Compound A is a primary alcohol whose formula is C₃H₈O. When c...

Question

Compound A is a primary alcohol whose formula is C₃H₈O. When compound A is heated with strong acid, it dehydrates to form compound B (C₃H₆). When compound A is oxidized, compound C (C₃H₆O) forms. Draw the condensed structural formulas and give the IUPAC names for compounds A, B, and C. (12.3, 12.4)

Accepted Answer

Welcome back, everyone. Let's take a look at our next problem. Consider compound X an unbranched primary alcohol with a molecular formula of C 4h 10 0. When compound X undergoes a dehydration reaction heated to 180 degrees with an H two. So four catalyst compound yc 4h 8 is produced. On the other hand, when compound X undergoes an oxidation reaction, compound ZC 4h 80 is formed based on the information given predict the condensed structural formulas of compounds. Xy and Z and provide their Iupac names. On first glance, this definitely looks a little bit intimidating. Have all we have is the molecular formulas in a general description of our first compound. And we have to predict what happens when we have a dehydration reaction and an oxidation reaction. But if we take this step by step, we'll be able to work through all of these things. So we know we're starting with an alcohol. So our compound X we know is unbranched and it's a primary alcohol. So we recall that a primary alcohol is one in which the carbon that has the alcohol group has only a single alkyl group attached to it So actually, I'm going to erase the alcohol group, that carbon will then have two hydrogens on it and an alcohol group. So let's think about what this structure could be. We have that molecular formula of C 4h 10 0. So we know that we have four carbons in our main chain. So we'll draw and it's unbranched. So we'll draw a condensed structural formula. So we have C see CC connected by lines to represent those bonds. I've left a space between each one for the hydrogens. We know it's a primary alcohol. So the alcohol group must be on one of the terminal carbons. It can't be in the middle because that would make it a secondary alcohol because those carbons in the middle have two alky groups on them. So it doesn't matter which side we put the alcohol group on, that's a symmetrical molecule and would be the same either way. So I'm just adding that oh group to the last carbon. And we know, so we have our four carbons, there's our one oxygen and one hydrogen. So what do we have left? We have C 4h 10 0 will subtract an oh group and we'll have left C 4h 9. So we've got our four hydrogens. We just need to place nine hydrogens. Well, the simplest way to do that is if we fill in what we would expect to be a number of hydrogens and then just count and make sure it adds up to nine. So that first carbon will have three hydrogens as the terminal carbon, carbon two will have two hydrogens, carbon 32 and carbon four as well. Two hydrogens. So we have ch three ch two, ch two C h2o H and that gives us three plus two, is five plus two, is seven plus two, is a nine. So we match our molecular formula. It's an unbranched primary alcohol. So this is compound X, we've written in the form of a condensed structural formula and we just need to give it its Iupac name. So we have four carbons. So we know our root must be butane. And with that alcohol group on the end, we add the ending ol butanol and we just need to indicate the location of the alcohol group. So we'll put one hyphen butanol. So that's our correct I name there are structural formula for compound X. So now we need to move on to compound Y. So compound Y occurs after we have a dehydration reaction where we've applied heat and in the presence of an acid. So there's protons available. Well, under these conditions, and alcohol will undergo a dehydration reaction to produce an alkene to produce that the alcohol group loses its hydroxy group and the neighboring carbon loses its hydrogen. So if we look at our structural formula, we have CH three ch two, ch two. Here's our neighboring carbon or excuse me, I should have stopped one carbon earlier, my third carbon is a neighboring ar carbon. So I have CH three CH two C and then I'm going to draw the HS on here as bonds and then my CH two which has an O group on it. And we're going to remove the hydroxy group from carbon four, a hydrogen from its neighboring carbon. Remember, this is a primary alcohol with only one neighboring carbon. So there's no decision about which carbon the hydrogen will be removed from. You'll have a loss of water. Again, this is in conditions where we have heat and we have protons and the end result will be in our keen. So we have CH three C two. This is now ach we've lost one hydrogen and a double bond is formed between carbon three and carbon four and this is now ch two. So we still have four carbons. So let's check this against the molecular formula. We know that compound four is C 4h 8. We've lost the oxygen. We still have four carbons. Let's add up our hydrogens to double check. We have three plus two is five plus one, is six plus two more, is eight. So that's correct. It matches our molecular formula. So again, for compound Y, that's our structural formula with that double bond between carbons three or excuse me, in the way we've drawn it, it's carbons three and four. But when it comes to naming it, will name our main chain still butane with its four carbons, now that it has a double bond, it will be in Beine. But when we need to locate the bond, we want to number it so that our double bond has the lowest possible number. So instead of numbering left to right, we will number it from right to left. So we have carbon one as the double bond to carbon two. So always keep that in mind in terms of sometimes we're tempted to number from left to right, but we want to number this so that that double bond or the importance substituent has the lowest possible number. So in this case, we have one beine as our Iupac name. So we've gone from one butanol to one, but now we just have compound Z left and we know that compound Z is a result of oxidation of our alcohol. So we have oxidation and it's important to note that this is the oxidation of a primary alcohol because that has sort of two steps. It initially gets oxidized to an aldehyde and then it can undergo complete oxidation into a carboxylic acid. This is however, a two step process and let's recall the format of these two compounds. So an aldehyde is in the form of our group bonded to a carbon carbon. And that carbon carbon has just a hydrogen bonded to it. On the other side, a carboxylic acid, on the other hand, is the R group carbonyl carbon. And then an oh, which means that if our compound is oxidized to an aldehyde. So just the first step of this process, it will only have a single oxygen. It's oxidized to a carboxylic acid. It should have two oxygens. But when we look at the molecular formula of compound, Z it's given to us as C 4h 80. So therefore, we'll cross out carboxylic acid because that should, would have two oxygens. So we know it hasn't been completely oxidized, just gone through this first step. Also good to note that when we have a secondary alcohol, it can become oxidized to a ketone but can't go any further because there are no more without a hydrogen on the neighboring carbon, we have no more hydrogens, they can go no further. And tertiary alcohols won't react at all in an oxidation reaction. So we know that compound Z must be an aldehyde. So let's look at our chain here again, we have that four carbon chain. So we know that the carbon that had the alcohol group previously is now an aldehyde. So that last carbon on the right now has a double bond to an oxygen and then just a hydrogen attached. So fill in our other hydrogens on our other carbons. So H three H two and H two and let's double check it against its molecular formula. FC 4h 80, we have the one oxygen, four carbons and let's count our hydrogens three plus two is five plus two, is seven plus one on the aldehyde is eight. So that is correct. We've got the correct condensed structure. We just need a name. So again, we've got that butane root. So we have butane, it's an aldehyde. So now the ending changes to a L Butan, now it needs its locator and just as with the others, we start from the end that gives it the lowest number on that functional group. So it's one Butan now as its I UAC name. So once again, that alcohol, that primary alcohol gets oxidized in just the first step to an aldehyde. We have our structure with CH three CH two, CH two and then that's C double Bondo with the hydrogen and our iupac name for compound Z of one button. Now, so again, our steps are compound X is a four carbon alcohol, one butanol compound Y a one button. So it's been dehydrated to form an alkane and compound Z which has been oxidized to form an aldehyde is one Butan. Now see you in the next video.

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