Constitutional isomers of C4H8O2 | Carboxylic acid & Ester – Dr K
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Constitutional isomers of C4H8O2 | Carboxylic acid & Ester – Dr K

October 25, 2019


We’re gonna draw out all possible
carboxylic acid and ester isomers for C4H8O2 in this video. Before we begin, how many isomers do you think we will get? Stay tuned until the end to
compare notes. Let’s begin. Quick recap. Carboxylic acid has a carboxyl
functional group which is C double bond O connected to OH, whereas ester is
C double bond Oo connected to OC. The difference between carboxylic acid and ester is carboxylic acid has a H connected to that O, whereas ester has a C connected to O. Another thing to note is that for carboxylic acid that COOH
group which is called carboxyl group, it’s always a carbon number 1, so, let’s
start with carboxylic acid. Since we have four carbons, let’s place the COOH group
at the beginning and then we will add in the remaining carbons in a row like this.
That’s our first carboxylic acid isomer having a 4 carbon chain. 4 carbon
chain because we have 4 carbon straight in a row. Now instead of placing all 4
carbons in a row we can actually place 3 carbons in a row and then stick the
fourth carbon like this on carbon 2. That will give us another isomer.
What if we were to stick that extra carbon on carbon 3? It’s actually the
same as our first structure. Notice it’s a 4 carbon chain, so, it looks like we are
done drawing the carboxylic acids. Let’s fill in with the hydrogens and here we go the two structures for carboxylic acid. Let’s move on to ester. For ester,
the functional group is COOC, so, let’s start by placing that at the
beginning. We’ve already used up 2 carbon and 2 oxygens so far, so, for the
remaining two carbons we are going to place them after that COOC
group like this so that will give us a 1+3 combo. Retaining the 1 carbon at
carbonyl group, let’s shorten the 3 carbon to 2 carbon and we’ll place the
third carbon like this. That will give us another isomer for
ester. Now, since we can’t shorten the chain after O anymore, let’s move on to
two carbons on the carbonyl group like this. So this gives us a 2+2 combo. Looks
like we can’t shorten the chain, either chain anymore, so, let’s move on to 3
carbon on the carbonyl group and this will give us a 3+1 combo.
Since we can’t shorten either chain and we need at least one carbon after that
oxygen, because we want to draw ester, so, it looks like we are done. So, we fill up
these structures with remaining hydrogens and it will give us these. Keep
in mind that each carbon makes four bonds, so, make sure that you check to
make sure that every structure has 8 hydrogens, no more and no less. Quick
recap on how we drew all 4 isomers for ester. We started with 1 carbon on the
carbonyl group and then we placed 3 carbon of the oxygen. Since we can
shorten the three carbon chain to two carbon, we then drew the second isomer
with the two carbon and extra carbon sticking on C1. Since we’ve exhausted
this combo, we then move on to 2 carbon on the carbonyl group. Because we can’t
shorten the chain anymore we then move on to 3 carbon on the carbonyl group. We
can’t have 4 carbon on the carbonyl group for this formula because that would have
given us a carboxylic acid instead of esters. So using this systematic way, we
know that we have drawn out all possible isomers for ester. To recap, we have drawn out a total of 2 carboxylic acids and 4 esters for C4H8O2. A quick
disclaimer though the numbering that I used in these structures, they are not
meant for naming, it’s just to help us see how many carbons that we have on each
side of the structure, so, don’t get confused. If you’re interested in drawing
other isomers aside from carboxylic acid and esters for C4H8O2, to let me know
in the comments and I’ll post a video for it. Hope the video was helpful somehow. Do subscribe and thanks for watching!

Only registered users can comment.

  1. Helped me understand drawing isomers in an easy way.Useful for nmr problems.Thanks a lot for posting the video on yuutube.

  2. I did not understand the third isomer of carboxylic acid . Can u plz explain it properly plz . Otherwise it is very effective

  3. really it help me a lot but there is one error in video in ester second chain as it shows 7 carbon but it will be 8 carbon. the ch will be CH2

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