When the lone pair of an amino group $\ce{R-NH2}$ is involved in resonance, its basicity decreases. But, between conjugation with carbonyl group $\ce{R}=\ce{R'CO}$ or resonance with benzene $\ce{R} = \ce{Ph}$, which one will reduce the basicity of amino group to a greater extent?
Answer
Measures for acid and base strength
First, you need to be clear about 'strong' and 'weak' acids. In order to measure the strength of an acid relative to water and find out how effective a proton donor it is, you must look at the equilibrium constant for the reaction
\begin{equation} \ce{HA (aq) + H2O(l) <=> H3O+ (aq) + A- (aq)} \end{equation}
The position of equilibrium is measured by the equilibrium constant for this reaction $K_{\text{eq}}$.
\begin{equation} K_{\text{eq}} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{H2O}][\ce{HA}]} \end{equation}
The concentration of water remains essentially constant (at $55.56 \, \text{mol} \, \text{dm}^{–3}$) with dilute solutions of acids wherever the equilibrium may be and a new equilibrium constant, $K_{\text{a}}$, is defined and called the acidity constant.
\begin{equation} K_{\text{a}} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]} \end{equation}
Like $\text{pH}$, this is also expressed in a logarithmic form, $\text{p}K_{\text{a}}$.
\begin{equation} \text{p}K_{\text{a}} = -\log (K_{\text{a}}) \end{equation}
Because of the minus sign in this definition, the lower the $\text{p}K_{\text{a}}$, the larger the equilibrium constant, $K_{\text{a}}$, is and hence the stronger the acid. The $\text{p}K_{\text{a}}$ of the acid is the $\text{pH}$ where it is exactly half dissociated. At $\text{pH}$s above the $\text{p}K_{\text{a}}$, the acid $\ce{HA}$ exists as $\ce{A^{–}}$ in water; at $\text{pH}$s below the $\text{p}K_{\text{a}}$, it exists as undissociated $\ce{HA}$.
An acid's $\text{p}K_{\text{a}}$ depends on the stability of its conjugate base. $\ce{HCl}$ is a much stronger acid than acetic acid: the $\text{p}K_{\text{a}}$ of $\ce{HCl}$ is around $–7$ compared to $4.76$ for acetic acid. This tells you that in solution $K_{\text{a}}$ for hydrogen chloride is $10^{7} \, \text{mol} \, \text{dm}^{–3}$ whilst for acetic acid it is only $10^{–4.76} = 1.74 \cdot 10^{–5} \, \text{mol} \, \text{dm}^{–3}$. But, why are the equilibria so different? Why does hydrogen chloride fully dissociate but acetic acid do so only partially?
\begin{equation} \ce{HCl (aq) + H2O (l) <=>> H3O+ (aq) + Cl- (aq)} \quad \qquad \qquad K_{\text{a}} = 10^{7} \, \text{mol} \, \text{dm}^{–3} \\ \ce{CH3COOH (aq) + H2O (l) <<=> H3O+ (aq) + CH3COO- (aq)} \qquad K_{\text{a}} = 1.74 \cdot 10^{–5} \, \text{mol} \, \text{dm}^{–3} \\ \end{equation}
The answer must have something to do with the conjugate base $\ce{A^{–}}$ of each acid $\ce{HA}$, since this is the only thing that varies from one acid to another. In both the equilibria above, water acts as a base by accepting a proton from the acid. For the hydrochloric acid equilibrium in the reverse direction, the chloride ion is not a strong enough base to deprotonate the hydronium ion. Acetate, on the other hand, is easily protonated by $\ce{H3O+}$ to give neutral acetic acid, which means that acetate must be a stronger base than chloride ion. A nice thing to remember:
- The stronger the acid $\ce{HA}$, the weaker its conjugate base, $\ce{A^{–}}$
- The stronger the base $\ce{A^{–}}$, the weaker its conjugate acid $\ce{AH}$
So, the $\text{p}K_{\text{a}}$ is a good measure for the strength of an acid: the lower the $\text{p}K_{\text{a}}$ the stronger the acid.
But what about bases? To say something about the strength of a base it is convenient to take the following viewpoint: to what extent does a protonated base want to keep its proton or in other words: how big is the acid strength of its conjugate acid? For example if you want to know which is the stronger base - say, formate anion or acetylide anion - you look up the $\text{p}K_{\text{a}}$s for their conjugate acids. You find that the $\text{p}K_{\text{a}}$ for formic acid ($\ce{HCO2H}$) is $3.7$, whilst the $\text{p}K_{\text{a}}$ for acetylene is around $25$. This means that acetylene is much more reluctant to part with its proton, that is, acetylide is much more basic than formate. This is all very well for anions - you simply look up the $\text{p}K_{\text{a}}$ value for the neutral conjugate acid, but what if you want to know the basicity of ammonia? If you look up the $\text{p}K_{\text{a}}$ for ammonia you find a value around $33$ but this is the value for deprotonating neutral ammonia to give the amide ion, $\ce{NH2^{–}}$. If you want to know the basicity of ammonia, you must look up the $\text{p}K_{\text{a}}$ of its conjugate acid, the ammonium cation, $\ce{NH4+}$, protonated ammonia. Its $\text{p}K_{\text{a}}$ is $9.24$ which means that ammonia is a weaker base than hydroxide - the $\text{p}K_{\text{a}}$ for water (the conjugate acid of hydroxide) is $15.74$. In conclusion: the higher the $\text{p}K_{\text{a}}$ of the conjugate acid the stronger the base.
The answer to your question
The $\text{p}K_{\text{a}}$ of the conjugate acid of aniline is $4.6$. So, since this value is much lower than the $9.24$ you had for ammonia, it is much less basic than ammonia, as is expected since the lone pair of aniline is delocalized into the phenyl ring. But for amides the $\text{p}K_{\text{a}}$ values of the conjugate acid are typically smaller than $0$, so an amide is much less basic than aniline. This is to be expected because the carbonyl group contains an electronegative oxygen atom and thus has a lower lying LUMO than the phenyl ring which can interact more efficiently with the nitrogen's lone pair, thus spreading the negative charge more evenly among the conjugated groups and leading to a higher stabilization of the molecule.
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