Among the following compounds most basic amino acid is:

We begin by recalling the general idea of basicity in amino acids. The basic strength of an amino acid depends upon the availability of a lone pair of electrons on a nitrogen atom that can readily accept a proton, $$\text{H}^+.$$ The more easily this lone pair can be donated, the higher will be the basicity of the molecule.

Now, let us examine each of the four given amino acids in turn, always keeping in mind the presence (or absence) of additional nitrogen-containing groups that can provide extra basic sites beyond the normal $$-NH_{2}$$ of the amino acid backbone.

For clarity, we write the structural fragments of the side chains:

• Lysine has the side chain $$-CH_{2}-CH_{2}-CH_{2}-CH_{2}-NH_{2}.$$ Thus, in addition to the backbone $$-NH_{2}$$ group, lysine possesses another terminal $$-NH_{2}$$ group on the side chain, giving it two basic nitrogens.

• Asparagine has the side chain $$-CH_{2}-CONH_{2}.$$ The amide nitrogen in $$-CONH_{2}$$ is involved in resonance with the adjacent carbonyl group $$C=O,$$ which withdraws electron density and reduces the availability of its lone pair. Hence, this nitrogen is a poor base.

• Serine has the side chain $$-CH_{2}-OH.$$ There is no additional nitrogen present; the -OH group is neutral and actually exhibits weak acidity rather than basicity. Therefore, serine has only the single backbone $$-NH_{2}$$ and is not strongly basic.

• Histidine contains an imidazole ring with two nitrogens. One of these nitrogens has a lone pair that can accept a proton, making histidine noticeably basic. However, quantitative pKa data show that the side-chain imidazole nitrogen has a pKa of about $$6.0,$$ whereas the terminal $$-NH_{2}$$ of lysine’s side chain has a pKa of about $$10.5.$$ A larger pKa for the conjugate acid corresponds to a stronger base.

We can summarise the essential trend by writing:

$$\text{Basicity} \propto \text{p}K_a(\text{conjugate acid})$$

and then substituting the approximate numerical values:

$$\begin{aligned} \text{Lysine : } & pK_a \approx 10.5 \\ \text{Histidine : } & pK_a \approx 6.0 \\ \text{Asparagine : } & pK_a \text{ very low (amide; non-basic)} \\ \text{Serine : } & \text{no extra basic site} \end{aligned}$$

Since $$10.5 > 6.0,$$ the side-chain amine of lysine is able to accept a proton much more readily than the imidazole nitrogen of histidine. Asparagine and serine lack strongly basic nitrogens altogether.

Therefore, lysine exhibits the greatest overall basicity among the four options.

Hence, the correct answer is Option A.