Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
Aniline is nitrated with a mixture of concentrated nitric and sulphuric acids. A real nitrating mixture also holds a trace of free amine, which reacts very quickly. So the figures below are a simplified model of what happens. In strong acid, aniline is largely turned into the anilinium ion. The of the anilinium ion is . Take the pH of the medium as for this calculation, although a real nitrating mixture is much more acidic. NCERT gives the products of direct nitration as 2 percent ortho, 47 percent meta and 51 percent para. Take of aniline to be nitrated and ignore the loss by oxidation. All nitroanilines have . Find the ratio of anilinium ion to aniline at pH 1.0 and the percentage of the aniline that is in the form of the ion. Then find the amount of the meta product and its mass.
The amino group activates the ring and directs to the ortho and para positions. Bromine water gives 2,4,6-tribromoaniline. In strong acid, aniline becomes the anilinium ion, which directs meta. An acetyl group protects the amine: acetylate, react, hydrolyse. The yields of the steps multiply. The ratio of the ion to the free amine is .
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. Ratio. .
Step 2. Percent as ion. . Almost all of the aniline is the anilinium ion, which directs meta.
Step 3. Meta product. .
Step 4. Its mass. .
Answer: , , and .
Why the meta product is not 99.97 percent. The small amount of free aniline reacts very much faster than the ion, because its ring is highly activated. The para product comes from that free amine, and so the product mixture is about half meta and half para. The calculation also does not include the oxidation of aniline by nitric acid.
Check. The three amounts add to .
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