Draw a Lewis structure, and classify each of the following com...

Question

Draw a Lewis structure, and classify each of the following compounds. The possible classifications are as follows:

alcohol

ether

ketone

aldehyde

carboxylic acid

alkene

(a) CH₂CHCHO

(b) CH₃CH₂CH(OH)CH₃

(c) CH₃COCH₂CH₃

Accepted Answer

Hey everyone, Let's solve this problem. It says draw the lewis structures of the following organic compounds classify each as an alkaline Aldo hide ketone. Car box silic acid, ether or alcohol. Okay, so let's quickly review what these types of organic compounds look like. Okay, so al keen we know has a carbon double bonded to a carbon and Aldo Hyde has the formula C. H. O. And it will look like this carbon double bonded to that oxygen bonded to a hydrogen. A ketone is a carbon steel group. So it'll which is a C double bond O. And you will see our groups on either side of that carbon. So R C double double bonnet or sorry, single bond to carbon double bonded to oxygen bonded to another carbon group or our group. Car box silic acid is similar except instead of an R group, we have an alcohol or an O. H. Rather I don't want to mix up the functional groups, but we have the O. H group on the other side. So that will look like this, C 00 H in the condensed formula. And when it's drawn out it looks like this. Okay, and ether is R O R. Okay, and alcohol is R O H. Okay, So it says to draw the Lewis structures 1st. So let's quickly review how to do that. So first we count the total number of valence electrons so that we can have the proper amount on our structure and that's the valence electrons for each atom. Right? We add them all up. Then we draw a sigma bond structure. Using those electrons kind of get a rough draft of the connectivity. We have the atom with the highest bond preference in the center. And then we complete the octet with our leftover electrons by adding them as lone pairs or forming double bonds to complete octet. It's then lastly we have to count calculate any formal charges. Ah This is usually done when you're figuring out if you need to form double bonds and we do that by adding the total number of valence electrons for that Adam minus the number of bonds touching that atom plus the non bonding electrons. Which would be lone pairs. Right. I like to call those touching electrons. So you could say total valence electrons minus touching electrons. It's the same thing. That's the reason why they say the number of bonds because a bond has two electrons but only one electron from that bond is touching that adam quote unquote touching. Okay, so let's draw the lewis structures structures and then we'll classify them. Okay, so first we have CH three, CHOHCH 3. This is 26 valence electrons if you add them all up, which I do, I'll show you the first one just so you know, we have four from carbon three from the three hydrogen is four from carbon one from hydrogen, six from oxygen, one from hydrogen. Four from carbon and three from hydrogen. Okay, that's how we get our 26. Now, let's draw out our structure. So we have CH three C H. O. H. And it has the parentheses. So I know it's attached to that same Carbon and then CH three. So drawing all these Sigma bonds, we get 22 electrons which means we have four left. And that means we can add two lone pairs and which atoms here needs some electrons. Well all the carbons have four sigma bonds so they're satisfied that oxygen would be very happy with two more lone pairs. So now it has reached its octet and it has a formal charge of zero because it has 123456 electrons touching it. And oxygen likes to have six electrons. Right? That's its valence electron number from the group, the group number. So there's our lewis structure and let's classify this compound. So do you see any functional groups on it? Yes. Even in the name I C O H. Right, so this O H is an alcohol as we reviewed from our list. So this is an alcohol. Okay, Cool, Let's move on to the next one. We have CH three CH. O. And if you add up their group numbers you get 18 valence electrons let's draw out the structure. So we have c. h. three Bonded to another carbon bonded to an oxygen and a hydrogen. Okay, so when we do this we have 12 valence electrons. Which means we need six more and that would be three lone pairs. So I see oxygen only has one electron touching it. So we'll add two lone pairs there. And this carbon also does not have a full octet. So we'll add to their to fill its octet. But even though I'm feeling that carbon octet, what's the formal charge of that carbon? It has carbon likes to have four electrons and it has 12345 touching it. So it has a formal charge of negative one. And oxygen only has six electrons in its octet. So it's definitely not happy. So carbon is going to share those electrons with the oxygen in the form of a double bond. So that will look like this C double bond oh bonded to H And as I mentioned before, when you C C H O. That will be what functional group C H O will be in Aldea Hood. Okay, so compound B is an alga hide and notice how it's on the end of the compound. That's always how it is. Out to hide. Alright, and the last one we have ch two, CH two, CHCOCH 3. And if we add this up we get 28 valence electrons. Let's draw it out. See see H two Bonnie to C H c O c H three. Okay. And once we draw all those sigma bonds, we've used 20 electrons. That means I have eight left which is four lone pairs. So let's add those lone pairs. Well again, oxygen is pretty bare. So let's add two lone pairs there. And then this carbon, its octet is not full. So we can add a lone pair there. And same with this carbon octet is not full. So we'll just put them there for now. But as you can probably tell, this is not their final place. Right? This carbon does not have a full octet and this carbon has a formal charge of negative one. Right? 12345 electrons touching it. It likes to have four. So that's a negative charge of negative one. Or formal charge of negative one. And same with this carbon, right, Three sigma bonds and a lone pair is a negative formal charge. This oxygen has an incomplete octet. So the carbon and the oxygen are going to need some electrons from those carbons. So they will share their electrons giving us this structure. So we have a c a carbon double bonded to another carbon bonded to a carbon eel group which has the carbonell group has two R groups on either side. Right? The methyl group and then that other carbon. So we have a ketone on our hands. And the double bond is also a functional group. Right? This is an alkaline. So we have an elk in and a keto. All right. I hope this video was helpful. And thanks for watching

Draw a Lewis structure, and classify each of the following compounds. The possible classifications are as follows:
alcohol
ether
ketone
aldehyde
carboxylic acid
alkene
(a) CH₂CHCHO
(b) CH₃CH₂CH(OH)CH₃
(c) CH₃COCH₂CH₃

Key Concepts

Lewis Structure

Functional Groups

Molecular Classification

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