[darkdevil]

Greetings,

I performed the following amide synthesis reaction. I expected to get two product as seen below, but the MALDI result showed an intense peak of 475m/z after I purified my product with flash chromatorgraphy. Can any of you think of any side reactions/products that can be form which led to a peak of 475m/z ?? I repeated 2 times for the synthesis, and 475 is always the dominant peak.



I am sure some reactions must have taken place, because my starting material is blue in color and the final product is red. The red substance is soluble in ethyl acetate, acetone. while the starting material does not dissolve in any solvent due to the intermolecular hydrogen bondings.

Any ideas will be greatly appreciated! Thank you so much!

[discodermolide]

Cannot see your attachment!

[darkdevil]

Cannot see your attachment!

Im sorry! it's ok now

[orgopete]

Can you give complete details? There isn't enough information to determine whether another reactant contributed to the 475.

[darkdevil]

Can you give complete details? There isn't enough information to determine whether another reactant contributed to the 475.

HI,! I didnt add solvent for this reactions

I then work up the reaction with addition of chloroform and aqueous sodium bicarbonate (to remove unreacted acid bromide), followed by extraction with chloroform. I then precipitated the red solids with addition of hexane. The last step was flash chromatography to obtain an orange red product for its mass spec data.

[willug]

What does the NMR look like? Mass spec is not a good means of characterisation unless you have all of the other data to go with it.

Also, did you just mix the indigo with the butanoyl bromide? You might want to think about the what the HBr gets up to if not.

[MOTOBALL]

Your [M+H]⁺ at m/z 475 is +72 amu greater than that of the expected product (MW 402)

Could this be due to the addition of CH₃CH₂CH₂(CO)H derived from the acyl bromide ??

Since the chromophore of indigo is the grouping shown below (apparently, the benzene rings do not contribute)

                         O=C---C=C---C=O

the addition cannot be across the >C=C< bond.

Can you post the MALDI spectrum, and was this determined on a recrystallized sample or directly on the material recovered by chromatography ??

[darkdevil]

What does the NMR look like? Mass spec is not a good means of characterisation unless you have all of the other data to go with it.

Also, did you just mix the indigo with the butanoyl bromide? You might want to think about the what the HBr gets up to if not.

Hi, Thank you for your information. I think I did it wrongly during the reaction. I just mixed the the acid bromide and indigo and let the reaction to proceed. I read some information about the Schotten-Baumann acylation, that I should have slowly added a base to slowly remove the HBr.

For my method,  I just mixed the the acid bromide and indigo, and then after 1 day I worked up my reaction mixture with NaHCO3. But is it just the same except I would have a lower yield? but not a completely different product? Thank you

[darkdevil]

Your [M+H]⁺ at m/z 475 is +72 amu greater than that of the expected product (MW 402)

Could this be due to the addition of CH₃CH₂CH₂(CO)H derived from the acyl bromide ??

Since the chromophore of indigo is the grouping shown below (apparently, the benzene rings do not contribute)

                         O=C---C=C---C=O

the addition cannot be across the >C=C< bond.

Can you post the MALDI spectrum, and was this determined on a recrystallized sample or directly on the material recovered by chromatography ??

Hi, Thank you for your reply

The Maldi spectrum has only 1 very intense peak at 475m/z. (I may post it later)
If I add  CH₃CH₂CH₂(CO)H  to 402m/z , it will give 474m/z, But is  CH₃CH₂CH₂(CO)H  an aldehyde? and if I add to the molecule with 402m/z, I need to first remove 1 H from  CH₃CH₂CH₂(CO)H  and another H from the 402m/z molecule to get to a nearest 472m/z which does not match the M = 474m/z ([M+H]⁺ m/z)

I tried to get an aldol addition product to work with a 474m/z like the following:
but I dont think this will happen right?? haha 

[MOTOBALL]

You should be aware of some MS basics,

1) The ESI, APCI and MALDI techniques of ionization (not electron impact which gives M^+.) generate [M+H]⁺ in the positive ion mode.

2) The Nitrogen Rule: if the N atom count is EVEN (N = 0, 2, 4, etc), then molecular weight (MW) is EVEN; therefore, an even MW requires an even N atom count.
Similarly, if the N atom count is ODD (N = 1, 3, 5, etc) then the MW is ODD.

The expected product has MW 402, and would  give [M+H]⁺ at m/z 403; your product gives [M+H]⁺ at m/z 475 (+72 shift) and therefore has MW 474.

This MW 474 is even, therefore N count is even (0, 2, 4 etc); given this higher mass and the structure of indigo, it is very likely that N = 2 rather than N = 0.

Now revisit your last post, but don't confuse MW (M) with molecular ion [M+H]⁺.

 

[darkdevil]

Your [M+H]⁺ at m/z 475 is +72 amu greater than that of the expected product (MW 402)

Could this be due to the addition of CH₃CH₂CH₂(CO)H derived from the acyl bromide ??

Since the chromophore of indigo is the grouping shown below (apparently, the benzene rings do not contribute)

                         O=C---C=C---C=O

the addition cannot be across the >C=C< bond.

Can you post the MALDI spectrum, and was this determined on a recrystallized sample or directly on the material recovered by chromatography ??

Thanks MOTOBALL for pointing out the mistakesss in my post. The m/z in MALDI should be 475m/z and the target compound MW should be 474.
Im so sorry for that! The below is the corrected version:

Hi, Thank you for your reply

The Maldi spectrum has only 1 very intense peak at 475m/z. (I may post it later)
If I add  CH₃CH₂CH₂(CO)H  to 402 MW , it will give a MW of 473 , But is  CH₃CH₂CH₂(CO)H  an aldehyde? and if it is added to the molecule with 402 MW, I need to first remove 1 H from  CH₃CH₂CH₂(CO)H  and another H from the 402 MW molecule to get to a nearest MW of 473 and a 474 m/z which does not match the spectral data

I tried to get an aldol addition product to work with a 475m/z (or 474 MW) like the following:
but I dont think this will happen right?? haha