stephen reduction $(SnCl_2/HCl)$ which typically reduces nitriles to imines,and $NaBH_4$ is not the standard reagent to complete this specific conversion to a primary amine.
Therefore,options $A$ and $C$ are not standard methods for this preparation.

(D) The reaction sequence is as follows:
$CH_3-COOH + NH_3 \rightarrow CH_3-COONH_4 (X)$ (Ammonium acetate)
$CH_3-COONH_4 \xrightarrow{\Delta} CH_3-CONH_2 (Y) + H_2O$ (Acetamide)
$CH_3-CONH_2 \xrightarrow{LiAlH_4} CH_3-CH_2-NH_2 (Z)$ (Ethylamine)
Thus,$Z$ is $CH_3-CH_2-NH_2$ (Ethylamine).

(A) Acetophenone reacts with hydroxylamine $(NH_2OH)$ in an acidic medium to form acetophenone oxime $(X)$.
The reduction of this oxime with sodium and ethanol $(Na/C_2H_5OH)$ yields $1$-phenylethanamine $(Y)$.
Reaction:
$C_6H_5-C(=O)-CH_3 + NH_2OH \xrightarrow{H^{+}} C_6H_5-C(=NOH)-CH_3 (X) + H_2O$
$C_6H_5-C(=NOH)-CH_3 \xrightarrow{Na/C_2H_5OH} C_6H_5-CH(NH_2)-CH_3 (Y)$

(B) The reaction of $CH_3CN$ (acetonitrile) with $Na + C_2H_5OH$ is a reduction reaction known as the $Mendius$ reduction.
In this reaction,the nitrile group $(-CN)$ is reduced to a primary amine $(-CH_2NH_2)$.
The chemical equation is: $CH_3CN + 4[H] \xrightarrow{Na/C_2H_5OH} CH_3CH_2NH_2$.
Thus,the product $x$ is $CH_3CH_2NH_2$ (ethylamine).

(B) The reaction sequence is as follows:
$1.$ $CH_3-CH_2-COOH + SOCl_2 \rightarrow CH_3-CH_2-COCl + SO_2 + HCl$ (Formation of acid chloride $[X]$)
$2.$ $CH_3-CH_2-COCl + NH_3 \xrightarrow{\Delta} CH_3-CH_2-CONH_2 + HCl$ (Formation of amide $[Y]$)
$3.$ $CH_3-CH_2-CONH_2 + Br_2 + 4KOH \rightarrow CH_3-CH_2-NH_2 + K_2CO_3 + 2KBr + 2H_2O$ (Hoffmann Bromamide Degradation to form amine $[Z]$)
Thus,$[Z]$ is $CH_3-CH_2-NH_2$.

(C) The reaction sequence is as follows:
$1$. Compound $A$ ($C_6H_5COOH$,benzoic acid) reacts with aqueous ammonia $(NH_3)$ followed by heating to form compound $B$ ($C_6H_5CONH_2$,benzamide).
$2$. Benzamide $(C_6H_5CONH_2)$ undergoes the Hofmann bromamide degradation reaction with $Br_2$ and $KOH$ to form compound $C$ ($C_6H_5NH_2$,aniline).
$3$. The molecular formula of aniline is $C_6H_7N$,which matches the given formula.
Therefore,the correct sequence is $A = C_6H_5COOH$,$B = C_6H_5CONH_2$,and $C = C_6H_5NH_2$.

(C) Step $1$: $CH_3-CH_2-I$ reacts with $NaCN$ to form $CH_3-CH_2-CN$ $(A)$ via nucleophilic substitution.
Step $2$: Partial hydrolysis of $CH_3-CH_2-CN$ $(A)$ in the presence of $\Theta OH$ yields $CH_3-CH_2-CONH_2$ $(B)$ (propanamide).
Step $3$: The reaction of $CH_3-CH_2-CONH_2$ $(B)$ with $Br_2$ and $NaOH$ is the Hofmann bromamide degradation reaction,which converts an amide into a primary amine with one carbon atom less.
Step $4$: The product $C$ is $CH_3-CH_2-NH_2$ (ethanamine).

