BR orgo Section 6 passage 7 question 44

[Student1331]

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BR orgo Section 6 passage 7 question 44
Treatment of Compound X with sodium borohydride (NaBH4) would yield which of the following?

A - An optically active diacid
B - An optically inactive diacid
C- An optically active hexa-ol
D- An optically inactive hexa-ol

The book says the correct answer is C, but would the answer be D because the compound would turn meso - thus optically inactive?

 

[MedPR]

What is compound X?

 

[Student1331]

Looks like ribose but has six carbons (see in fisher projection)

http://www.google.com/imgres?q=ribo...nw=84&start=0&ndsp=30&ved=1t:429,r:0,s:0,i:67

 

[syoung]

Looks like ribose but has six carbons (see in fisher projection)

http://www.google.com/imgres?q=ribo...nw=84&start=0&ndsp=30&ved=1t:429,r:0,s:0,i:67

http://www.scribd.com/api_user_11797_hamzas_pecial/d/7237703-Carbohydrates
says that NaBH4 reduction of a sugar simply reduces the carbonyl carbon on the aldehyde side, so you would still get chiral centers resulting in active hexa-ol.

 

[Student1331]

Right, but then the molecule becomes meso, so wouldnt that make in not optically active?

 

[syoung]

Well are you saying it's a allose sugar molecule?
http://images.search.yahoo.com/r/_y.../imagesF09/L17/Energy%20Metabolism%20MPs.html

Yeah, the thing would be very meso if it were D-Allose, and i think meso is not opt active...

 

[Student1331]

Yep, allose (didnt know the compound name). Seems like the book is wrong here

 

[syoung]

Sounds like from all the research I'm trying to find. meso achiral compounds are not optically active, more so in this case 6 carbon sugar version of ribose is allose and when you reduce with NaBH4 it turns to D-allitol which is a meso achiral compound