How many isomers can be formed from C7H16?

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Discussion Overview

The discussion revolves around the molecular formula C7H16 and the identification of its possible isomers. Participants explore methods for determining isomer counts, the nature of isomers, and the implications of structural variations, including the consideration of optical isomers.

Discussion Character

  • Exploratory
  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that there are nine isomers of C7H16, while others claim there are more, leading to disagreement on the total count.
  • One participant suggests a method for identifying isomers by visualizing structures and checking for duplicates.
  • Another participant mentions that C7H16 follows the formula CnH2n+2, indicating it is a saturated acyclic alkane.
  • There is a discussion about the absence of optical isomers due to the lack of chiral centers in the molecule, with some participants clarifying the conditions for chirality.
  • Participants propose drawing structures to identify isomers, including variations with different alkyl groups such as methyl and ethyl.
  • A formula for calculating unsaturation in hydrocarbons is introduced, but its relevance to the discussion of C7H16 is debated.
  • Some participants discuss the naming conventions for isomers, particularly regarding 2-ethylpentane and 3-methylhexane, indicating a potential overlap in structural representation.

Areas of Agreement / Disagreement

There is no consensus on the total number of isomers for C7H16, with participants presenting conflicting views on whether it is nine or more. The discussion remains unresolved regarding the inclusion of optical isomers and the correct naming of certain structures.

Contextual Notes

Participants express uncertainty about the mathematical methods for determining isomers and the implications of structural variations. The discussion includes assumptions about the nature of isomers and the conditions under which they are counted.

garytse86
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I was given a molecular formula C7H16 and asked to find all possible isomers. I know there are nine because the question said it. But I am just wondering how to find all isomers correctly without repeating myself. Is there a way to do this mathematically or a general rule that makes it easier?
 
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hmm i don't really know if there's a mathematical formula (probably would be a waste of time), but if you think you're repeating yourself just try flipping the structure in your head, then try to match it with any of the structures you have already written down, if none match it's a new isomer :smile:. you could also number the carbons as you would if you were naming the molecule, but if you haven't covered that yet then just stick with trying to visualize it.
 
There's a lot more than nine.
 
Notice that C7H16 is Cn H2n +2 ?

That means it can only be a straight chain of atoms (not a cycloalkene) and it may not be unsaturated. The rest is for you to figure out. I don't know of any formulas :)


Basically start with heptane, and start taking of a C atom from the end and place it somewhere else, making sure you don't make a mirror image. Every time you take one off and finally you should come to 9 different forms.
 
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Originally posted by Chemicalsuperfreak
There's a lot more than nine.
There are only nine!
 
Maybe you can figure out a formule from the following data:

Code: 
Pentane C5H12  3 isomers 
Hexane  C6H14  5 isomers 
Heptane C7H16  9 isomers 
Octane  C8H18  18 isomers 
Nonane  C9H20  35 isomers 
Decane  C10H22 75 isomers
 
Originally posted by Monique
There are only nine!
Whoops. Sorry. Just saw the C7 and not the H16.
 
how come they don't count optical isomers? if they did C7H16 would have 11 isomers correct?
 
There are no optical isomers, since there are no double bonds in the molecule.. molecules are able to rotate freely around their axes. *edit* I guess I am talking about geometric isomers (cis/trans)

Well, no optical isomers either (L/D), since you would need different sidechains right? You cannot have a chimeric atom when all groups are identical.. I guess?

I drew all 9 of them out.. in order for a C atom to be chiral, it needs to have 4 groups attached.. there are a few.. no, no enantiomers.
 
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  • #10
Originally posted by Monique
There are no optical isomers, since there are no double bonds in the molecule.. molecules are able to rotate freely around their axes. *edit* I guess I am talking about geometric isomers (cis/trans)

Well, no optical isomers either (L/D), since you would need different sidechains right? You cannot have a chimeric atom when all groups are identical.. I guess?

