Write the structures of the following compounds:
a. Butyldiethylammonium bromide

Question

Write the structures of the following compounds:

a. Butyldiethylammonium bromide

Accepted Answer

Hi, everyone. Welcome back. Our next question says, draw the structure of py diprop ammonium bromide. We have four answer choices. A through D each has a condensed structural formula and we'll look at them in more detail in a minute. But as I look at them overall, I see that each has a fairly long carbon chain, just an alkyl chains, carbons and hydrogens until we get to the end where it has a nitrogen with either one or two hydrogens and then a bromine ion, both the nitrogen and bromine are ions, some with positive, some with negative ions, negative charges, excuse me. And then I see that in each case, my nitrogen also has 23 carbon chains attached to it. So we'll again, we'll look at each one more closely as we work our way through the name. So we start with the fact that we have an ammonium bromide. So this means our compound is a salt. It's an ionic compound. So we have an anion and a cion. So first of all, our bromide is a halogen halogens have seven valence electrons and therefore, are extremely happy to take on an extra electron to fill that shell. So our anion, it's pretty easy to say is BR minus. That one's quite straightforward halogens are pretty much always going to be anions with a negative one charge. Then my ammonium, that means I have a nitrogen with four bonds to it and nitrogen with its five valence electrons when it has four bonds, that means it will carry a positive charge. So that makes sense since we're still looking for ac so we have it in our ammonium. So what are the four bonds to the nitrogen? Well, one thing when we're looking at this name, we're often used to thinking of branches as being attached to carbons. But in this case, when we have an ammonium, these are branches, these are structure or functional groups attached to the nitrogen of our ammonium ion. So again, we're looking for four things bonded to that nitrogen, we can also look at our answer choices. Uh Before we even start that process, we know that bromine has to be our anion. And when I look at those answer choices, two of them have bromine with a positive charge and nitrogen with a negative charge. So now that we've identified our cion anion, we can eliminate two answer choices right off the bat. So choice B has at the end of the molecule NH minus BRP. So it has the wrong charges on the ions. And choice D has NH two minus BR plus at the end of the molecule. So again, it has the wrong charges. So choices B and D are incorrect as they have the wrong charges, the wrong an in and CM. So were down to A and C. So now let's think about what's bonded to our nitrogen. When we look at the name here, we have pal diprop. So Penil here bracket, not a little carrot means a five carbon alkyl chain. And then of course, diprop means we have two of them. And there are two, three carbon alky chains. Well, right there, we have three things attached to our nitrogen. The pil and the two propol, there's nothing else in our name. So we know that the fourth bond must be to a hydrogen. So we have NH and then our three alky chains. Well, again, without even starting to count carbons, look at those chains, we can eliminate an answer choice because choice C at the end of the long carbon chain, we have NH two plus BR minus. So the BR minus is correct. But this nitrogen can only take one hydrogen. It has three alkyl chains. So choice C is incorrect. It's given by nitrogen. Too many bonds. So, too many bonds on the nitrogen. So if I'm on a test, I'm trying to get through quickly. I've already got my answer choice. A I've eliminated the other three, but let's be a little more thorough and check our chains here. So our structure has a long carbon chain going from left to right ending in that NH plus VR minus. So we see we have the correct NH plus VR minus. We just need to make sure there's five carbons in that chain and indeed +12345. So there's our pencil chain correctly and then the nitrogen should have 23 carbon chains. And we see indeed on top of it, ch two ch, two, ch three and attach below it ch two ch two ch three. So choice A is our answer here for the structure of pal diprop ammonium bromide. We have five carbons, then NH plus BR minus and the nitrogen also has +23 carbon chains attached to it. See you in the next video.

Write the structures of the following compounds:
a. Butyldiethylammonium bromide

Verified video answer for a similar problem:

Key Concepts

Quaternary Ammonium Compounds

Ionic Compounds

Chemical Structure Representation

Write the structures of the following compounds:a. Butyldiethylammonium bromide

Verified video answer for a similar problem:

Key Concepts

Quaternary Ammonium Compounds

Ionic Compounds

Chemical Structure Representation

Write the structures of the following compounds:
a. Butyldiethylammonium bromide