Your cart is empty.
Practice each question, then open the answer or worked solution.
Best method for preparing primary amines from alkyl halides without changing the number of carbon atoms in the chain is
Ans: (B)
The reaction of transformation of primary alkyl halides to primary amines using potassium phthalimide is the Gabriel phthalimide reaction.
Which of the following compounds will not undergo azo coupling reaction with benzene diazonium chloride?
Ans: (D)
Weak electrophile such as Diazonium cation readily reacts with electron-rich compounds which are having electron-rich compounds such as hydroxyl group, amino group. They don’t react with electron-withdrawing groups like nitro groups. Therefore, nitrobenzene will not undergo an azo coupling reaction with benzene diazonium chloride.
Which of the following compounds is the weakest Bronsted base?
(A)
(B)
(C)
(D)
Ans: (D)
Since phenol is the strongest acid among the four options listed above, it is the weakest Brönsted base. The stronger the acid, the weaker the conjugate base.
Amines have a strong tendency to accept electrons thus they are a strong bronsted base while phenol is the strongest acid among all, therefore as per the relation of conjugative strong acids and weak bases, phenol is the weakest base.
Among the following amines’ the strongest Bronsted base is
(A)
(B) $\mathrm{NH}_{3}$
(C)
(D)
Ans: (D)
Pyrrolidine is the strongest of two bases because the lone pair of nitrogen does not involve sin resonance, and the presence of two alkyl basic compounds increases the basic strength among the given four compounds.
The correct decreasing order of basic strength of the following species is
$\mathrm{H}_{2} \mathrm{O}, \mathrm{NH}_{3}, \mathrm{OH}^{-}, \mathrm{NH}_{2}^{-}$
Ans: (A)
The basic strength of an atom is determined by its electron-donating capacity; in this case, the amide is the most basic due to the presence of a negative charge and two lone pairs of electrons on the nitrogen atom.
Which of the following should be most volatile?
(i) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}$
(ii) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}$
(iii)
(iv) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}$
Ans: (B)
Among primary, secondary, and tertiary amines, tertiary amines are the most volatile compounds as they don’t have any strong intermolecular H-bonding between N-H like the primary and secondary amines have.
So due to weak dipole-dipole interactions, (B) will be the most volatile.
Which of the following methods of preparation of amines will not give the same number of carbon atoms in the chain of amines as in the reactant?
Ans: (D)
Only treatment of amide with $\mathrm{Br}_{2}$ in aqueous $\mathrm{NaOH}$ solution yields an amine with fewer carbon atoms than the reactant, whereas $\mathrm{RCONH}_{2} \mathrm{Br}_{2} / \mathrm{NaOH} \mathrm{RNH}_{2}$ all other reactions yield an amine with the same number of carbon atoms as the reactant.