(A) The reaction proceeds in two steps:
$1$. Ethyl chloride $(C_2H_5Cl)$ reacts with potassium cyanide $(KCN)$ to form ethyl cyanide (propanenitrile): $C_2H_5Cl + KCN \rightarrow C_2H_5CN + KCl$.
$2$. The reduction of ethyl cyanide $(C_2H_5CN)$ with sodium $(Na)$ and alcohol $(C_2H_5OH)$ (Mendius reduction) yields propylamine ($C_2H_5CH_2NH_2$ or $CH_3CH_2CH_2NH_2$): $C_2H_5CN + 4[H] \xrightarrow{Na/C_2H_5OH} CH_3CH_2CH_2NH_2$.

(D) The conversion of an amide $(R-CONH_2)$ to an amine $(R-CH_2NH_2)$ with the same number of carbon atoms is best achieved by reduction using $LiAlH_4$ in ether.
The reaction is as follows:
$C_6H_5-CH(CH_3)-CONH_2 \xrightarrow{LiAlH_4 / \text{ether}} C_6H_5-CH(CH_3)-CH_2NH_2$
Option $B$ ($Br_2$ in aqueous $NaOH$) is used for the Hofmann bromamide degradation reaction,which converts an amide to an amine with one carbon atom less than the parent amide.

(C) The reaction sequence is as follows:
$1$. $m$-bromobenzoic acid reacts with $SOCl_2$ to form $m$-bromobenzoyl chloride $(B)$.
$2$. $m$-bromobenzoyl chloride reacts with $NH_3$ to form $m$-bromobenzamide $(C)$.
$3$. $m$-bromobenzamide undergoes the Hofmann bromamide degradation reaction with $NaOH$ and $Br_2$ to form $m$-bromoaniline $(D)$.
Therefore,the final product $D$ is $m$-bromoaniline.

(A) The reduction of nitroalkanes $(R-NO_2)$ yields primary amines $(R-NH_2)$.
For example,$CH_3CH_2NO_2 + 6[H] \xrightarrow{Sn/HCl} CH_3CH_2NH_2 + 2H_2O$.
Option $B$ is an alkyl nitrite,which on reduction gives alcohol and ammonia.
Option $C$ is azobenzene,which on reduction gives aniline (a primary amine),but nitroalkanes are the standard textbook example for primary amine preparation via reduction.
Option $D$ is an isocyanide,which on reduction gives a secondary amine $(R-NH-CH_3)$.

$(i)$ $C_2H_5-Cl + NH_3 (\text{ethanolic}) \xrightarrow{\Delta} C_2H_5-NH_2 + HCl$
$(ii)$ $C_6H_5-CH_2-Cl + NH_3 \rightarrow C_6H_5-CH_2-NH_2 + HCl$
$C_6H_5-CH_2-NH_2 + 2CH_3-Cl \rightarrow C_6H_5-CH_2-N(CH_3)_2 + 2HCl$

(N/A) $(i)$ $CH_3-CH_2-Cl$ $\xrightarrow{Ethanolic \ NaCN} CH_3-CH_2-CN$ $\xrightarrow{H_2/Ni \ \text{or} \ LiAlH_4} CH_3-CH_2-CH_2-NH_2$
$(ii)$ $C_6H_5-CH_2-Cl$ $\xrightarrow{Ethanolic \ NaCN} C_6H_5-CH_2-CN$ $\xrightarrow{H_2/Ni \ \text{or} \ LiAlH_4} C_6H_5-CH_2-CH_2-NH_2$

(N/A) $(i)$ Propanamine contains three carbon atoms. Hoffmann bromamide degradation reaction involves the loss of one carbon atom from the amide group. Hence,the starting amide must contain four carbon atoms. The structure and $IUPAC$ name of the amide are:
$CH_3-CH_2-CH_2-CONH_2$ (Butanamide)
$(ii)$ Benzamide is an aromatic amide containing seven carbon atoms. Hoffmann degradation of benzamide removes the carbonyl carbon,resulting in an aromatic primary amine containing six carbon atoms $(C_6H_5NH_2)$. The product is aniline,also known as benzenamine.

(N/A) Gabriel phthalimide synthesis is used for the preparation of aliphatic primary amines. It involves nucleophilic substitution $(S_{N}2)$ of alkyl halides by the anion formed by the phthalimide.
However,aryl halides do not undergo nucleophilic substitution with the anion formed by the phthalimide because the carbon-halogen bond in aryl halides has partial double bond character and the electron-rich benzene ring repels the nucleophile.
Hence,aromatic primary amines cannot be prepared by this process.

