DIBALL
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How to find out, which of 2 products are more likely to form?
So if I have a reaction, where 2 products can be formed and only 1mol of the reactant is given to 1mol of the other.
Example:
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An alkane with a chlorine group and on the carbon next to it is a ketone,
which are reacting with MeNH2:
Cl O
. ..
Rn-C-C-C-C-Rn + CH3NH2 →
Predicted products:
Cl CH3N
. ..
1. Rn-C-C-C-C-Rn + H2O
Imine formation
OR
CH3NH O
. ..
2. Rn-C-C-C-C-Rn + HCl
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Sorry for the crappy "painting" (just imagine the hydrogen atoms) and I know that both reactions can happen on the same molecule as well. Did I miss
something? Everything correct?
I know that I can use the rate equation, but since I don´t have a table/graph for this reaction, I can´t get the rate constants. Or can I?
My second thought was to compare the pKA values and see which one is more basic or acidic compared to the reactant and the winner was the
halogen group.
So can I use the pKA´s to find out, what product will form more likely?? Stronger base + stronger acid = more/faster reactions between
those compared to the weaker base?
Or is there an easier and better way to find out, without doing the actual experiment?
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theAngryLittleBunny
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I would say the second one would happen, yeah I'am pretty sure, because the first one is reversable, the second one is not. That way you can also make
cathinones, where you have a alpha halo propiophenone and react it with an amine.
But it often also depends on which haloketone you use, if you have hydroxide ions around, some haloketones make the so called Faworski rearrangement,
where they rearrange into a carboxylic acid, maybe some haloketones would do that with amines around too. So yeah, to sum up, I'am pretty sure it's
the first one, or mayyyybe something completely different, I hope that was helpful.
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DIBALL
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Thanks!
Sadly this isn´t what I was hoping for... I want to do a gringard reaction with the imine, but the Cl should stay. Isn´t there a formula for such
reactions with many possible products? I mean some formula, that tells how likely one group reacts with the other? Also I´m not sure, if the Cl might
be a problem during the gringard reaction? If I use 1mol magnesium for 1mol of the substance (which should be the gringard reagent) then there
shouldn´t be a problem or can some other side reaction occur with the halogen group, when I react it with the gringard reagent?
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theAngryLittleBunny
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Oh you wanted to make an imin and make a grignard reaction with that, that won't really happen tho. A halogen next to a carbonyl is completely
different from a normal halogen group. The carbon attached to the halogen is much more electrophile because of the carbonyl next to it, so the
grignard reagent will definitly attach there. I'am honestly not even sure if a alpha chloro imine exists.
And I don't really think such a formula exists, every group influences the molecule, so it's extremly complicated, and you need computer programs
which calculate stuff like that. I would suggest you to tell the people here what you wanna make, and they can help you to find another way. I know,
chemistry is pretty harsh sometimes.
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DIBALL
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Could I solve this problem, if I turn the halogen with NaOH to an alcohol and attach a protective group? Or is there still some "electrophilic" issue,
if the resulting ester is next to an imine? I thought about adding benzyl alcohol as a protection group and then hydrogenate it. If I get rid of the
H+ of the benzyl alcohol with NaOH, I probably can swap it with the halogen without turning it to an alcohol first?
Can you suggest a good program for such calculations?
[Edited on 2-5-2018 by DIBALL]
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JJay
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I am only speculating here, but I think it really depends on how fast water is removed from the reaction. Product 1 forms much faster than product 2
(right?), but the reaction is reversible, so if water is allowed to hang around and sufficient heat/pressure are maintained, product 2 might be the
dominant product. If the reaction conditions are mild and water is removed as it forms, I'd expect product 1 to be the dominant product. The best way
to be really sure is to do some experiments and find out.
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theAngryLittleBunny
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Quote: Originally posted by DIBALL  | Could I solve this problem, if I turn the halogen with NaOH to an alcohol and attach a protective group? Or is there still some "electrophilic" issue,
if the resulting ester is next to an imine? I thought about adding benzyl alcohol as a protection group and then hydrogenate it. If I get rid of the
H+ of the benzyl alcohol with NaOH, I probably can swap it with the halogen without turning it to an alcohol first?
Can you suggest a good program for such calculations?
[Edited on 2-5-2018 by DIBALL] |
If you touch this stuff with NaOH, it will make a Faworski rearrangement. A halogen next to a carboneö behaves completely different then a normal
halogen group. The base won't replace the chloride, it will deprotonate the carbon on the other side of the carbonyl, making the faworski
rearrangement.
And I think JJays approach has also it's problems, because I'am sure the imine is still going to attach the Cl Group of another molecule, making some
weird cyclic product.
[Edited on 2-5-2018 by theAngryLittleBunny]
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JJay
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As in any organic reaction, I do think there will be side products (including polymers and others not mentioned); the question is how much of these
side products will form. Alkyl halides with lighter atomic weight halides don't alkylate amines easily (nor imines, right?).... I'm not sure why this
is being referred to as "my" approach, but anyway....
I will say that I know of at least one synthesis that is actually very easy to do despite rather widespread claims that it is impossible due to [much
slower] competing reactions. In organic chemistry, there are *always* competing reactions. I would take claims that such and such reaction is
impossible due to competing reactions with a grain of salt unless there is a clear theoretical argument (with mechanisms and reaction rate
information) or experimental evidence offered as well.
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