Compound X is a secondary alcohol whose formula is C₃H₈O. When compound X is heated with strong acid, it dehydrates to form compound Y (C₃H₆). When compound X is oxidized, compound Z (C₃H₆O) forms, which cannot be oxidized further. Draw the condensed structural formulas and give the IUPAC names for compounds X, Y, and Z.

Question

Accepted Answer

Welcome back everyone. Let's take a look at our next problem. It says compound M is a secondary alcohol with a formula of C 4h 10 0 compound M can undergo dehydration by heating with a strong acid to produce compound N with a formula of C 4h 8 compound M can also undergo oxidation to produce compound L with a formula of C 4h 80. What are the IAC names? And condensed structural formulas for the compounds? Mn and L? So this one looks a little scary at first three different compounds. The only information we're given, it seems at first is the molecular formulas. But if we work our way through this problem step by step, it kind of works out like a logic problem where we eliminate possibilities to get down to the right answer. So let's start with our initial compound. What do we know about it? So we have compound M, we know that it has a molecular formula of C 4h 10 0 and that it's a secondary alcohol. So what does that mean? What is a secondary alcohol? What's one where the carbon bound to the oxygen of the alcohol group has two alkyl groups on it. So the format will be R and then our carbon with an oh group carbon will have one hydrogen and then a second, our prime group. So that's all that means referring to that carbon that's in the oxygen bond. So we know that it must be on an internal carbon. It can't be on a terminal carbon and we have four carbons. So that makes it much easier. We're not dealing with a big long chain. There's only four of them. Uh They can be arranged in different arrangements. So let's think through what's possible. So, first of all, let's start with the simplest arrangement. All four Carmens in a row. So we'll draw, we're going to be looking at a condensed structural formula. So we'll draw c leave a space for hydrogens a line for a bond, another carbon and do the same thing until we have four carbons in a row. Again, we know that Carmen's one and four are not possibilities. I, no, with a little arrow to each of them. And that's because adding a, a, an alcohol group to either of those would result in a primary alcohol. So we know those can be eliminated from our answer. So we know that our alcohol group has to go on one of the internal carbons. So we'll put it here on carbon two for lying down in oh group. What about the possibility of putting it on carbon three? Well, look at this carbon chain, a alcohol, an alcohol group on carbon three is just, it's the exact same molecule. If you rotated it over, it would be the same. So there's only one arrangement possible with all four carbons in the chain. Now, what about a case where there might be a branch? The four carbons don't all have to be in, in a row. So let's imagine a structure where we have three carbons in the main chain. Well, if we put a carbon on a methyl group on either end, that just makes a four carbon chain, so let's add a carbon to the middle carbon. So we have this t shaped structure. So a propane chain with a methyl group in the middle. Well, again, we can't use any of the terminal carbons for alcohol group because that will make a primary alcohol. So we'd have to put it on the middle carbon to make isopropanol. But look at this middle Carmen, it has bonds to three other cars. So this would be a tertiary alcohol, not a secondary alcohol. So this branch structure is not a possibility, it's going to cross it out with a big red X. So as I said, it's again, it's like a logic problem. We eliminate all these possibilities. We've only got one possible structure for a secondary alcohol with four carbons. So last thing we need to do to get a condensed structural formula is fill in the hydrogens we write hydrogens and carbons in groups rather than delineating every single bond. That's why it's a condensed formula. So just go along starting with carbon one, it's a terminal carbon with no substituents. So it gets three hydrogens. Now we get to carbon two, it has the alcohol substituent and it's internal. So it just gets one hydrogen. Carbon three is internal with no substituents. So it gets two hydrogens and carbon four is terminal with no substituents. So it gets three hydrogens. Does this fit our molecular formula? We've got four carbons, we've got one oxygen, we should have 2, 10 hydrogens, excuse me. 10. So let's add them up in the main chain. Three plus one is four plus two, is six plus three, is nine and there's one more hydrogen on the alcohol group. So that is 10 hydrogens. So we've got a structural formula that fits our parameters. All we need is its Iupac name. So we have a four carbon chain. So that's going to be a butane route with an alcohol group. So it will be Butanol with an OL ending. We just need to locate the alcohol. It's on carbon two. So we have two hyphen butanol as our Iupac name. So we're done with compound M we got its structural formula, ch three single bond to ach which has the oh substituent single bond to CH two, single bond to CH three and its Iupac name of Tuinal cross off one out of three. So now we move on to compound N and here N is produced by the dehydration of compound M. So what happens when we do a dehydration? Um and alcohol, so we have a dehydration with, by heating with a strong acid dehydration, we remove a water molecule. So in the case of an alcohol, we remove the alcohol group and then we remove a hydrogen from one of the adjacent carbons. water is removed and we form a double bond between the carbon that