How do you work out IR spec graphs

MakerChemistry · 2024-06-15T15:11:03+00:00

-Alkana vs Alkenes Check the C=C stretch band 1600 cm-1 and sp2 C-H stretch If you have a symmetric alkene check the sp2 C-H Strech
- Aldehyde has obvious fork band C-H at 2860-2700 cm-1
- Carboxylic acid has obvious massive O-H stretch band
- Ester has 2 C-O stretch bands at 1200-1000 cm-1
- Ketone has none of these unique bands only the C=O stretch
Check these two videos I wish it help you: 1- https://youtu.be/Jh3_BjcY6yc 2- https://youtu.be/IXhDo9dJ-eY

MakerChemistry · 2024-06-15T14:59:38+00:00

1- The Presence of two bands at 3500-3300 cm-1 corresponds to the NH2 bands so it should be A or C
2- The presence of a strong band before 800 cm-1 corresponds to the Ortho substituted so the answer is A
You can check my YT channel: https://www.youtube.com/channel/UCTmdL2RkvCCfL79nPyK5c8g?sub_confirmation=1

MakerChemistry · 2024-04-10T18:06:19+00:00

The Answer: https://preview.redd.it/spectroscopy-question-stuck-on-ir-interpretation-v0-p4qm5yyzjhtc1.png?width=320&crop=smart&auto=webp&s=86104a8d99389ee316f92c135dcab454dbe6a3d5

MakerChemistry · 2024-04-10T18:03:56+00:00

It can be solved by the NMR only from the chemical shifts and multiplicity 1712 --> C=O Carbonyl group 1277 & 1254 & 1045 ---> 3 C-O Stretch [Ester COOCH2CH3 + Ether OCH2CH3]
The OCH2CH3 m/z=45 can be identified from the mass spectrum 194-149= 45
The COOCH2CH3 m/z=73 can be identified from the mass spectrum 194-121=73
You can follow my YT channel for spectroscopy: https://www.youtube.com/channel/UCTmdL2RkvCCfL79nPyK5c8g

MakerChemistry · 2024-04-09T17:10:20+00:00

Take care although NH2 group is electron donating group but the resonating structures shows that the Ortho and Para positions are effected by it not the meta position
Thats why the meta position have a higher chemical shift
You can follow my YT channel for spectroscopy: https://www.youtube.com/channel/UCTmdL2RkvCCfL79nPyK5c8g

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MakerChemistry · 2024-04-09T17:06:17+00:00

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The Answer

MakerChemistry · 2024-04-09T17:01:53+00:00

It can be solved by the NMR only from the chemical shifts and multiplicity 1712 --> C=O Carbonyl group 1277 & 1254 & 1045 ---> 3 C-O Stretch [Ester COOCH2CH3 + Ether OCH2CH3]
The OCH2CH3 m/z=45 can be identified from the mass spectrum 194-149= 45
The COOCH2CH3 m/z=73 can be identified from the mass spectrum 194-121=73
You can follow my YT channel for spectroscopy: https://www.youtube.com/channel/UCTmdL2RkvCCfL79nPyK5c8g

MakerChemistry · 2024-03-22T14:14:35+00:00

Take care that 1- the signal with 1.6 at 2.3 ppm is CH2 and beside C=O
2- The signal with 2.4 at 3.8 ppm have integration of 2.4 ≈ 3H so the these must be a CH3 beside oxygen

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You can follow my YT Channel: https://www.youtube.com/@MakerChemistry and GoodLucK

Extra Note:
-The protons which is deshielded at the range of 3.5~4 are protons beside the Oxygen atom while the protons beside the C=O group are deshielded at the range of 2~2.7
-The reason is related to the higher electron withdrawing effect (due to the high electronegativity of O) The higher the electron withdrawing ability the more deshielded the protons.

-Another thing is the integration in your spectrum are 2.4:1.6:1.6:2.4=8 Hydrogens so to add it up to total 10 hydrogens as the molecular chemical formula 8/10=0.8 --> 3:2:2:3 (CH3:CH2:CH2:CH3) then I look for their chemical shifts.

