Draw both condensed and line structures corresponding to the f...

Question

Draw both condensed and line structures corresponding to the following IUPAC names and label each carbon as primary, secondary, tertiary, or quaternary.

c. 2,2,4-Trimethylpentane

Accepted Answer

All right. Hi, everyone. So this question says to sketch a line structure and condense structure for the compound 334 tri methyl hexane classify the carbons in the structure as primary secondary tertiary or ordinary. So first, let's go ahead and get started with the line structure or the skeletal structure. Recalled, skeletal structures are representations of organic molecules in which carbon carbon bonds are simply shown as lines, meaning that carbon atoms are not actually shown rather they're implied because all corners and open ends are designated to be carbon right now. Other atoms like nitrogen, oxygen and chlorine have to be drawn in. But hydrogen atoms attached to carbons are implied because they are expected to fill however many bonds carbon is lacking. So let's go ahead and get started by drawing the line structure of 334 triethyl hexane. The phrase Hexane implies that the parent chain or the longest carbon chain is six carbons long. So I'm first going to start by drawing a line that has a total of six corners and open ends, right? And for ease of reference, I'm going to go ahead and label them So now we can talk about substituents. The phrase 334 triethyl implies that there are three methyl substituents present in this molecule or three C three groups. Two of them are on carbon three of the molecule itself, whereas one of them is on carbon four of namely the parent chain. So on carbon three, there will be two lines emerging from it to indicate the two metal groups, whereas on carbon four, there will be only one line. So now we have the line structure 4334 tri methyl hexane. But in this case, I'm actually going to go ahead and rotate the structure so that it looks a little bit similar too how it appears in your answer choices. Now it's the same molecule regardless of how you choose to dry it. But for the sake of cons consistency, I'm just going to jar it in a slightly different way. So from here, we can talk about classifying each carbon as primary secondary tertiary or ordinary. Now recall that each designation is based on how many other carbon atoms are attached to the carbon atom that you're focusing on. Right. So primary carbons only have one other carbon connected to it directly. Secondary carbons have two other carbons. Tertiary carbons have three other carbons and ordinary carbons have four other cars attached directly to it. So in this case, if we consider carbon one of our structure here, notice how carbon one is only connected to one other carbon, namely carbon two of the parent chain. Therefore, it is going to be primary and also carbon six also has one bond to only one other carbon, meaning that it also is going to be primary. And likewise all methyl substituents here are primary considering that they only have one bond to another carbon. So the three methyl groups are primary as well. Now carbon two on the parent chain. However, carbon two here has two bonds, two other carbons, namely carbons, one in three of the parent chain. So it is secondary as is carbon five of the parent chain because notice how carbon five has two bonds to other carbons as well. So it is also secondary. Now, carbon three in the parent chain is connected to four other carbons when you consider two from the parent chain and the two methyl substituents. So carbon three here is actually ordinary whereas carbon four has three bonds to other carbons making it tertiary. So now I can focus my attention on the condensed structure. Recall that the condensed structure shows how the atoms in a given compound are connected without actually showing the lines to indicate covalent bonds. Now, carbon and hydrogen atoms are grouped together, for example, ch three CH two or CH whereas groups in parentheses can be designated or actually can refer to either branches on a chain or repeating CH two units within a chain. So if I scroll down here recall that the number of hydrogens is actually implied according to how many bonds, carbon already has, right? Because this compound is an all cane recall that carbon prefers to have four covalent bonds to other atoms. Now, depending on its designation, whether it's primary secondary et cetera, we can infer how many hydrogens that carbon needs to ensure that it has four covalent bonds. So in our all here, primary carbons have only one bond to other carbons. This means that three hydrogens are needed to make sure that that carbon has all four covalent bonds that it prefers to have secondary carbons. Therefore require two hydrogens. Tertiary carbons require one hydrogen and ordinary carbons require zero hydrogen because they already have all four bonds necessary. So in my condensed formula, carbon one here COVID one is going to be written as AC three, whereas carbon two being secondary is written as a sh two. Now, carbon three on the other hand has no hydrogens, right. So that is simply ac but next to carbon three in my condensed formula goes, the identity of any substituents present in parentheses, right, because there are two methyl groups on carbon three right next to carbon three in my parent chain in my condensed structure, I'm going to write the age three in parenthesis. However, because I have two of them on the same carbon outside of the parentheses goes a subscript of two and similarly, right, carbon three or carbon four, excuse me being tertiary requires one hydrogen. So that's ac but it also has a metal group. So next to carbon four in my condensed structure goes a CH three written in parentheses. Lastly, we have carbons five and six of the parent chain. Carbon five is secondary. So that is a CH two, whereas carbon six is primary. So that is ach three and there you have it, here is the complete lined and condensed structures for the compound 334 tri methyl hexane and also their designations as either primary, secondary, tertiary or ordinary. So if you stuck around to the end of this video, thank you so very much for watching and I hope you found this helpful.

Draw both condensed and line structures corresponding to the following IUPAC names and label each carbon as primary, secondary, tertiary, or quaternary.
c. 2,2,4-Trimethylpentane

Key Concepts

IUPAC Nomenclature

Condensed and Line Structures

Carbon Classification

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