# Does elimination occur when bromonapthalene is reacted with sodium amide in potassium hydroxide?

What should be the product of this reaction? Some books show that an elimination reaction would happen since $$\ce{NH2-}$$ is a strong enough base to take away a hydrogen on a benzene ring and create a triple bond. I've never seen that before and why would the stable ring react?

• Are you trying to say that the ring would not react because of loss of aromatic stability? Jun 15 at 1:28
• @M.L regarding your latest edit, try to avoid using MathJax in title filed as the search engines can't read it and thus it makes a post harder to find. Jun 15 at 4:08
• Have you learnt about benzyne mechanism? If not, I would suggest you to read about that first. Books would generally have an overview of benzyne first before showing it on napthalene. Your reaction is similar to that.
– TRC
Jun 15 at 6:28

Elimination does occur, but only to form an intermediate. As TRC has mentioned above, this involves a benzyne mechanism.

$$\ce{NH2+}$$ is a very strong base; it is the conjugate base of ammonia, which itself is already a base.

Under such strong conditions, the aromaticity of the benzene ring can be broken with the $$\ce{NH2+}$$ eliminating a hydrogen from the aryl halide and forming ammonia. This leaves a negative charge on the arene, and naturally to resolve this negative charge, the halide, being a good leaving group, leaves the benzene ring as an anion.

This hydrogen is always eliminated alpha from the aryl halide, meaning to one of the carbons next to it.

Two of the carbons of the ring, one at the halide position and one alpha to it, now have no substituents and form a triple bond between themselves. This is an aryne.

This Aryne is too reactive to remain for long, with a very strained triple bond and no aromaticity.

The aryne contains two locations, on each end of the triple bond, where the carbon has no substituents. Therefore, each end of the triple bond is now a prime target for an addition reaction.

Therefore, the ammonia attacks the aryne at one end of the triple bond, forming an brief intermediate before one of the hydrogens from ammonia undergoes a hydride shift to add to the other end of the triple bond. Net result is a substitution reaction. An amine group is substituted for a halide group.

However, because of how both ends of the triple bond can be attacked, the amine group may be at the same position as the original halide, or it may be at one of the alpha positions relative to it.

Libretexts has a good page on the mechanism which I shall link.

https://chem.libretexts.org/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Benzyne_Mechanism