Practice free Amines (Chemistry) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
The correct order of increasing basicity in aqueous solution is.
$ C_6H_5 NH_2 $ is weaker than $NH_3$ and basicity of amines in aqueous solution is$ 2 ^\circ \gt 3 ^ \circ \gt 1 ^ \circ $
The order of basic strength among the Following amines in the Vapour phase (non - aqueous) Solution is.
In non aqueous solvents the base strength increases as the magnitude of + 1 - effect increases, $ 3 ^\circ \gt 2 ^\circ \gt 1 ^\circ $
Dye test can be used to distinguish between
Only aromatic primary amines gives dye test.
Deamination of benzenediazonium chloride can be carried out with
Deamination of benzenediazonium chloride can be carried out using hypophosphorous acid (H_3PO_2). This reaction involves the replacement of the diazo group (N_2^+) with a hydrogen atom, thus converting the diazonium salt back to benzene.
Arrange the following amines in order of increasing basicity
n - pentylamine (I)
Sec-pentyl amine (II)
iso - pentylamine (III)
tert - pentylamine (IV).
As the steric hindrance increases from (I)+ (IV) the basicity decreases, so, increasing order of basicity becomes IV < III < II < I.
Which of the following statement is correct ?
Aniline is a weaker base than ethylamine because the lone pair of electrons on the nitrogen in aniline is delocalized into the benzene ring, making it less available for protonation. In ethylamine, the lone pair on nitrogen is more available, making it a stronger base.
Benzylamine may be prepared by.............
When a primary amine reacts with chloroform in ethanolic KOH, then the product is............
When a primary amine reacts with chloroform in the presence of ethanolic KOH, the product is an isocyanide. This reaction is known as the Carbylamine reaction or Hofmann's isocyanide test. The reaction can be represented as: $$ R-NH_2 + CHCl_3 + 3KOH ightarrow R-NC + 3KCl + 3H_2O $$
Each question given below contains statement - 1 (Assertion) and Statement - 2 (Reason). Each queastion has 4 choices (1), (2), (3) and (4). out of which only one is correct choose the correct option as under : Statement - 1: Aniline is less basic than P - toludine. Statement - 2: P- toludine is more basic than Aniline due to electron donating group - $CH_3$ .
Aniline (C6H5NH2) is indeed less basic than p-toluidine (C6H4CH3NH2). The methyl group (-CH3) in p-toluidine is an electron-donating group that increases the electron density on the nitrogen atom, making it more willing to donate a pair of electrons. This increased electron density makes p-toluidine more basic compared to aniline. Both statements are true, and the reason provided in Statement 2 correctly explains Statement 1.
Each question given below contains statement - 1 (Assertion) and Statement - 2 (Reason). Each queastion has 4 choices (1), (2), (3) and (4). out of which only one is correct choose the correct option as under : Statement - 1: Primary aliphatic amine forms highly stable alkyldiazonium salt. Statement - 2: Benzenediazonium chloride is easily soluble in water while Benzenediazonium fluoroborateis insolubel in water.
Primary aliphatic amines do not form highly stable alkyldiazonium salts. These salts are generally unstable and decompose easily, unlike aromatic diazonium salts, which are more stable. Statement 1 is false. Statement 2 is true because benzenediazonium chloride is soluble in water due to its ionic nature, while benzenediazonium fluoroborate is insoluble in water due to the presence of the large and less polarizable BF4- anion.
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