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 $AgCN$.