(N/A) Reduction of nitro compounds: Nitro compounds are reduced to amines by passing hydrogen gas in the presence of finely divided nickel,palladium,or platinum,and also by reduction with metals in an acidic medium.
Reduction with iron scrap and hydrochloric acid is preferred because the $FeCl_2$ formed gets hydrolysed to release hydrochloric acid during the reaction. Thus,only a small amount of hydrochloric acid is required to initiate the reaction.
Nitroalkanes can also be similarly reduced to the corresponding alkanamines.
$CH_3CH_2NO_2 \xrightarrow[\text{ethanol}]{H_2 / Pd} CH_3CH_2NH_2$
Additionally,nitrobenzene can be reduced to aniline as follows:
$C_6H_5NO_2 \xrightarrow[\text{ethanol}]{H_2 / Pd} C_6H_5NH_2$
$C_6H_5NO_2 \xrightarrow{Sn + HCl \text{ or } Fe + HCl} C_6H_5NH_2$

(N/A) The carbon-halogen bond in alkyl or benzyl halides can be easily cleaved by a nucleophile. Hence,an alkyl or benzyl halide on reaction with an ethanolic solution of ammonia undergoes a nucleophilic substitution reaction in which the halogen atom is replaced by an amino $(-NH_{2})$ group. This process of cleavage of the $C-X$ bond by an ammonia molecule is known as ammonolysis.
The reaction is carried out in a sealed tube at $373 \ K$. The primary amine thus obtained behaves as a nucleophile and can further react with alkyl halide to form secondary and tertiary amines,and finally a quaternary ammonium salt.
The free amine can be obtained from the ammonium salt by treatment with a strong base: $R-NH_{3}^{+}X^{-} + NaOH \rightarrow R-NH_{2} + H_{2}O + Na^{+}X^{-}$.
Ammonolysis has the disadvantage of yielding a mixture of primary,secondary,and tertiary amines and also a quaternary ammonium salt.
However,a primary amine is obtained as a major product by taking a large excess of ammonia.
The order of reactivity of halides with amines is $R-I > R-Br > R-Cl$.

Reduction of nitriles: Nitriles on reduction with lithium aluminium hydride $(LiAlH_{4})$ or catalytic hydrogenation produce primary amines. This reaction is used for the ascent of the amine series,i.e.,for the preparation of amines containing one carbon atom more than the starting nitrile.
Reduction of amides: The amides on reduction with lithium aluminium hydride yield primary amines.
$R-C \equiv N \xrightarrow{H_{2}/Ni \text{ or } Na(Hg)/C_{2}H_{5}OH} R-CH_{2}-NH_{2}$
$CH_{3}-C \equiv N \xrightarrow{H_{2}/Ni \text{ or } Na(Hg)/C_{2}H_{5}OH} CH_{3}-CH_{2}-NH_{2}$
$R-CONH_{2} \xrightarrow{LiAlH_{4} / H_{2}O} R-CH_{2}-NH_{2}$
$CH_{3}-CONH_{2} \xrightarrow{LiAlH_{4} / H_{2}O} CH_{3}-CH_{2}-NH_{2}$

(N/A) Gabriel synthesis is used for the preparation of primary amines.
Phthalimide on treatment with ethanolic potassium hydroxide $(KOH)$ forms the potassium salt of phthalimide,which on heating with an alkyl halide $(R-X)$ followed by alkaline hydrolysis produces the corresponding primary amine $(R-NH_2)$.
The reaction sequence is as follows:
$1$. Phthalimide reacts with $KOH$ to form potassium phthalimide.
$2$. Potassium phthalimide reacts with alkyl halide $(R-X)$ to form $N$-alkylphthalimide.
$3$. $N$-alkylphthalimide undergoes alkaline hydrolysis to yield the primary amine $(R-NH_2)$ and phthalic acid (as a salt).
Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.

Hoffmann developed a method for the preparation of primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide.
In this degradation reaction,migration of an alkyl or aryl group takes place from the carbonyl carbon of the amide to the nitrogen atom.
The amine so formed contains one carbon atom less than that present in the amide.
The general reaction is:
$R-CONH_2 + Br_2 + 4NaOH \rightarrow R-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$
Example:
$CH_3CONH_2 + Br_2 + 4NaOH \rightarrow CH_3NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$
(Ethanamide to Methanamine)

(N/A) The conversion of primary aromatic amines into diazonium salts is known as diazotization. $C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278 \ K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$.
Benzenediazonium chloride is prepared by the reaction of aniline with nitrous acid at $273-278 \ K$. Nitrous acid is produced in the reaction mixture by the reaction of sodium nitrite with hydrochloric acid.
Due to its instability,the diazonium salt is not generally stored and is used immediately after its preparation. Examples include benzenediazonium chloride and benzenediazonium hydrogen sulphate.

