A compound with the formula C₄H₈O is synthesized from 2-methyl-1-propanol and oxidizes easily to give a carboxylic acid. Draw the condensed structural formula and give the IUPAC name for the compound.

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Welcome back, everyone. Here's my next problem. Compound X with a molecular formula of C six H 12 0 can be formed from two ethyl, one butanol compound X is also easily oxidized to yield a carboxylic acid. Using this information, determine the condensed structural formula of compound X and provide its Iupac name. We're given four different answer choices. A through D with four different structural formulas and four different names. And we'll describe these and look at them in more detail as we work through what this compound must be and compare it to our answer choices. So first, let's look at the fact that we know that we began with an alcohol and it was oxidized. So I'm drawing an arrow with an O and brackets to sum wise oxidation two form compound X which was further oxidized to form a carboxylic acid. So when we see this two step oxidation process with an alcohol getting completely oxidized to carboxylic acid, let's recall that primary alcohols can be completely oxidized to carboxylic acids going through a first oxidation to become an aldehyde. Secondary alcohols can also be oxidized to form a ketone, but that can't go any further because the carbon carbon doesn't have any more hydrogens on it that could be removed in an oxidation reaction. So we know our compound X must be an aldehyde. So we'll just take a quick look at our answer choices to see if we can rule any out on that basis alone. So choice. A, we have a carbon groups at the end of a four carbon chain. So we have co it's got an H attached to it, making an aldehyde and then there's an ethyl group attached to the carbon adjacent to our aldehyde group. So A is definitely a possibility. Choice B also has an aldehyde group. At the end of a four carbon chain, it has an ethyl group attached to the carbon two carbons away from our aldehyde group. So very similar structures, the ethyl group just attached to a different carbon. When we look at choice C, we have a four carbon chain. But that carbonel group, which I'll highlight in red and B I highlighted my aldehyde in blue is internal to that chain. So it has bonds to two other carbons. So this is not an aldehyde, this is a ketone and therefore cannot be our answer. So cross that choice C since a ketone cannot be oxidized to form a carboxylic acid and looking at choice D I have a six carbon chain with a carbonel group in the middle on the third carbon from the right. So again, it has a carbon on either side, no hydrogen. So this is also a ketone highlighted those carbon groups and there our, our group carbons on either side in red. So C and D cannot possibly be our answer. So now let's work through which of the two structural formulas left are our answer. So to understand how the carbons are attached to each other, let's look at our original alcohol. We have two ethyl, one butanol. So we know we must have four carbons. So I'll draw a condensed structural formula. So four CS with lines between them room for the hydrogens. There, I have the alcohol group on carbon one. Since we have one butanol, I normally might draw that on the left side since we often think left to right. But my answer choices have the aldehyde group on the left side, we can number our chain starting in either direction. We just know that the alcohol and the aldehyde will be on the first carbon. So just to make it easier to compare structures, I will call the carbon, the carbon on the right carbon one and then we'll call the and so then we will not be numbering right to left. So that la that carbon number one on the left has our alcohol group. So we'll write, oh, there. Now we just need to locate our substituent. If this carbon on the left on the right and excuse me is carbon one, then the next one over to the left is carbon two. And we know we have our ethyl group from our name, it's two ethyl so that I'll write a line down from that carbon two. And I'll put that ch two ch three ethyl group on that carbon. So now I just need to fill in the hydrogens. Carbon two has three bonds already with that ethyl branch. So it has just one hydrogen carbon one which has the alcohol group has just two bonds, one to oxygen and one to carbon. So let me just move over that. Oh group, I need to add two hydrogens there. So ch two oh and then carbon three, it is just an internal carbon with no substituents. So it has two hydrogens and carbon four is a terminal carbon with no substituents. So it's three. So IC H three ch two ch with the ethyl group on it and then C H2O H and as we expect this is a primary carbon, the carbon bonded to the alcohol group has only one other carbon attached to it. So that's our alcohol. So this will be oxidized to form our compound X which will be the aldehyde. So that would be CH three ch two, it's not very good too c that has the ethyl group C two C three. Then our fourth carbon to the right which is carbon one is now an aldehyde instead of an alcohol. So we have carbon double bonded to an oxygen and then it has a hydrogen attached to it, giving it four bonds. So there is our compound. And what would be the name here? Well, we have the aldehyde group as the most important. So we have a Butan now four carbons ending in aldehyde. So that would be choice A which says the name is two ethyl button. Now, choice B, the name is three methyl Pentin. Now and that is because the ethyl group being attached to carbon two means there is a five carbon chain. In Choice B starting with the aldehyde, 12345, going down the branch and leaving a methyl group aside. But that has the ethyl group attached to the wrong carbon because this is carbon three when you start with the aldehyde being carbon one. So choice B, our branch is attached in the wrong place. So our answer choice here is choice A which has a structural formula ch three CH two CHC double bond to an oh and then on that carbon two has our ethyl group ch two ch three. Its name is two ethyl button. Now, now we do need to double check one more thing, make sure that it matches the molecular formula. So we have 1234 carbons in the chain two in the branch, six carbons, one oxygen in our carbon group, double check our hydrogens. So in our main chain, we have three plus two plus one. So six plus one more on the aldehyde is seven then on our ethyl branch seven plus two is nine plus three is 12. So 12 hydrogens our molecular formula matches. So our answer choice is definitely choice. A see you in the next video.

A compound with the formula C₄H₈O is synthesized from 2-methyl-1-propanol and oxidizes easily to give a carboxylic acid. Draw the condensed structural formula and give the IUPAC name for the compound.

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A compound with the formula C4H8O is synthesized from 2-methyl-1-propanol and oxidizes easily to give a carboxylic acid. Draw the condensed structural formula and give the IUPAC name for the compound.

Verified video answer for a similar problem:

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A compound with the formula C₄H₈O is synthesized from 2-methyl-1-propanol and oxidizes easily to give a carboxylic acid. Draw the condensed structural formula and give the IUPAC name for the compound.