[kapital]

If I want to draw  (2S)-2-methyltetrahydro-2H-pyran
(chair conformation), how can I now, is methyl grup axial or equatorial?
(I know both(two)conformations exist, one is more stable, but I woud like to know hot to trasform S,R nomenclature in axial and equatorial)


thanx

[discodermolide]

If I want to draw  (2S)-2-methyltetrahydro-2H-pyran
(chair conformation), how can I now, is methyl grup axial or equatorial?
(I know both(two)conformations exist, one is more stable, but I woud like to know hot to trasform S,R nomenclature in axial and equatorial)


thanx

The methyl group is equatorial, when you ring flip it becomes axial. The equatorial isomer is preferred.
See my picture.

[kapital]

OK, but how do you know that methyl group in the first conformer is equatorial?

[discodermolide]

OK, but how do you know that methyl group in the first conformer is equatorial?

Because if it were axial it would be the (R) isomer!

[kapital]

Yes, but I woud like to know how can you tell that. I know that stereochemistry is shwon with black or dashed bonds, but here this is not the case. So how can you know that?

[discodermolide]

Yes, but I woud like to know how can you tell that. I know that stereochemistry is shwon with black or dashed bonds, but here this is not the case. So how can you know that?

Work out the absolute configuration of both compounds using the Cahn-Ingold-Prelog rules and you will see. See pictures.

[kapital]

Yes, but what I woud like to know, is how do you know that this two structures are equivalent?

[discodermolide]

Yes, but what I woud like to know, is how do you know that this two structures are equivalent?

In The right structure the methyl group points up, in the left structure the methyl group is pointing up, i.e. equatorial.
If you can make a model you will see this. Look at the absolute configuration using the Cahn-Ingold-Prelog system i.e. (R) or (S) and you will see they are the same molecule.
Perhaps this picture will help?

[fledarmus]

I see what Kapital is saying. Without the dashed and solid bonds, there is really no way of telling in the chair conformation view of the tetrahydropyran whether the bond to oxygen is supposed to be going into the paper or out of it. To be truly accurate, in the chair drawing, the bond to O should be a dashed bond, and the bond to the ring carbon should be a wedged bond. The equatorial methyl and the hydrogen are in the plane of the paper.

[Jasim]

Looking at the ring straight on, with the dark wedge to the methyl that indicates the methyl group is coming out of the page, likewise the dashed line on the hydrogen indicates the hydrogen is going into the page.

You can't tell anything regarding the conformation from the drawing that is looking at the ring as a hexagon. The drawing on the left is showing stereochemistry for that conformation, the drawing on the right is not.