Preparing inorganic and organic compounds

Six preparations

A preparation has a plan: react, separate, purify, weigh. Each step has a reason. The yield tells how well it went:

The theoretical mass comes from the equation and the limiting reagent, the one that is used up first.

Mohr's salt, $\text{FeSO}_{4}(\text{NH}_{4})_{2}\text{SO}_{4}\cdot 6\text{H}_{2}\text{O}$

Mohr's salt is a double salt. It is stable in air, whereas iron(II) sulphate alone slowly oxidises, so it is a primary standard.

Steps.

1. Dissolve about 7 g of iron(II) sulphate crystals, $\text{FeSO}_{4}\cdot 7\text{H}_{2}\text{O}$, in a little water with a few drops of dilute sulphuric acid.
2. Dissolve about 3.5 g of ammonium sulphate in a little water. The two amounts are almost equimolar.
3. Mix the solutions. Heat to concentrate, but not to dryness.
4. Let the solution cool slowly. Light green crystals form. Filter and dry between filter papers.

$\text{FeSO}_{4}\cdot 7\text{H}_{2}\text{O}\ +\ (\text{NH}_{4})_{2}\text{SO}_{4}\ \rightarrow \ \text{FeSO}_{4}(\text{NH}_{4})_{2}\text{SO}_{4}\cdot 6\text{H}_{2}\text{O}\ +\ \text{H}_{2}\text{O}$

The acid stops hydrolysis and oxidation of the iron(II). Heating to dryness would oxidise the iron(II) and give a brown crust. Slow cooling gives larger crystals.

Potash alum, $\text{KAl}(\text{SO}_{4})_{2}\cdot 12\text{H}_{2}\text{O}$

Aluminium scrap dissolves in potassium hydroxide, and sulphuric acid then makes the alum.

$2\text{Al}\ +\ 2\text{KOH}\ +\ 6\text{H}_{2}\text{O}\ \rightarrow \ 2\text{K}[\text{Al}(\text{OH})_{4}]\ +\ 3\text{H}_{2}$

$2\text{K}[\text{Al}(\text{OH})_{4}]\ +\ 4\text{H}_{2}\text{SO}_{4}\ \rightarrow \ 2\text{KAl}(\text{SO}_{4})_{2}\ +\ 8\text{H}_{2}\text{O}$

Steps. Warm the scrap with potassium hydroxide solution until the bubbles of hydrogen stop. Filter off the residue. Add dilute sulphuric acid. A white precipitate of aluminium hydroxide forms first, then it dissolves. Concentrate, then cool in ice. Colourless octahedral crystals form. The $M$ of the dodecahydrate is $474\ \text{g/mol}$.

Acetanilide, $\text{C}_{6}\text{H}_{5}\text{NHCOCH}_{3}$

Acetanilide is made by acetylation of aniline. The $\text{-}\text{NH}_{2}$ group is changed to the amide group.

$\text{C}_{6}\text{H}_{5}\text{NH}_{2}\ +\ (\text{CH}_{3}\text{CO})_{2}\text{O}\ \rightarrow \ \text{C}_{6}\text{H}_{5}\text{NHCOCH}_{3}\ +\ \text{CH}_{3}\text{COOH}$

Steps. Warm aniline with acetic anhydride, glacial acetic acid and a pinch of zinc dust. The zinc dust stops the aniline from oxidising and turning dark. Pour the hot mixture into cold water. Acetanilide separates as a solid. Filter, wash with cold water, and recrystallise from hot water or ethanol. The melting point is $114\ ^\circ\text{C}$.

p-Nitroacetanilide

Direct nitration of aniline gives oxidation products and a large amount of the meta isomer, because the amino group is protonated in strong acid. We first protect it as the amide. The acetamido group is much less activating, resists oxidation and directs to the para position, with a little ortho product.

$\text{C}_{6}\text{H}_{5}\text{NHCOCH}_{3}\ +\ \text{HNO}_{3}\ \rightarrow \ \text{p}\text{-}\text{O}_{2}\text{N}\text{-}\text{C}_{6}\text{H}_{4}\text{-}\text{NHCOCH}_{3}\ +\ \text{H}_{2}\text{O}$ (concentrated sulphuric acid present)

Steps. Dissolve acetanilide in glacial acetic acid and concentrated sulphuric acid in an ice bath. Add the cooled mixture of concentrated nitric and sulphuric acids drop by drop. Keep the temperature below about 293 K, since heat gives dinitration and oxidation. Leave for a short time, then pour onto crushed ice. Filter and wash. Recrystallise from ethanol. The para isomer separates as pale yellow crystals, with a melting point of about $216\ ^\circ\text{C}$. The ortho isomer is more soluble in ethanol, so it stays in the solution. Hydrolysis of the product gives p-nitroaniline.

