Properties of Amines Questions in English

(C) In an aqueous solution, the basic strength of ethyl-substituted amines is determined by the interplay of three factors: the inductive effect ($+I$ effect), the solvation effect (hydrogen bonding), and steric hindrance.
For ethyl-substituted amines, the secondary amine $(2^\circ)$ is the most basic due to the optimal balance of these factors.
The tertiary amine $(3^\circ)$ is less basic than the secondary amine due to steric hindrance, but more basic than the primary amine $(1^\circ)$.
Thus, the correct order of basic strength is $(C_2H_5)_2 NH > (C_2H_5)_3 N > C_2H_5NH_2 > NH_3$.

(A) $1$. Ammonolysis of benzyl chloride $(C_6H_5CH_2Cl)$ involves the nucleophilic substitution of the chlorine atom by an ammonia molecule, resulting in the formation of benzylamine $(C_6H_5CH_2NH_2)$.
$2$. Benzylamine is a primary amine. When it reacts with two moles of methyl chloride $(CH_3Cl)$, the two hydrogen atoms attached to the nitrogen atom are replaced by two methyl groups through nucleophilic substitution.
$3$. The final product formed is $C_6H_5CH_2N(CH_3)_2$.
$4$. According to $IUPAC$ nomenclature, this compound is named $N$, $N$-dimethylphenylmethanamine.

(C) Benzenediazonium fluoroborate $(C_6H_5N_2^+BF_4^-)$ is unique among diazonium salts because it is water-insoluble and relatively stable at room temperature, which allows it to be dried and stored.
Upon heating, it undergoes thermal decomposition to yield fluorobenzene, a process known as the Schiemann reaction.

(A) The chemical equation for the Hoffmann bromamide degradation reaction is: $RCONH_{2} + Br_{2} + 4NaOH \rightarrow RNH_{2} + Na_{2}CO_{3} + 2NaBr + 2H_{2}O$.
$(A)$ is correct because the stoichiometry of the reaction requires $4$ moles of $NaOH$ and $1$ mole of $Br_{2}$ for every mole of primary amide.
$(B)$ is false because alkyl amides (primary amides) readily undergo this reaction.
$(C)$ is correct because this reaction is a standard laboratory method for the synthesis of primary amines.
$(D)$ is false because secondary amides $(RCONHR')$ lack the necessary $-NH_{2}$ group required for the rearrangement mechanism.
$(E)$ is correct because the by-products formed are indeed $Na_{2}CO_{3}$, $NaBr$, and $H_{2}O$.

(A) The reaction is an electrophilic aromatic substitution (diazo coupling).
The reactivity is determined by the electron density on the aromatic ring, which is enhanced by electron-donating groups like $-N(Me)_2$.
Steric hindrance at the ortho position to the $-N(Me)_2$ group significantly reduces the reactivity by preventing the approach of the electrophile.
Molecule $P$ has no ortho substituents, molecule $Q$ has one ortho methyl group, and molecule $R$ has two ortho methyl groups.
Therefore, the steric hindrance increases in the order $P < Q < R$, which means the reactivity decreases in the order $P > Q > R$.
Wait, let us re-evaluate: $P$ has no ortho substituents, $Q$ has one ortho methyl group, and $R$ has two ortho methyl groups. The reactivity order is $P > Q > R$.

(D) $1$. Aniline reacts with $Br_2/H_2O$ to form $2,4,6$-tribromoaniline as the major product $(A)$.
$2$. $2,4,6$-tribromoaniline reacts with $NaNO_2/HCl$ at $273-278 \ K$ to form the diazonium salt, $2,4,6$-tribromobenzenediazonium chloride $(B)$.
$3$. The reaction of the diazonium salt with $HBF_4$ followed by $NaNO_2/Cu, \Delta$ is a variation of the Balz-Schiemann or Sandmeyer-type reaction where the diazonium group $(-N_2^+Cl^-)$ is replaced by a nitro group $(-NO_2)$.
$4$. Thus, the final product $C$ is $1,3,5$-tribromo$-2-$nitrobenzene.

