Practical Organic Chemistry

The organic compound $X$ is $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2$ (aniline). Here's a concise breakdown of the solution.

We're using the Kjeldahl method to find the nitrogen content of compound $X$.

(A) Moles of acid reacted with ammonia

• Total $\mathrm{H}_2 \mathrm{SO}_4$ added ; $0.060 \mathrm{~L} \times 0.1 \mathrm{M} =0.006 \mathrm{~mol}$.

• NaOH used in back-titration : 0.020 l $\times 0.1 \mathrm{M}=0.002 \mathrm{~mol}$.

• Unreacted $\mathrm{H}_2 \mathrm{SO}_4$ (from 2 NaOH : $1 \mathrm{H}_2 \mathrm{SO}_4$ ) : $0.002 \mathrm{~mol} / 2=0.001 \mathrm{~mol}$.

• $\mathrm{H}_2 \mathrm{SO}_4$ reacted with $\mathrm{NH}_3: 0.006 \mathrm{~mol}$ $0.001 \mathrm{~mol}=0.005 \mathrm{~mol}$.

(B) Mass of nitrogen in compound $X$

• Moles of $\mathrm{NH}_3$ (from $2 \mathrm{NH}_3$ :

$ \left.1 \mathrm{H}_2 \mathrm{SO}_4\right): 0.005 \mathrm{~mol} \times 2=0.010 \mathrm{~mol} . $

• Mass of nitrogen ( $\mathrm{N}=14.01 \mathrm{~g} / \mathrm{mol}$ ) : $0.010 \mathrm{~mol} \times 14.01 \mathrm{~g} / \mathrm{mol}=0.1401 \mathrm{~g}$.

(C) Percentage of nitrogen in compound $X$.

• Sample mass $=0.933 \mathrm{~g}$

• $\% \mathrm{~N}=(0.1401 \mathrm{~g} / 0.933 \mathrm{~g}) \times 100 \%$

$ \approx 15.01 \% . $

(D) Compare with options

• Aniline $\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2\right)$ : Contains one N .

Molar mass $\approx 93 \mathrm{~g} / \mathrm{mol} . \% \mathrm{~N}=(14 / 93) \times 100 \% \approx 15.05 \%$.

• Other options (benzylamine, n-propylamine, acetamide) have significantly different nitrogen percentages (approx. 13.08\%, 23.73%, $23.73 \%$ respectively).

• Since compound $X$ has approximately $15.01 \%$ nitrogen, it matches Aniline.

First, when the sodium fusion extract reacts with iron(II) sulfate, a chemical reaction happens that forms a compound called sodium ferrocyanide.

$ \begin{aligned} &6 \mathrm{NaCN} + \mathrm{FeSO}_4 \longrightarrow \mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] + \mathrm{Na}_2 \mathrm{SO}_4 \end{aligned} $

Next, when you add concentrated $\mathrm{H}_2 \mathrm{SO}_4$, the iron(II) changes to iron(III). This iron(III) reacts with sodium ferrocyanide to make a blue colored compound called Prussian blue.

$ 3 \mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] + 4 \mathrm{Fe}^{3+} \xrightarrow{x \mathrm{H}_2 \mathrm{O}} \mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3 \cdot x \mathrm{H}_2 \mathrm{O} $

The blue color (Prussian blue) shows that nitrogen is present in the organic compound.

Kjeldahl's method can estimate nitrogen only in organic compounds where nitrogen is present in amino or amide forms and not in aromatic rings, azo groups, or nitro groups.

Compound I (aniline) and compound(V) (benzylamine) is suitable.

$n$-pentane ion

$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3$ one is aromatic and other is straight chain compound. Therefore, it can be separated by simple distillation.

Hence, option (b) is the correct answer.

While estimating nitrogen by Kjeldahl's method, the organic compound is heated with conc. $\mathrm{H}_2 \mathrm{SO}_4$, so that nitrogen in the organic compound gets converted to $\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4$.

The carbon and nitrogen present in the organic compound of fusion with sodium metal give sodium cyanide (NaCN) soluble in water. This is converted into sodium ferrocyanide by the addition of sufficient quantity of ferrous sulphate. Ferric ions generated during the process react with ferrocyanide to form the prussian blue precipitate of ferric ferrocyanide. The prussian blue colour is due to formation of ferric ferrocyanide, $\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_4 \cdot X_2 \mathrm{O}$.

$ \text { (i) } \mathrm{Fe}^{2+}+6 \stackrel{\ominus}{\mathrm{C}} \mathrm{~N} \longrightarrow\begin{gathered} {\left[\mathrm{Fe}^{2+}(\stackrel{-6}{\mathrm{CN}})_6\right]^{4-}} \\ \text { Hexa cyanoferrate (II) } \end{gathered} $

$ \text { (ii) } \mathrm{Fe}^{2+} \xrightarrow[\mathrm{H}_2 \mathrm{SO}_4]{\text { Conc. }} \mathrm{Fe}^{3+}+e^{-} $

$ \text { (iii) }\left[\mathrm{Fe}^{2+}(\stackrel{6-}{\mathrm{CN}})_6\right]^{4-}+4 \mathrm{Fe}^{3+} \longrightarrow\left[\mathrm{Fe}^{2+}(\stackrel{6-}{\mathrm{C}})_6\right]^{4-} \cdot \mathrm{H}_2 \mathrm{O} $