Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Snobbibjorn on March 25, 2023, 08:36:54 PM
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Hi
I'm currently writing a school paper where i need to analyse a synthesis of 2,4Dichlorophenoxyacetic acid from 1941.
I have a good general idea of why the synthesis makes sense but some parts of it confuses me, mostly the part about why NaOH is added my current thinking is to neutralise the acid created from the reaktion and potentially as a catalyst. What i'm most confused about is the acidification where it seperates the liquid, i get that it's because of polarity but why does the warm water not split it?
I would love if anyone had some input on my two "questions" and on the entire synthesis in general.
Hope someone is able to help, here is a link to the synthesis ( i'm only looking at the one making 2,4Dichlorophenoxyacetic acid):
https://pubs.acs.org/doi/pdf/10.1021/ja01851a601
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I assume the NaOH is to deprotonate the phenol; the phenoxide is the nucleophile. I suppose in acid conditions you could get esterification. Addition of acid separates the liquid because the sodium salt initially formed is soluble in water, but the undissociated acid form is not.
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Thanks for the answer, i'm still a little confused about the seperation. How would the NaCl react with the HCl?
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There is no reaction between NaCl and HCl
The Phenol is converted with NaOH to Phenolat and the Chloraceticacid to Chloracetate.
ROH + NaOH => RONa + H2O
ClCH2COOH + NaOH => ClCH2COONa + H2O
So the Chlorine and the Phenolate react in an nucleophile Substitution.
RO- + ClCH2COO- => Cl- + ROCH2COO-
The acidification with HCl gives the free acid
ROCH2COO- + HCl => ROCH2COOH + Cl-
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So is it correctly understood that when the acidification happens, the ROCH2COOH becomes the phase described as the heavy oil that crystalises, and then there is another phase a water phase with: water, Cl- and Na+/NaCl?
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Correct understood