[THC]

Q1: How would you prepare benzoic acid from 2-phenylethanol?
A1: Ar-CH2-CH2OH -> Ar-CH=CH2 (with H2SO4 - elimination) -> Ar-COOH (with KMnO4 - cleavage of double bonds)
Is that feasible?

Q2: Predict the product of the following transformation: Propylene oxide + HBr -> ?
Propylene oxide is: http://en.wikipedia.org/wiki/Propylene_oxide
A2: Is the product 1-brompropan-2-ol?

Thanks.

[macman104]

Seems feasible to me, but KMnO4 will also oxidize alkyl groups (including your ethanol) to your benzoic acid.

[THC]

Wouldn't KMnO4 oxidize Ar-CH2CH2OH to Ar-CH2CHO or Ar-CH2COOH?

Also, could you check Q2?

Thank you.

[azmanam]

#2 is wrong.  what are the differences in epoxide-opening mechanisms under acidic or basic conditions?

[THC]

#2 is wrong.  what are the differences in epoxide-opening mechanisms under acidic or basic conditions?

I didn't know... but I checked Wikipedia: http://en.wikipedia.org/wiki/Epoxide#Reactions
The product is 2-bromopropan-1-ol, right?
Also, wikipedia says: "Under basic conditions, the nucleophile attacks the least substituted carbon, in accordance with standard SN2 nuclephilic addition reaction process." Isn't R1 the least substituted carbon? If R2 is the most substituted carbon?

[azmanam]

The product is 2-bromopropan-1-ol, right?

yes.  Wanna discuss stereochemistry?

Under basic conditions, the nucleophile attacks the least substituted carbon

true.

Isn't R1 the least substituted carbon?

Well, I don't know what you label as R¹ in your starting material.  If we're referring to the wikipedia page, the carbon atom with R² attached is the most substituted in the acidic mechanism, and the carbon atom with R¹ attached is the more substituted in the basic mechanism.

So why do you suspect 2-bromopropanol is the product?  That would imply the nucleophile is attacking the more sterically hindered carbon atom. 

[THC]

Thank you

The product is 2-bromopropan-1-ol, right?

yes.  Wanna discuss stereochemistry?

It's a net anti reaction. I think I need more information before I can decide whether it's R or S.

Well, I don't know what you label as R¹ in your starting material.  If we're referring to the wikipedia page, the carbon atom with R² attached is the most substituted in the acidic mechanism, and the carbon atom with R¹ attached is the more substituted in the basic mechanism.

So why do you suspect 2-bromopropanol is the product?  That would imply the nucleophile is attacking the more sterically hindered carbon atom. 

Yes, I was referring to the wiki article. I just got a bit confused since R1 != R2, but I think I'm with you now.

[azmanam]

I think I need more information

Stereochemistry was not assigned in the problem, assume the starting material is a mixture of both enantiomers.

[THC]

I think I need more information

Stereochemistry was not assigned in the problem, assume the starting material is a mixture of both enantiomers.

Then the product is a racemic mixture.

Could you also help me with Q1?

[azmanam]

I think your answer to #1 is fine, but there is one slight technical detail missing from your second step.  The modification of one reaction condition changes the chemical outcome of the product.  Do you know what I'm talking about and what the two different conditions/products are?

[THC]

I think your answer to #1 is fine, but there is one slight technical detail missing from your second step.  The modification of one reaction condition changes the chemical outcome of the product.  Do you know what I'm talking about and what the two different conditions/products are?

Ah, of course. It's KMnO4 in acidic solution - if it was in basic solution, the product would be a 1,2-diol. But what about the solution macman104 proposed? That would be a lot easier.

[azmanam]

no... my book says both products occur under basic conditions.  but something else is modulated to provide either the diol or the di-acid.

As for the other proposal, I'm unfamiliar with that reactivity of KMnO₄, but that doesn't mean it wouldn't work.  I don't see that specific example in my undergrad organic text, but again that doesn't mean it wouldn't work.

[THC]

no... my book says both products occur under basic conditions.  but something else is modulated to provide either the diol or the di-acid.

As for the other proposal, I'm unfamiliar with that reactivity of KMnO₄, but that doesn't mean it wouldn't work.  I don't see that specific example in my undergrad organic text, but again that doesn't mean it wouldn't work.

That's weird. "Fundamentals of organic chemistry" (McMurry), pp. 113-114 says that alkene + KMnO4 in basic solution yields a diol, while alkene + KMnO4 in acidic solution cleaves the double bond and 1) if the double bond is tetrasubsituted, the products are ketones 2) if a H is present on the double bond, the product is a carboxylic acid and 3) if two H are present, CO2 is formed.
What does your book says?

[azmanam]

Ok, then stick with your text book, as that's the same textbook your instructor will be grading you from

My text says keeping the reaction cold and dilute favors diol, and concentrated and hot favors cleaved products (ketone/acid - depending).

Good work, though.

[THC]

Ok, then stick with your text book, as that's the same textbook your instructor will be grading you from

My text says keeping the reaction cold and dilute favors diol, and concentrated and hot favors cleaved products (ketone/acid - depending).

Good work, though.

I'll ask my instructor, then
Thanks.