[mass]

Can someone take me through how to do it. I don't get it at all, I have read and read for ever and still don't get it.

[sjb]

OK, here is a simple chiral molecule, butan-2-ol to be exact.



Take us through e.g. the protocol at http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch07/ch7-6.html, so we can begin to understand where your problems lie.

[mass]

OK, here is a simple chiral molecule, butan-2-ol to be exact.



Take us through e.g. the protocol at http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch07/ch7-6.html, so we can begin to understand where your problems lie.

ok basically I know that you are meant to put them in order of increasing atom size. So the H would go where the CH3 went right. The highest priority is H0, then CH2CH2 and then CH3. But I dont get where to place the molecules and so I always end up getting an R system.

[sjb]

OK.

Here are the two enantiomers of butan-2-ol


I appreciate it's hard on a forum like this, but can you show how you're moving the groups around to "always end up getting an R system"

Can you visualise what happens if you tip the hydrogen away from you?

Do you have a molecular model kit? Sometimes it's easier to play around with one of these. Remember, you can't break any of the bonds, so moving for instance the ethyl group by exchanging it with the hydroxyl such that the ethyl now has the dash and the hydroxyl a flat bond is not allowed.


S

[mass]

OK.

Here are the two enantiomers of butan-2-ol


I appreciate it's hard on a forum like this, but can you show how you're moving the groups around to "always end up getting an R system"

Can you visualise what happens if you tip the hydrogen away from you?

Do you have a molecular model kit? Sometimes it's easier to play around with one of these. Remember, you can't break any of the bonds, so moving for instance the ethyl group by exchanging it with the hydroxyl such that the ethyl now has the dash and the hydroxyl a flat bond is not allowed.


S

well the first enantiomer is s right?. OH to CH3CH2 to CH3. The other one is R.

I need just one enantiomer to show what I mean, finding it hard to explain.

[sjb]

Just one enantiomer?

OK, how about the enantiomer of Pencillin G I showed you when we were discussing double bond equivalents, over at http://www.chemicalforums.com/index.php?topic=25196.msg95257#msg95257  ?

How would you assign the chiral centres there?

[english]

You always start with the highest priority group (typically the group connected to the chiral center with a high molecular weight atom), then move in a clockwise or counterclockwise fashion from that atom to the second highest priority group.

In doing so the highest priority group must be coming out at you (a "wedge" line) and the lowest priority group must be going away from you (a "dashed" line). 

An easy way of looking at it is by always moving from the highest priority group to the second highest priority group; if the highest priority group is not a "wedge," or in some cases the lowest priority group is not a "dash," then you have to think in opposites.

Using the former example given,



The highest priority is OH, and the lowest priority is H.  You have to move clockwise or counterclockwise by going from OH to CH3CH2, the second highest priority group.  However, since H is not a dash as it should be (lowest priority, review statements above) whatever notation you deem the chiral center the answer will be opposite to that.

Moving from OH to CH3CH3 yields an S conformation (it looks like S), yet it is actually R because the H is not a dash as it should be.

 

A rule of thumb is when moving from the highest priority to second highest priority, you can always move over a dashed bond or solid bond (just a line) on your way but never over a wedge bond.

[Structuralist]

This may hepl

[mass]

g english, I kind of get it better now, except I am only confused where and mainly  I have to put the dash and wedge and what atoms I have to do this for.

For example. the atom you have used:
where would I have to put the H, wouldn't I have to move it as its the lowest priority and with the fact there is a wedge coming out. Thats where my confusion lies for these molecules, do I need to rearrange the molecules when they are done wrongly (e.g. highest priority having a dashed line instead of a wedge)

[mass]

An easy way of looking at it is by always moving from the highest priority group to the second highest priority group; if the highest priority group is not a "wedge," or in some cases the lowest priority group is not a "dash," then you have to think in opposites.





Right I have finally got it, I have used this rule and it seems to be working, except one last question. What happens if both are drawn wrong, i.e. the highest priority is not a wedge and the lowest priority is not a dash?.

