smaerd
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Alkaline Hydrolysis
Okay so let me know if I understand this right.
In Hydrolysis a compound is placed into water and the parent part of the molecule gets the H+ ion from the water, and the other part gets the
hydroxy(OH-) ion. Thus seperating the two parts from each other.
Now in Alkaline Hydrolysis is the same principle but uses a base as a catalyst. Would NaOH be a good catalyst for this? I am imagining it would be
nearly perfect.
So after the two compounds have been seperated into the aqueas solution, then is it commonplace to use a solvent to retrieve/extract the desired
portion? Or am I misunderstanding the procedure.
Thanks for baring with me, I'm still learning, slowly but surely.
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mnick12
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Could you be more specific? It is really hard to understand what you are asking unless you have a specific example. Also I dont think NaOH would be a
catalyst as it would be consumed in the reaction, but like I said unless you give a rection you are thinking about it is hard to tell.
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Nicodem
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Quote: Originally posted by smaerd  | | In Hydrolysis a compound is placed into water and the parent part of the molecule gets the H+ ion from the water, and the other part gets the
hydroxy(OH-) ion. Thus seperating the two parts from each other. |
No, it is much more involved than that. The exact mechanism depends on what hydrolysis you talk about. For example, PCl<sub>5</sub> will
hydrolyse by a nucleophilic substitution with H<sub>2</sub>O being the attacking nucleophile and chloride being the leaving group. The
intermediate is deprotonted (by H<sub>2</sub>O or any other base), followed by the elimination of HCl. If the attacking nucleophile is the
hydroxide (basic hydrolysis), the mechanism is the same except for the deprotonation step which is skipped. So as you see, even such a simple
hydrolysis includes a nucleophilic substitution, elimination and proton transfers. Acidic compounds hydrolyse by reacting with
H<sub>2</sub>O having the role of a base, followed by deprotonation and elimination (ZrCl<sub>4</sub>,
BiCl<sub>3</sub>, etc.). Some transition metal complexes hydrolyse by ligand exchange where H<sub>2</sub>O or hydroxide
exchanges some other ligand. Functional groups with Pi-bonds hydrolyse trough nucleophilic addition followed by E1 elimination SOCl2, POCl3, RCOOR,
RCN, etc.). Here proton transfers also have an important role and the attacking nucleophile is either a hydroxide (basic hydrolysis) or
H<sub>2</sub>O (acidic hydrolysis). In acidic hydrolysis the Pi-bond first gets protonated so that the system becomes electrophilic enough
to react with such a poor nucleophile like H<sub>2</sub>O (hence the acid catalysis). The net mechanism is thus a retro-condensation.
| Quote: | | Now in Alkaline Hydrolysis is the same principle but uses a base as a catalyst. Would NaOH be a good catalyst for this? I am imagining it would be
nearly perfect. |
As you were told the hydroxide has no catalytic role in hydrolysis reactions. If you look at the net reaction equation you can see why it is not a
catalyst, but a reactant.
| Quote: | | So after the two compounds have been seperated into the aqueas solution, then is it commonplace to use a solvent to retrieve/extract the desired
portion? Or am I misunderstanding the procedure. |
Commonplace for what hydrolysis? You mention no specific case.
| Quote: | | Thanks for baring with me, I'm still learning, slowly but surely. |
We have the Beginnings section for pedagogic discussions.
…there is a human touch of the cultist “believer” in every theorist that he must struggle against as being
unworthy of the scientist. Some of the greatest men of science have publicly repudiated a theory which earlier they hotly defended. In this lies their
scientific temper, not in the scientific defense of the theory. - Weston La Barre (Ghost Dance, 1972)
Read the The ScienceMadness Guidelines!
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Nicodem
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Thread Moved
15-3-2010 at 11:46 |
smaerd
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I need to stop reading/skipping chapters in my books, haha. What you said makes a lot more sense from the pieces and parts I can put together, than
what I was thinking. Thanks for the explanation, I have a lot more studying to do.
I'll certainly be posting more in the beginning's section where I certainly belong .
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