I drew all 9 of them out.. in order for a C atom to be chiral, it needs to have 4 groups attached.. there are a few.. no, no enantiomers.

butyl, ethyl, methyl, and hydrogen. Four different groups. You CAN have more than nine isomers, if you count stereoisomers and not just structural isomers.
 
  • #12
3-methylhexane and 2,3-dimethylpentane each have a chiral carbon.
 
  • #13
You are correct, my err!
 
  • #14
That means it can only be a straight chain of atoms (not a cycloalkene) and it may not be unsaturated. The rest is for you to figure out. I don't know of any formulas :)
Erm...

The isomers don't have to be straight chained, (if they did then there would be only one isomer), and since it is in the general formula CnH2n+2 then all of the isomers will be saturated.
Sorry for being so nit picky.

Anyway, the best way to spot isomers at a glance is to do many many examples. At least that's what my chemistry teacher made me do. It's also best if you learn to spot them now, that way when you get on to cis/trans and then other more complicated molecules (with benzene rings in, cis/trans and some halides thrown in for fun) you can spot the possibilities a lot easier.
 
  • #15
Originally posted by lavalamp
Erm...

The isomers don't have to be straight chained, (if they did then there would be only one isomer), and since it is in the general formula CnH2n+2 then all of the isomers will be saturated.
Sorry for being so nit picky.

He means acyclic.
 
  • #16
That makes more sense. I blame the fact that I was tired.
 
  • #17
2-ethylpentane

Correct me if i´m wrong please, but isn´t there a 2-ethylpentane isomer?




Rui.
 
  • #18


Originally posted by RuiMonteiro
Correct me if i´m wrong please, but isn´t there a 2-ethylpentane isomer?




Rui.

You mean 3-methylhexane?
 
  • #19
No, i mean 2-ethylpentane.



Rui.
 
  • #20
Code: 
  C
  |
  C
  |
C-C-C-C-C
The above is the molecule that you (RuiMonteiro) named (2-ethylpentane).

It just so happens that it is also the same as this molecule (3-methylhexane).

Code: 
  C
  |
C-C-C-C-C-C
Chemists prefer to name the molecule after the longest chaain in the molecule (ie: hexane) and so it's proper name is 3-methylhexane, although 2-ethylpentane is not technically wrong.
 
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  • #21
That´s right, i knew i was missing something cause nobody talked about that isomer lol, thanks for clarifying. :)




Rui.
 
  • #22
The simplest way to find all the isomers is to draw the the possible ways to arrange it as a straight chain, then ways to arrange with a methyl, then ways with several methyls, then an ethyl, then several ethyls and so on. I think 1 ethyl is as high as she goes.
 
  • #23
With any hyrdrocarbon there is a way to narrow down your possible isomers. I always use this formula to start a problem of this kind. This formula is used to find the unsaturation of the molecule which in turn can help you figure out how many rings, double or triple bonds are in the structure.

(2C+2-H-X+N)/2

You just multiply the number of Carbons by 2 add 2 subtract the number of hydrogens and halides (X= Cl, I, Br) and add the number of nitrogens. Divide all of that by 2. NOTE: the number of oxygens does not factor into the unsaturation

Here's an example of how to find unsaturation:

C6H9Cl

(2(6) + 2 - 9 - 1)/2 = 2

How to interpret your results:
An unsaturation of o indicates no unsaturation and therefore the molecule has no rings, or double or triple bonds
An unsaturation of 1 indicates that there is either a ring or a double bond
An unsaturation of 2 indicates there is either a ring and a double bond, 2 double bonds or a triple bond
etc...

We now know that the example will have isomers with 1 triple bond, 2 double bonds, and ones with a ring and a double bond.

This formula is very useful and a great tool for organic chemistry.
 
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  • #24
(2C+2-H-X+N)/2
I'm always wondering how chemists figered out this formula. [?]
 
  • #25
Originally posted by KL Kam
I'm always wondering how chemists figered out this formula. [?]

