Leah here from leah4sci.com and in this video we're going to work through some nomenclature
practice problems. If you're not comfortable with nomenclature or you feel you could benefit
from a review, pause this video and check out my in-debt naming tutorial video series
along with the nomenclature practice quiz linked below or by going to leah4sci.com/naming.
Early in the semester professors will start by drawing molecules with a lot of the atoms showing
as I have here. But, the more advanced you get, the bigger the molecules, the more overwhelming
so I strongly urge you to check out the skeleton structure tutorial linked below
and learn how to convert these molecules to line structure. In this case I hav 1,2,3,4,5.
So let's re-draw this as a simple 5-carbon chain where carbon 3 has a CH3 attached and an OH.
I know this looks scary at first but once you're comfortable with it,
isn't it so much easier to see exactly what's going on with this line structure? I'm gonna
argue, yes, so let's take the simple version and name it. But first we need to number it. And with
the alcohol coming out of this carbon, by having the same number of carbons for the right and left,
it doesn't matter if I choose to number this from the right side or the left side because I'll get
the same exact name. To name molecules like this, I'm going to use the puzzle piece approach that I
teach in the tutorial linked below, starting with 5 carbons in the parent for a first name of pent.
Only single bonds for a last name of ane. I have a one carbon substituent coming off of carbon 3
so I'll call this 3-methyl where 3 tells me where it is located, meth tells me that it's only one
carbon and yl tells me that it's a substituent. The functional group on this molecule is an OH
or an alcohol, 3 tells me it's an alcohol sitting on carbon 3. And ol for the suffix tells me that I
have an alcohol functional group. Now that we have the pieces, let's put the name together
starting with the prefix of 3 methyl, then we grab the first and last name for pentane
and finally add the suffix 3-ol. Some professors will accept this as your final answer
but in the case of a functional group where you have a number indicating where it is,
you can drag that number out in front of the parent name to give us 3-methyl-3-pentane. And
then, because we have a vowel following another vowel, we actually drop that "e" to add that "ol"
for a final name of 3-methyl-3-pentanol. Both of these answers are correct, you're more likely
to see this but find out which your professor prefers to make sure you don't lose any points.
Once again we're given a molecule where all of the atoms are drawn out and even though
it still looks simple, I urge you to get into the habit of drawing this as a line structure,
trust me it will save you a lot of confusion later. So what do we have? If I number this,
not yet for the sake of naming, just for the sake of making sure I redraw the same number of atoms,
I can even copy the shape and draw it exactly the same way. Carbons 1 and 2, go down to 3, up for 4,
right for 5, down for 6, a double bond between 1 and 2, between 4 and 5. All the hydrogens
disappear, the molecules are so much simpler, and now we're ready to name. We numbered it here, just
to see what's going on but now we wanna verify. This molecule is not symmetrical so it does matter
where we start, and in the case of having 2 double bonds, we want to start with the total lowest set
of numbers, which means if I put a carbon 1 on the left, I hit the pibond at carbon 1,
but if I start it at right, I would hit the pibond at carbon number 2 which means I have to stick to
the left. 1,2,3,4,5,6. I have 6 carbons on my parent chain giving me a first name of hex.
I have two double bonds which will give me a last name of ene, but I have to specify. I have a
double bond between carbons 1 and 2 but it starts at 1. I have another double bond between 4 and 5
but it starts at 4, which means I put 1 for the first pibond comma, 4 for the second pibond, dash,
di to tell me that I have two, and ene for the pibonds. Again, we have ene for the pibonds,
di because I have to specify 2 and then 1 and 4 tells me where the pibonds begin on the molecule
and that's it! I don't have any functional group or any substituent so I'm ready to put the name
together. Just like with the alcohol, we can take these two numbers and drag them out to the front
which gives me 1,4 but unlike the alcohol where we had a vowel following a vowel.
