Below is the written transcript of my YouTube tutorial video Naming Branched Chain Alkanes.
Leah here, from leah4sci.com and in this video, I will show you the IUPAC rules for naming branched alkanes. When naming branched alkanes you get to add on the prefix portion to the naming puzzle. Remember a branch is simply another carbon group on the molecule that doesn’t flow with your straight chain.
Here are few rules that I’d like to use when naming branched alkane,
First, identify your parent chain. This is the carbon chain you can trace without lifting your highlighter. Next is you number your chain so that your constituents gets the lowest number. Next, you want to name your parent and substituents using the format introduced in the first video when adding the following details : You have to include the number for each substituents telling where it shows up on the parent chain, use the prefix that identifies the number of carbon and then end with the letter ‘yl’ to show that it’s a substituent. And last but not the least, tell how many of each substituent you have.
If you have a single substituent, you don’t need to number because it is understood to be 1. If you have more than 1, include the prefix di for 2, tri 3, tetra and penta 5.
Let’s start with this simple example: I like to mark up my chain as I put the puzzle pieces together because this way I know what I named and what I have yet to name. we highlight the longest carbon chain which is our parent chain, and number from the side that gives me the lowest number for my substituent. I have the option to number from the right giving me, 1, 2 and the substituent of 3 or naming from the left which gives me 1, 2 and a substituent of 2. Since 2 is lower than 3, I have to number from the left , this gives me a parent chain of 4 carbons and a first name of but. Only single bonds gives me the last name of “ane”.
For my substituent, I have one carbon on carbon number 2. I use the number 2 since it’s the second carbon on the parent chain, myth to show that it’s a single carbon and “yl” to show that it’s a substituent. Putting the name together, this is my prefix, this is my first , “ane” is my last name, giving this molecule the name of 2-methylbutane.
Now let’s try adding a second group to this molecule,. We’ll approach it the same way, highlight the parent chain and then numbers so that you have the lowest number for the substituents. One more time I number from the left for a total of 4 carbons. 4 carbons gives me a first name of “but”, single bonds gives me a last name of “ane” . and now we have 2 substituents both coming off of carbon 2. Since the substituents are the same, they’re both methyls, I write 2 comma 2 dimethyl. The reason I include 2 number 2s is that both substituents are on number 2 and the di tells me there 2 of the methyl group. You have to include both the numbers and the di, this way I know how many I have and exactly where each one shows up.
Notice the way I structured the name of the substituents, you use a comma when you have a number followed by another number and a dash when you have a number and letter or a letter and a number. It this case I have a 2 comma 2 and then the second 2 gets a dash to dimethyl. Putting the name together, I have 2,2-dimethylbutane and notice there is no space between the substituent and the parent chain but rather it’s all one word.
Now let’s look at an example that has two different substituents. I start by highlighting the parent chain and then numbering from the side that gives me the lowest number for the first substituent I come across. This means if I start from the right, I get 1, 2 ,3 for my first group. Or if I start from the left, I get 1, 2. Since 2 is lower than 3, I start counting from the left for a total of 7, giving me a first name of hept. Since I have single bonds, I have “ane” and that’s my parent name. Now let’s look at the substituents. On carbon number 2 I have a single carbon substituent, this is a 2-methyl, and on carbon number 5 I have a 2 carbon substituent which gives me 5-ethyl. With more than 1 substituent, you order them in alphabetical order, so we’ll compare M to E, since E comes before M, I get a final name of 5-ethyl-2methylheptane.
And now for a tricky example, at first glance it appears that my parent chain only has 5 carbons, however don’t go by what appears to be the longest chain. Remember that carbon to carbon bonds can rotate and so your chain can actually be slightly twist. In fact the longest carbon chain actually extends downward for a total of 6 carbons. I’ve adopted the junction rule in physics to help you easily identify your longest carbon chain. Instead of counting 1,2, 3 going this way, going that way, going that way, this can potentially take a long time, instead identify your junction or the one point that has multiple chains coming out of it and then simply identify your longest pieces. Notice that to the right, I have 2, to the left I have 2, but going down I have 3. This tells me that my parent chain has to include the branch going down. And I have 2 to the right and to the left, I can arbitrarily choose 1 for the longest carbon chain.
Now that I have identified the longest carbon chain, I continue with the standard naming. Starting from the left I have a total of 6 carbons for a first name of “hex”, single bonds gives me an “ane” and a 2 carbon substituents on carbon 3 gives me 3-ethyl. Putting the name together I have 3-ethylhexane.
Let’s do one last tricky example by naming this molecule. Don’t be fooled by the obvious which appears to be a carbon chain of 6 when in fact I have a much longer carbon chain. Applying the junction rule, I have a branch here, here, here and here. Now, let’s not get carried away, let’s only take the branches that are not obvious at first glance to be our longest carbon chain. And so the one tricky branch to look at is right here. To the right, I have 2 carbons. Counting upwards I have 3 carbons and counting to the left, I also have 3 carbons. I verify this with a highlighter tick, where I start highlighting at one end and make sure I can get all the way to the other end without having to lift my highlighter. And so we get a total of one, two, three, four, five, six, seven carbons in my longest carbon chain. I have the option of numbering from the left which gives me my first substituent of 2 or numbering from the top which also gives me the first substituent at 2. So we have to number it one more step and see which one has more substituents for lower number. From the left, I have another substituent at 3, from the top I don’t have a substituent at 3 and so I have to count from the left.
This is where the puzzle idea comes in handy. If you try to name this as you see it, you will likely to make a mistake. But if you mark off each piece individually by naming that aspect of the puzzle, the entire thing will come together nicely. I have seven carbons giving me a first name of “hept”, I have only single bonds giving me “ane”. And now let’s look at substituents. I have the methyl group on carbon 2, 3 and 6 and so I name this 2 comma 3 comma 6 trimethyl. I also have a 2 carbon substituent on number 4 which gives me 4-ethyl and that’s it! Every piece of my molecule is marked and numbered, every puzzle piece is named and now we put the puzzle together. Your substituents are named in alphabetical order but we don’t look at the prefix “tri”, tri is just an identifier, but the actual letter to look at is M compared to E. since E comes before M, you get a final name of 4-ethyl-2,3,6-trimethylheptane.
In the next video, we’ll look at how to name alkanes that has branches on their branches.
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