Consider a straight chain hydrocarbon. Each carbon has two hydrogens, except for the ones at the end, which each have an additional hydrogen. So the number of H is 2n+2, where n is the number of carbon atoms, follow? You can than arrange it in anyway you want and it will still work. Halogen atoms will take up the place of a hydrogen atom. So bromopropane has three carbons, and one bromine, so it would have 8 hydrogens just given the carbons but it really has only seven, due to the bromine, so you subract the number of halogens (-X). Oxygen won't affect it at all, it just spaces out the molecule, it's kind of like a filler. Nitrogen is kind of like oxygen, except it's got one more bond for a hydrogen (or something else as the case may be), so you add a hydrogen for every nitrogen. So your final equation is 2n+2-X+N for how many hydrogens you should have. This, minus actual number of hydrogens you actually have, divided by two (each pi bond takes the place of two hydrogens, or the ring forming bond takes the place of those two protons at the end of the chain) gives the degree of unsaturation.
 
  • #26
Originally posted by Chemicalsuperfreak
Consider a straight chain hydrocarbon. Each carbon has two hydrogens, except for the ones at the end, which each have an additional hydrogen. So the number of H is 2n+2, where n is the number of carbon atoms, follow? You can than arrange it in anyway you want and it will still work. Halogen atoms will take up the place of a hydrogen atom. So bromopropane has three carbons, and one bromine, so it would have 8 hydrogens just given the carbons but it really has only seven, due to the bromine, so you subract the number of halogens (-X). Oxygen won't affect it at all, it just spaces out the molecule, it's kind of like a filler. Nitrogen is kind of like oxygen, except it's got one more bond for a hydrogen (or something else as the case may be), so you add a hydrogen for every nitrogen. So your final equation is 2n+2-X+N for how many hydrogens you should have. This, minus actual number of hydrogens you actually have, divided by two (each pi bond takes the place of two hydrogens, or the ring forming bond takes the place of those two protons at the end of the chain) gives the degree of unsaturation.
Thank you very much for answering! It was difficult to understand the part about oxygen and nitrogen, but after drawing out some structures of molecules, I could grasp the idea.

While considering some molecules, I drew one with triple bond accidentally, that was a molecule with one C\equiv N but no double bond. After applying the formula, I got 2 as the answer.
Then I consider another molecule, H-C\equiv C-C\equiv C-CH_3, and applied the formula, the answer was 4. For H-C\equiv C-C\equiv -C-CH=CH_2, I got 5. Is it because we can treat a triple bond as two double bonds if we move one of the pi bonds from the triple bond to a single bond?

If that's true, next time when we apply this formula to find out how many unsatuated compounds, should we consider triple bonds as well ? For example, draw all possible isomers of C11H11N. After applying the formula, we got 7. The possible structure may contain either 3 triple bonds, 1 double bond, or 2 triple bonds, 3 double bonds, or 1 triple bond, 5 double bonds, or 7 double bonds.

(It seems that it is correct)
 
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  • #27
it could also be a variety of ring structures.
 
  • #28
Originally posted by KL Kam
Is it because we can treat a triple bond as two double bonds if we move one of the pi bonds from the triple bond to a single bond?

(It seems that it is correct)

Yes, degrees of unsaturation can either be a ring or a pi bond. Triple bonds have two pi bonds so they count as two degrees of unsaturation.
 
  • #29
n-heptane
2-methylhexane
3-methylhexane
2,2-dimethylpentane
3,3-dimethylpentane
2,3-dimethylpentane
2,4-dimethylpentane
3-ethylpentane
2,2,3-trimethylbutane

Should be 9.

tks!
 
  • #30
What about this formula?

(2^n) + 1, where n = 0 when number of carbon atoms = 4

So, C7H16 would have:

(2^3)+1 = 9 isomers.

The only exception I know of this formula is when C=10, then the formula says the no. of isomers is 75 when actually it is 65.
 

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