Here we have a consonant followed by a consonant which means we wanna add
a vowel in. And that gives me 1,4-hexadiene , hex diene doesn't read well. Add that a,
giving us 1,4-hexadiene. If you haven't been following the puzzle piece approach up to now,
I strongly recommend use it for this molecule. And if you don't know that approach, pause this video,
go watch my naming videos, then come back and see if this becomes a little easier.
The first thing we want to do is identify our parent chain which does not only the
longest carbon chain but also because we have a carboxylic acid it has to include the carboxylic
acid. So let's start here and we have 1,2,3,4. Now at this junction you might be wondering do
I go this way or do I go that way? And the way to figure that out is by the junction rule which
I also teach in the naming series, and this simply counting to see which one has more.
1,2,3 as the longest, or 1,2,3,4 the longest wins and I keep going to the left. 5,6,7,8. Now we know
we have 8 carbons in the parent chain but let's number it just to verify. The carboxylic acid
is an automatic number 1 because it's the highest priority functional group. 2,3,4,5,6,7,8.
I have 8 carbons in my parent chain giving me a first name of Oct.
I have only a single bond giving me the last name of "ane". I have already identified my functional
group as a carboxylic acid which gives me a suffix of oic acid, two separate words. I'm` not going
to include 1 for the position of carboxylic acid because as a highest priority functional group it
is automatically understood to be on the first carbon meaning it cannot be on something that
isn't the first carbon unless we have two carboxylic acids in this case one is on the
first and one is on the last. But I didn't make it that crazier. We have our functional group,
let's identify all the substituents. I like to mark them off as I name them to make sure
I didn't miss anything. On carbon 3 I have an OH which is an alcohol functional group
except that in the presence of carboxylic acid where the carboxyl is higher priority, the oxygen
is not the highest priority functional group and that means it does not get name with a suffix,
instead, I like to think of it as getting demoted to a substituent which would give me
a hydroxy group on carbon 3 where hydroxy tells us alcohol as a substituent rather than alcohol as a
functional group. On carbon 4 I have a four carbon substituent but there are now 4 carbons in a row
so it's not just a butyl group instead it's got that iso tail at the end reminding me that this is
an isobutyl group. An isomer, a butyl specifically with that beth tail at the end. On carbon 5 I have
a chlorine which gives me 5 chloro. On carbon 6 and 7 I have the same group, the methyl group.
That means I have to include the number 6 and the number 7 to tell me where they show up,
dash, di to tell me that I have 2 and methyl for the group. So the substituents on 6
and 7 are 6,7-dimethyl. And this is why I color it up. Notice that every single group has been
circled and therefore accounted for, that means nothing is gonna be left out so I wanna put the
name together.I need to start with my substituents but I first have to put them in alphabetical order
so let's see what we have to compare. I have an H for 3 hydroxy, I have an i for isobutyl,
I have a C for chloro and I have an M for dimethyl. This is a trick. Anytime you have
more than one and you have to name it di, tri, tetra, those don't count, only the M. But iso
is an exception to that role where you do count the i for iso. Taking this into consideration,
we'll start with the C for chloro, then the H for hydroxy, i for iso and M for methyl. That gives
me 5-chloro-3-hydroxy-4-isobutyl-6,7-dimethyl and we're running out of space but we still have some
name to go so let's squish them together and then just like with the alcohol if we
have ene ending with an e, ene ending with a vowel and oic acid starting with a vowel, we drop that e
to give us octanoic acid where acid is a separate word. So our final name is
5-chloro-3-hydroxy-4-isobutyl-6,7-dimethyloctanoic acid. Phew! that's a mouthful.
If you followed up to here and you want even more practice, or if you first want to review
the naming series including the puzzle piece approach in everything I showed you here,
first make sure you give this video a thumbs up, subscribe to my channel and hit that bell icon so
you don't miss out on any new videos then catch the entire series along with the naming practice
quiz on my website leah4sci.com/naming. Again that's leah4sci.com/naming.