TBR (New Book) Orgo Section VII Question 16 Errata

[plukfelder2017]

Hi Folks,

-I am not sure if any of you folks have bought the new TBR books. I had a quick question on #16 in Section VII of the Orgo Books under the Carbohydrates Review Questions. In this question, they ask you which structure represents the C6 deoxy form of L-mannose.

-I understand the logic which states that mannose is a C2 epimer of d-glucopyranose. I also understand that d-glucopryanose has all of its bulky groups in the equatorial position. This means that d-mannose would have all of its bulky groups in the equatorial position other than C2 where the hydroxyl group would be in the axial position.

TBR, however, asks for the chair confirmation of L-mannose and I wanted to confirm that L-mannose would essentially have all substituents in opposite config relative to D-mannose. If so, this would invalidate their answer.

Thanks,
plukfelder2017

@BerkReviewTeach (if you could provide some guidance I would appreciate it as well.)
***PS: your new books rock.

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[aalamruad]

Hi Folks,

-I am not sure if any of you folks have bought the new TBR books. I had a quick question on #16 in Section VII of the Orgo Books under the Carbohydrates Review Questions. In this question, they ask you which structure represents the C6 deoxy form of L-mannose.

-I understand the logic which states that mannose is a C2 epimer of d-glucopyranose. I also understand that d-glucopryanose has all of its bulky groups in the equatorial position. This means that d-mannose would have all of its bulky groups in the equatorial position other than C2 where the hydroxyl group would be in the axial position.

TBR, however, asks for the chair confirmation of L-mannose and I wanted to confirm that L-mannose would essentially have all substituents in opposite config relative to D-mannose. If so, this would invalidate their answer.

Thanks,
plukfelder2017

@BerkReviewTeach (if you could provide some guidance I would appreciate it as well.)
***PS: your new books rock.

Nope, they'd end up equatorial in L-mannose too (except for the C2 hydroxyl group). I think you're assuming that L-mannose should have all OH's axial except for C2 which should be equatorial, but that's not the case. It would just flip to its other chair conformation to make all but one OH's equatorial, making the C2 OH axial.

Look at this picture for example. All OH's are equatorial in both:

Edit: picture didn't work. Just type "L glucose ring" into google and look at the first picture

The Fischer projection, though, has all the OH's on opposite sides:

 

[plukfelder2017]

Ah. I completely forgot that molecules interconvert between chair-boat-reverse chair.

Awesome answer. Thanks for the clarification.

 

[BerkReviewTeach]

Sorry for being late to the party, but "what walloobi said"! That was an eloquent answer walloobi.

plukfelder2017, you have a great understanding of the topic here, and just happened to forget about ring-flipping. That is probably because we emphasize that all of the chiral centers change when going from one enantiomer to another (thus all of the axials become equatorial and all of the equaltorials become axial). But you can also think of it as D-deoxymannose being the mirror image of L-deoxymannose, in which case equatorial substituents will remain equatorial and axial substituents will remain axial, just the mirror image version.

From a learning standpoint, you took the perfect perspective on this question of inspecting axial versus equatorial compared to glucopyranose. That is a fast and efficient way to answer questions like these.

On a related note, thanks for posting a question from the new organic chemistry book. We are so very proud of that book and are very excited to see its impact on our homestudy students. Keep up the excellent work.

 

[plukfelder2017]

Thank you for the guidance @BerkReviewTeach. I will make sure to pepper this forum with additional questions as I run across them. I have also acquired your new Psychology/Sociology book and should be asking questions out of there as well.