Mix Examples of Amines Questions in English

(A) The chemical reaction for the preparation of acetanilide from aniline is: $C_6H_5NH_2 + (CH_3CO)_2O \rightarrow C_6H_5NHCOCH_3 + CH_3COOH$.
$1$. Calculate the moles of aniline used:
$n_{\text{aniline}} = \frac{\text{mass}}{\text{molar mass}} = \frac{9.3 \ g}{93 \ g/mol} = 0.1 \ mol$.
$2$. Since the stoichiometry of the reaction is $1:1$,the theoretical yield of acetanilide is $0.1 \ mol$.
$3$. Calculate the actual moles of acetanilide obtained:
$n_{\text{acetanilide}} = \frac{11 \ g}{135 \ g/mol} \approx 0.08148 \ mol$.
$4$. Calculate the percentage yield:
$\text{Percentage yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100 = \left( \frac{0.08148 \ mol}{0.1 \ mol} \right) \times 100 = 81.48 \% \approx 81.5 \%$.

(C) $1$. The starting material is $4-$nitrotoluene $(C_7H_7NO_2)$,which has a molar mass of $(7 \times 12) + (7 \times 1) + 14 + (2 \times 16) = 84 + 7 + 14 + 32 = 137 \ g \ mol^{-1}$.
$2$. Reduction of $4-$nitrotoluene with $Sn/HCl$ gives $p-$toluidine ($A$,$C_7H_9N$).
$3$. Acetylation of $p-$toluidine with $(CH_3CO)_2O$ gives $N-(p-tolyl)acetamide$ $(C_9H_{11}NO)$.
$4$. Bromination of $N-(p-tolyl)acetamide$ with $Br_2/AcOH$ gives $2-bromo-4-methylacetanilide$ ($B$,$C_9H_{10}BrNO$).
$5$. The molar mass of $B$ $(C_9H_{10}BrNO)$ is $(9 \times 12) + (10 \times 1) + 14 + 80 + 16 = 108 + 10 + 14 + 80 + 16 = 228 \ g \ mol^{-1}$.
$6$. Moles of $4-$nitrotoluene $= \frac{137 \times 10^{-3} \ g}{137 \ g \ mol^{-1}} = 0.001 \ mol$.
$7$. Since the stoichiometry is $1:1$,moles of $B = 0.001 \ mol$.
$8$. Mass of $B = 0.001 \ mol \times 228 \ g \ mol^{-1} = 0.228 \ g = 228 \ mg$.

(C) Step $1$: $Nitrobenzene$ reacts with $Br_2/FeBr_3$ to give $m-bromonitrobenzene$.
Step $2$: Reduction with $Sn/HCl$ converts the $-NO_2$ group to $-NH_2$,yielding $m-bromoaniline$.
Step $3$: $pH$ neutralisation is performed.
Step $4$: $m-bromoaniline$ reacts with $Br_2/H_2O$ to give $2,4,6-tribromo-3-bromoaniline$ (which is $2,3,4,6-tetrabromoaniline$).
Step $5$: Diazotization with $NaNO_2/HBr$ at $0-5^{\circ}C$ converts $-NH_2$ to $-N_2^+Br^-$.
Step $6$: Sandmeyer reaction with $CuBr/NaBr$ replaces the diazonium group with a $Br$ atom.
The final product is $1,2,3,4,5-pentabromobenzene$ (or $1,2,3,4,6-pentabromobenzene$ depending on numbering,but it contains $5$ bromine atoms).
Thus,the number of $Br$ atoms in the final product $(P)$ is $5$.

(C) $1$. The molecular formula $C_8H_9ON$ and the reaction with $Br_2 / HO^{(-)}$ (Hofmann bromamide degradation) suggest that $A$ is an amide.
$2$. The product $B$ is an amine,which is soluble in dilute acid.
$3$. The reaction sequence $B$ $\xrightarrow{NaNO_2/HCl} C$ $\xrightarrow{CuCN} D$ $\xrightarrow{H_3O^+} E$ is the standard conversion of an amine to a carboxylic acid via a diazonium salt and nitrile.
$4$. $E$ is also obtained from the hydrolysis of $A$,confirming $A$ is an amide.
$5$. $E$ on oxidation with $KMnO_4$ gives $F$. If $E$ is $p$-toluic acid,oxidation gives terephthalic acid $(F)$,which has two types of hydrogen atoms (aromatic and carboxylic).
$6$. Based on the reaction path,$A$ must be $p$-methylbenzamide.

(A) $1$. Reaction $A$: Benzyl bromide $(C_6H_5CH_2Br)$ reacts with $AgCN$ to form benzyl isocyanide $(C_6H_5CH_2NC)$. This is because $AgCN$ is a covalent compound,and the nucleophilic attack occurs through the nitrogen atom of the ambident $CN^-$ ion.
$2$. Reaction $B$: Benzylamine $(C_6H_5CH_2NH_2)$ undergoes the carbylamine reaction with $CHCl_3$ and $KOH$ (or $NaOH$) to form benzyl isocyanide $(C_6H_5CH_2NC)$.
$3$. Reaction $C$: Bromobenzene does not undergo nucleophilic substitution with $AgCN$ under normal conditions due to the partial double bond character of the $C-Br$ bond.
$4$. Reaction $D$: Aniline $(C_6H_5NH_2)$ undergoes the carbylamine reaction to form phenyl isocyanide $(C_6H_5NC)$,not benzyl isocyanide.
$5$. Reaction $E$: $2-$Phenylethyl bromide reacts with $KCN$ to form $3-$phenylpropanenitrile $(C_6H_5CH_2CH_2CN)$.
Therefore,benzyl isocyanide is obtained from reactions $A$ and $B$.

