[thesurgeon]

Well the first reaction is an alcohol condensation to an ether and the second is an esterification where the alcohol keeps the same configuration.. Why am I wrong?

[Hunter2]

Normaly you to get propene CH₃-CH=CH₂ what can add a second alcohol. I would say C will be main product.

The second willchange th stereochemistry. Probably you can have racemate.

[thesurgeon]

I didn't quite understand why the first one isn't an alcohol condensation... If you think the answer is C can you please tell me how is it possible to get it?

For the second question , the Oxygen of the alcohol attacks and no attack occurs on the chiral center , so why would we expect a racemic mixture?
Thanks.

[mjc123]

Look more closely at the stereochemistry of the alcohol. Don't be misled by how the bond is drawn.

[Hunter2]

Diisopropyl ether cannot synthesized in this way. Its possible to use 2-Brom-Propane and the Sodium isopropylate.

With Sulfuric acid mainly water is eliminated CH₃-CH⁺-CH₃ is unstable and loose von H⁺ of one of the Methyl-groups. This active  species can attack a second alcohol. I think the teacher gave also hint where to look.

[thesurgeon]

Thank you Hunter, I guess the section number was also misleading since it's the ether synthesis. Now I understand it's not a primary alcohol thus ether synthesis isn't favored. Thank you again, sir.

Mjc123, I tried my best to understand the stereochemistry of the alcohol but I didn't really understand how to proceed. I just know that the stereochemistry doesn't change in this reaction , however correct me if I'm wrong but if I choose answer C , it's the R configuration just like the one in the product? Thank you!

[Hunter2]

mjc123said it already. It is the way how it is drawn on the paper. I would choose the other one. With the Racemate I am wrong.

[kriggy]

Diisopropyl ether cannot synthesized in this way. Its possible to use 2-Brom-Propane and the Sodium isopropylate.

With Sulfuric acid mainly water is eliminated CH₃-CH⁺-CH₃ is unstable and loose von H⁺ of one of the Methyl-groups. This active  species can attack a second alcohol. I think the teacher gave also hint where to look.

Maybe I dont understand but I fail to see how CH₃-CH=CH₂ would form the product C. I agree with the elimination of water but I dont see how the other molecule of isopropanol could react at the methyl group with propene?
Is it thet the propene reacts with CH₃-CH⁺-CH₃ and then there is an addition of water?

[orgopete]

I can understand how C could be formed in problem 7, however I wouldn't ask this question. It is my opinion that if one wanted to form 4-methyl-2-pentanol from isopropyl alcohol, that most chemists would be unable to successfully carry out that reaction in the lab. I think it more likely they would get other products, such as propene or diisopropyl ether. If diisopropyl ether were not one of the choices, then I could predict C.

For problem 8, I'd say this is wrong. I agree with the poster, the S-alcolohol will give the S-ester. I don't see how R-configuration fits in here in any way. Am I missing something?

[mjc123]

Is it the S-ester? Look at the orientation of the wedge. Where is the H?

[Dan]

Is it the S-ester? Look at the orientation of the wedge. Where is the H?

I think it's ambiguous. It's an unusual way to draw it. If it was meant this way, then I'd say it is an unnecessary and confusing trap for students.

As a little experiment, I put different variations of wedges and dashes into ChemDraw and asked it to name the compounds (see below).

As you can see from the automatically generated names, ChemDraw does pay attention to the direction of solid wedges (as mjc123 is suggesting), but not dashed wedges.

[orgopete]

Kudos Dan for the 'real solution'. I didn't think of that possibility. This implies if the hydrogen were added to avoid ambiguity, it would have a second wedge to the carbon. I've never seen that drawn. I think this should be reported to ChemDraw.

[Dan]

Kudos Dan for the 'real solution'.

It was mjc123, not me who spotted it.