Hydrolysis of ester: Follow the Oxygen...

[SaintJude]

Ok, so I've seen a few FL question, where it asks: What would happen if you radioactively labeled the oxygen...blah blah..

If you radioactively-labeled the oxygen in the water nucleophile (e.g. O-18) in the following 3 reactions, would you see the labeled oxygen only in the resulting carboxylate product?

I think so b/c the OH group of the alcohol products are actually formed form the 'OR leaving group of the ester. ?

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[syoung]

well, i would say that yes, the new product does incorporate the O18 from H2O, because that is what's attacking the carbonyl carbon and kicking off the ROH.
But not all the water would be used up, since it's the solvent.
But as it says, the acid catalyzed is reversible, so you could end up with an O18 in your product side as well. However in the base catalyzed version, it would not reverse back towards the reactants, so O18 is only on product side (carboxylic acid).

 

[SaintJude]

Ah, yes, very good point!

B/c an acid-catalyzed reaction is reversible, O18 will show up in the original ester compound and the carboxylic acid. But in the non-reversible base-catalyzed reaction, the O18 will (theoretically) only show up only in the carboxylic acid.

Thank you so much syoung. I hope this will pay off b/c it looks like the April test takers have encountered good amount of orgo.

 

[folktale]

well, i would say that yes, the new product does incorporate the O18 from H2O, because that is what's attacking the carbonyl carbon and kicking off the ROH.
But not all the water would be used up, since it's the solvent.
But as it says, the acid catalyzed is reversible, so you could end up with an O18 in your product side as well. However in the base catalyzed version, it would not reverse back towards the reactants, so O18 is only on product side (carboxylic acid).

Wouldn't it only kick of the -OR? Then it will just get protonated by H+ in solvent?

 

[SaintJude]

Wouldn't it only kick of the -OR? Then it will just get protonated by H+ in solvent?

The OR group gets protonated before it gets kicked off. So, as you can see, an alcohol is what leaves the tetrahedral intermediate

 

[folktale]

The OR group gets protonated before it gets kicked off. So, as you can see, an alcohol is what leaves the tetrahedral intermediate

 

[circulus vitios]

Did you memorize the ester hydrolysis mechanism? I've never understood how some mechanisms 'make sense' e.g. why can't the methoxide just deprotonate the water and kick itself off in one step versus an outside water deprotonating the water, then the methoxide protonating itself with a free-floating proton, etc. I don't see how or why that's favorable.

 

[syoung]

Did you memorize the ester hydrolysis mechanism? I've never understood how some mechanisms 'make sense' e.g. why can't the methoxide just deprotonate the water and kick itself off in one step versus an outside water deprotonating the water, then the methoxide protonating itself with a free-floating proton, etc. I don't see how or why that's favorable.

Not too sure about your question but here goes.
Well here's the thing, is it CH3COOCH3 in methoxide or methanol? Or is it in water? I'm assuming it's in water and methanol, and NOT methoxide because the mechanism has H+.

My thinking goes: the "methoxide" can't kick itself off, because you are in acidic conditions. RO- is a strong base, and could never "leave". But if you protonate it, it can leave as ROH easy!

 

[circulus vitios]

Not too sure about your question but here goes.
Well here's the thing, is it CH3COOCH3 in methoxide or methanol? Or is it in water? I'm assuming it's in water and methanol, and NOT methoxide because the mechanism has H+.

My thinking goes: the "methoxide" can't kick itself off, because you are in acidic conditions. RO- is a strong base, and could never "leave". But if you protonate it, it can leave as ROH easy!

It's in acidic conditions. I've always been horrible at o-chem so this probably makes no sense, but why doesn't this happen?

Step 1: Water leaves because positive formal charge on oxygen isn't preferential and lone pair on oxygen forms double bond.
Step 2: Methoxide deprotonates hydroxyl group, MeO kicked off.

 

[syoung]

Like I said earlier, MeO- CANNOT be kicked off under acidic conditions, only basic. It must be R-OHMe+ before it can leave as MeOH
Secondly, what you are trying to show is way too much to occur in "one" step.

 

[circulus vitios]

Like I said earlier, MeO- CANNOT be kicked off under acidic conditions, only basic. It must be R-OHMe+ before it can leave as MeOH
Secondly, what you are trying to show is way too much to occur in "one" step.

I combined steps. It is R-OHMe+ before it leaves.

Maybe I'm better off just memorizing the mechanism.

 

[syoung]

Oh I think I understand your question now.
You want the Methoxy O to deprotonate the H2O+ on the compound and then get kicked off?

 

[circulus vitios]

Oh I think I understand your question now.
You want the Methoxy O to deprotonate the H2O+ on the compound and then get kicked off?

I now see that my mechanism won't work.

 

[amandawilson]

Well, we learned in our chemistry subject about Hydrolysis. It is a chemical process in which a water molecule is added to a substance resulting in the split of that substance into two parts. One fragment of the target molecule (or parent molecule) gains a hydrogen ion (H+) from the split water molecule. The other portion of the target molecule collects the hydroxyl group (OH&#8722 of the split water molecule. In effect an acid and a base are formed. Well, with technological advances we have today, its no wonder to see a vehicle that don't required filling up at fuel stations. Hydrogen fuel cell vehicles seem the perfect evolution beyond electric cars and gas-powered cars. There are two methods of powering an electric car, though the bits like the motor and so forth are always the same. Car makers can use a battery pack, which is the dominant method, or a hydrogen fuel cell stack. Batteries store electricity and then have to be recharged. Hydrolysis process are being applied here that is the reason why an electric current produced that is used to power the car. You may look for more at: Hydrogen fuel cell vehicles making slow, steady progress.