For each molecular formula, draw all the possible constitution...

Question

For each molecular formula, draw all the possible constitutional isomers of alcohols with that formula. Give the IUPAC name for each alcohol.

(d) C₃H₄O

Accepted Answer

Hey everyone, let's solve this problem. It says draw all possible constitutional ice members for the following molecular formula of an alcohol C five H 12 0. Open chain compounds only. Right? You pack names for each of the ice hammers. Okay, so it's sort of a two part question first drawing the ice hammers and then naming them. Well, what our constitutional customers? That just means all the structures will have the same molecular formula, but they could be connected differently. The atoms can be connected in different orders. Right. Well, a few clues about what our structure will look like from what the problem gives us is that it says it's open chain compounds only, which means there are no rings. And it says that we have an alcohol. So we know that oxygen is going to be in an O. H. Group. So one of those hydrogen also has to be attached to that oxygen. Okay, so all the structures are going to have that O. H group. Any other clues about our structure? Is it going to have a double bond or a triple bond? Well, we can check by calculating the index of hydrogen deficiency. Right. Which that formula is, I. H. D. Equals two. N plus two -H. divided by two where N. Is the number of carbons and h you guessed it is the number of hydrogen and you would also add one for religions but we don't have any allergens in this case. So that's okay, so let's calculate R I. H. D. First to know if we have any un substituted carbons. So we have I. H. D. Equals two times five. Right? five carbons plus 2 -12 hydrogen five times two is 10 plus two is 12. 12 minus 12 is zero. So we know we have no rings but also no double bonds etcetera. So our carbon is totally are totally sub substituted. Okay so that makes it a little bit easier. So we have five carbons. So first we can start with a five carbon straight chain and see how many possibilities of that we can do. So if we have 12345 carbons we can add the O. H. group on the end. That's one option. We can also add 12345. The O. H. Group on the second carbon 12345. Or we can add it on the middle carbon. If I add it to this next carbon or this next carbon, that would be the same as these two structures. Right? Just flipped around. So these are our three structures are three summers we can have with a five carbon straight chain. So what are the impact names for these? Well they're all going to have the parent name Pen Tan, Right because we have five carbon chains. And if we start this carbon as 12345 then we'll have Pen Tan one All pen tan to all & Penton three All. Okay and again if we had four all or five all that would be the same as one or two just flipped around. So we don't need to draw those. Okay, so that's with a five carbon parent chain. But again, we can add different connectivity of the atoms. So how about we try a four carbon chain and maybe we can add the other carbon and hydrogen as say, a methyl group. So if we have 1, 2, 3, 4 carbon chain, I can add, let's say a methyl group to carbon to And oxygen to carbon or sorry, an alcohol group to carbon one. And now let's say I take this structure and keep it the same. And then I can move the alcohol to different carbons. Right? So it's easier to try and just change one thing first. So keep these substitue in groups, keep all of them the same except one. So you can try the different combinations of that before you change something else. So now we have the same thing except our alcohol has moved to carbon too. Or we can move the alcohol to carbon three. And that metal is still staying on carbon 21234. Methyl on carbon to alcohol and carbon three. And we can also do the same thing with alcohol on the opposite end. Right? Because it's still a different relationship than this end. Right? Because now there's one carbon in between. So if I tried to move the methyl group. Well, if I move the methyl group to four, that would just be a five carbon chain, which is not what we're going for. And if I move the method to three, well, we would have those same combinations of structures here. It would just be flipped around. Right? So these our these are our four carbon chain summers. So the names of these are two metal Butin. Right now we have four carbons Butin one all in same parent chain. Butin Butin all but now we have two metal Beauty to or sorry, tan to all Alcohol groups get the priority in numbering. So instead of saying two methyl Beauty and all, we're going to switch the numbering to go this way 1234. Right, alcohol's get priority. So this one will actually be three methanol Butin to all. Does that make sense? In our last one? Again, we're numbering our priority with the alcohol. I forgot a carbon here. 12,340. H. Okay, so we have three metal Butin one. All. All right. Now let's see if we could do three carbon chain and see what we have. Now, a three carbon chain is going to be symmetrical. Right? So there are less combinations of different things that we can do. So, if we have 123. Well, if I added a methyl group to either end, that would just make a longer chain. Right? So that that's what I meant by having less possibilities. Right, we can only add two carbons to this central carbon, which leaves our alcohol to be added to one of these end carbons. And again, like I was saying about the symmetry. If you add it to either end, it's going to be the same structure. So we can only have this one summer with a three carbon parent chain. Okay, So if we number starting with our alcohol 123, this will be 22, two, comma 2 dime ethel pro pan all. Oops pro pan one all. Okay, that's it for this problem. I hope this video is helpful and thanks for watching.

For each molecular formula, draw all the possible constitutional isomers of alcohols with that formula. Give the IUPAC name for each alcohol.
(d) C₃H₄O

Key Concepts

Constitutional Isomers

IUPAC Nomenclature

Functional Groups

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