#tautomerism

[ibalkisai]

‏‎#tautomerism #tautomerism #keto #enol ‎‏ (في ‏‎Lagos, Nigeria‎‏) https://www.instagram.com/p/Bx3OwHYloST/?igshid=13ep7098m1pfj

[eleonorau3o6]

tautomerism methods and theories

Download tautomerism methods and theories

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[science-of-anthropology]

Carbonyl Alpha Substitution Reactions

Alright, so, I'm kinda skipping around the order here a little bit because apparently there's just something about this I didn't get for whatever reason.  I got a not awesome grade on the homework for these, and not awesome grades on the quiz, and so I really want to make sure I try to get to them before the test.  I haven't been tested on them, and the final is 40% this material from the last three chapters....except I lost my notes packet for ch. 24 and that makes me =******( because all the reactions I got right were in the first half that I made into flash cards, and the ones I missed were in the second half that I didn't make into flash cards (on my phone...awesome app called Study Droid.  Speaking of apps, one of the girls in my class loved this iphone-only app: https://itunes.apple.com/us/app/reagents/id453336174?mt=8  I don't have an iphone, so I don't personally know if it's good or not.).  

ANYWAY, ok, so what *are* these reactions? If you'll recall, we talked about the way that an e- withdrawing group (usually something containing O) will cause the alpha carbon to be vulnerable to certain types of reactions.  One of these reactions is an acid/base reaction.  The H is lost, but because of the stabilizing effect of the surrounding atom, there's a lot of resonance.  So the double bond can jump from place to place.  Remember, the more resonance, the more stability.  When that H is lost, the resulting ion is called a Enolate ion.  Your teacher, like mine, may ask you to draw this on a test, so let me explain it.  

An enolate ion is an ion that is able to perform something called keto-enol tautomerism, due to its lost electron.  At first, the actual ion itself is missing an H.  So: O-C-C-R, with dotted double bonds on those first two bonds indicating that the electrons of the double bond shift between those two spots.  Like all electron movement, this is really fast, so we can't just say "it hangs out here".  When these ions accept protons, something weird happens.  The movement of the double bond causes the O to have a very, very negative charge.  A lot of the time it doesn't have the double bond that normally makes it less negative, so it will be all like PROTONS OMNOMNOM, and grab them.  What you get is this: HO-CH=CH-R.  That isn't stable.  That double bond is just way too close to the O, and so basically the double bond and the H trade places and we get: O=CH-CH2-R  This movement switches from the ROH - the enol - to the keto form, hence the name keto-enol tautomers.  This is very stable, and this happens really quickly.  You won't be able to isolate that first form, although these do exist in equilibrium (where the keto form overwhelmingly predominates.).  But you can't do it without spare Hs.  Which is kind of funny when you consider that to get it to do this in the first place you have to remove the H.  It's almost like...why bother? Well, that's a good question.  I used to think it was just because teachers liked making us memorize reams of information that had no use whatsoever, and that keto-enol tautomer was a cool word, but it turns out that's not exactly right.  

That electron movement allows us to DO STUFF to them.  Take, for example, acid promoted bromination.  The reagents: Br2/HC2H3O2 (<-- when you see something written with an H in front like that it's a giant neon sign saying that that thing is an acid.).  The movement of the H from the alpha carbon allows us to sneak a Br on there when no one's looking and you end up with an acid halide.  USEFUL THINGS.  For example, you can expose them to Pyridine and heat to turn them into beta unsaturated ketones: O=C-CH=CH2.  The tautomerism doesn't happen here because the double bond is attached to the beta C and not the alpha C.  

Adding more groups onto a ring, as long as they are e- w/drawing, promotes stability.  So a highly substituted enol is more stable, just like more highly substituted alkenes are more stable.  

Apparently this is a good way to make a C=C bond.  WHO KNEW? ;) 

(Side note, my dumbass neighbor's kid will NOT stop screeching its aggravating little lungs out.  It's 9:30 at night, what is your baby doing up still? >.< PUT IT TO BED SO I CAN STUDY BETTER.)  

Iodoform test - So, when you mix a ketone, a halide, and a base a long mechanism happens and you get: O=C-O-  + CHX3.  This is actually really useful.  You can attach all sorts of things to that hanging negatively charged O.  When you use Iodine as the halide, your leftover product is CHI3 and it's a QA test for methyl ketones.  Now, my notes have this as being performed on a benzene ring substituate, but *says* methyl ketones (which is anything where the ketone is on the 2 C).  I'm thinking it isn't the stabilizing effect of the ring that causes this.  

The other important thing to note about this reaction is that the last step of the mechanism is a COOH + CX3, negatively charged.  The negative charge steals the H from the COOH, resulting in the more stable of the two being negatively charged.  I'm telling you this because I've been asked a lot about this reaction's last two steps.  So my teacher gives the reactant, the X/OH reagents, and then asks for the COOH, adds another arrow, and asks for the haloform.  Keep this in mind when studying.  

