Sciencemadness Discussion Board - weird reactions

weird reactions

CrossxD - 25-1-2018 at 04:10

Hi guys I have 3 problematic reactions can you help me? please

in the frist one: radical substitution - Br will substitute α (on ethyl and closer to benzene) hydrogen or other??

2: I dont have any idea what can happen

3: will there work addition-elimination mechanism (even when i have I^- and not M^-???

Dr.Bob - 25-1-2018 at 05:58

Why don't you propose some ideas and then we can review them. Or explain what the pros and cons are of each possible reaction or product. Most people here don't want to do the work for other people without them trying first.

ie, in 1) the alpha proton may substitute more due to X and less due to Y, but the beta position will be more reactive due to z.

CrossxD - 25-1-2018 at 06:24

ok thanks,

I think that in 2nd case palladium is good for attracting hydrogens and next to nitrogen (in cycle) is saturated C-C bond so it might rip off those hydrogens and make more stable aromatic heterocycle.....?
and in third case i think it should work because on meta carbon should be parcial positive carge, am I wrong?

CxD

LearnedAmateur - 25-1-2018 at 06:47

Quote: Originally posted by CrossxD

ok thanks,

Pd/C is used as a catalyst for reduction using H2, try have a look at all of the atoms around the molecule and see which ones can be reduced. Also consider that aromatic systems cannot be reduced via this method at low pressure (is there any indication as to whether it is high or low?), and that a carbon atom saturated with hydrogen cannot be reduced further.

For number 3, I’ll give you two hints. Nucleophilic aromatic substitution, and Schiff base. Technically three, if I tell you that these two occur separately.

[Edited on 25-1-2018 by LearnedAmateur]

CrossxD - 25-1-2018 at 07:06

I know that saturated carbon cannot be reduced but i dont have H2 in that reaction.... so I thought that palladium would dehydrogenate at high temperature and make double bond and h2 and that bond would participate in cojugated double bond system to stabilize it

number 3... I get it

i forgot about Schiff there will be imine right?

CxD

Sigmatropic - 25-1-2018 at 09:11

The 3rd example is perhaps proceeding through a benzyne derivative. Noting that the acetyl group is in the metaposition it does not allow for a resonance stabilized meisenheimer complex (a.k.a. Sigma complex), although it could proceed through one, but it simply wouldn't be stabilized by resonance.
http://www.ochempal.org/index.php/alphabetical/a-b/benzyne-m...
http://www.ochempal.org/index.php/alphabetical/s-t/snar-mech...

[Edited on 25-1-2018 by Sigmatropic]

Melgar - 25-1-2018 at 19:08

Pd/C implies hydrogenating conditions. I guess that's one of those things you learn with experience. There just isn't much use for Pd/C without hydrogen present.

For the first one, the conditions clearly indicate this will be a free radical reaction. However, what usually happens is bromine will pull off a hydrogen to form HBr, then the other bromine will occupy its place. I guess I'm not sure if that counts as a substitution or not.

[Edited on 1/26/18 by Melgar]

Cryolite. - 25-1-2018 at 19:14

Quote: Originally posted by Melgar

Pd/C implies hydrogenating conditions. I guess that's one of those things you learn with experience. There just isn't much use for Pd/C without hydrogen present.

Incorrect. Note that in the reaction drawn above the temperature is extremely high and no hydrogen is given as a reagent. This could not possibly be a hydrogenation reaction.

Instead, at high temperatures palladium on carbon functions as a dehydrogenating reagent. It converts cyclohexenones to phenols, and (usefully in this case) dihydropyridines to pyridines. Thus, the product of the corresponding reaction is the corresponding isoquinoline plus a mole of hydrogen.

[Edited on 26-1-2018 by Cryolite.]

Melgar - 3-2-2018 at 02:11

I have literally never seen a single reaction in literature that was done with Pd/C without hydrogen.

But I guess there's a first time for everything.