[english]

Right I have finally got it, I have used this rule and it seems to be working, except one last question. What happens if both are drawn wrong, i.e. the highest priority is not a wedge and the lowest priority is not a dash?.

You rotate the molecule until its stereochemistry suits your liking (use models to your advantage)...

Take the following example (Et = ethyl)...where the highest priority is an OH and the second highest priority is the Et group.  The OH however is not initally a "wedge" and the H (being the lowest priority) is not initially a "dash."



The structure is S because the lowest priority is not a "dash" even after this particular rotation...even though the highest priority is a "wedge."  Of course, if you could be bothered enough, you could rotate it to the point where OH was a wedge and H was a dash, but this maneuver isn't easy without a model.

Use models!

[christina]

You always start with the highest priority group (typically the group connected to the chiral center with a high molecular weight atom), then move in a clockwise or counterclockwise fashion from that atom to the second highest priority group.

In doing so the highest priority group must be coming out at you (a "wedge" line) and the lowest priority group must be going away from you (a "dashed" line). 

An easy way of looking at it is by always moving from the highest priority group to the second highest priority group; if the highest priority group is not a "wedge," or in some cases the lowest priority group is not a "dash," then you have to think in opposites.

Using the former example given,



The highest priority is OH, and the lowest priority is H.  You have to move clockwise or counterclockwise by going from OH to CH3CH2, the second highest priority group.  However, since H is not a dash as it should be (lowest priority, review statements above) whatever notation you deem the chiral center the answer will be opposite to that.

Moving from OH to CH3CH3 yields an S conformation (it looks like S), yet it is actually R because the H is not a dash as it should be.

 

A rule of thumb is when moving from the highest priority to second highest priority, you can always move over a dashed bond or solid bond (just a line) on your way but never over a wedge bond.

I have a question...
OH=> 17.0074 g/mol
CH2CH3=> 29.0613 g/mol

so how is it that you say that OH is a higher priority group
______________________________________________________________
My take on this problem of yours:

I learned it as, if the lowest priority group is in the wrong place( the H or etc.) you make a switch on paper, but after doing that you have to make sure everytime you make a switch you switch the configuration in your head of course...

like for example the molecule here below: (from above)


you make one switch for the Hydrogen and thus it goes on the top.

But what is the configuration now? It is R but since you made 1 switch it is the opposite thus, S.
If you make more than one switch remember to also remember how many switches you do make so you can figure out what is the actual configuration(don't forget).

[sjb]

I have a question...
OH=> 17.0074 g/mol
CH2CH3=> 29.0613 g/mol

so how is it that you say that OH is a higher priority group

It has little to do with molecular weight, more to do with atomic number of the atoms directly (or further afield if necessary) attached to the chiral centre. You have ₈O, and ₆C, so the oxygen is higher priority.

In the alkyl chains, you have 2 x C directly attached to the centre, but one is only attached to 3 x ₁H, the other 2 x ₁H and 1 x ₆C, hence the latter is higher priority.

My take on this problem of yours:

I learned it as, if the lowest priority group is in the wrong place( the H or etc.) you make a switch on paper, but after doing that you have to make sure everytime you make a switch you switch the configuration in your head of course...

like for example the molecule here below: (from above)


you make one switch for the Hydrogen and thus it goes on the top.

But what is the configuration now? It is R but since you made 1 switch it is the opposite thus, S.
If you make more than one switch remember to also remember how many switches you do make so you can figure out what is the actual configuration(don't forget).

There is that option as well. I usually forget how many switches I've made, so don't like it and is why I was trying to veer away from that way to do it.

S

edit damn forum formatting tags 

[mass]

It's easier to use the whole its opposite rule rather than rotating in your head surely?

[sjb]

It's easier to use the whole its opposite rule rather than rotating in your head surely?

For some people, perhaps, just not me

There's always the danger you may try and skip steps and "always end up getting an R system"