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Practice each question, then open the answer or worked solution.
Which of the following is a 3° amine?
Ans: (B) Triethylamine
The structure of the given amines are shown below.
(A)
(B)
(C)
(D)
The correct IUPAC name for $\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{NH}-\mathrm{CH}_{3}$ is
Ans: Correct answer: (D)
Amongst the following, the strongest base in an aqueous medium is.
Ans: (C)
Because of the electron-releasing nature of the alkyl group, it (R) pushes electrons towards nitrogen, making the unshared electron pair more available for sharing with the acid's proton. Furthermore, the +I effect of the alkyl group stabilizes the substituted ammonium ion formed from the amine by dispersing the positive charge.
As a result, alkylamines are more powerful bases than ammonia.
As a result, the basic nature of aliphatic amines should increase as the number of alkyl groups increases.
Which of the following is the weakest Bronsted base?
(A)
(B)
(C)
(D) $\mathrm{CH}_{3} \mathrm{NH}_{2}$
Ans: (A)
The $\mathrm{NH}_{2}$ group is directly attached to the benzene ring in aniline and other aryl amines. As a result, the unshared electron pair on the nitrogen atom is conjugated with the benzene ring, making it less available for protonation.
Benzylamine may be alkylated as shown in the following equation:
$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{NH}_{2}+\mathrm{R}-\mathrm{X} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{NHR}$
Which of the following alkyl halides is best suited for this reaction through $\text{S}{{\text{N}}^{\text{1}}}$ mechanism?
Ans: (C)
SN1 reaction: A nucleophilic reaction that occurs in two steps, first is the bond-breaking step and the second is the production of the carbocation. The stability of carbocation formed in the second step determines the rate of reactivity of reactant toward $\text{S}{{\text{N}}^{\text{1}}}$ reaction. Here, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{Br}$,
In the process of ionization, removal of bromine, a stable Benzyl carbocation is produced. Therefore, it is best suited for reaction through the $\text{S}{{\text{N}}^{\text{1}}}$ mechanism.
Which of the following reagents would not be a good choice for reducing an aryl nitro compound to an amine?
Ans: (B)
Lithium aluminum hydride in ether is a strong reducing agent that donates its $\mathrm{H}^{-}$ hydride ion to any $\mathrm{C}=\mathrm{O}$ containing a functional group. In addition to $\mathrm{LiAlH}_{4}$ to aryl nitro compounds, no reaction will be observed, the desired products of amines will not be produced, rather it will form diazobenzene products.
$2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2} \stackrel{\text { LiAlH }_{4} / \text { ether }}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}=\mathrm{N}-\mathrm{C}_{6} \mathrm{H}_{5}$
In order to prepare a primary amine from an alkyl halide with simultaneous addition of one $\mathrm{CH}_{2}$ group in the carbon chain, the reagent used as a source of nitrogen is
Ans: (C)
Potassium cyanide, as cyanide on reduction with sodium metal in alcohol, produces amine with increased carbon atoms.
$\mathrm{R}-\mathrm{X} \stackrel{\mathrm{KCN}}{\stackrel{-\mathrm{KX}}{\longrightarrow}} \mathrm{R}-\mathrm{CN} \stackrel{\mathrm{Na} / \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}{\longrightarrow} \mathrm{R}-\mathrm{CH}_{2} \mathrm{NH}_{2}$
The source of nitrogen in Gabriel synthesis of amines is.
Ans: (D)
Gabriel synthesis reaction:
Gabriel’s synthesis is a method for producing primary amines. When phthalimide is treated with ethanolic potassium hydroxide, it forms a potassium salt of phthalimide, which when heated with an alkyl halide and then alkaline hydrolyzed yields the corresponding primary amine. This method cannot produce aromatic primary amines because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.
Amongst the given set of reactants, the most appropriate for preparing $2^{\circ}$ amine is.
Ans: ( C)
$\underset{1^{\circ} \text { Amine }}{\mathrm{R}-\mathrm{NH}_{2}}+\mathrm{O}=\mathrm{CHR} \underset{\text { Aldehyde }}{\stackrel{\text { Reductive }}{\text { amination }} \longrightarrow \mathrm{R}-\mathrm{N}=\mathrm{CHR} \frac{\mathrm{H}_{2} / \mathrm{Pt}}{\text { Reduction }} \underset{2^{\circ} \text { Amine }}{\mathrm{RNH}}-\mathrm{CH}_{2} \mathrm{R}}$
The best reagent for converting 2-phenylpro-panamide into-2-phenyl- propanamine is.
Ans: (D)
Lithium aluminium hydride in ether is a strong reducing agent that donates its hydride ion to any $\mathrm{C}=\mathrm{O}$ containing functional groups into corresponding reduced compounds. Here, the amide group is converted to an amine functional group.
$\mathrm{LiAlH}_{4}$ in ether is the best reagent for converting 2-phenylpropanamide to 2-phenylpropanamine.