[aisha]

Hi...

Been spending weeks and weeks trying to figure out the correct mechanism for this isomerization. Apparently, heating the 5-cholesten-3-one with oxalic acid in ethanol leads to the formation of the isomer 4-cholesten-3-one. Problem is...I can't find anywhere why this specific rearrangement occurs nor how to draw up a mechanism of the acid-catayzed conversion of 5-cholesten-3-one to 4-cholesten-3-one.

I have tried several mechanisms, the following is what I thought might be happening:

1. The H+ of the acid attacks the oxygen of the C=O of 5-cholesten-3-one, leaving a positive charge on the oxygen, electrons from the double bond feed the oxygen to form a single bond O-H (alcohol), at the same time H2O deprotonates a hydrogen from carbon 4, forming another double bond between carbons 3 and 4. Then uncertain of what the rest involves.

Please help...anyone, and anything is much appreciated

[azmanam]

The H+ of the acid attacks the oxygen

Typically we talk of the atom with the electron density doing the 'attacking' - I would have said the lone pair on the carbonyl oxygen attacks the proton from the acid...  but I get your point.

I can envision 2 ways to get to the product, but I don't have any proof that either is correct.

We'll start where you leave off.  You propose an intermediate with an carbon-oxygen single bond at carbon 3 (oxygen has hydrogen attached), a double bond between carbons 3 and 4, single bond between carbons 4 and 5, and the original double bond between carbon atoms 5 and 6, right?  The next step in my mechanism has the electrons from the lone pair on oxygen reforming the double bond between the carbon and the oxygen (this will put the positive charge back on oxygen).  Two more arrows to the product.  Can you take it from there?

The other mechanism I can propose neglects the oxygen atom all together.  If you had just an isolated alkene and some acid lying around - say HBr - do you know what reaction would occur, and by what mechanism?  My second proposal involves the first step of this acid/alkene mechanism leading to a particularly stable carbocation.  The second step deviates from the typical acid/alkene mechanism.  Instead of the conjugate base acting as a nucleophile, I could see that conjugate base acting as a base to remove a proton to get to the product directly.

I don't know which - if either - is correct.  Someone around here probably does though

[macman104]

This page seems to indicate it is an oppenauer oxidation.  But there are some references there of the original research that may point you in the right direction.

[azmanam]

The Oppenauer is when you start with the alcohol version.

http://www.orgsyn.org/orgsyn/prep.asp?prep=cv3p0207

I don't suspect that's what's happening here, because it's not a formal oxidation.  The double bond is just moving into conjugation.

[movies]

Yeah, there is no oxidation here.

Azmanam, your first proposal is on the right track.  The second one would involve a carbon atom that does not possess a full octet, so it would be quite high energy!

Oxalic acid is not strong enough to substantially protonate an alkene, you really need something like sulfuric acid for that.

[azmanam]

a carbon atom that does not possess a full octet

True, but carbocations are prevalent in acid promoted alkene mechanisms - and a tertiary carbocation would be the most stable.  I'm not proposing an isolable intermediate, but sorta the first half of an alkene addition and the last half of an E1.

Oxalic acid is not strong enough to substantially protonate an alkene

That I buy.  And that makes the above explanation moot, as the carbocation wouldn't form in significant quantities.

Looks like aisha was on the right track the whole time.

[movies]

The point I was trying to make is given the choice between a mechanism that has an intermediate where a relatively electronegative atom does not have a full octet and a mechanism where that doesn't happen, you should choose the latter! 

[aisha]

hey... so i have my intermediate intermediate with an carbon-oxygen single bond at carbon 3 (oxygen has hydrogen attached), a double bond between carbons 3 and 4, single bond between carbons 4 and 5, and the original double bond between carbon atoms 5 and 6. What would be the point of feeding electrons fromt the oxygen back to form a double bond (this gives back the original positively charged oxygen). What i am trying to figure out is that if this intermediate i am proposing is right then how do i get rid of the double bond between the carbon atoms 5 and 6? and how do i then go back to forming the C=O ??

[azmanam]

how do i then go back to forming the C=O

by 'feeding electrons from the oxygen back to form a double bond'

how do i get rid of the double bond between the carbon atoms 5 and 6

Start by feeding those electrons from oxygen, that would give a carbon with 5 electrons (not cool), so you'll need to move some electrons away from that carbon.  Move those in the right way and you'll do everything you set out to.

[AISHA/aishabashir3@hotmai]

ok will set out to do that. thanks mate

[aisha]

so... i fed those electrons from oxygen to form the C=O bond, whilst water (i guess) deprotonates the H (which is part of the O-H bond). I then moved the electrons in the double bond at C3 and C4 to the C4 to C5 position (where i want it to be- p.s. don't know if this is do-able???) then I got the rid of the double bond at the C5 to C6 position by adding H3O+ (to give a single bond) ... does this sound good?? ( I don't know why, but I keep thinking that perhaps an enolate intermediate is formed; only cos it seems to be going from a beta, gamma unsaturated ketone to an alpha, beta unsaturated carbon)...hmmm...please help

p.s..does anybody know how i can follow this isomerization (other than TLC) apparently spectrophotometrically there is an absorption max at 240 nm for the 4-cholesten-3-one...but what would i observe for 5-cholesten3-one??

[azmanam]

Yes, all that sounds fine.  An enolate is a carbon oxygen single bond, carbon carbon double bond, and a negative charge on the oxygen (C=C-O^-), so this is not technically an enolate.  The chemistry is similar, though.  The alpha carbon atom is the nucleophile.  However, in this system, there's another double bond in conjugation (C=C-C=C-O-H).  This makes the gamma carbon nucleophilic.  It is definitely plausible for the lone pair on oxygen to reform a carbon-oxygen double bond and move electrons to the alpha-beta carbon atoms.  The electrons in the double bond of the beta-gamma carbon can then pick up a proton from H₃O⁺ as you mentioned.  The oxygen atom is probably still protonated at this point.  Then a water molecule acts as a nucleophile and picks up that proton from the positively charged oxygen atom.  Now your double bond is in conjugation - which should make the system lower in energy and therefore favorable.

I don't know much about the properties of steroid systems, so unfortunately I can't help you much with following.  You might be able to follow by NMR or GC, though.

[movies]

I would guess that these compounds might be too heavy for GC, but I may be wrong.  LCMS would be another alternative, but not everyone has one of those.

You could monitor by IR because the carbonyl stretch would change from ketone to enone!

[aisha]

IR sounds good...thank u movies