Draw the condensed structural or line-angle formula, if cyclic...

Question

Draw the condensed structural or line-angle formula, if cyclic, for each of the following: (12.3)

d. 3,5-dimethylhexanal

Accepted Answer

Hi, everyone. Welcome back. Let's take a look at our next problem. It says sketch the following the condensed structural formula or line angle formula for cyclic compound of 223 triethyl hepton. And we have four different answer choices A through D with four different condensed structural formulas. So we aren't even worrying about cyclic compounds here. And we'll describe them in a little more detail as we think about our compound. So let's say I'm on a test, I want to work efficiently rather than working through all the steps of drawing the structure. I often can take a quick kind of overview, glance at the name of my compound and rule out some or perhaps three of my answer choices and get to my answer without even the somewhat time consuming steps of writing the whole structure myself. So let's look at this name real quick and see what we can gather at, you know, right away, which is the whole purpose of these names to help us quickly get a sense of the structure of our compound. So we have a Hepton now, so that a L ending means we know we must have an aldehyde, that prefix hept means we have seven carbons in a chain. And the 223 triethyl tells us we should see three methyl groups, two of which are on the same carbon. So that's a pretty distinctive shape compound, especially those two methyl groups on the same compound and the aldehyde group. So using that, let's see what we can eliminate and see if we can take sort of a shortcut to the answer. So let's look at choice. A choice. A we see this long carbon chain if we count them, 1234567, got our seven carbons. We have our aldehyde group on the first carbon, there are two methyl groups, not three and they're not on the same carbon. So we're missing our pair of methyl on the same carbon. So choice A is incorrect. We've only got two methyl groups. So we've already eliminated one. Then I took a choice B we have 1234567 carbons. We have an aldehyde group on our first carbon. And then we see the next carbon. After that, we've got two methyl, the next carbon. After that one, we've got three methyl groups with two on the same carbon. So choice B is looking pretty promising. Choice C we've got our seven carbon chain aldehyde on carbon one, we do have three methyl groups, but they are on three different carbons. 23 and four. So wrong carbons for our metal groups across South. Choice C then choice D seven carbon chain aldehyde group on carbon one, but we have only one methyl group on carbon four. So definitely not correct. So you can see how that name lets us quickly rule out three of our answer choices. We've already got our answer. Choice B because none of the others can be correct. So, on a test, that's a very quick and efficient way to your answer of choice B because this is a practice problem, stay tuned if you want to see the actual process for getting to this structure yourself, if you had a short answer question rather than multiple choice, for instance. So as we said, we have a seven carbon chain. So I would draw seven CS with lines linking them. Since this is a condensed structural formula, I'm leaving space after each C to fill in the right number of hydrogens. So I've got 1234567. As we know we have an aldehyde, there's no number in front of the hap anal. And indeed, since I have this chain that aldehyde needs to be on a terminal carbon, so we'll put it over here on the left on carbon one, add a double bond to an O on that carbon and add the hydrogen that makes it an aldehyde. So that takes care of my main chain. And then I have three methyl groups two on carbon two since it's 223 tri methyl and one on carbon three. So since my aldehyde group must be in carbon one, then I just number 12 and three. So on carbon two, I have two methyl groups. So let's write a line above and a line below with C three on each one. And then I have one on carbon three. So I'll draw a line up to ach three group on carbon three. Now, the last thing I have to do is just fill in the right number of hydrogens on each carbon. Carbon one has four bonds. Already don't add anything else there. Carbon two has four bonds since it has two substituents. Carbon three has one substituent. So it still needs one hydrogen. Carbons. 45 and six are internal carbons with no substituents. So they all get two hydrogens and then carbon seven needs three hydrogens at the end. So we have HC double Bondo then C with two methyl groups, then ch with CH three, then CH H two, C two, C two and C three, which does match that structure in choice B. Choice B. Definitely our correct answer for the condensed structural formula of 223 tri methyl hepton. L see you in the next video.

Struggling with GOB Chemistry?

Draw the condensed structural or line-angle formula, if cyclic, for each of the following: (12.3)

d. 3,5-dimethylhexanal

Watch next

Naming Aldehydes practice set

Struggling with GOB Chemistry?

Draw the condensed structural or line-angle formula, if cyclic, for each of the following: (12.3)d. 3,5-dimethylhexanal

Watch next

Naming Aldehydes practice set

Draw the condensed structural or line-angle formula, if cyclic, for each of the following: (12.3)

d. 3,5-dimethylhexanal