(B) The Hoffmann bromamide degradation reaction $(Br_2/KOH)$ is specific to primary amides $(R-CONH_2)$, converting them into primary amines $(R-NH_2)$.
Gabriel phthalimide synthesis is also used for the preparation of primary amines $(R-NH_2)$ from primary alkyl halides $(R-X)$.
Therefore, we need to identify the number of primary amides $(R-CONH_2)$ in the given set.
Let's analyze the structures provided in the image:
$1$. $C_6H_5CONH_2$ (Benzamide) - Primary amide.
$2$. $C_6H_5CH_2CONH_2$ ($2$-Phenylacetamide) - Primary amide.
$3$. $CH_3CONH_2$ (Acetamide) - Primary amide.
$4$. $C_6H_{11}CONHCH_2CH_3$ - Secondary amide (Does not undergo Hoffmann degradation).
$5$. $(CH_3)_3CCONHCH_3$ - Secondary amide (Does not undergo Hoffmann degradation).
$6$. $C_6H_{11}CONH_2$ (Cyclohexanecarboxamide) - Primary amide.
Counting the primary amides, we have compounds $1$, $2$, $3$, and $6$.
Thus, there are $4$ such compounds.

(D) Statement $I$ is false: The reaction of benzamide with $Br_2$ and $NaOH$ (Hofmann bromamide degradation) produces aniline, not benzylamine.
Statement $II$ is true: In the strong acidic medium of nitration $(H_2SO_4)$, aniline gets protonated to form anilinium ion $(-NH_3^+)$, which is meta-directing. Hence, $m$-nitroaniline is formed in a significant amount due to the deactivating and meta-directing nature of the $-NH_3^+$ group.

(C) conjugate acid is formed by the protonation of a base. $A$ stronger conjugate acid is formed from a weaker base.
Among the given substituted anilines, the base strength depends on the electron-donating or electron-withdrawing nature of the substituents.
The $-NO_2$ group in $4$-nitroaniline is a strong electron-withdrawing group via resonance and induction, which significantly reduces the availability of the lone pair on the nitrogen atom, thus making it the weakest base.
Since the strength of a conjugate acid is inversely proportional to the strength of its parent base, the protonated form of $4$-nitroaniline is the strongest conjugate acid.

(C) The boiling point of a compound depends on the strength of intermolecular forces.
$n-C_4H_{10}$ $(C)$ is an alkane and possesses only weak van der Waals forces, resulting in the lowest boiling point.
$C_2H_5NHC_2H_5$ $(D)$ is a secondary amine; it exhibits hydrogen bonding, but the extent is less than that of a primary amine due to steric hindrance.
$n-C_4H_9NH_2$ $(B)$ is a primary amine, which has stronger hydrogen bonding compared to the secondary amine $(D)$.
$n-C_4H_9OH$ $(A)$ is an alcohol, which forms the strongest hydrogen bonds due to the high electronegativity of oxygen compared to nitrogen, resulting in the highest boiling point.
Therefore, the increasing order of boiling points is $C < D < B < A$.

(A) The reaction sequence involves the acid-catalyzed rearrangement of diazoaminobenzene to $p$-aminoazobenzene.
$1$. The starting material is diazoaminobenzene $(C_{12}H_{11}N_3)$.
$2$. Upon treatment with $HCl$ and aniline, it undergoes a rearrangement reaction known as the diazoamino-to-aminoazo rearrangement.
$3$. The yellow product $(X)$ formed is $p$-aminoazobenzene, which has the chemical formula $C_{12}H_{11}N_3$.
$4$. The molar mass of $C_{12}H_{11}N_3$ is $(12 \times 12) + (11 \times 1) + (3 \times 14) = 144 + 11 + 42 = 197$ g/mol.
$5$. The percentage of nitrogen in the product is $\frac{\text{Total mass of N}}{\text{Molar mass of X}} \times 100 = \frac{42}{197} \times 100 \approx 21.32\%$.
$6$. Rounding to the nearest integer, we get $21\%$.

(B) Reaction $1$: $C_2H_5Cl + AgCN \rightarrow C_2H_5NC$ (Ethyl isocyanide, which is a foul-smelling compound).
Reaction $2$: This involves the Hofmann bromamide degradation followed by the carbylamine reaction.
Step $1$: $C_2H_5CONH_2 \xrightarrow{Br_2, NaOH} C_2H_5NH_2$ ($Y$ is ethylamine).
Step $2$: $C_2H_5NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} C_2H_5NC + 3KCl + 3H_2O$ (Carbylamine reaction).
Thus, $Z$ is ethyl isocyanide $(C_2H_5NC)$ and $X$ is $AgCN$.