Alpha Bromination of RCOOH, the Hell-Volhard-Zelinskii rxn - Is it me, or would it be helpful if ochemists stopped naming reactions after themselves and started naming them after what they do? For instance, you could call this the "make an acid bromide from a COOH" reaction and that'd be more useful than knowing who came up with it.  Cause, frankly, I don't care.  I might be a little annoyed at you for coming up with yet another thing for college ochem students to memorize.  See? I'm giving you bitchface, Hell-Volhard-Zelinskii: 

So in a NOT AT ALL pleasing turn of events, somehow adding the bitchface picture deleted a bunch of stuff I wrote afterwards.  So I'm going to write it again, but it'll be shorter and not as amusing.  NOT NICE BITCHFACE.  

Ok, real quick: COOH with Br2, PBr3/H2O yields a Bromine attached to the alpha carbon.  This is then used for other stuff, yay!

Ok, moving on.  Next we have Alkylation of Enolate Anions.  We're back to Enolates.  So you put a ketone in a base, then allow it to undergo Sn2, and you end up with a ketone on a longer carbon chain.  The way you do this is that you add an RX molecule during the Sn2 portion of the reaction.  The reactivity of the X groups is: 

Tosylate > -I > -Br > -Cl

The reactivity of the R groups:

H3C- > RCH2 -

Vinylic (C=C) and Aryl (aromatic) halides are unreactive.  So you can't tack a benzene ring onto something with this.  

Malonic Ester Synthesis - You should definitely learn this.  I was tested on this, and I didn't know it, and i lost a slew of points.  Looking at it now it makes me mad because i know this reaction really well, but I somehow neglected to add the *name* of it to my memory.  Do not be fooled like I was - this is not a way to MAKE esters, this is a way to synthesize something FROM esters.  

This reaction is used to stick extra R groups onto a diester.  Really quickly since we haven't done anything on Esters yet, this is an ester:

R-O,O=C-R

It's basically a ketone and an ether on the same C.  A DIester is two esters on the same molecule.  For this synth reaction we're using a malonic ester, which if you think back to that list of molecule names I made a few entries ago, is a 3 carbon molecule.  So it's ester-CH2-ester.  If you do 1.NaOEt/EtOH 2. RX 3. H2O/heat, what happens is that the R group from the X gets tacked onto the C in between the two ketones, and then one of the esters gets removed and hydrolyzed to a COOH.  

REALLY IMPORTANT NOTE!!!!  The 3rd step isn't necessary! You can skip it if you'd like to keep the two esters on the molecules instead of getting a COOH.  This is, in fact, sometimes preferable.  So sometimes it will be left off of questions, and sometimes it won't.  

The chemistry behind understand why this happens is really in line with what we've already been talking about.  Remember e- withdrawing O? Yeah, me too! Well, that C that's between two esters is extra positive because it's between four O atoms.  This means it's really susceptible to losing those H's in an acid-like reaction.  The NaOEt (EtOH is ethanol - grain alcohol - and it's just a solvent.) steals the H atom, leaving the carbon with a negative charge.  The R group that's on the halide loves that, so it attaches there.  This can, in fact, be done twice to remove both H's if you don't do step 3 (hydrolysis).  So you can actually tack 2 R groups onto that middle molecule.  

I could have, in fact, drawn this all out on a board for you without even looking at my notes and would have done it on the test...had I remembered the term Malonic ester synth.  Not gonna forget THAT one any time soon.  :/  Watch, now it won't be on the test tomorrow...  ;) 

Acetoacetic Ester Synthesis - This is another one that isn't actually creating an ester, it's using acetoacetic ester to create something else.  Ok, so the difference between this one and the previous one is that this one isn't a diester, it's just an ester.  My teacher told use that you can tell the difference by looking at the molecule of the starting material.  If you can see acetate in it, it's this one.  if not, it's the other one.  

The reagents are the same as the previous reaction, 1.NaOEt/EtOH 2. RX 3. H2O/Heat.  Because I'm tired I'm going to shorten this and say, it does pretty much the exact same thing as the previous reaction.  The difference is that when you do the hydrolysis portion at the end, you end up with a ketone and not a COOH, and the ester is gone.  The end of the molecule ends up staying as acetate, and the ester is removed.  So if you had 3 Os in the starting material and you elect to do step 3, you'll end up with the acetate ketone staying behind and 1 O on the molecule.  If you skip step three, you also can add two R groups to that poor, abused C.  

Ok you guys, there's actually one more reaction in this chapter but it's almost 11 and I need to go to sleep or i won't be able to get up for work tomorrow.  I'll add it tomorrow, I promise.  :)  

[chemcool]

Question on enol Content

Out of acetone and acetyl acetone, which one has the highest enol content in water and why?

Answer:

Acetyl acetone will have high enol content in water because of it having an active methylene groups which has hydrogen atoms more acidic than the ones in acetone and also because that enol is stabilised by intra molecular hydrogen bonding.