4

I understood the concept of hemiacetals quite well, but have considerable trouble remembering which are called $\alpha$ or $\beta$ respectively.
Is there some concept/mnemonic/common reaction/something else to faciliate remembering this convention?

caconyrn
  • 667
  • 1
  • 4
  • 18
  • 1
    The short answer is no, not really. Or, more precisely, not that I know of. Probably because $\alpha$ and $\beta$ depends on whether the sugar is a D or an L. I simply learned that for D, $\alpha$ means down. Then everything else logically follows. – Linear Christmas Jun 19 '16 at 17:23

2 Answers2

5

When you draw the Haworth projection, the alpha anomer is written with the -OH group pointing downwards (see the image), when you write the symbol alpha, you start by writing it downwards. For the beta anomer the -OH group is written upwards and when you write the symbol beta, you start writing it upwards. This mnemonic applies only to D-monosaccharides. For L-monosaccharides the opposite applies.

Actually, a better definition is that the alpha isomer has the -OH group pointing in the direction opposite to the -CH2OH group (on carbon 5), the opposite applies for the beta isomer, therefore for L-monosaccharides the -CH2OH group should point downwards and the alpha isomer will have the -OH group pointing upwards.enter image description here

EJC
  • 14,302
  • 15
  • 67
  • 157
3

When I was studying for my organic chemistry class last semester, I ran across an ACS published paper that provides several useful mnemonics for interconverting Fischer and Haworth projections of monosaccharides. I believe number four is what you are looking for specifically:

  1. Start the ring with the anomeric carbon on the right, and continue to draw the remaining atoms in the ring in a clockwise direction, ending with the ring oxygen at the top of the drawing.

  2. Hydroxyl groups on the right (Dextro-) are drawn Down; those on the left (Levo-) Levitate up.

  3. The last -$\ce{CH2OH}$ group is drawn beLow for an L-sugar; above for a D-sugar.

  4. The hydroxyl group attached to the anomeric carbon and the last -$\ce{CH2OH}$ group are drown Both on the same side of the ring for the $\beta$-anomer; for the $\alpha$-anomer each is Alone.

ringo
  • 24,013
  • 6
  • 87
  • 135