easy for ochem people

[BoneMental]

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Blah, some ochem things I am struggling with. Blah! I'd appreciate some halp.

1)Can someone tell me why a tertiary alkyl chloride would undergo Sn1 with a Bromide ion? Isn't the chloride a much weaker leaver group? Wouldn't there be no reaction?

2) When assigning R/S on a wedge/dash/dotted line representation, I thought that following rules worked:
-Assign priorities to your substituents
-When lowest priority group is in the back, trace your substituents from highest to lowest. Clockwise is R, counterclockwise is S
-When lowest priority group is not in the back, trace your substituents from highest to lowest. Clockwise is S, counterclockwise is R.

This usually works....but sometimes it doesn't. Why is this?

Thank you, kind people. :]

 

[ridethecliche]

1) Leaving groups can easily leave Tertiary carbons as the carbocation is stable. So they spontaneously dissociate and the other nuc can attack. There will be a reaction.

2) I suck at stereochemistry so I'll let someone else answer that.

 

[loveoforganic]

Your second rule is "lowest priority in front" not "lower priority not in back"

 

[sonnybobiche]

Lowest priority group goes in the back, then trace your substituents from highest to lowest priority. Clockwise is R.

The best way to deal with having a lowest priority group in the front is to ROTATE THE MOLECULE IN YOUR HEAD and trace your substituents from highest to lowest priority. Clockwise is R.

 

[orgohacks]

Blah, some ochem things I am struggling with. Blah! I'd appreciate some halp.

[snip]

This usually works....but sometimes it doesn't. Why is this?

Thank you, kind people. :]

It sometimes doesn't work because your lowest priority group is on the SIDE (i.e. not a dash or wedge). In these cases it may fail. If you have the spacial skills, try to rotate it in your head (imagine looking at the molecule from the side). If not, you can always try using the "single swap rule" to put it in the back first, and then figure out whether it's R or S.

http://masterorganicchemistry.com/2011/01/24/the-single-swap-rule/

hope this helps - James

 

[plzNOCarribbean]

....when the lowest priority is not in the back and IN THE PLANE (solid line, not dashed or wedged) name the compound as you normally would and then flip the designation. For instance, if the chiral center was (R) when you named it but the lowest priority was not in the back, instead it were on a wedge or in the plane of the board, you would simply flip the designation to (S). This always holds true.

 

[Catburr]

Here's how I figure out R and S, if the molecule is too complicated for my mental rotating skills (doesn't take a very complicated molecule to beat my skills, admittedly) without having to worry about whether my lowest priority substituent is in the plane or the front or the side or whatever...

If the lowest priority substituent isn't already in the back, then find it and SWITCH it with whatever is in the back. Then go about the clockwise/counterclockwise numbering. But whatever you get, R or S, the ORIGINAL molecule (before you did the switch) has the opposite stereochemistry.

I just noticed that someone mentioned the single swap rule already. So uh, I second that, I guess. Hopefully my explanation of it is useful to someone...

 

[BerkReviewTeach]

....when the lowest priority is not in the back and IN THE PLANE (solid line, not dashed or wedged) name the compound as you normally would and then flip the designation. For instance, if the chiral center was (R) when you named it but the lowest priority was not in the back, instead it were on a wedge or in the plane of the board, you would simply flip the designation to (S). This always holds true.

Actually, when the #4 priority is in the plane, it doesn't always work as you mentioned. In fact, it works half of the time. A better approach is to (1) exchange #4 for whatever substituent is in back, (2) determine whether the altered structure you created by placing #4 in back is R or S, and (3) take the opposite of what you just determined for the altered structure and assign that to the original structure.

The idea is that whenever you exchange two substituents on a stereocarbon, the configuration changes. Switching #4 and the substituent in back changes the configuration, so whatever you get is inverted from the original structure.

Edit: just acknowledging that Catburr had already mentioned this and I was late to the party and replied before reading thier response.

 

[SmurfinUSA]

stick your thumb down the direction of the H and let your hand curl in the order of substituents, and if your right hand works, R if your left hand works its S

haha works everytime.

 

[BerkReviewTeach]

stick your thumb down the direction of the H and let your hand curl in the order of substituents, and if your right hand works, R if your left hand works its S

haha works everytime.

You must be taking the BR class.

 

[AnastomoSteve]

Your second rule is "lowest priority in front" not "lower priority not in back"

👍

If the lowest priority is coming out of the plane (wedge) you need to reverse the configuration. The rest of your logic seems to be correct.