[discodermolide]

Compare with my drawing of this molecule.
The t-Butyl group cannot be in this position, axial due to unfavorable 1,3-diaxial interactions with the hydrogens.
The t-Butyl group must be in an equatorial position.

[discodermolide]

here are two pictures.
Cyclohex 01 is mine, cyclohex 02 is yours
Drawn with ChemDraw 3d, not energy minimized.

[Twickel]

Can someone please check that the one I have just done is correct?

Would be much appreciated.

[discodermolide]

Can someone please check that the one I have just done is correct?

Would be much appreciated.

Here is my interpretation.
Wedged bonds point up, hashed bonds point down. Start by drawing the cyclohexane ring as I do then put in the substituents according to the previous sentence.
Could you please elaborate just what it is that you do not understand?

[Twickel]

My groups are in the same postion as yours, just that they are shifted by one carbon, was what I did correct?

I do not understand how to draw the trans linked cyclohexanes

Thanks for the help.

[discodermolide]

My groups are in the same postion as yours, just that they are shifted by one carbon, was what I did correct?

I do not understand how to draw the trans linked cyclohexanes

Thanks for the help.

Your structure is not correct the methyl and the chlorine are both "up" in your structure the methyl group is "down" and the chlorine is down in your structure it should be up.

I don't understand what you mean by"trans linked cyclohexanes".

Start off by drawing the ring as I have done. See picture. Then put the substituents on it.
As I said before solid bolds mean the substitiuent is "up* wedged bonds it is "down", this applies to the first structure of latest question. Have a look at the second set of pictures and study the relationship between the various bonds in both structure representations.

[Twickel]

I thought that, because all the bonds are pointing in the same direction, then yo just have to be consistent when drawing the chair ( iin this make all the groups be below hydrogens)

Looking at the diagram you drew ( the latest one) all the cls are above the hydrogens, which is what I thought I did for the example I posted

Ahh, I see what I did, the tert butyl group is not above the H

Does it matter which carbon I start iwth on the chair. aka can I make any carbon number 1?

I always start with the bottom right corner being carbon one on the ring, but working like that I get the tert butyl group axial and thats not good.... can I just do a ring flip

[Twickel]

If I drew that, can I just do a ring flip and get the correct answer?

[discodermolide]

No I'm sorry your answer is wrong, ring flip will not help.
Again look at my pictures, and compare them with yours.
The other structure is a decalin. I've drawn it for you.
Have you got any molecular models? Or can you get some?

[Twickel]

Thank you, I do have a set and started building some, just one question, when you draw your chair conformations, which carbon the chair do you denote as number 1? I thought the top most center carbon on the hexagon corresponding to the bottom right corner of the chair as 1 worked all the time, obviously it does not.

[discodermolide]

Thank you, I do have a set and started building some, just one question, when you draw your chair conformations, which carbon the chair do you denote as number 1? I thought the top most center carbon on the hexagon corresponding to the bottom right corner of the chair as 1 worked all the time, obviously it does not.

It does not matter which is carbon number 1 on the unsubstituted ring.
I just know that a t-Butyl group will always be equatorial due to it's size, it also prevents ring flip. So if you have a t-Butyl group there are only 3 positions it can go on the run and be "up" i.e. equatorial.

[Twickel]

hmm, ok I knew that I just need to stop over complicating things maybe.

I  thought that the CH3 needs to be equatorial as well, or is that not considered bulky?

[discodermolide]

hmm, ok I knew that I just need to stop over complicating things maybe.

I  thought that the CH3 needs to be equatorial as well, or is that not considered bulky?

Sure it does if there is not a t-Butly group present. If the methyl group happens to be axial the ring can flip then it's equatorial.
Try and keep it simple, you got that right