A few questions regarding sn2

[510586]

So I know that you can't do sn2 on a bridgehead carbon on a bycyclo compound, but can you do it on all the other carbons? Also, a tertiary halide with a bad leaving group cannot undergo sn2 but what about a very great leaving group such as I?

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[zEkii]

Just to clarify some points that are important. "Also, a tertiary halide with a bad leaving group cannot undergo sn2"... A halide is an excellent leaving group, the problem is that it won't undergo SN2 because of its halide attachment to a carbon to three other carbons. I is the most ideal leaving group for SN2, in order to do a SN2 reaction, the Halide must be on a primary or methyl group to continue.

The reason why you can't put it on a bycyclo compound is because of the tertiary compound. You might be able to do a SN2 on another source of compound, but that is pushing it... Especially since it is tertiary, thus, you must look for the nucleophile, if it is strong or not.

Make sure you read up on the criteria about SN2 vs E2's!

 

[510586]

Just to clarify some points that are important. "Also, a tertiary halide with a bad leaving group cannot undergo sn2"... A halide is an excellent leaving group, the problem is that it won't undergo SN2 because of its halide attachment to a carbon to three other carbons. I is the most ideal leaving group for SN2, in order to do a SN2 reaction, the Halide must be on a primary or methyl group to continue.

The reason why you can't put it on a bycyclo compound is because of the tertiary compound. You might be able to do a SN2 on another source of compound, but that is pushing it... Especially since it is tertiary, thus, you must look for the nucleophile, if it is strong or not.

Make sure you read up on the criteria about SN2 vs E2's!

I think destroyer said that Fluorine was not a good leaving group.

 

[zEkii]

I think destroyer said that Fluorine was not a good leaving group.

Here is a better schematic on what they mean: I>Br>Cl. All of these are good leaving groups, fluorine is relatively not a good leaving group compared to those three.

 

[510586]

Here is a better schematic on what they mean: I>Br>Cl. All of these are good leaving groups, fluorine is relatively not a good leaving group compared to those three.

Oh ok yea the answer to the question in the back was tertiary halides with a bad lg cannot do leaving group so I was wondering if there were exceptions like a super good nucleophile. Doubt it's important. Thanks!

 

[DAT Destroyer]

So I know that you can't do sn2 on a bridgehead carbon on a bycyclo compound, but can you do it on all the other carbons? Also, a tertiary halide with a bad leaving group cannot undergo sn2 but what about a very great leaving group such as I?

interesting questions. Yes....a bicyclo compound may react by an SN2 reaction at another carbon distinct from the bridgehead. Leaving groups at the bridgehead carbon do NOT do SN1 nor SN2. Lets see why..... No SN2......too hindered to do the backside attack...... No Sn1 can occur either.....catbocations need to be sp2...planar......the molecule is too RIGID and cannot flatten out. Now for your next question......A tertiary halide or pseudo-halide such as Tosylate cannot do SN2 on a tertiary halide.....Even though they are great leaving groups... the nucleophile is still unable to attack the antibonding orbital that lies at the rear of the carbon.

Hope this helps.

Dr. Romano

 

[510586]

interesting questions. Yes....a bicyclo compound may react by an SN2 reaction at another carbon distinct from the bridgehead. Leaving groups at the bridgehead carbon do NOT do SN1 nor SN2. Lets see why..... No SN2......too hindered to do the backside attack...... No Sn1 can occur either.....catbocations need to be sp2...planar......the molecule is too RIGID and cannot flatten out. Now for your next question......A tertiary halide or pseudo-halide such as Tosylate cannot do SN2 on a tertiary halide.....Even though they are great leaving groups... the nucleophile is still unable to attack the antibonding orbital that lies at the rear of the carbon.

Hope this helps.

Dr. Romano

Thank you once again Dr. Romano!