$(A)$. Diethyl amine (secondary amine) and Ethyl amine (primary amine) can be distinguished using the Carbylamine test $(CHCl_3 + KOH, \Delta)$; only primary amines react to form an isocyanide with a foul smell $(A-II)$.
$B$. Acetaldehyde (aldehyde) and Acetone (ketone) can be distinguished using Tollens' test (Ammonical silver nitrate); only acetaldehyde reacts to form a silver mirror $(B-IV)$.
$C$. Ethanol and Phenol can be distinguished using Neutral $FeCl_3$; phenol gives a characteristic violet color with neutral $FeCl_3$ $(C-III)$.
$D$. Benzoic acid and Cinnamic acid can be distinguished using Bromine water; cinnamic acid contains a carbon-carbon double bond (unsaturation) and decolorizes bromine water, whereas benzoic acid does not $(D-I)$.
Therefore, the correct matching is $A-II, B-IV, C-III, D-I$.

(C) Let's analyze each statement:
$(A)$ Benzylamine $(C_6H_5CH_2NH_2)$ is a stronger base than aniline $(C_6H_5NH_2)$ because in benzylamine,the lone pair on nitrogen is not involved in resonance with the benzene ring,whereas in aniline,it is delocalized. Thus,statement $(A)$ is correct.
$(B)$ Gabriel phthalimide synthesis is used for the preparation of primary aliphatic amines. It cannot be used for the preparation of aromatic primary amines because aryl halides do not undergo nucleophilic substitution with the phthalimide anion. Thus,statement $(B)$ is incorrect.
$(C)$ The Hofmann bromamide degradation reaction is used to convert primary amides $(RCONH_2)$ into primary amines $(RNH_2)$. The reaction shown involves phenylacetamide $(C_6H_5CH_2CONH_2)$,which would yield benzylamine. However,the product is labeled as a 'primary aromatic amine',which is incorrect as the amine formed is a primary aliphatic amine. Thus,statement $(C)$ is incorrect.
$(D)$ The diazotization of $4-$nitroaniline followed by hydrolysis gives $4-$nitrophenol. $4-$Nitrophenol is acidic and will dissolve in $NaOH$. Thus,statement $(D)$ is correct.
Therefore,the incorrect statements are $(B)$ and $(C)$.

(C) The reaction sequence starting from propanoic acid $(CH_3CH_2COOH)$ is as follows:
$1$) Reaction with $NH_3/\Delta$ gives propanamide $(CH_3CH_2CONH_2)$.
$2$) Hofmann bromamide degradation with $NaOH/Br_2$ converts propanamide to ethanamine $(CH_3CH_2NH_2)$.
$3$) Reaction with nitrous acid ($HNO_2$,from $NaNO_2/HCl$) at $0^\circ C$ converts ethanamine to ethanol $(CH_3CH_2OH)$.
$4$) The reaction of ethanol with benzenediazonium chloride $(C_6H_5N_2Cl)$ at $0^\circ C$ is not a standard textbook reaction for producing the given options. However,if the question intended to ask for the product after step $(iii)$,the product is ethanol. Given the options provided,there is no correct match for the final product. Assuming a potential error in the question's options or reagents,the most logical intermediate product is ethanol,which is not listed. Therefore,the question is flawed.

(B) The starting material is $p$-nitrotoluene $(CH_3-C_6H_4-NO_2)$.
$(i)$ Reduction of $-NO_2$ to $-NH_2$ gives $p$-toluidine $(CH_3-C_6H_4-NH_2)$.
(ii) Acetylation of the amino group with acetic anhydride gives $p$-methylacetanilide $(CH_3-C_6H_4-NHCOCH_3)$.
(iii) Bromination occurs at the ortho position relative to the $-NHCOCH_3$ group,yielding $2$-bromo-$4$-methylacetanilide.
(iv) Hydrolysis of the acetamido group gives $2$-bromo-$4$-methylaniline $(C_7H_8BrN)$.
This product $(P)$ is $2$-bromo-$4$-methylaniline. Its molar mass is $(7 \times 12) + (8 \times 1) + 80 + 14 = 84 + 8 + 80 + 14 = 186 \text{ g/mol}$.
In Carius analysis,$1 \text{ mole}$ of $P$ produces $1 \text{ mole}$ of $\text{AgBr}$.
Moles of $P = 1.0 \text{ g} / 186 \text{ g/mol} \approx 0.005376 \text{ mol}$.
Moles of $\text{AgBr} = 0.005376 \text{ mol}$.
Mass of $\text{AgBr} = 0.005376 \times (108 + 80) = 0.005376 \times 188 \approx 1.01 \text{ g}$.
The nearest integer is $1$.

(B) $1$. $C_2H_6 \xrightarrow{Cl_2, UV \text{ light}} C_2H_5Cl$ ($X$ is ethyl chloride).
$2$. $C_2H_5Cl \xrightarrow{NH_3} C_2H_5NH_2$ ($Y$ is ethylamine).
$3$. $C_2H_5NH_2 \xrightarrow{(i) NaNO_2/HCl, (ii) H_2O} C_2H_5OH$.
Diazotization of ethylamine followed by hydrolysis yields ethanol $(Z)$.