[DannyBoi]

Hi again,

Our class has not gone into reactions deeply yet but this question stumped me:



This is my second and last attempt:



I figured the sulfate ion in solution isn't too good of a nucleophile because it's the anion of a strong acid? But I'm not sure about the charge or the chirality. Any clarification would be great!

[DannyBoi]

Ok now I am thinking that it's actually a hydroxy group attached and that's how the charge is neutral, and the hydrogen cation is just reintroduced into the solvent? So this is what I would guess is the product, but I'm still not sure about the chirality. Is it unimportant in this question whether the OH group is toward or behind the ring?

[DannyBoi]

(deleted)

[zeropoint]

So, the first thing to do with this sort of reaction is almost always to push electrons to form the carbonation (positively charged carbon). This gives you an ' vacancy' where you can put something else (ie. a nucleophile). The acid there just as a source of H+. Hopefully you can piece it together from this.

I have a problem with the 'optically active' question however. I think that what the person is intending to ask is if you get a racemic mixture or not. Even if the mixture is totally racemic, it's a mixture of two optically active components, so the question is misleading.

[DannyBoi]

I see what you mean, well I submitted my second attempt as shown and it was correct. I figured it's not optically active because it has a plane of symmetry, although I'm not completely sure about this.

[contra]

There are no chiral centers in that product, so it would not be considered optically active. Using a plane of symmetry to test is another good method.

[spirochete]

You're correct about the connectivity of the final product.  The charged intermediate you initially drew gets depronated by something else to give a neutral product.  If you're not sure, 99% of the time you're drawing a neutral major product.

Addition of the sulfate ion, as you initially drew, will always be a reversable process because sulfate is a good leaving group.  In contrast, once the water addition loses a proton the addition is irreversible because OH- (hydroxide) is a terrible leaving group.

Also, that reaction almost certainly gives a horrible mixture of diastereomers, neither of which are chiral.  Technically there are two stereocenters in that compound, that is atoms where a flip in configuration can result in stereoisomers.  In this case flipping the configuration of one relative to the other will give a diastereomer.  There are however NO asymmetric centers in that molecule.  Another example of a stereocenter that is not an asymmetric center is an SP2 carbon where switching the configuration of an R group would result in Cis/Trans or E/Z isomerism.

Are they really asking you to predict the major diastereomer, or just predict the major regioisomer?

[TSClimber]

I might be missing something here but how can you have an -OH attack under acidic conditions? any ^-OH would be quenched by the acid surly...

[stewie griffin]

@TSClimber
There is no hydroxide in the reaction. The nucleophile is a water molecule which then loses a proton so that, in the end, an OH group is connected to a carbon.
@spirochete
"once the water addition loses a proton the addition is irreversible" Not quite; the reaction actually is reversible. The resultant hydroxyl product can still pick up a proton from the acid present and then undergo E1 elimination back to the starting material. In fact, dehydration of alcohols (usually catalyzed by either sulfuric or phosphoric acid) to give alkenes is a common undergrad lab.
I would argue there are two reasons why the product is the hydroxyl and not the sulfonic acid. 1) Since water is the solvent (and sulfuric is most likely present in catalytic amounts) it is far more likely that a water molecule will attack the intermediate carbocation. 2) Even if the sulfonic acid is formed, water can do a simple "ligand exchange" reaction at the sulfur resulting in formation of the hydroxyl product.

[spirochete]

Under these conditions it seems to be irreversable, if for no other reason than the fact that it's so thermodynamically favored in excess water.

[Dan]

Under these conditions it seems to be irreversable, if for no other reason than the fact that it's so thermodynamically favored in excess water.

I don't understand how you can claim that this hydration reaction is irreversible. The excess of water pushes the equilibrium to the alcohol, but acid catalysed addition of water across a C=C double bond is still reversible.

the addition is irreversible because OH- (hydroxide) is a terrible leaving group.

^-OH is not the leaving group for the reverse reaction, H₂O is the leaving group - and a good leaving group.

Here's a question for you - if you dissolved the tertiary alcohol product in H₂¹⁸O with H₂SO₄ would you expect to recover labelled alcohol?

[voidSetup]

Pretty sure this is definitely reversible. If you actually do the reverse reaction (dehydration) you have to remove water from the system (usually by distillation i believe) to push the equilibrium forward.