Practical Organic Chemistry

The Carius method is used to determine the composition of sulphur and halogen in an organic compound.

The reaction is,

$ 2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{Na}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O} $

Moles of $\mathrm{H}_2 \mathrm{SO}_4$ neutralised $=\frac{0.030}{2}=$

0.015 moles

$ \begin{aligned} \text { Initial moles of } \mathrm{H}_2 \mathrm{SO}_4 & =0.050 \mathrm{~L} \times 0.5 \mathrm{~mL} \\ & =0.025 \mathrm{~mol} \end{aligned} $

Moles of $\mathrm{H}_2 \mathrm{SO}_4$ reacted

$ =0.025-0.015=0.010 \text { mole } $

Moles of $\mathrm{NH}_3$ produced

$ =2 \times 0.010=0.020 \text { mole } $

Mass of $\mathrm{N}=0.020 \times 14=0.28 \mathrm{~g}$

$ \% \text { of } N=\frac{0.28}{\text { Mass of compound }} \times 100 $

$\Rightarrow$ Mass of compound

$ =\frac{0.28 \times 100}{56}=0.50 \mathrm{~g} \text { or } 500 \mathrm{mg} $

Distillation under reduced pressure method is used to purify liquid with high boiling points and liquids that decompose at or below their boiling point.

This technique lower the boiling point of liquid by reducing the pressure.

In column chromatography, compounds with lower adsorption (or weaker interactions with the stationary phase) elute earlier than those that are more strongly adsorbed.

Given the degree of adsorption:

$D > B > C > A$

Compound D is most strongly adsorbed and will therefore come out last.

Compound A, being the least adsorbed, will come out first.

Now, let’s review the options:

Option A: "D comes out first from the column." – This is incorrect because D has the highest retention.

Option B: "A comes out first from the column." – This is correct as A is the least adsorbed.

Option C: "C comes out after B from the column." – This is incorrect since the elution order will be A first, then C, followed by B, and lastly D.

Option D: "B comes out after D from the column." – This is incorrect because D, being the most adsorbed, should elute last.

Thus, the correct answer is:

Option B: A comes out first from the column.

In the test of halogens from sodium extract, a small amount of conc. $\mathrm{HNO}_3$ is added. This is done to remove $\mathrm{NaCN}$ and $\mathrm{Na}_2 \mathrm{~S}$, so that they do not interfere with the test for halogen. Therefore, dilute nitric acid is added before testing halogens to expel all the gases if evolved.

$\begin{gathered} \mathrm{Na}+\mathrm{C}+\mathrm{N} \stackrel{\Delta}{\longrightarrow} \operatorname{NaCN}(X) \quad \text{(if N is present)}\\ 2 \mathrm{Na}+\mathrm{S} \stackrel{\Delta}{\longrightarrow} \mathrm{Na}_2 \mathrm{~S}(Y) \quad \text{(if S is present)}\end{gathered}$

$\mathrm{Sodium\,extract + \mathop {conc.\,HN{O_3}}\limits_{(Z)} \buildrel {} \over \longrightarrow \mathop {\text{Halogen identity}}\limits_{[Cl,|Br|,I]}}$