Kjeldahl's method cannot be used to estimate nitrogen for which of the following compounds?

First, recall the principle of Kjeldahl’s method. In this method, the compound is digested with concentrated $$\text{H}_2\text{SO}_4$$, which converts all nitrogen present in the compound into ammonium sulphate $$\left((\text{NH}_4)_2\text{SO}_4\right)$$. The general reaction written symbolically is

$$\text{Organic nitrogen} \;+\; \text{H}_2\text{SO}_4 \;\xrightarrow{\text{heat}}\; (\text{NH}_4)_2\text{SO}_4$$

After digestion, the reaction mixture is made alkaline with excess $$\text{NaOH}$$. This liberates ammonia gas:

$$(\text{NH}_4)_2\text{SO}_4 + 2\,\text{NaOH} \;\longrightarrow\; 2\,\text{NH}_3 \uparrow + \text{Na}_2\text{SO}_4 + 2\,\text{H}_2\text{O}$$

The evolved $$\text{NH}_3$$ is then absorbed in a known excess of standard acid and finally estimated by back-titration. Thus, the success of Kjeldahl’s method depends on the ability of hot concentrated $$\text{H}_2\text{SO}_4$$ to convert every form of nitrogen in the sample into $$\text{NH}_4^+$$.

However, if nitrogen is present in such a way that it cannot be reduced to $$\text{NH}_4^+$$ under these conditions, the method fails. Specifically, nitrogen present in its oxidised form (such as the nitro group $$\left(-\text{NO}_2\right)$$ or the nitroso group $$\left(-\text{NO}\right)$$) resists conversion to ammonium ions during sulphuric-acid digestion. Because of this resistance, the total nitrogen is not quantitatively transformed into $$\text{NH}_3$$ and the estimation becomes inaccurate.

Now we examine each option to see how the nitrogen is bonded:

Option A: $$\text{C}_6\text{H}_5\text{NH}_2$$ is an aniline molecule where nitrogen is present as a primary amine $$(-\text{NH}_2)$$. In concentrated $$\text{H}_2\text{SO}_4$$, an amine nitrogen is easily protonated and ultimately converted to ammonium sulphate. So, Kjeldahl’s method works here.

Option B: $$\text{CH}_3\text{CH}_2{-}\text{C}\equiv\text{N}$$ is propionitrile, containing a cyano group $$(-\text{C}\equiv\text{N})$$. Nitrile nitrogen is in a reduced state and, on digestion with hot $$\text{H}_2\text{SO}_4$$, it is hydrolysed to give ammonium sulphate. Thus, Kjeldahl’s method also works for nitriles.

Option C: $$\text{C}_6\text{H}_5\text{NO}_2$$ is nitrobenzene. Here nitrogen is present in the nitro group $$(-\text{NO}_2)$$ where its oxidation state is $$+5$$. Because this nitrogen is already highly oxidised, concentrated $$\text{H}_2\text{SO}_4$$ cannot further oxidise it to $$\text{NH}_4^+$$. Instead, the nitro group remains largely unaffected or sometimes even forms $$\text{NO}_2$$ or $$\text{NO}$$ gases that escape. Therefore, nitro-nitrogen is not estimated by Kjeldahl’s method.

Option D: $$\text{NH}_2{-}\text{CO}{-}\text{NH}_2$$ (urea) contains two amide nitrogens $$(-\text{NH}_2)$$. Amide nitrogens are also reduced forms of nitrogen and get converted to ammonium sulphate during digestion, so Kjeldahl’s method works here as well.

From the above discussion, only the compound with a nitro group, namely $$\text{C}_6\text{H}_5\text{NO}_2$$ (Option C), cannot be analysed by the Kjeldahl procedure.

Hence, the correct answer is Option C.