[orgo]anti/gauch

[ippie]

which is most stable?
not #1!!! be careful!!!
the answer is #2 because of intramolecular hydrogen bonding.
BUT WHY NOT #3???

THIS PROBLEM IS FROM DESTROYER ORGANIC ODYSSEY.
DO YOU THINK ORGANIC ODYSSEY PROBLEMS ARE GOOD???
SO MANY HARD QUESTIONS AND INSUFFICIENT EXPLANATION.

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[doc3232]

good question. I learned this like 5 days ago.
So O-chem is completely random. This question should not be on the DAT because it is too complicated. Like you said the trick is Hydrogen bonding, so you want OH's close but not too close. As close as you said (#3) would make it gauche and hence makes it too unstable (and hence the hydrogen bonding does not make up for the instability from the gauche positioning). #2 is just right.
I tried my best to explain.

 

[doc3232]

From what I know. All the O-chem is basic, very basic. nothing is tricky.
If you understand the concepts then your ahead of everyone else.

 

[joh020340]

good question. I learned this like 5 days ago.
So O-chem is completely random. This question should not be on the DAT because it is too complicated. Like you said the trick is Hydrogen bonding, so you want OH's close but not too close. As close as you said (#3) would make it gauche and hence makes it too unstable (and hence the hydrogen bonding does not make up for the instability from the gauche positioning). #2 is just right.
I tried my best to explain.

#3 is eclipsed, not gauche. 2 is gauche and 1 is anti.

 

[doc3232]

It has been a while, sorry. Thats what I meant.
The eclipsed makes it too unstable and the Hydrogen bonding does not make up for it.
Rule of thumb: Eclipsed is always too unstable.

 

[RockandRoll24]

Steric Strain causes the eclipsed confirmation to be the most unstable.

 

[Holopeachy1]

which is most stable?
not #1!!! be careful!!!
the answer is #2 because of intramolecular hydrogen bonding.
BUT WHY NOT #3???

THIS PROBLEM IS FROM DESTROYER ORGANIC ODYSSEY.
DO YOU THINK ORGANIC ODYSSEY PROBLEMS ARE GOOD???
SO MANY HARD QUESTIONS AND INSUFFICIENT EXPLANATION.

I know im replying super late, but if you buy a molecular model and build it, you can clearly see how through steric hindrance #3 would be unstable. The molecular models help you with most of the basic stereochemistry. Its really good!

 

[joonkimdds]

I am surprised to find out that #1 isn't the most stable.
I think stability order should be 1>2>3 but I guess I am wrong if TC is correct.

 

[Zerconia2921]

which is most stable?
not #1!!! be careful!!!
the answer is #2 because of intramolecular hydrogen bonding.
BUT WHY NOT #3???

THIS PROBLEM IS FROM DESTROYER ORGANIC ODYSSEY.
DO YOU THINK ORGANIC ODYSSEY PROBLEMS ARE GOOD???
SO MANY HARD QUESTIONS AND INSUFFICIENT EXPLANATION.

Are you sure there isnt a typo. Anti is always the most stabel and it dosent matter if there H-bonds the bulky group in the #2 puts hinderance.

 

[doc3232]

Are you sure there isn't a typo. Anti is always the most stable and it doesn't matter if there H-bonds the bulky group in the #2 puts hindrance.

The one thing I know about O-chem is that it is never an always rule (to an extent). I am not sure if I am right and I would love to be corrected but H-bonds are very stabilizing.

 

[eatkabab]

it is definitely number 2 in this case. H-bonding is the strongest intermolecular bond, so strong that weak covalent bonds and H-bonds are both given basically the same bond stregnth, about 155KJ/mol. So imagine that the H-bond between the two alcohols are just about if not just as strong as the covalent bond between the carbons. additionally, the H bonding radius is more flexible than a covalent bond radius. this is what allows for the gauche configuration. if H bonding radi were as strict as covalent bonding radi, the answer would be number 1.

I dunno if this is such a complicated problem for the DAT. I feel like methyl/hydride shifts are more complicated than this. you could answer this question with basic bio knowledge and VERY basic Ochem knowledge. but yes, when you start thinking too much along the lines of Ochem, you could easily choose number 3. so you basically need to know when to think smart and when to think dumb...? dunno if that made any sense

 

[Brunella]

That's strange...I would think that the anti conformation (#1) is the most stable.