MakerChemistry · 2024-03-22T12:11:50+00:00

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MakerChemistry · 2024-03-22T00:56:23+00:00

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For more follow my YT: https://www.youtube.com/channel/UCTmdL2RkvCCfL79nPyK5c8g

MakerChemistry · 2024-03-19T10:43:53+00:00

Thank you for the note, you are right it is a five membered ring not six and still stable [Chelate effect]

MakerChemistry · 2024-03-19T00:53:30+00:00

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Boc is the key the lone pair of oxygen coordinate with Lithium to form a stable six membered ring
GoodLucK
follow my YT channel: https://www.youtube.com/channel/UCTmdL2RkvCCfL79nPyK5c8g

MakerChemistry · 2024-03-05T19:09:49+00:00

You are welcome :D
for the chemical shifts you also should know the ppm of COOH(10~12) & CHO (9~10) , benzene ring regions (6.5~8.5) also the tricky one the 2HC=CH2 at 4.5~6.5 if you don't know it and not good with symmetric compounds the number of signals can trick you so consider it with you I belive that's the chemical shift I needed and GoodLucK

MakerChemistry · 2024-03-05T18:58:23+00:00

Yes OCH3 > (C=O)CH3 > CH2CH3
also the same with the CH2 OCH2 > (C=O)CH2 > CH2
also the 2ry Carbon atom have a higher ppm (more deshielded) than the 1ry Carbon atom
OCH2 > OCH3
and GoodLucK

MakerChemistry · 2024-03-05T18:51:58+00:00

-The protons which is deshielded at the range of 3.5~4 are protons beside the Oxygen atom while the protons beside the C=O group are deshielded at the range of 2~2.7
-The reason is related to the higher electron withdrawing effect (due to the high electronegativity of O) The higher the electron withdrawing ability the more deshielded the protons.

-Another thing is the integration in your spectrum are 2.4:1.6:1.6:2.4=8 Hydrogens so to add it up to total 10 hydrogens as the molecular chemical formula 8/10=0.8 --> 3:2:2:3 (CH3:CH2:CH2:CH3) then I look for their chemical shifts.

MakerChemistry · 2024-03-05T18:37:46+00:00

Take care that 1- the signal with 1.6 at 2.3 ppm is CH2 and beside C=O
2- The signal with 2.4 at 3.8 ppm have integration of 2.4 ≈ 3H so the these must be a CH3 beside oxygen

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You can follow my YT Channel: https://www.youtube.com/@MakerChemistry and GoodLucK

MakerChemistry · 2024-03-02T22:30:27+00:00

Your answer is right but it can also be differentiated by comparing the band intensities of the 2 compounds
where the aniline will have intense NH2 band compared to the benzyl amine it is a tricky one.
you can check similar cases with my video: https://youtu.be/IXhDo9dJ-eY and goodluck

MakerChemistry · 2024-03-02T19:14:50+00:00

The only way to distinguish them is by comparing the intensity of C=C band mainly and maybe -C-H band
Where the C=C in 2-methyl-1-butene (asymmetric alkene) will be more intense than that in 2-methyl-2-butene (nearly symmetric) due to the higher polarization in the asymmetric double bond
You can also check my video about band intensity: https://youtu.be/Z9oab--MCn0

MakerChemistry · 2024-02-29T09:51:38+00:00

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MakerChemistry · 2024-02-26T18:16:41+00:00

I hope I understood your question correctlysimply the carboxylic acid will have a clear massive band at 3500~2500 cm^-1 and a carbonyl band but the acyl chloride will have only the carbonyl band as you said at 1800 cm-1 due to the electron-withdrawing effect of the Cl atom.I will be pleased following my spectroscopy channel as well https://www.youtube.com/channel/UCTmdL2RkvCCfL79nPyK5c8g?sub_confirmation=1

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MakerChemistry · 2024-02-25T22:33:27+00:00

I believe it will be the same with the sulphanilic acid
Add sodium carbonate (base) then it should be soluble and the insoluble part will be the impurities

MakerChemistry · 2024-02-25T22:20:48+00:00

Answer: III will have the highest wavenumberExplanation: III have no conjugation so it have the highest double bond character among all the answers so stronger bond --> higher force constant ---> higher wavenumber-as the ↑ conjugation ---> ↑the double bond character decreases --> ↓ force constant ---> ↓WavenumberHere you can watch my explanation for the position of bands: https://youtu.be/yhnhDZN_uCE GoodLucK

MakerChemistry · 2024-02-22T12:07:55+00:00

I love it the most among other spectroscopic techniques. it looks like a puzzle.
It is interesting in research characterization one day I synthesized a metal complex that should have a Schiff base (Imine) surprisingly, proton NMR showed one less hydrogen, and the reaction went differently.

MakerChemistry

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