Kap FL 2 2015, C/P Q23

[StarFall]

Spoiler alert: Question 23 from Chem/Phys section of Kaplan FL #2.



At first, I thought sugars are a-anomer when the -OH group on Carbon 1 (anomeric carbon) is axial and b-anomers are when -OH group on Carbon 1 are equitorial, leading me to Kaplans correct answer.

But I though of the wrong designation (even though I got the question correct). See link http://www.chem.ucla.edu/harding/ec_tutorials/tutorial08.pdf
How I learned to designate it was compare C1 OH group and C6 orientations.
a-anomer = OH on C1 opposite of C6 orientation (C1 axial or equitorial down, C6 up OR C1 up, C6 down)
b-anomer = OH on C1 same as C6 orientation (C1 down, C6 down OR C1 up, C6 up)

1. Which designation is correct? I'm assuming Kaplan is wrong.
2. II is Beta, III is alpha, how would you label I and IV?
3. Relevant thread with same question but still unsure: http://forums.studentdoctor.net/threads/sugars-question.1018919/

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[amy_k]

It is true that beta form is when the anomeric OH on c1 is cis to c6, and alpha form is trans. For IV and V i would compare the anomeric carbon c1 to the c5, which has the OH group on, since alpha and beta assigns based on the conformation of c1 to the reference carbon, that which is furthest away. So IV is beta and V is alpha.

 

[amy_k]

If you look up alpha-d-fructopyranose they have the OH group pointing down, though, which leads me to believe if there is no c6 configuration to compare then alpha is with the OH axial, which makes sense because alpha fructose in furanose form has the OH below the plane of the ring.

 

[StarFall]

It is true that beta form is when the anomeric OH on c1 is cis to c6, and alpha form is trans. For IV and V i would compare the anomeric carbon c1 to the c5, which has the OH group on, since alpha and beta assigns based on the conformation of c1 to the reference carbon, that which is furthest away. So IV is beta and V is alpha.

If you look up alpha-d-fructopyranose they have the OH group pointing down, though, which leads me to believe if there is no c6 configuration to compare then alpha is with the OH axial, which makes sense because alpha fructose in furanose form has the OH below the plane of the ring.

It seems like alpha for d-fructopyranose is when the OH group on the anomeric carbon is in the same direction as C3 and C4's OH groups. Wouldn't IV be alpha and V be beta then? http://upload.wikimedia.org/wikiped...Haworth.svg/1226px-D-Fructose_Haworth.svg.png
This would make V beta as the OH on the anomeric carbon is up and C3/C4 OHs are down (trans).

But this seems to go against the alpha = trans for C1 and C5 and beta = cis for C1 and C5 rule..
If you change the perspective and compare C2 OH group with C1/anomeric carbon's OH group, then it follows the alpha/trans, beta/cis rule. But I don't know how I feel about that "rule" in comparing C2 and C1..

II is beta and III is alpha for sure, correct?