Classify the following amine as 1º, 2º, 3º or 4° (primary, secondary, tertiary, quaternary).
(CH3)4N+
A. primary
B. secondary
D. quarternary
C. tertiary

D. Quaternary

Explanation:

I'm going to attach a picture that can help you understand how amines are classified!

I'll explain how amines are classified below. You can skip it if you feel that you already understand that!

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So we know that amines are composed of nitrogen, at least one R group, and hydrogens. The way that we categorize amines as primary, secondary, tertiary, or quarternary is by counting the number of R groups.

Now, how do we identify the R groups?

Well, basically anything that isn't a hydrogen. You will see that the nitrogen in an amine standardly only makes 3 bonds. If all those three bonds are with a hydrogen, then it will be ammonia. If it is a primary amine, then one of those hydrogens will be replaced with another compound (like in this case). This other compound is called an R group. R groups will change based on the amine. A secondary amine will have two hydrogens replaced with groups. A tertiary amine will have all three hydrogens replaced with R groups.

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Now, by this logic, we can just count the number of R groups in and we should get our answer! We don't see any hydrogens. That means we can immediately rule out both A. and B.

Well, let's see. It looks like is our R group. The subscript for it is also 4.

But we've used up our 3 bond spots, how is a quaternary amine possible? And that is why you see that your nitrogen has a positive charge in . Since the nitrogen is bonded with more groups than it usually does, the electrons of nitrogen and therefore the electrical charge will be shared between more compounds and elements, leading to less charge for the nitrogen. This will cause it to have a positive charge. Making your amine an ion will allow for 4 bonds.

That is why even though nitrogen usually only has 3 bonds, we can still classify as a quaternary amine.

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The last part got a little lengthy, but I hope I was able to explain this question in-depth! If you had any trouble understanding what I wrote, feel free to leave a comment. I will reply as soon as possible!

you need to give more information

the answer is c. nacl