Sandmeyer Reaction
Sandmeyer Reaction
Aniline synthesis to set up the Sandmeyer reaction
We have already learned how to nitrate an aromatic ring with nitric and sulfuric acid. A reduction of a nitro group on an aromatic ring to an amino group with fewer nitrogen-oxygen bonds is accomplished by reducing it with iron filings and hydrochloric acid followed by hydroxide. The meta-directing nitro group is now an o,p-directing amino group. Remember that aminobenzene is commonly called aniline.
9. Show the two reaction steps needed to turn propyl benzene into 4-aminopropylbenzene.
Amino groups on an aromatic ring can be turned into the world’s greatest leaving group, N2, by reacting it with sodium nitrite (NaNO2) and HCl. Once this happens, the leaving group is so good, it is only stable below 10°C. It can be replaced with many different types of nucleophiles.
The substitution reaction is called a Sandmeyer reaction if the nucleophile is a Cu+ salt like cuprous cyanide (CuCN), cuprous chloride (CuCl), or cuprous bromide (CuBr).
Sandmeyer Reaction
Many different nucleophilic substitutions can occur. With acid and water, we get a phenol. With HBF4, we get an aromatic fluoride. With potassium iodide (KI), we get an aromatic iodide. With hypophosphorous acid (H3PO2), the N2 is replaced with a proton to reform the benzene ring.
10. Write in the reagents to complete the following Sandmeyer and Sandmeyer-like reactions.
11. Draw in all reagents and products in the following reaction sequence. Pay careful attention if groups are ortho-para directing or meta-directing. Notice how groups sometimes direct to one location and then are changed into a different type of directing group.
Friedel-Crafts on aniline
Suppose you wanted to make o- or p-methylaniline from aniline using a Friedel-Crafts alkylation. You might think you could react methyl chloride (CH3Cl) with aluminum trichloride (AlCl3) with aniline. You would find that this reaction does not work.
The reason it does not work is that AlCl3 is a good Lewis acid and wants electrons. The amino NH2- group of aniline has a negative, lone pair of electrons on it that can attack the AlCl3 Lewis acid. This makes a positively charged quaternary ammonium salt that is a deactivating, meta-directing group. The reaction would, therefore, not proceed at all.
Like many times before in human history, when we hit a roadblock, we have figured out a way around it. In order to do a Friedel-Crafts alkylation on aniline to make o- or p-methylaniline, we need to turn the amino NH2 group into something that is still electron donating and o,p-directing, but does not have such a reactive negative lone pair of electrons on it. The way we do that is we can turn the amino group into an amido group. To do this, we add an acid chloride like acetyl chloride to the aniline making N-phenylacetamide. The amido group has a lone pair of electrons on the nitrogen atom, but they are not as concentrated on the nitrogen atom as they were in aniline. The lone pair is spread out towards the oxygen atom of the amide through resonance. In fact, the electrons are now spread out enough that they will not react with the AlCl3 Lewis acid. There is a lone pair of electrons on the nitrogen atom, so the amido group is still an o,p-director.
This amido compound can now undergo a Friedel-Crafts reaction.
This amido compound can now undergo a Friedel-Crafts reaction.
In this roundabout way, the Friedel-Crafts reaction of an amino benzene can occur. The amino group is transformed into an amido group, the Friedel-Crafts reaction occurs, then the amido group is hydrolyzed to an amino group once again.
12.
a) Why is the following Friedel-Crafts reaction NOT a good way to make isopropyl aniline from aniline?
b) What is a better synthesis of isopropyl aniline?
Answers
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12.
a) The lone pair of electrons on the amino group would react with the Lewis Acid, AlCl3 instead of the AlCl3 helping pull off the chloride of isopropyl chloride.
b)
