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I'm reading this IR spectra and trying to determine if the large broad band at 3388 is alcohol from an intermediate or from water in the sample. The details of the reaction, conditions, and mechanism can be found below if it's helpful.

enter image description here

I attempted to run a double aldol condensation of p-anisaldehyde and acetone using 2 mL of 3 M NaOH, 0.3 mL of p-anisaldehyde, and 1.6 mL of 0.625 M acetone in ethanol. I let the reaction run for 30 minutes at room temperature and recrystallized the product.

I drew all of the steps I think the reaction mechanism could have stopped that have an alcohol here:

enter image description here

Thanks for any advice.

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The main working rule of a good FTIR spectrum is that everything must be bone dry! The absorbance in IR is additive just like Beer's law, i.e. your spectra is the sum of water traces in your sample as well as your suspected alcohol. There is no simple way to deconvolute it. Dry your product thoroughly in a vacuum dessicator for at least a day and re-run the IR. If your big -OH band vanishes, it means your product was not dry. If it persists, your product likely has an hydroxide group.

KBr must be oven dried well before use as well.

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  • $\begingroup$ Thanks for the advice! I really appreciate it. $\endgroup$ – Cade Apr 18 at 3:10
  • $\begingroup$ Also check the melting point before getting puzzled by spectra of all kind. $\endgroup$ – Alchimista Apr 18 at 7:05
  • $\begingroup$ @Alchimista Thanks for the tip. The melting point is about 1.5 deg C less than the final product of the double aldol condensation [Bis(4-methoxybenzylidene)acetone], but my guess is that's due to minor impurities or my partner wasn't paying careful attention. I suppose there's a chance some of the intermediates with alcohol were in the sample, but I'm not sure the likelihood of that. $\endgroup$ – Cade Apr 18 at 21:50

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