Aniline yellow

Aniline yellow is p-aminoazobenzene, $\text{C}_{6}\text{H}_{5}\text{-}\text{N}=\text{N}\text{-}\text{C}_{6}\text{H}_{4}\text{-}\text{NH}_{2}$, a yellow azo dye. Aniline is diazotised with sodium nitrite and hydrochloric acid at 273 to 278 K. The cold diazonium salt reacts with aniline in a weakly acidic solution. At first diazoaminobenzene, $\text{C}_{6}\text{H}_{5}\text{-}\text{N}=\text{N}\text{-}\text{NH}\text{-}\text{C}_{6}\text{H}_{5}$, forms. Warm with aniline hydrochloride, and it rearranges to p-aminoazobenzene. Filter the yellow crystals.

$\text{C}_{6}\text{H}_{5}\text{N}_{2}^{+}\text{Cl}^{-}\ +\ \text{C}_{6}\text{H}_{5}\text{NH}_{2}\ \rightarrow \ \text{C}_{6}\text{H}_{5}\text{-}\text{N}=\text{N}\text{-}\text{C}_{6}\text{H}_{4}\text{-}\text{NH}_{2}\ +\ \text{HCl}$

The temperature is kept low because the diazonium salt breaks down above about 283 K, giving phenol and nitrogen.

Iodoform

Iodoform, $\text{CHI}_{3}$, is made from ethanol or propanone with iodine and sodium hydroxide.

$\text{CH}_{3}\text{COCH}_{3}\ +\ 3\text{I}_{2}\ +\ 4\text{NaOH}\ \rightarrow \ \text{CHI}_{3}\ +\ \text{CH}_{3}\text{COONa}\ +\ 3\text{NaI}\ +\ 3\text{H}_{2}\text{O}$

$\text{CH}_{3}\text{CH}_{2}\text{OH}\ +\ 4\text{I}_{2}\ +\ 6\text{NaOH}\ \rightarrow \ \text{CHI}_{3}\ +\ \text{HCOONa}\ +\ 5\text{NaI}\ +\ 5\text{H}_{2}\text{O}$

Steps. Dissolve iodine in potassium iodide solution, add the ethanol or propanone, and then sodium hydroxide until the iodine colour fades. Warm gently to about 333 K. Pale yellow crystals with a strong medicinal smell separate on cooling. Wash with water and recrystallise from ethanol. The melting point is about $119\ ^\circ\text{C}$.

Set 4.5 g of aniline and 6 g of acetic anhydride. The aniline is the limiting reagent. Now reduce the acetic anhydride to 3 g. The line flattens, because the anhydride is now limiting.

Practise with new numbers

Try it in Python

The program finds the limiting reagent for acetanilide and the percentage yield.

Common mistakes

• Heating Mohr's salt solution to dryness. The iron(II) oxidises.
• Forgetting the dilute sulphuric acid in the Mohr's salt solution.
• Nitrating aniline directly. Acetylate it first.
• Letting the nitration warm up. Dinitration and oxidation follow.
• Using the mass of the excess reagent to find the theoretical yield. Use the limiting reagent.
• A yield above 100 %. The product is still wet or impure.

Remember these

• Mohr's salt: pale green, stable double salt. Potash alum: colourless octahedral crystals.
• Acetanilide ($135\ \text{g/mol}$) protects the amine. p-Nitroacetanilide is the main product, pale yellow.
• Aniline yellow is p-aminoazobenzene, made at 273 to 278 K.
• Iodoform: three iodine molecules and four sodium hydroxide for each propanone. Yellow, medicinal smell.
• Yield is mass obtained over theoretical mass, from the limiting reagent.

References

• NCERT Chemistry Lab Manual for Class 12, the preparation of inorganic compounds and of organic compounds.
• Vogel's Textbook of Practical Organic Chemistry, the preparations of acetanilide, p-nitroacetanilide and iodoform.

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 dilute sulphuric acid added when the solution of Mohr's salt is made?

2. Why is aniline acetylated before it is nitrated?

3. Why is the diazotisation of aniline done at 273 to 278 K?

4. How many moles of iodine and of sodium hydroxide does 0.10 mol of propanone need for iodoform?