(D) The reduction of nitriles $(R-CN)$ to primary amines $(R-CH_2NH_2)$ can be achieved using various reducing agents such as $LiAlH_4$,catalytic hydrogenation $(H_2 / Ni)$,or the Mendius reduction using sodium and alcohol $(Na / C_2H_5OH)$.
Therefore,all the listed reagents are effective for this conversion.

(N/A) The conversion can be achieved in three steps:
$1$. $Hofmann$ bromamide degradation: Cyclohexanecarboxamide reacts with $Br_2/KOH$ to form cyclohexanamine.
$2$. Carbylamine reaction: Cyclohexanamine reacts with $CHCl_3/KOH$ to form cyclohexyl isocyanide.
$3$. Reduction: Cyclohexyl isocyanide is reduced using $H_2/Pd$ to yield $N$-methylcyclohexanamine.

(A) The reaction proceeds in two steps:
$1$. The starting material is $2-(2-chloroethyl)cyclohexanone$. Treatment with $KCN$ leads to a nucleophilic substitution reaction where the $Cl^-$ ion is replaced by the $CN^-$ group,forming $A$,which is $3-(2-oxocyclohexyl)propanenitrile$.
$2$. The subsequent reduction of the nitrile group $(-CN)$ using $H_2/Pd$ (catalytic hydrogenation) reduces the nitrile to a primary amine $(-CH_2NH_2)$,forming $B$,which is $2-(3-aminopropyl)cyclohexanone$.

(B) The conversion of a nitrile $(R-CN)$ to a primary amine $(R-CH_2-NH_2)$ involves the reduction of the cyano group.
$LiAlH_4$ (Lithium aluminium hydride) is a strong reducing agent capable of reducing nitriles to primary amines.
The reaction is:
$CH_3-CH_2-C \equiv N \xrightarrow{LiAlH_4} CH_3-CH_2-CH_2-NH_2$
Therefore,the correct reagent is $LiAlH_4$.

(A) Gabriel phthalimide synthesis is used for the preparation of primary $(1^{\circ})$ aliphatic amines.
In this reaction,potassium phthalimide reacts with an alkyl halide to form an $N$-alkylphthalimide,which upon hydrolysis yields a primary aliphatic amine.
Aromatic amines (like aniline) cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the phthalimide anion under these conditions.
Among the given options,benzylamine $(C_6H_5CH_2NH_2)$ is a primary aliphatic amine and can be prepared by this method using benzyl chloride as the alkylating agent.

(B) The reaction of an amide with bromine $(Br_2)$ in the presence of an aqueous or ethanolic solution of a strong base like sodium hydroxide $(NaOH)$ is known as the Hofmann bromamide degradation reaction.
In this reaction,the amide is converted into a primary amine containing one carbon atom less than the original amide.
The reaction for propanamide $(CH_3CH_2CONH_2)$ is:
$CH_3CH_2CONH_2 + Br_2 + 4NaOH \rightarrow CH_3CH_2NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$
As shown in the reaction,propanamide ($3$ carbons) yields ethylamine ($2$ carbons).

(A) Gabriel phthalimide synthesis is used for the preparation of primary $(1^{\circ})$ aliphatic amines.
$1$. $(CH_3)_2CH-CH_2-NH_2$ is a primary aliphatic amine. It can be synthesized by this method.
$2$. $CH_3CH_2NH_2$ is a primary aliphatic amine. It can be synthesized by this method.
$3$. $C_6H_5-CH_2-NH_2$ (benzylamine) is a primary amine where the $-NH_2$ group is attached to an aliphatic carbon. It can be synthesized by this method.
$4$. $C_6H_5-NH_2$ (aniline) is an aromatic amine. Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.
Thus,there are $3$ such amines that can be synthesized by this method.

