How to assign configuration to a newman projection?

[skyisblue]

---

Members don't see this ad.

How do you assign configuration to a newman projection without using models??

There's an attachment. Thanks to everyone being so patient with me.👍

 

[Streetwolf]

Start with the carbon backbone going vertical, then assign horizontal spaces based on what you should have - R or S.

 

[skyisblue]

Start with the carbon backbone going vertical, then assign horizontal spaces based on what you should have - R or S.

Sorry for being slow, but I don't understand...

 

[Lonely Sol]

What are the answers, is it
A. R
B. S

 

[skyisblue]

What are the answers, is it
A. R
B. S

The answer is:

A. S
B. R

Help!

 

[Streetwolf]

Oh sorry I was thinking Fischer.

For A, you want H in the back, so you look at it from the left side. The Cl is on the top, the rest of the molecule is on the left, and the CH2 is sorta down. So you start on top, go left, then go down. That's counterclockwise = S.

For B, you want H in the back, so again we'll look from the left. Remember we want the carbon in the back (the one in the front is achiral). We have to rotate it in our minds so that the H is on the right side of the drawing. When we do that, we have the OH on top and the CH2 jutting out towards us (from our left side perspective) and a bit down. The rest of the molecule is to our right. So the OH is first, the rest of the molecule second, the CH2 third. Start at the top, then the right, then the bottom. It's clockwise = R.

 

[skyisblue]

Oh sorry I was thinking Fischer.

For A, you want H in the back, so you look at it from the left side. The Cl is on the top, the rest of the molecule is on the left, and the CH2 is sorta down. So you start on top, go left, then go down. That's counterclockwise = S.

For B, you want H in the back, so again we'll look from the left. Remember we want the carbon in the back (the one in the front is achiral). We have to rotate it in our minds so that the H is on the right side of the drawing. When we do that, we have the OH on top and the CH2 jutting out towards us (from our left side perspective) and a bit down. The rest of the molecule is to our right. So the OH is first, the rest of the molecule second, the CH2 third. Start at the top, then the right, then the bottom. It's clockwise = R.

I guess there is just no other way to do this except to visualize it.....

Streetwolf do you mind taking a look at my other two questions??

Thanks!

 

[Lonely Sol]

Oh sorry I was thinking Fischer.

For A, you want H in the back, so you look at it from the left side. The Cl is on the top, the rest of the molecule is on the left, and the CH2 is sorta down. So you start on top, go left, then go down. That's counterclockwise = S.

For B, you want H in the back, so again we'll look from the left. Remember we want the carbon in the back (the one in the front is achiral). We have to rotate it in our minds so that the H is on the right side of the drawing. When we do that, we have the OH on top and the CH2 jutting out towards us (from our left side perspective) and a bit down. The rest of the molecule is to our right. So the OH is first, the rest of the molecule second, the CH2 third. Start at the top, then the right, then the bottom. It's clockwise = R.

Yea it makes sense after you said that, I started looking at it like a 3D structure in my mind, this is where PAT comes in handy, loll!

 

[BodybldgDoc]

Yea it makes sense after you said that, I started looking at it like a 3D structure in my mind, this is where PAT comes in handy, loll!

yes I see it too!!!! Hooray for PAT skillzzz lol

 

[Molar Man]

Hi,
i think this is the easiest way to do this type of problem. get a piece of paper and draw out what i type so its clear.

you have to switch the newmans to fischers. this can be done as follows:
the c facing you in the newman is the top carbon in the fischer, the rear facing carbon in the newman is the bottom carbon in the fischer.
to convert:1)take the front C from the newman and transfer the groups attached to it (as they are-in the same places) to the fischer. 2) the top two groups on the back carbon on the newman have to be switched[rotated] when put on the newman(to visualize: revolve it 180 degrees using the bottom facing group as the axis, so what was right is now left and what was on the left is now on the right). ok this is a bit confusing written out so heres an example :

for (a) the completed fischer (should have two parallel horizontal lines with one vertical perpendicular line cutting through both of them) has on the top carbon pointing up: Cl ,right: H, Left:CH3. On the bottom carbon: pointing down:H, Right: H, Left: CH3.

For (B) top Carbon: Up:H, Right:CH3, Left:CH3. Bottom Carbon: pointing down:H, Right:CH3, Left: OH.

now you just assign R or S for the fischer, if you need help with that i can post 2 easy rules for you as well.

i hope this was helpful to you, i got this technique from the destroyer. Remember if you are confused just draw it out, the 3 seconds itll take you to get a definite answer will be well worth the(possibly) confused 10 seconds it would have taken you. with a bit of practice you can do these problems very quickly.

 

[skyisblue]

Hi,
i think this is the easiest way to do this type of problem. get a piece of paper and draw out what i type so its clear.

you have to switch the newmans to fischers. this can be done as follows:
the c facing you in the newman is the top carbon in the fischer, the rear facing carbon in the newman is the bottom carbon in the fischer.
to convert:1)take the front C from the newman and transfer the groups attached to it (as they are-in the same places) to the fischer. 2) the top two groups on the back carbon on the newman have to be switched[rotated] when put on the newman(to visualize: revolve it 180 degrees using the bottom facing group as the axis, so what was right is now left and what was on the left is now on the right). ok this is a bit confusing written out so heres an example :

for (a) the completed fischer (should have two parallel horizontal lines with one vertical perpendicular line cutting through both of them) has on the top carbon pointing up: Cl ,right: H, Left:CH3. On the bottom carbon: pointing down:H, Right: H, Left: CH3.

For (B) top Carbon: Up:H, Right:CH3, Left:CH3. Bottom Carbon: pointing down:H, Right:CH3, Left: OH.

now you just assign R or S for the fischer, if you need help with that i can post 2 easy rules for you as well.

i hope this was helpful to you, i got this technique from the destroyer. Remember if you are confused just draw it out, the 3 seconds itll take you to get a definite answer will be well worth the(possibly) confused 10 seconds it would have taken you. with a bit of practice you can do these problems very quickly.

Can you post those rules for fishcher? Thank you!👍