(D) $1$. Quaternary ammonium salts have the general formula $[R_4N]^+X^-$.
$2$. The four $C-N$ bonds are covalent bonds.
$3$. One of these $C-N$ bonds is formed by the donation of a lone pair from the nitrogen atom to the fourth alkyl group, which is a coordinate (dative) bond.
$4$. The electrostatic attraction between the quaternary ammonium cation $[R_4N]^+$ and the anion $X^-$ constitutes an ionic bond.
$5$. Therefore, all three types of bonds are present.

(C) $1$. The reaction described is the Carbylamine reaction.
$2$. Primary amines $(R-NH_2)$ react with chloroform $(CHCl_3)$ and alcoholic $KOH$ to form alkyl isocyanides (carbylamines), which have a foul smell.
$3$. Secondary $(R_2NH)$ and tertiary $(R_3N)$ amines do not undergo the Carbylamine reaction.
$4$. Methanamine $(CH_3NH_2)$, Propan$-2-$amine $(CH_3CH(NH_2)CH_3)$, and Ethanamine $(CH_3CH_2NH_2)$ are primary amines.
$5$. $N, N-Diethylethanamine$ $((C_2H_5)_3N)$ is a tertiary amine and therefore does not give this test.

(B) $1$. The reaction of primary amines with ethanolic $KOH$ and $CHCl_3$ is known as the Carbylamine reaction.
$2$. This reaction produces isocyanides (carbylamines), which have a characteristic foul smell.
$3$. Secondary and tertiary amines do not undergo the Carbylamine reaction because they lack the necessary hydrogen atoms on the nitrogen to form the isocyanide group.
$4$. Benzenamine $(C_6H_5NH_2)$, Propan$-2-$amine $(CH_3CH(NH_2)CH_3)$, and Phenylmethanamine $(C_6H_5CH_2NH_2)$ are primary amines and will give the test.
$5$. $N, N - \text{Dimethylaniline}$ is a tertiary amine $(C_6H_5N(CH_3)_2)$ and will not produce a foul smell.

(D) Step $1$: Analyze the structure of amines. Primary amines $(R-NH_2)$ have two hydrogen atoms attached to the nitrogen, allowing for intermolecular hydrogen bonding.
Step $2$: Tertiary amines $(R_3N)$ have no hydrogen atoms attached to the nitrogen, so they cannot form intermolecular hydrogen bonds.
Step $3$: Tertiary butylamine is $(CH_3)_3C-NH_2$, which is a primary amine because the nitrogen is attached to only one carbon atom.
Step $4$: iso-propylamine is $(CH_3)_2CH-NH_2$, which is a primary amine because the nitrogen is attached to only one carbon atom. Therefore, the statement that iso-propylamine is a secondary amine is incorrect.

(D) $1$. Primary $(R-NH_2)$ and secondary $(R_2NH)$ amines contain $N-H$ bonds, which allow for inter-molecular hydrogen bonding.
$2$. Tertiary amines $(R_3N)$ do not have any hydrogen atoms attached to the nitrogen atom.
$3$. Due to the absence of $N-H$ bonds, tertiary amines cannot form inter-molecular hydrogen bonds.
$4$. Consequently, they have weaker inter-molecular forces compared to primary and secondary amines, resulting in lower boiling points.

(B) Step $1$: In the aqueous phase, the basicity of aliphatic amines depends on the combined effect of the inductive effect $(+I)$, solvation effect, and steric hindrance.
Step $2$: For methyl-substituted amines, the order is $(CH_3)_2NH > CH_3NH_2 > NH_3$.
Step $3$: $C_6H_5NH_2$ (aniline) is the least basic because the lone pair on the nitrogen atom is involved in resonance with the benzene ring, making it less available for protonation.
Step $4$: Combining these, the decreasing order of basic strength is $(CH_3)_2NH > CH_3NH_2 > NH_3 > C_6H_5NH_2$.

(B) $1$. Ethyl amine $(CH_3CH_2NH_2)$ reacts with excess methyl iodide $(CH_3I)$ via nucleophilic substitution ($S_N2$ reactions).
$2$. The primary amine reacts to form a secondary amine: $CH_3CH_2NH_2 + CH_3I \rightarrow CH_3CH_2NH(CH_3) + HI$.
$3$. The secondary amine reacts further to form a tertiary amine: $CH_3CH_2NH(CH_3) + CH_3I \rightarrow CH_3CH_2N(CH_3)_2 + HI$.
$4$. The tertiary amine reacts with the remaining excess methyl iodide to form a quaternary ammonium salt: $CH_3CH_2N(CH_3)_2 + CH_3I \rightarrow [CH_3CH_2N(CH_3)_3]^+I^-$.
$5$. The final product is ethyltrimethylammonium iodide.