(C) Nitrobenzene can be reduced to aniline using various reducing agents:
$1.$ $Sn/HCl$: This is a standard reduction method for nitro compounds to amines.
$2.$ $Fe/HCl$: This is also a standard reduction method for nitro compounds to amines.
$3.$ $Zn/HCl$: This is a strong reducing agent that can reduce nitrobenzene to aniline.
$4.$ $H_2-Pd$: Catalytic hydrogenation effectively reduces the nitro group to an amino group.
$5.$ $H_2-$ Raney Nickel: Catalytic hydrogenation using Raney Nickel is also effective for this reduction.
$Sn/NH_4OH$ is not a standard reagent for this specific reduction.
Therefore,the reagents that can convert nitrobenzene into aniline are $I, III, IV, V,$ and $VI$.
The total number of such reagents is $5$.

(A) $1$. The reaction of benzene with conc. $HNO_3$ and conc. $H_2SO_4$ (nitration) yields $A = \text{Nitrobenzene}$.
$2$. Nitrobenzene is a meta-directing group. Therefore,chlorination of nitrobenzene using $Cl_2$ and anhydrous $AlCl_3$ gives $B = m\text{-Chloronitrobenzene}$.
$3$. The reduction of the nitro group $(-NO_2)$ in $m\text{-Chloronitrobenzene}$ using $Fe/HCl$ yields $C = m\text{-Chloroaniline}$.

(D) Gabriel phthalimide synthesis involves the nucleophilic substitution of an alkyl halide by the phthalimide anion via an $S_N2$ mechanism.
Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution reactions under these conditions due to the partial double bond character of the $C-X$ bond and the instability of the phenyl cation.
Therefore,both Assertion $(A)$ and Reason $(R)$ are true,and Reason $(R)$ is the correct explanation of Assertion $(A).$

(B) The reaction shown is the $Gabriel \ phthalimide \ synthesis$.
In this reaction,potassium phthalimide reacts with benzyl bromide $(C_6H_5CH_2Br)$ to form $N$-benzylphthalimide.
Subsequent hydrolysis with $OH^-/H_2O$ yields benzylamine $(C_6H_5CH_2NH_2)$ and phthalic acid as the products.
Therefore,the major product $A$ is benzylamine.

(A) The conversion of propan$-1-$ol $(CH_3CH_2CH_2OH)$ to $n-$butylamine $(CH_3CH_2CH_2CH_2NH_2)$ requires increasing the carbon chain length by one carbon atom.
Step $1$: Propan$-1-$ol reacts with $SOCl_2$ to form $1-$chloropropane $(CH_3CH_2CH_2Cl)$.
Step $2$: $1-$chloropropane reacts with $KCN$ (nucleophilic substitution) to form butanenitrile $(CH_3CH_2CH_2CN)$.
Step $3$: Butanenitrile is reduced using $H_2/Ni$ or $Na(Hg)/C_2H_5OH$ (Mendius reduction) to form $n-$butylamine $(CH_3CH_2CH_2CH_2NH_2)$.

(C) The reaction sequence described is the Gabriel phthalimide synthesis. \\
$1$. Compound $A$ is phthalic acid $(C_{6}H_{4}(COOH)_{2})$. \\
$2$. Reaction of phthalic acid with $NH_{3}$ followed by heating gives phthalimide $(C_{8}H_{5}NO_{2})$,which is compound $C$. The intermediate $B$ is phthalamide $(C_{6}H_{4}(CONH_{2})_{2})$. \\
$3$. Phthalimide $(C)$ reacts with ethanolic $KOH$ to form potassium phthalimide. \\
$4$. Potassium phthalimide reacts with alkyl chloride $(R-Cl)$ via $S_{N}2$ mechanism to form $N$-alkylphthalimide. \\
$5$. Hydrolysis of $N$-alkylphthalimide with alkali gives a primary amine $(RNH_{2})$ and phthalic acid. \\
Thus,compound $A$ is phthalic acid.

(A) The reaction sequence is as follows:
$1$. The starting material $A$ reacts with $NH_3$ to form an amide $(C_8H_8ClNO)$.
$2$. The amide then undergoes the Hofmann bromamide degradation reaction with $Br_2$ and $NaOH$ to form an amine.
$3$. Looking at the final product,which is $3-chloro-4-methylaniline$ (or a similar substituted aniline derivative),we can trace back the structure.
$4$. The final product has a $Cl$ group at the ortho position and a $CH_3$ group at the meta position relative to the $NH_2$ group.
$5$. The starting material $A$ must be $4-chloro-2-methylbenzoyl$ chloride,which upon reaction with $NH_3$ gives $4-chloro-2-methylbenzamide$,and subsequent Hofmann degradation yields $3-chloro-4-methylaniline$.