 

[americanpierg]

it is definitely number 2 in this case. H-bonding is the strongest intermolecular bond, so strong that weak covalent bonds and H-bonds are both given basically the same bond stregnth, about 155KJ/mol. So imagine that the H-bond between the two alcohols are just about if not just as strong as the covalent bond between the carbons. additionally, the H bonding radius is more flexible than a covalent bond radius. this is what allows for the gauche configuration. if H bonding radi were as strict as covalent bonding radi, the answer would be number 1.

I dunno if this is such a complicated problem for the DAT. I feel like methyl/hydride shifts are more complicated than this. you could answer this question with basic bio knowledge and VERY basic Ochem knowledge. but yes, when you start thinking too much along the lines of Ochem, you could easily choose number 3. so you basically need to know when to think smart and when to think dumb...? dunno if that made any sense

What you said doesn't make any sense at all. It's easy to make up explanations when you know the correct answer. Why does the "flexibility?" of a Hydrogen bond radius and covalent bond radius have anything to do with this? The bonding raidus of a substituent group differs from group to group, hence two methyl groups causes strain next to each other in gauche conformation such as for butane, while the hydrogens don't, is because their distance is less than the sum of their van der wals radii. In order to solve this, you'd have to know the amount of energy removed from hydrogen bonding of the O and H attached to the other O (which, I can tell you for sure off the top of my head, is no where near 155kj/mol), the van der wal radii of the OH groups, the bonding distances between the OH-H (which in itself changes depending on other factors such as temperature) and C-C molecules, and the energy change from going from anti to gauche and weigh the energy changes against each other.

This isnt exactly common sense. Neither is the fact that the two OH groups even form hydrogen bonds in that structure. Hydride and methly shifts are simple concepts...form a carbocation...move a hydrogen/methyl group to make a more stable carbocation. This isn't something that follows the simple newman projection concepts. Either that, or I guess I didn't know I was supposed to memorize all of the bonding distances, energies, and the know how.

If we removed the two hydrogens opposite each hydroxyl group, and replaced them with methyls, would the gauche conformation still be the most stable?

 

[doc3232]

What you said doesn't make any sense at all. It's easy to make up explanations when you know the correct answer. Why does the "flexibility?" of a Hydrogen bond radius and covalent bond radius have anything to do with this? The bonding raidus of a substituent group differs from group to group, hence two methyl groups causes strain next to each other in gauche conformation such as for butane, while the hydrogens don't, is because their distance is less than the sum of their van der wals radii. In order to solve this, you'd have to know the amount of energy removed from hydrogen bonding of the O and H attached to the other O (which, I can tell you for sure off the top of my head, is no where near 155kj/mol), the van der wal radii of the OH groups, the bonding distances between the OH-H (which in itself changes depending on other factors such as temperature) and C-C molecules, and the energy change from going from anti to gauche and weigh the energy changes against each other.

This isnt exactly common sense. Neither is the fact that the two OH groups even form hydrogen bonds in that structure. Hydride and methly shifts are simple concepts...form a carbocation...move a hydrogen/methyl group to make a more stable carbocation. This isn't something that follows the simple newman projection concepts. Either that, or I guess I didn't know I was supposed to memorize all of the bonding distances, energies, and the know how.

If we removed the two hydrogens opposite each hydroxyl group, and replaced them with methyls, would the gauche conformation still be the most stable?

You are really overanalyzing it.

 

[kema]

What you said doesn't make any sense at all. It's easy to make up explanations when you know the correct answer. Why does the "flexibility?" of a Hydrogen bond radius and covalent bond radius have anything to do with this? The bonding raidus of a substituent group differs from group to group, hence two methyl groups causes strain next to each other in gauche conformation such as for butane, while the hydrogens don't, is because their distance is less than the sum of their van der wals radii. In order to solve this, you'd have to know the amount of energy removed from hydrogen bonding of the O and H attached to the other O (which, I can tell you for sure off the top of my head, is no where near 155kj/mol), the van der wal radii of the OH groups, the bonding distances between the OH-H (which in itself changes depending on other factors such as temperature) and C-C molecules, and the energy change from going from anti to gauche and weigh the energy changes against each other.

This isnt exactly common sense. Neither is the fact that the two OH groups even form hydrogen bonds in that structure. Hydride and methly shifts are simple concepts...form a carbocation...move a hydrogen/methyl group to make a more stable carbocation. This isn't something that follows the simple newman projection concepts. Either that, or I guess I didn't know I was supposed to memorize all of the bonding distances, energies, and the know how.

If we removed the two hydrogens opposite each hydroxyl group, and replaced them with methyls, would the gauche conformation still be the most stable?

I read your lengthy response to the question but I must ask, why is the Anti not the most stable still? Isnt it the least sterically hindered formation of the three please explain further, thank you.