(A) $1$. The reaction described is the carbylamine reaction, which is a characteristic test for primary $(1^\circ)$ amines.
$2$. In this reaction, a primary amine reacts with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ to form an isocyanide (carbylamine), which has a foul smell.
$3$. Among the given options, $\text{Benzenamine}$ (aniline, $C_6H_5NH_2$) is a primary aromatic amine.
$4$. $N-\text{methylaniline}$ and $N,N-\text{Dimethylaniline}$ are secondary and tertiary amines, respectively, and do not undergo the carbylamine reaction.
$5$. $\text{Ethylethanamine}$ (diethylamine) is a secondary aliphatic amine and also does not undergo this reaction.
$6$. Therefore, $\text{Benzenamine}$ is the correct answer.

(B) Step $1$: Hofmann's exhaustive alkylation involves the reaction of a primary amine with an excess of alkyl halide to form a mixture of secondary amine, tertiary amine, and finally a quaternary ammonium salt.
Step $2$: Option $(A)$ is correct as the final product is a quaternary ammonium salt.
Step $3$: Option $(B)$ is incorrect because the reaction of $1 \text{ mole}$ of primary amine with $2 \text{ moles}$ of alkyl halide produces a tertiary amine, not a tetraalkyl ammonium halide. Tetraalkyl ammonium halide requires $4 \text{ moles}$ of alkyl halide.
Step $4$: Option $(C)$ is correct as the product distribution depends on the molar ratio of reactants.
Step $5$: Option $(D)$ is correct as the reaction releases $HX$, which is neutralized by a base like $\text{NaHCO}_3$ or excess amine to drive the reaction forward.

(C) The reaction of an amide with $Br_2$ and aqueous $KOH$ is known as the $Hofmann$ bromamide degradation reaction.
In this reaction, the amide group $(-CONH_2)$ is converted into a primary amine $(-NH_2)$ with the loss of one carbon atom.
Propionamide is $CH_3CH_2CONH_2$.
Reaction: $CH_3CH_2CONH_2 + Br_2 + 4KOH \rightarrow CH_3CH_2NH_2 + K_2CO_3 + 2KBr + 2H_2O$.
The product obtained is ethylamine $(CH_3CH_2NH_2)$.

(A) $1$. The reaction is the Hofmann elimination of a quaternary ammonium salt.
$2$. Step $(i)$ involves the exchange of the halide ion with the hydroxide ion using moist $Ag_2O$ $(AgOH)$, forming the quaternary ammonium hydroxide: $A + AgOH \rightarrow [CH_3CH_2N^+(CH_3)_3]OH^- + AgX$.
$3$. Step (ii) is the thermal decomposition (Hofmann elimination) of the hydroxide. According to the Hofmann rule, the less substituted alkene is formed. Here, $CH_2=CH_2$ (ethene) is formed from the ethyl group, and the remaining part is $CH_3CH_2N(CH_3)_2$.
$4$. Thus, the substrate '$A$' must be $Ethyltrimethylammonium$ halide.

(B) $1$. In a strong acidic medium, the lone pair of electrons on the nitrogen atom of aniline is protonated by the acid.
$2$. This results in the formation of the anilinium ion $(-NH_3^+)$.
$3$. The $-NH_3^+$ group is strongly electron-withdrawing due to its positive charge and exerts a $-I$ effect.
$4$. Consequently, it deactivates the benzene ring and directs the incoming electrophile to the meta position.

(D) Step $1$: The basic strength of alkylamines in the aqueous phase depends on three main factors: inductive effect, solvation effect, and steric hindrance.
Step $2$: The inductive effect ($+I$ effect) increases the electron density on the nitrogen atom, making it more basic.
Step $3$: The solvation effect (stabilization of the conjugate acid by hydrogen bonding with water) stabilizes the cation; more substituted amines have less hydrogen bonding.
Step $4$: Steric hindrance (crowding around the nitrogen atom) hinders the approach of the proton.
Step $5$: Hyperconjugation effect does not play a significant role in determining the basicity of alkylamines in the aqueous phase. Therefore, the correct option is $D$.