[sharbeldam]

Is it true that atoms pointing upwards are always wedged? so if i look at the right carbon it should be R but since the hydrogen is wedged, it is S? that's the way to do it?

Thanks alot in advance

[sharbeldam]

Anyone? please

[AWK]

Both carbon atoms show the same configuration

[sharbeldam]

Thank you AWK, but can you tell me if the way i did it is true? i can consider the upward atoms wedged? always?

[AWK]

You can change the surrounding of carbon atom into a spatial view or typical Fischer view. For both cases, rules exist that allow you to assign the proper configuration unambiguously.

[sharbeldam]

to simplify this question, without changing the view of this molecule, can i consider this molecule ins the picture haworth projection which means that any upward atom is wedged and figure out the configuration simply by doing that?

[AWK]

Rules for configuration in Haworth projection works correctly for normal sugars only. They do not work for deoxysugars, for example.
You should show you attempt, according to Forum rules.

[sharbeldam]

I'm really sorry If I seem dense.
So since I cant consider this molecule a haworth projection, may I ask and if i get this question in the exam to figure out if the carbon is R or S. can you show me simple steps how can you tell if it's clockwise or counterclockwise? knowing that the 4th priority need to be going inside (Dashed)

I'd really appreciate it, thank you

[AWK]

First of all, rotate your picture by 180 degrees around the axis perpendicular to the plane of your picture. You can see that the configuration of both carbon atom is the same.
Then assign priorities for, eg left carbon atom - Cl>CHCl>CH₂>H. Now turn the carbon atom with its substituents so that the hydrogen atom is behind the carbon atom. Then look only at the priority of the three substituents over the carbon atom - if they are arranged clockwise - the configuration is R; if on the contrary - S.

[sharbeldam]

Thanks alot.

[hollytara]

to answer the wedged bond aspect of your question.... 

The figure you show is a Haworth projection.  It is a form of perspective drawing that portrays the ring as planar (which cyclopropane is, but other cycloalkanes like cyclohexane are not - still the Haworth portrays all rings as planar).  The ring is always drawn horizontal.  Substituents on the ring are shown with bonds that are vertical - they either go up above the ring or down below the ring.  Haworth projections are useful in that they show cis/trans relationships very clearly, but they do not show the true conformation of the ring or the substituents. 

If you convert a Haworth projection to a drawing with wedged bonds/ hashed bonds, you are changing your perspective from the side of the ring to above or below the ring.  It is easiest to change to above the ring.  In that case, the drawing will have all substituents above the ring as wedged bonds, because they come toward you when you are above the ring!  All substituents below the ring now go away from you so they are hashed bonds. 

It is possible to imagine changing your perspective to below the ring as drawn in the Haworth projection.  Then below the ring = wedges, above the ring = hashes.  But this takes a little more experise in visualization to make sure you get it right.