(B) The correct option is $B$. The conversion of $2-$phenylpropanamide into $1-$phenylethylamine involves the loss of a carbonyl carbon atom,which is characteristic of the Hofmann bromamide degradation reaction.
The reaction is as follows:
$CH_{3}CH(Ph)CONH_{2} + Br_{2} + 4NaOH \rightarrow CH_{3}CH(Ph)NH_{2} + Na_{2}CO_{3} + 2NaBr + 2H_{2}O$
In this reaction,the amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide. The alkyl or aryl group migrates from the carbonyl carbon to the nitrogen atom,resulting in the formation of a primary amine with one carbon atom less than the starting amide.

(C) The given reaction is the Hoffmann bromamide degradation reaction.
In this reaction,an amide $(RCONH_2)$ reacts with bromine $(Br_2)$ in the presence of a strong base like potassium hydroxide $(KOH)$ to form a primary amine $(RNH_2)$ with one carbon atom less than the starting amide.
The reaction for benzamide is:
$C_6H_5CONH_2 + Br_2 + 4KOH \rightarrow C_6H_5NH_2 + K_2CO_3 + 2KBr + 2H_2O$
Here,benzamide $(C_6H_5CONH_2)$ is converted into aniline $(C_6H_5NH_2)$.
Therefore,the major product is aniline.

(C) The starting material is $5-oxo-5-phenylpentanamide$ $(Ph-CO-(CH_2)_3-CONH_2)$.
Step $1$: Clemmensen reduction $(Zn/Hg, HCl)$ specifically reduces the ketone group to a methylene group $(-CH_2-)$ while leaving the amide group intact.
This yields $5-phenylpentanamide$ $(Ph-(CH_2)_4-CONH_2)$.
Step $2$ and $3$: Treatment with $LiAlH_4$ followed by acidic workup $(H_3O^+)$ reduces the amide group $(-CONH_2)$ to a primary amine $(-CH_2NH_2)$.
The final product is $5-phenylpentan-1-amine$ $(Ph-(CH_2)_5-NH_2)$.
Thus,the correct option is $C$.

(D) The reaction is a Hofmann bromamide degradation,which converts an amide $(-CONH_2)$ into a primary amine $(-NH_2)$.
In the given molecule,the $-CONH_2$ group is converted to $-NH_2$ by the action of $Br_2/NaOH$.
The resulting intermediate is an amino ester,which undergoes intramolecular cyclization (nucleophilic acyl substitution) to form a cyclic amide,known as a lactam.
The final product is $isoindolin-1-one$.

(A) Gabriel Phthalimide synthesis is used for the preparation of primary aliphatic amines. It involves the reaction of potassium phthalimide with an alkyl halide followed by alkaline hydrolysis.
It cannot be used for the preparation of aromatic amines because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.
Therefore,only those amines with the formula $C_8H_{11}N$ that have the $-NH_2$ group attached to an aliphatic carbon (i.e.,primary aliphatic amines) can be synthesized by this method.
The isomers of $C_8H_{11}N$ that are primary aliphatic amines are:
$1$. $C_6H_5-CH_2-CH_2-NH_2$ ($2$-phenylethanamine)
$2$. $C_6H_5-CH(CH_3)-NH_2$ ($1$-phenylethanamine)
$3$. $o-CH_3-C_6H_4-CH_2-NH_2$ ($2$-methylbenzylamine)
$4$. $m-CH_3-C_6H_4-CH_2-NH_2$ ($3$-methylbenzylamine)
$5$. $p-CH_3-C_6H_4-CH_2-NH_2$ ($4$-methylbenzylamine)
There are $5$ such isomers that can be synthesized by this method.

(B) $1$. Phthalimide reacts with $KOH$ to form potassium phthalimide.
$2$. Potassium phthalimide then undergoes an $S_N2$ reaction with benzyl chloride to form $N$-benzylphthalimide (product $P$).
$3$. The structure of $N$-benzylphthalimide contains a benzene ring ($3 \ \pi$ bonds),another benzene ring from the benzyl group ($3 \ \pi$ bonds),and two carbonyl groups ($C=O$,each having $1 \ \pi$ bond).
$4$. Total $\pi$ bonds = $3$ (phthalimide ring) + $3$ (benzyl ring) + $2$ ($C=O$ bonds) = $8$.