had the, the alcohol group and adjacent carbon. So I'm going to sketch out how this will work with a line structure just because it's a little quicker remembering that we need a condensed structural formula at the end. So let's draw our initial alcohol that four carbon chain with an oh on carbon two. And we need to think about the two adjacent carbons to our alcohol or not our alcohol and the carbon with the alcohol. So our two are groups, do they each have a hydrogen that could possibly be removed? And yes, they do. Carbon one is CH three. So there's definitely a hydrogen there. Carbon three is CH two. So it has a hydrogen as well. So for a dehydration reaction, we're going to remove the oh and we can remove one of the two hydrogens that gives two possibilities for our alkene. It will be an alkene with AC C double bond. If I remove the hydrogen on carbon one. So I'll put C one, I will form a double bond between carbons one and two. And if I remove the hydrogen from carbon three, so, but C three, the double bond will form between carbons two and three. So how do I know which will be my answer? Well, when we're looking at the formation of alkenes in this way, the most substituted alkene is the major product. So let's look at our two alkenes. When I have my alkene from, with a double bond from carbon one to carbon two, the only substituent is a single ethyl group. When I have my double bond between carbon two and carbon three, I've got a methyl substituent on each carbon. So carbon three, removing the hydrogen, carbon three. So having a double bond between two and three is going to be the major project as it is the most substituted alkene. So what's going to be our structural formula here? Well, let's write it out. We have the line structure to go on. So we can just draw that out. So we write out our carbons and our bonds and write the CHS as groups. So let's start with carbon one. We have CH three. It's a terminal carbon with just a single bond. Then we have a line drawing it to carbon two. Carbon two is in a double bond. So it has one bond to carbon 12 bonds to carbon three. So one leftover needs one hydrogen. Then we have a double bond, two, carbon, three, carbon three also has a double bond and a single bond to other carbons. So just one hydrogen and then we're going to run out of space. So let me move that structure a little bit. And then we just have a single bond to carbon four, which will be ch three being a terminal carbon with no substituents. So now we just have to name this with its proper Iupac name or carbon chain with a double bond will be a butane since we still have that bet root there with four carbons, and we just have to locate the double bond. It goes from carbon 2 to 3. We use the lowest number. So it's two hyphen beine. So now we've worked our way through compound N. Our strut formula is CH three, single bond to CH double bond to ch single bond to CH three. It's our condensed structural formula and our Iupac name is two beine. So we're two thirds of the way through. Now, we need to compound L oh and before we move on quick double check of our structural, our molecular formula here. So we've said that compound and had a molecular formula of C 4h 8. So looking at our structural formula, we have four carbons, let's add up our hydrogens three plus one is four plus one, is five plus three, is eight. So that checks matches the given molecular formula, move on to compound. Hell compound L is produced after the oxidation of compound L. So we have an oxidation of a secondary alcohol. Why is it important to specify that it's secondary oxidation of a primary alcohol produces first an aldehyde and then it can undergo a second oxidation reaction to produce carboxylic acid. Oxidation of a secondary alcohol produces a ketone. It then can't go any farther because there's no more hydrogens on that carbon carbon. So you can't have any further oxidation. There's no more hydrogens to take away. And then if you have a tertiary alcohol, it can't undergo any oxidation reaction at all. Since there's no hydrogens to start out with on the carbonyl carbon or on the alcohol carbon, excuse me. So we know our alcohol will be transformed into a ketone by the reaction. So we can just go straight from our alcohol's structural formula. So we have C three. Now our alcohol group was on Carmen two. So that alcohol group, so we have a single bond to carbon two is now transformed into a ketone. So we have instead a Carbone group so that carbon has a double bond to an oxygen. It's also lost its hydrogen that was there. So no hydrogens on that carbon, it has a single bond to carbon three. So ch two and then a single bond two, carbon four, CH three. So our condensed structural formula here has that ketone on carbon two. So let's double check that with our molecular formula. Our molecular formula for L is C 4h 80. Well, we know we have the one oxygen four carbon. Still let's check our eight hydrogens. So adding them up three plus two is five plus three is eight. So that checks, we just need its Iupac name. So again, that root butt for the four carbons, it's a ketone. So we will call it but and no that one ending and we just have to locate the carbon carbon. It's on carbon two. So two hyphen Banon. So now we've got compound L compound L condensed structural formula of CH three, single bonded to AC which has a double bond to an O that C is single bonded to CH two which is single bonded to CH three and its I AAC name is two bean know. So there we have it all three compounds, Mn and L they condensed structural formulas and their Iupac names. See you in the next video.

Verified video answer for a similar problem:

Key Concepts

Secondary Alcohols

Dehydration Reaction

Oxidation of Alcohols