What happens if naoh is substituted for nh3

A tertiary amine is one which has three alkyl groups attached to the nitrogen. The final stage! The triethylamine reacts with bromoethane to give tetraethylammonium bromide - a quaternary ammonium salt one in which all four hydrogens have been replaced by alkyl groups. This time there isn't any hydrogen left on the nitrogen to be removed. The reaction stops here. What do you actually get if you react bromoethane with ammonia?

Whatever you do, you get a mixture of all of the products including both amines and their salts shown on this page. To get mainly the quaternary ammonium salt, you can use a large excess of bromoethane. If you look at the reactions going on, each one needs additional bromoethane.

If you provide enough, then the chances are that the reaction will go to completion, given enough time. On the other hand, if you use a very large excess of ammonia, the chances are always greatest that a bromoethane molecule will hit an ammonia molecule rather than one of the amines being formed.

That will help to prevent the formation of secondary etc amines. Jim Clark Chemguide. Making a Primary Amine The reaction happens in two stages. This gets so complicated that it is dealt with on a separate page.

You will find a link at the bottom of this page. The facts of the reactions are exactly the same as with primary halogenoalkanes. The halogenoalkane is heated in a sealed tube with a solution of ammonia in ethanol. This time the slow step of the reaction only involves one species - the halogenoalkane.

It is known as an S N 1 reaction. There is a second stage exactly as with primary halogenoalkanes. An ammonium ion is formed, together with an amine. It is very unlikely that any of the current UK-based syllabuses for 16 - 18 year olds will ask you about this. In the extremely unlikely event that you will ever need it, secondary halogenoalkanes use both an S N 2 mechanism and an S N 1.

Make sure you understand what happens with primary and tertiary halogenoalkanes, and then adapt it for secondary ones should ever need to.

The reaction of primary halogenoalkanes with ammonia. Use the BACK button on your browser to return to this page. The facts The halogenoalkane is heated with a concentrated solution of ammonia in ethanol. We'll talk about the reaction using 1-bromoethane as a typical primary halogenoalkane. The more ammonia there is in the mixture, the more the forward reaction is favoured.

In terms of the chemistry, this is exactly the same as the last example - all that differs are the colors. Unfortunately, these are not quite so straightforward. The color of the tetrachlorocuprate II ion is almost always seen mixed with that of the original hexaaqua ion.

What you normally see is:. The reaction taking place is reversible, and you get a mixture of colors due to both of the complex ions. You may find the color of the tetrachlorocuprate II ion variously described as olive-green or yellow.

Adding water to the green solution, replaces the chloro ligands by water molecules again, and returns the solution to blue. Water molecules and ammonia molecules are very similar in size, and so there is no change in co-ordination this time. Unfortunately, the reactions aren't quite so straightforward to describe.

Ammonia solution can react with hexaaqua metal ions in two quite distinct ways, because it can act as a base as well as a ligand. If you add a small amount of ammonia solution you get precipitates of the metal hydroxide - the ammonia is acting as a base.

In some cases, these precipitates redissolve when you add more ammonia to give solutions in which a ligand exchange reaction has occurred. In the diagrams below, both steps are shown, but we are only going to consider the chemistry of the overall ligand exchange reaction.

The precipitates dissolve because of a complicated series of equilibrium shifts, and we shan't worry about that for the moment. Notice that the four ammonias all lie in one plane, with the water molecules above and below. What you see in a test tube is:. The color of the deep blue complex is so strong that this reaction is used as a sensitive test for copper II ions in solution.

Even if you try to reverse the change by adding large amounts of water to the equilibrium, the strength of the deep blue even highly diluted always masks the pale blue of the aqua ion. The straw colored solution formed changes color very rapidly on standing to a deep reddish brown.

However, that is a quite separate reaction, and is not a part of the ligand exchange reaction.