Sulphur and the halogens in an organic compound

One extract, many tests

In lesson 1 we made the sodium fusion extract. It can answer more than one question. Sulphur is now a sulphide ion. A halogen is a halide ion. Let us test for them one at a time. Use a fresh portion of the extract for each test.

Testing for sulphur

Test 1: sodium nitroprusside. Add a few drops of freshly prepared sodium nitroprusside solution, $\text{Na}_{2}[\text{Fe}(\text{CN})_{5}\text{NO}]$, to 1 mL of the extract. A violet or purple colour shows sulphide.

$\text{S}^{2-}\ +\ [\text{Fe}(\text{CN})_{5}\text{NO}]^{2-}\ \rightarrow \ [\text{Fe}(\text{CN})_{5}\text{NOS}]^{4-}$

Test 2: lead acetate. Acidify 1 mL of the extract with acetic acid. Add a few drops of lead acetate solution. A black precipitate of lead sulphide shows sulphur.

$\text{S}^{2-}\ +\ \text{Pb}^{2+}\ \rightarrow \ \text{PbS}$

We use acetic acid, not hydrochloric acid. Acetic acid is strong enough to neutralise the alkali of the extract. Hydrochloric acid would give white lead chloride, which could hide the black.

When nitrogen and sulphur are both present

During the fusion, carbon, nitrogen and sulphur can make sodium thiocyanate, $\text{NaSCN}$. Then there is little free cyanide or free sulphide. The nitrogen test gives no Prussian blue. The test for the thiocyanate ion is a blood red colour with iron(III) chloride:

$3\text{SCN}^{-}\ +\ \text{Fe}^{3+}\ \rightarrow \ \text{Fe}(\text{SCN})_{3}$

Acidify the extract with dilute hydrochloric acid first, then add a few drops of iron(III) chloride. (The red species in solution is really a complex such as $[\text{Fe}(\text{SCN})]^{2+}$, but books write $\text{Fe}(\text{SCN})_{3}$.)

If the sodium is in excess, the thiocyanate breaks down: $\text{NaSCN}\ +\ 2\text{Na}\ \rightarrow \ \text{NaCN}\ +\ \text{Na}_{2}\text{S}$. Then the extract gives Prussian blue and the violet nitroprusside colour, and not the blood red.

Testing for the halogens

Steps.

1. Take about 2 mL of the extract in a test tube.
2. Add a few drops of concentrated nitric acid. Boil gently for a few minutes in a fume cupboard.
3. Cool. Add silver nitrate solution.

Why boil with nitric acid? If nitrogen or sulphur is present, the extract holds cyanide or sulphide ions. These would give white silver cyanide or black silver sulphide and confuse the halogen test. Boiling with the acid drives them off as gases:

$\text{NaCN}\ +\ \text{HNO}_{3}\ \rightarrow \ \text{NaNO}_{3}\ +\ \text{HCN}$

$\text{Na}_{2}\text{S}\ +\ 2\text{HNO}_{3}\ \rightarrow \ 2\text{NaNO}_{3}\ +\ \text{H}_{2}\text{S}$

Both gases are poisonous. Work in a fume cupboard. The acid also destroys the alkali of the extract. Without it, silver nitrate would give a brown precipitate of silver oxide.

What you see.

• A white, curdy precipitate that dissolves in dilute ammonia solution: chlorine. $\text{Ag}^{+}\ +\ \text{Cl}^{-}\ \rightarrow \ \text{AgCl}$
• A pale yellow precipitate that dissolves only slightly in dilute ammonia solution: bromine.
• A yellow precipitate that does not dissolve in ammonia solution: iodine.

The layer test. Acidify a portion of the extract with dilute sulphuric acid and add a little chlorine water and carbon tetrachloride. Shake. Bromide gives an orange-brown colour in the lower layer. Iodide gives a violet one. Chlorine water frees the halogen, and the organic solvent dissolves it. Add the chlorine water drop by drop. Too much takes the colour away again.

Fluorine. Silver fluoride is soluble, so silver nitrate cannot detect fluorine.

Choose bromine and a precipitate of 0.50 g. Now choose chlorine. Which halogen has the larger share of its silver halide as halogen? Iodine is heaviest, so the line for iodine rises most.

Practise with new numbers

Try it in Python

This program does the same sum as the simulation. It finds the percentage of a halogen from the mass of its silver halide.

Common mistakes

• Skipping the nitric acid. Cyanide and sulphide give wrong precipitates.
• Using hydrochloric acid to acidify before silver nitrate. It adds chloride ions and gives a false white precipitate.
• Mixing up the colours of the silver halides. White is chloride, pale yellow is bromide, deep yellow is iodide.
• Calling a blood red colour a nitrogen test. It shows that both nitrogen and sulphur are present as thiocyanate.
• Using hydrochloric acid in the lead acetate test. The white lead chloride may hide the black sulphide.

Remember these

• Sulphur: violet with sodium nitroprusside, or black lead sulphide with lead acetate and acetic acid.
• Nitrogen and sulphur together: sodium thiocyanate gives blood red with iron(III) chloride.
• Halogens: boil with nitric acid first, then silver nitrate. White (dissolves in ammonia), pale yellow (sparingly), yellow (does not).
• Silver fluoride is soluble, so fluorine does not respond.

References

• NCERT Chemistry Class 11, the chapter Organic Chemistry: Some Basic Principles and Techniques.
• Vogel's Textbook of Practical Organic Chemistry, the chapters on the detection of elements.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Why is the extract boiled with nitric acid before silver nitrate is added?

2. A white precipitate forms with silver nitrate. It dissolves completely in dilute ammonia solution. Which halogen is present?

3. An extract gives a blood red colour with iron(III) chloride and no Prussian blue. What does it show?

4. Why can a silver nitrate test not show fluorine?