[pdffree]

can anyone guide me to draw the mechanism of the following attached file?

[azmanam]

start by adding acid.  which site is likely to be protonated first... how does that change the reactivity of the compound?

[pdffree]

double bond o  and what acid can be used? HNO3/H2so4?

[azmanam]

Doesn't really matter. If I were actually trying to do this reaction, I would probably use catalytic PPTS in dichloromethane as a solvent... but honestly, dilute HCl or H₂SO₄ would work, too.

HNO₃/H₂SO₄ is probably a bit harsh, but it would get the job done.

Yes, the O of the C=O bond will act as a nucleophile and will attack H⁺ as the electrophile.  What happens next?  What do you think is the nucleophile is of the next step?  The electrophile?

[pdffree]

.O-

[pdffree]

and after that H3o+ so I will get "OH", right?

[azmanam]

no...

Here's what we have so far.  The O of the C=O double bond has been protonated by acid, so it will have a bond to hydrogen and a full positive charge.

It's hard to read that image, but is that H^- attacking the carbonyl carbon?  First, H^- is a rather rare nucleophile.  Second, you won't see any minus charges in an acidic mechanism.   BUT, you have correctly identified the electrophile in the next step.  Something will attack the C of the C=O double bond... but what?

To help, start making a list of all the competent nucleophiles you have present in the molecule/reaction mixture.


(BIG HINT... the product of the reaction is given.  How can we get to the product from the intermediate we have drawn?)

[pdffree]

so is this what happens?

[azmanam]

huh?  where did you get those reagents?  The only reagent we've added so far is catalytic acid.

You're trying too hard.  We have the protonated molecule, the conjugate base of our catalytic acid, and unreactive solvent.  What are the nucleophiles in those compounds?

[pdffree]

is this what happens?

[azmanam]

yes.  well done.

one minor clarification... when the alcohol attacks the protonated carbonyl, the hemiacetal forms... BUT the oxygen within the 6-membered ring (the oxygen atom which came from the alcohol) is still protonated and carries a full positive charge.  Some conjugate base will need to come along and remove that proton to give you the neutral product.

I noticed your initial image was titled glucose.  This is not coincidental.  This is the mechanism by which straight-chain glucose cyclizes to glucopyranose.

[pdffree]

Thanks a lot!