Give a plausible explanation for each of the following:
$(i)$ Why are amines less acidic than alcohols of comparable molecular masses?
$(ii)$ Why do primary amines have higher boiling points than tertiary amines?
$(iii)$ Why are aliphatic amines stronger bases than aromatic amines?

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(N/A) $(i)$ Amines undergo deprotonation to give amide ions (amide anions).
$R-NH_2 \to R-NH^- + H^+$
Similarly,alcohol loses a proton to give an alkoxide ion.
$R-OH \to R-O^- + H^+$
In an amide ion,the negative charge is on the $N$ atom,whereas in an alkoxide ion,the negative charge is on the $O$ atom. Since $O$ is more electronegative than $N$,$O$ can accommodate the negative charge more easily than $N$. As a result,the amide ion is less stable than the alkoxide ion. Hence,amines are less acidic than alcohols of comparable molecular masses.
$(ii)$ In a molecule of a tertiary amine,there are no $H$ atoms attached to the nitrogen atom,whereas in primary amines,two hydrogen atoms are present. Due to the presence of $N-H$ bonds,primary amines undergo extensive intermolecular hydrogen bonding.
As a result,extra energy is required to separate the molecules of primary amines. Hence,primary amines have higher boiling points than tertiary amines.
$(iii)$ Due to the $-R$ (resonance) effect of the benzene ring,the lone pair electrons on the $N$ atom are delocalized into the ring in aromatic amines. Therefore,the electrons on the $N$ atom in aromatic amines are less available for donation. This explains why aliphatic amines are stronger bases than aromatic amines.

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Consider the following sequence of reactions:
Chlorobenzene $\xrightarrow[ii) CO_2, H_3O^{+}]{i) Mg, \text{dry ether}} \text{Benzoic acid}$ $\xrightarrow{NH_3, \Delta} \text{Benzamide (A)}$ $\xrightarrow{Br_2, NaOH} \text{Aniline (B)}$
$11.25 \ mg$ of chlorobenzene will produce $.......... \times 10^{-1} \ mg$ of product $B$.
(Consider the reactions result in complete conversion.)
[Given molar mass of $C, H, O, N$ and $Cl$ as $12, 1, 16, 14$ and $35.5 \ g \ mol^{-1}$ respectively]

The correct sequence of reagents for the following transformation is:

Match the following:
List-$I$ (Amine)List-$II$ ($pK_b$ value)
$A$. $N,N$-Dimethylaniline$I$. $9.30$
$B$. Aniline$II$. $8.92$
$C$. $N$-Ethylethanamine$III$. $9.38$
$D$. $N$-Methylaniline$IV$. $3.00$

The correct answer is

$A$ hydrocarbon $'A'$ $(C_4H_8)$ on reaction with $HCl$ gives a compound $'B'$ $(C_4H_9Cl)$,which on reaction with $1 \ mol$ of $NH_3$ gives compound $'C'$ $(C_4H_{11}N)$. On reacting with $NaNO_2$ and $HCl$ followed by treatment with water,compound $'C'$ yields an optically active alcohol,$'D'$. Ozonolysis of $'A'$ gives $2 \ mols$ of acetaldehyde. Identify compounds $'A'$ to $'D'$. Explain the reactions involved.

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Consider the following reaction sequence:
$p$-Nitrotoluene $\xrightarrow{P} Q$ $\xrightarrow{R} S$ $\xrightarrow{H_2O} T$
$S \xrightarrow{U} \text{Benzoic acid}$
Identify the correct statements from the following:
$(A) P = H_2/Pd, \text{ethanol}; R = NaNO_2/HCl; U = 1. H_3PO_2, 2. KMnO_4-KOH, \text{heat}$
$(B) P = Sn/HCl; R = HNO_2; S = p-\text{toluenediazonium chloride}$
$(C) S = p-\text{toluenediazonium chloride}; T = p-\text{cresol}; U = 1. CH_3CH_2OH, 2. KMnO_4-KOH, \text{heat}$
$(D) Q = p-\text{nitrobenzoic acid}; R = H_2/Pd, \text{ethanol}; T = p-\text{cresol}$
Which of the following combinations is correct?

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