Constitutional isomers of C4H10O | Alcohol & Ether – Dr K
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Constitutional isomers of C4H10O | Alcohol & Ether – Dr K

The question is asking to draw out all
constitutional isomers for C4H10O. Since this chemical formula fits in the
general formula of CnH2n+2O that means C4H10O can be an alcohol
which has an OH functional group. It can also be an ether which has an
oxygen sandwiched between two carbons. We’re gonna use a systematic way to draw
the isomers that way we won’t accidentally leave out any structures. So
let’s start by drawing a four carbon chain meaning four carbons connected
straight in a row and we place OH on the first carbon. That will give us our
first structure. There’s another place where we can move the OH group which
is on the second carbon. Now these are the two four carbon chain alcohols. If we
place a huge group on the third carbon you’ll be the same as the second
structure that we have drawn. Also if we were to move the OH all the way to
the end on the fourth carbon it will be actually the same as the first structure.
So it looks like we have exhausted all possibilities for the four carbon chain
alcohol. In that case let’s move on to three carbon chain. Since we have four
carbons total let’s place the extra carbon on carbon 2 like this. The
reason we don’t want to stick that fourth carbon at the end is because then
we will end up with four carbon chain rather than three carbon chain. So okay
now let’s stick the OH group on the first carbon like this. Keeping the same
three carbon chain, let’s move the OH to the second carbon
like that that give us another isomer. Now if we were to move the OH to the
third carbon it’ll be the same as placing OH on the first carbon. That
means we are now done with a three carbon chain alcohol since it’s not
possible to have a two carbon chain with four carbons. Let’s move on to drawing
ethers. We’ll do it the systematic way as well. we will start with one carbon
followed by oxygen and three remaining carbons. That’s one way of drawing ether
keeping one carbon before oxygen let’s see if there’s a different way we can
draw the three carbons at the back. In fact there is we can have two carbons
and then we stick the extra carbon on carbon one like that. Since there’s no
other way to drawing the three carbons at the back let’s just move on to
placing two carbons before the oxygen so we have two carbons oxygen and then two
remaining carbons since there is no other way of drawing the two carbons at
the back that’s all that we have here are the structures after placing in the
hydrogen’s we have a total of six isomers for C4H10O all where four of them
are alcohols and two (I mean THREE) of them ethers. I hope you find the video helpful. Do
subscribe. Thanks for watching and I’ll see you in the next video.


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