Fischer

[emminent]

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I don't understand how the C attached to the Br and COOH is R.

I ranked Br as 1; COOH as 2: and the C 3 as 3. And since H is on the side I reversed from R to S. Isn't a COOH a higher priority than C3 with the Cl and CH3 attached?

 

[TheDanger]

The picture isn't working for me.

 

[emminent]

Sorry about that. Should work now.

 

[jwan14]

I'm thinking that the Cl-CH-CH3 is second priority. Aren't priorities assigned based on weight? So you're basically comparing O to Cl and Cl would win out making that C the 2nd priority.

 

[TheDanger]

Hmm, I do agree that your priorities are correct. I think COOH gets a higher priority than C3 because COOH has 3 bonds to O, whereas C3 has 1 bond to Cl, 1 bond to H, and 1 bond to C.

When you have the lowest priority as a wedge in a Fischer diagram, I don't think you can simply switch between R and S. I think you have to swap the lowest priority molecule with a dashed molecule, and then determine the configuration. I'll attached a picture since my explanation may be confusing.

 

[TheDanger]

 

[TheDanger]

Ohhh snap actually...I'm not sure if my explanation is correct because if you swap 2 of the groups, you have to swap the other 2 as well, which would lead to S configuration. 😕 Is it possible the answer key has a mistake?

 

[emminent]

Lol I don't know. I guess we'll just wait for some more input. I think you might be right on the weight portion. Answer is C though.

 

[KyoPhan]

Priority will be based on atomic number of attachment.

Since the C3 has a Cl attached, and C1 has O attached, the C3 will be higher priority. It doesn't matter if there are 2, or even 3 O's attached. Whichever one has the single one with the highest atomic number attachment will take priority.

1) Br
2) C-Cl~~~
3) COOH

It looks like S, but since H is on horizontal, it will be R.

 

[jwan14]

Do you have Chad's videos? I'm pretty sure that he said it's not the sum total of what atoms are attached to the carbons but it's the first different atom. So going away from the central C you have the C of the COOH group and then the C of the ClCHCH3 group. So that's a tie because they are both carbons. Then you go to the next atom attached to those carbons. The next one attached to the COOH group is the O and the next one attached to the other group is the Cl. Now you compare those two. Cl is larger in size and this beats out the O according to stereo priorities. The C with the Cl is therefore the next priority group after the Br. That's how I learned it haha maybe I'm wrong but that's what I remember Chad teaching

 

[jwan14]

So you don't compare all three atoms attached you just compare the atoms where there you find the first difference. That's probably all hard to follow haha I couldn't think of a better way to explain

 

[TheDanger]

Looks like I have to rewatch Chad. Thanks for the correction, guys!