When do you consider H-bonding in cyclohexane/butane stability?

[Mantis Toboggin]

From what I understand the presence of substituents such as -OH allows for h-bonding with h groups.

For example if we have cis 1,3 cyclohexane, the 1,3 diaxial would be preferred to the 1,3 dieq due to H-bonding.

However if we consider something like R-2-methyl-1-butanol, the highest energy conformation is that which includes the CH2OH and C4 carbon eclipsed, whereas the SECOND highest energy conformation is where the two methyl groups are eclipsed.

Why wouldn't H-bonding occur between the OH there and the hydrogen of the methyl group in order to add stability? IN this case steric hindrance becomes a bigger factor, but what is the rule of thumb here?



Separate question: Berkley mentions in the case of trans 1,2 dimethyl cyclohexane there are two things to consider:

1.) Diequatorial having less strain than diaxial (the predominant factor for Keq being greater than 1)

2.) In opposition to point (1), Anti being more stable than gauche. The anti position belongs to the trans diaxial, the gauche belonging to the diequatorial. Is it only because they're on adjacent carbons that this entire comparison comes up?

My question is with regards to point 2, what exactly is the "anti" and "gauche" that they allude to here (equatorial has gauche whereas diaxial has anti, from what I gleaned).


Thanks a ton.

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[AureatEagle]

From what I understand the presence of substituents such as -OH allows for h-bonding with h groups.

For example if we have cis 1,3 cyclohexane, the 1,3 diaxial would be preferred to the 1,3 dieq due to H-bonding.

However if we consider something like R-2-methyl-1-butanol, the highest energy conformation is that which includes the CH2OH and C4 carbon eclipsed, whereas the SECOND highest energy conformation is where the two methyl groups are eclipsed.

Why wouldn't H-bonding occur between the OH there and the hydrogen of the methyl group in order to add stability? IN this case steric hindrance becomes a bigger factor, but what is the rule of thumb here?


Separate question: Berkley mentions in the case of trans 1,2 dimethyl cyclohexane there are two things to consider:

1.) Diequatorial having less strain than diaxial (the predominant factor for Keq being greater than 1)

2.) In opposition to point (1), Anti being more stable than gauche. The anti position belongs to the trans diaxial, the gauche belonging to the diequatorial. Is it only because they're on adjacent carbons that this entire comparison comes up?

My question is with regards to point 2, what exactly is the "anti" and "gauche" that they allude to here (equatorial has gauche whereas diaxial has anti, from what I gleaned).


Thanks a ton.

'
Just because you're on a ring doesn't mean the rules of H-bonding don't apply 🙂 There can't be hydrogen bonding between OH and H from a methyl group, because H-bonding can only take the form: X-H - - - - Y where X and Y are O, N, or F. Carbon isn't electronegative enough to polarize it's H to the extent that it can participate in hydrogen bonding.

Correct. When the molecule is 1,2-diaxial, the methyls are anti relative to each other, but both gauche relative to the ring. When the molecule is 1, 2- diequatorial, they're gauche relative to each other.