Give the molecular formula of each compound shown a. Pyridine
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Question

Give the molecular formula of each compound shown a. Pyridine

b. Pyrrolidine

c. Furan

Accepted Answer

Hey there, let's solve this problem. It says give a molecular formula that corresponds to each of the following structures. Structure one is two methyl paradyne. Structure two, we have two metal pyro ladin And structure three, we have two metal Furin. And the question is asking for a molecular formula, what is a molecular formula? Well, it simply states the number of each type of atom in the structure. Okay, so we could count directly from these line angle formulas but I would suggest doing that once you're more comfortable with recognizing the hidden carbons or the invisible hydrogen. But for now we're going to convert these line angle formulas two lewis structures just so we don't accidentally skip any of those invisible atoms. Okay, so important things to remember about line angle structures is that the intersection of two lines or the end of two lines symbolizes carbon atoms. Right. If it's unspecified, sometimes at the intersection we'll have a nitrogen or oxygen. But those are specified. Right? So if there's no other specified adam there, it's a carbon. Then similarly hydrogen are invisible. There's no line corner hydrogen ins so how do we know how many there are? Well, we just draw enough to fulfill carbons octave. Okay, and lastly this is pretty straightforward but just remember that 11 line equals a single bond. Right, Two electrons and two parallel lines represents a double bond or sigma and a pi bond for electron. Okay, so first we're going to replace the carbons. When we draw out our structure, then we're going to add the hydrogen is to fill the carbons octet and then we can add any lone pairs that are not drawn. Okay, so let's do this. We'll start with structure one a very good place to start. So we have our nitrogen and then I see carbons, right, all those corners in our base ring and then there's also a methyl group which is a six carbon out here on the end of that. So let's draw our ring with those carbons. So nitrogen 1, 2, 34 five and then six carbons. Okay, and now let's add the hydrogen that are in this. So we have to add our bonds right there not all sigma bonds. So we have a double bond at this carbon here, I guess that would be carbon five. And this carbon and in between the nitrogen and the first carbon. Okay, so now let's add the hydrogen is to complete the octet. So on the first carbon to the right of the nitrogen that's double bonded, there are already four bonds there. So there are no hydrogen, this carbon, so any carbon that's touching a double bond besides that one, since it has a methyl group on it, none of these carbons have other groups. So if there's a double bond touching it, there's going to be one hydrogen. Right? Because it has 1, 23 bonds already. So it only needs one more. Same with this carbon and it looks like all the carbons are touching a double bond. So they're all going to be in that similar situation. And then this methyl carbon we know our methyl group has three hydrogen. And so that's all of our hydrogen. Any lone pairs, yep, that nitrogen has 246 total electrons. So it needs two more electrons to fill its octet and also satisfy the formal charge 12345 electrons touching it. Ok, so like we said, the molecular formula is simply counting the number of types of atoms in the structure. So we usually start with the base adam, which is carbon and then hydrogen. And then any other atoms such as oxygen or nitrogen. So how many carbons do we have? 1, 2, 3, 456. We have 1234567 hydrogen and one nitrogen. So we don't usually write a subscript for the one. So that is our first molecular formula for our structure one. Okay, let's repeat this for structures two and three. So two methyl parola Dean, we have a five member ring starting with nitrogen at the top. Then we have our 1234 carbons at those other corners and a methyl group. So that's another carbon. And now let's add our hydrogen. So there are no double bonds in our ring. So all these carbons have two sigma bonds. Well, except the one with the metal group. Right, So the woman's without the methyl group will get to hydrogen. The carbon with the methyl group needs one more bond connected to it. Right? So that's four. And then our methyl group has three. Okay, this hydrogen is very tricky. I feel like it's easy to miss because that methyl group is attached to it and any loan pairs. Oh, sorry, I forgot. The hydrogen on the nitrogen. Now we have a lone pair as well on the nitrogen. All right. So now it has a complete octet. So let's count our atoms. We have carbon, hydrogen and nitrogen. So we have 12345 carbons one. Don't forget that hydrogen on the nitrogen. 123456789, 10, 11 hydrogen and one nitrogen. So we leave it just like that. And that is for structure number two. Lastly structure number three, we have two metal Furin. So we have our oxygen at the top two double bonds. Not in that location. I forgot my carbons. So, again, a five member ring with oxygen at the top and four carbons. And we have two double bonds in a missile group on that first carbon. So let's add our hydrogen. So the carbons that are again have a double bond here, we'll only get one hydrogen. And then this carbon does not get any hydrogen because it has that methyl group on it. Any lone pairs. Well, we know oxygen will have two lone pairs if it has two sigma bonds. Oh and our metal group has hydrogen, Right? Can't forget about that. So that's our structure. Let's count. So this time we have carbon, hydrogen and oxygen 12345 carbons, 123456 hydrogen and one oxygen. Okay, so that is it for this problem. I hope this video is helpful, and thanks for watching.

Give the molecular formula of each compound shown a. Pyridine b. Pyrrolidine c. Furan

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Give the molecular formula of each compound shown a. Pyridine
b. Pyrrolidine
c. Furan