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Hydrolysis chemistry

Help! Im confused I was watching machemguy and he said that for base hydrolysis the water splits the ester apart and then the Na+ present in the solution reacts with the products. But then why is there no water in the equation? And I know hydrolysis means splitting of water implying that there is water present but why’s it not in the equation??



Also this is a separate question but related I guess… for base hydrolysis why isn’t the 2nd Qs answer what I put on the whiteboard. Because the OH reacts with Na+ right?? Because in the 3rd question example it does react with Na + so I’m confused as to why the 2nd one doesn’t.
Is it because the q has told us how many products form, but in an exam If the q doesn’t tell me what example do I follow??
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Also for this question as well IMG_2044.jpeg


And this
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My confusion is annotated.
Plz help, I would be so grateful as my final exams are in June 😱
(edited 1 year ago)
1) Because the Hydroxide (OH-) does the hydrolyzing it splits the bond, this can be seen in the mechanism here: https://www.chemistrysteps.com/ester-hydrolysis-acid-and-base-catalyzed-mechanism/. So the equation wouldn't balance if water was included e.g.

NaOH + CH3CH2COOCH3 --> CH3CH2COO- Na+ + CH3OH
Left side: 1Na 3O 9H 4C
Right side: 1Na 3O 9H 4C

If you add water on the left handside the additional hydrogen and oxygen have no where to go on the right handside.

2) The carboxylate the part containing the C=O always forms the salt you can justify this in two different ways.

A) The Hydrogen on the carboxyilic acid (COOH) is acidic and therefore prone to be removed by either the base or as shown in the linked mechanism RO-. leading to ROH and RCOO- Na+

B) The negative charge (O-) is more stable as part of the carboxylate as it has a resonance form via the carbonyl (C=O). This is missing in the alcohol hence the negative charge prefers to be on the carboxylate (COO-) hence that part forms the salt.

This is why the answer you drew on the whiteboard is wrong. Remeber basic hydrolysis of esters always gives a carboxylate salt (RCOO- M+) and an alcohol ROH.

Questions 3,5 and 6 are special cases there are actually two different reactions happening.

Firstly an ester hydrolysis as discussed above and then secondly the phenol (benzene ring with an alcohol) is reacting with the NaOH.

RC6H4OH + NaOH --> RC6H4O- Na+ + H20

This occurs because phenol is a rare example of a (relatively) acidic alcohol, again you can justify the salt formation in two different ways:

A) The hydrogen on the Phenol is acidic (this is beause it is electron poor as its electrons are delocalised into the aromatic ring) and therefore prone to be removed by NaOH.

B) The negative charge CO- is stabilised by delocalisation into the aromatic ring, this can also be shown by resonance forms if you use the benzene structure with double bonds.

Hope that helps you out!
Reply 2
well, im not actually sure *why* it does that... but im gonna guess that the hydroxide nucleophilically attacks at the carbonyl, resulting in a nuc-add-elim that expells a deprotonated alcohol, and a carboxylic acid. The thing is, the charge can't be delocalised very well and it would be isolated onto the single oxygen ion. This makes it unstable, and by extension rly basic. As a result, it quickly yoinks another hydrogen from the the new carboxylic acid formed, which can delocalise the charge much better than oxyanion. Then that carboxylate forms a salt with the sodium.

BTW, the newly formed alcohol wouldn't rly want to get deprotonated and form sodium salt, as the newly proposed oxyanion would be too unstable - it rly doesn't want to exist, so its like nahh i'll just stay like that.

idk, im in yr 12, but i think amides can also be hydrolysed, but like idk if the question is even asking for that https://www.chemguide.co.uk/organicprops/amides/hydrolysis.html

eg, you can also deprotonate between the two carbonyls very easily since theyre rly acidic, but the questions not asking for that

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