Isomerism (Chemistry)

To draw all possible structures for C4H8O, the first step we need to do is to calculate the degree of unsaturation of the molecule.

Degree of Unsatuation of C4H8O = (C+O-H)/2 = (2*4+2-8*1)/2 = 1 (meaning it has 1 double bond)
[degree of C=2, H=1 and O=2]

My question is why can the degree of unsatuation be caculated by the above formula, which only works for compounds with double bond(s) or satuated compounds. Also why is the degree of Carbon assigned to be 2, H=1 and O=2 ?
 

I, Brian

I don't think it's possible to say "Why?" in Maths - because then you're in the realm of Mathematical Philosophy, which mathematicians apparently don't care to explain, and non-mathematicians cannot even begin to. :)
 
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Mathematically, i don't really know what to tell you.

Chemistry wise, I don't usually see this equation and terms used much anymore. but to answer your question, the c=2 because the highest bond for it is 2, and O is 2 because it double bonds with carbon. H is almost always 1 because it singally bonds with the rest of the singally bonded carbons.
 
Originally posted by PeteGt
the c=2 because the highest bond for it is 2, and O is 2 because it double bonds with carbon. H is almost always 1 because it singally bonds with the rest of the singally bonded carbons. [/B]
Carbon can form triple bond with nitrogen (-nitriles) and it can also form triple bond with carbon itself (alkyne)
 
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You are correct but the chemical you have listed is N-Butanone. Thus there are no triple bonds.
 
It doesn't necessarily belong to N-Butanone. Since C4H8O is only a moleuclar formula, after knowing it has 1 double bond, we know there are 4 possibilities.
1. Ketone (C=0)
2. Alkanol + C=C
3. Ether (-O-C-O-) + C=C
4. Aldehyde (C=OH)

In fact we can draw 15 different possible structures.

there are no triple bonds.
I get your point. As the formula (C+O-H)/2 can only be used to calculate how many double bonds it has, therefore there shouldn't be any triple bond.
 

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--- and, how many cyclics?
 
I forgot to take cyclics into consideration. :wink:
 
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Originally posted by KL Kam
It doesn't necessarily belong to N-Butanone. Since C4H8O is only a moleuclar formula, after knowing it has 1 double bond, we know there are 4 possibilities.
1. Ketone (C=0)
2. Alkanol + C=C
3. Ether (-O-C-O-) + C=C
4. Aldehyde (C=OH)

In fact we can draw 15 different possible structures.


I get your point. As the formula (C+O-H)/2 can only be used to calculate how many double bonds it has, therefore there shouldn't be any triple bond.
1.) Yeap it's a keytone alright, N-butanone

2.) Alkanol (proper alkene)-Can't be this. With a double bond between any of the carbons yields an H count > 8.

3.) Ether...nope, you have one Oxygen

4.) Aldehyde (watch how you write that, looks like an alcohol group with a double bond to carbon, violation of valence electrons. This is possible since an aldehyde and a ketone are the same thing. Just, an aldehyde occurs on the terminal carbon and has a c-h bond with it, while ketones have a bridging C with a =o and no c-h bonds.

Cyclics? Not possible by general valence bond theory, you would only have 6 or 7 hydrogens but not more.
 
Alkanol (proper alkene)-Can't be this. With a double bond between any of the carbons yields an H count > 8.
http://www.angelfire.com/freak2/antiwork/2.bmp [Broken]
Ether...nope, you have one Oxygen
Oops, ether should look like this, - C-O-C- , but not -O-C-O-. Sorry for the confusion.
http://www.angelfire.com/freak2/antiwork/3.bmp [Broken]
Cyclics? Not possible by general valence bond theory, you would only have 6 or 7 hydrogens but not more.
Yes, I think you're right. I tried that just now and couldn't attatch 8 hydrogens to the carbon ring
 
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Yeah, those look right, you're first one with the alkene, watch how you draw that, you've drawn a bond to a hydrogen :-)

But in everything you showed me there is a double bond with a carbon. As to why oxygen had it I dunno.

My field is not organic chemistry unforunately, but i'll toss it around the lab here tomorrow and find more info about that equation.

I'm a physical inorganic chemist.

Pete
 

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Cyclobutanol, methyl-cyclopropyl ether, cyclopropyl methanol --- you can play with the IUPAC names on your own --- tetramethylene oxide (a cyclic ether) --- getting the picture yet?
 
Yeah, those look right, you're first one with the alkene, watch how you draw that, you've drawn a bond to a hydrogen :-)
Let's call it KL Kam's compound then, which doesn't exist.
I'm a physical inorganic chemist.
Cool! In fact I only studied physical chemistry last school year (02-03), plus nomenclature of organic compounds(IUPAC) and part of the chapter about isomerism. I'll study organic chemistry and inorganic chemistry next school year.
Cyclobutanol, methyl-cyclopropyl ether, cyclopropyl methanol --- you can play with the IUPAC names on your own --- tetramethylene oxide (a cyclic ether) --- getting the picture yet?
C4H8O is an unsatuated compound after using ((C+O-H)/2) to calculate. But when I draw the sturcture you mentioned, it seems that they are correct too.
Why is it so?
http://www.angelfire.com/freak2/antiwork/structure.bmp [Broken]
By the way, I don't know the struture of tetramethylene oxide, could you please show me?
 
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"Cyclic ether:" swap oxygen for a methylene group in cyclopentane.
 
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Gosh, I need to stay away from organic chemistry. I look at all these compounds and think, damned, how did i miss that one???

Good luck with studying inorganic this year. 400 level Inorganic?

This structuring that we are looking at here by trying to identify isomers of C4H8O. That's why when i can, always identify through common place names. (IUPAC gets to be annoying too) Like caffiene, thank god a can just refer to it as caffiene and now give a lecture about the IUPAC. :-)
 

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THF, tetrahydrofuran, natural products chemistry (corn cobs), still explosions, Grignard rxns --- quite a few inorganic syntheses.
 
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what does THF have to do with inorganic chemsitry?

And what are you saying? :-p

Just seems like that post came from rigth field. lol :-)

Pete
 

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Can't put my hands on references, so, technically, this is hearsay; THF is/was a fairly popular solvent among the metal hydride crowd --- at least a few years back --- times change, and so do lab techniques --- never got into that business personally, and a quick check of Cotton & Wilkinson, Drago, and Greenwood & Earnshaw comes up blank --- mebbe I'm "worng."
 

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