Draw the structures of the ammonium ions formed when the amine...

Question

Draw the structures of the ammonium ions formed when the amines in Problem 16.30 are treated with acid.

a. N-Methylpentylamine

b. N-Ethylcyclobutylamine

c. p-Propylaniline

Accepted Answer

All right. Hi, everyone. So for this question, let's go ahead and consider the following amine. For part one, we have perchloroethylene. Part two is N methyl cyclops amine and part three is N butty ethyl amine. What are the structures of the ammonium ion produced from the reaction of each of the following means with an acid. So let's go ahead and talk about the reaction in question. Recall that a means can be represented by the general formula of NR three where R is an abbreviation for any a group or simply nitrogen or hydrogen. Excuse me. So when an amine reacts with an acid that will refer to as H A, then a Bronzy acid base reaction is going to take place because specifically a means are characteristically weak basis. And according to the braid acid based definition bases accept protons or as acids donate protons. So during this acid base reaction, the nitrogen of the amine is going to gain an extra proton. Now, during this process, the lone pair of electrons on nitrogen is lost, right? Because these electrons here are used to create a bond between nitrogen and the proton from the acid. So when we consider our product here, the the product of the protonation of this nitrogen, the nitrogen is going to have a positive charge due to the loss of the electron pair. And so this right here, the ion in which nitrogen has a positive charge is referred to as the ammonium ion. And of course, we have the conjugate base of the acid which in this case would be a negative. So for the purpose of answering this question, our task here is to find the structures of each of the three A means and then find the ammonium ion by adding a proton and a positive charge to each nitrogen. So let's go ahead and do that starting off with our first compound here, which is per Chloro Annele. Now, in this case, when it comes to drawing structures based on names, I always like to start from the right side and then add on substituents as I go towards the left. Because Adeline, if you recall is a very special type of amine in which there is an NH two group attached to a benzene ring, that would be the first thing I would draw here. Here's my benzene ring and so attached to it is an NH two group. So now we have pera chloral and le so this means that we have a chlorine atom attached to the pers position relative to the NH two group on the benzene ring. So if you recall recall that the pairs position is the carbon directly opposite the carbon in the benzene ring that contains the amino group. So here, if the amino group is on carbon one of the benzene ring, then the preposition is carbon for. So this means that chlorine and the amino groups are going to be on opposite sides of the bending ring. So now we have the structure of pero chloral annele, which means that we can now go ahead and draw the ammonium ion by simply adding an extra hydrogen to our nitrogen, which would make this NH three and then adding a positive charge on nitrogen. And with that, I can proceed to the second compound, which if I scroll up here is n methyl cyclopentolate. Now here cyclopedia amine refers to a five carbon ring or a cyclops ring that has an amino group attached to it. So that's nitrogen connecting to the five member ring. Now recall that and methyl indicates that there is a methyl group or ch three attached to the nitrogen directly. So this indicates that hydrogen or rather nitrogen would have one hydrogen atom attached to it and one lone pair of electrons because recall that nitrogen prefers to have three bonds and one lone pair of electrons. So to find the ammonium ion, I'm going to go ahead and remove my loan pair of electrons replacing that with a new bond to a second hydrogen atom. And so now that nitrogen has four bonds, it's going to have a positive charge which I will go ahead and depict. And lastly for the third part, I have N butyl ethyl of me. So for part three, here, I am going to have a nitrogen atom, I can move this a bit. I have a nitrogen atom in the center of my structure with 20 substituents on either side of it. One of them is an ethel group or simply a two carbon chain, right? So that's ch two ch three and the other substituent is a burial group or a four carbon substituent. So that's ch two ch two ch two ch three, right. So considering the ocho groups that we have nitrogen is going to require one more hydrogen and its loan pair of electrons. So now this is the structure of an bey eo amine, but the corresponding ammonium ion would require removing the lone pair of electrons on nitrogen and adding an extra bond to a second hydrogen atom. This would result in a positive charge on nitrogen which completes the ammonium ion and there you have it right here are the corresponding ammonium ions for all three amines discussed in this question. The bottom line here is to recall that nitrogen is a weak base or excuse me a means or a weak base because nitrogen has a loan pair of electrons which allows it to accept protons. So to draw the corresponding ammonium ion that lone pair of electrons is replaced by a new bond to hydrogen, thereby resulting in a positive charge on nitrogen. And so, with that being said, thank you so very much for watching. And I hope you found this helpful.

Draw the structures of the ammonium ions formed when the amines in Problem 16.30 are treated with acid.
a. N-Methylpentylamine
b. N-Ethylcyclobutylamine
c. p-Propylaniline

Key Concepts

Ammonium Ion Formation

Amines

Acid-Base Reactions

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