April 17, 2021, 11:36:41 PM
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Topic: vanillin reduction with NaBH4 to vanillyl alcohol  (Read 262 times)

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Offline xshadow

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vanillin reduction with NaBH4 to vanillyl alcohol
« on: March 25, 2021, 04:46:05 AM »
I have a doubt about this synthesis

The reaction is made under basic conditioon in  an aqueous /ethanol solution with NaOH.

After the adding of NaBH4 we get the vanillyl alcohol deprotonated ...so we use HCl in  order to protonate  it and getting the neutral vanillyl alcohol that with an ice bath it will precipitate


Now my question is why is this alcohol not soluble in  a mixtures  of H2O/EtOH and it will precipitate after a while??
because ,for example, example  ethanol is soluble in water...we get mixture of both of them. It will not precipitate!!

thanks!!

Offline luscofusco

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Re: vanillin reduction with NaBH4 to vanillyl alcohol
« Reply #1 on: March 25, 2021, 06:38:43 AM »
Im not sure but this makes sense to me: Ethanol, methanol or propanol (low molecular mass) are soluble in water because they can form hydrogen bonds H2O-OH but regarding vanillin alcohol, it has less electron density around the OH group because of the phenyl group so it can't form that H2O-OH bond and that's why it's not soluble.

Offline rolnor

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Re: vanillin reduction with NaBH4 to vanillyl alcohol
« Reply #2 on: March 25, 2021, 07:04:50 AM »
The benzene-ring makes the product lipophilic:
https://en.wikipedia.org/wiki/Lipophilicity

Offline Babcock_Hall

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Re: vanillin reduction with NaBH4 to vanillyl alcohol
« Reply #3 on: March 25, 2021, 10:50:59 AM »
Benzoic acid is quite insoluble in water at low temperatures, despite having a carboxylic acid group.

Offline xshadow

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Re: vanillin reduction with NaBH4 to vanillyl alcohol
« Reply #4 on: March 26, 2021, 07:19:18 PM »
thanks :)

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