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What do we get when we joint two α- glucoses via an α1,4 bond?

  1. Is it alpha maltose or beta maltose?
  2. In other words, if we cleave amylopectin with amylase (α-amylase or β-amylase), which form of maltose is the ultimate result?

I think Maltose, which has an axial anomeric hydroxy group, will be the answer to (1) and (2), because α-glucose has has axial anomeric hydroxy group; am I right? However, I could not identify which form of maltose has axial anomeric hydroxy group.

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1 Answer 1

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The anomeric carbon of of the second glucose unit of maltose interconverts between the alpha and the beta form in aqueous solution (just like glucose). The stereochemistry of the anomeric carbon in the glycosidic bond, on the other hand, is fixed.

For some sugars, you can isolate one or the other stereoisomer in the solid form. Glucose, for example, shows a large difference in solubility between the alpha and the beta form. If you crystallize glucose at room temperature, you get crystals of alpha glucose (details). In this crystal structure of maltose, the main form is the alpha isomer, with about 20% in the beta form. In solution, the ratio of alpha and beta will be different (and there will be some open form species as well), and will depend on the temperature.

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  • $\begingroup$ Thank you!! So, is it correct to understand that for (1), both α-maltose and β-maltose can be made, and for (2), both α-maltose and β-maltose can be made? $\endgroup$ Commented May 16, 2023 at 13:18
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    $\begingroup$ @BlueVarious Yes, and usually you don't have a choice. Enzymes might make one rather than the other, but then they interconvert anyway and reach an equilibrium mixture (of alpha, beta and open form) after a while. $\endgroup$
    – Karsten
    Commented May 16, 2023 at 13:24
  • $\begingroup$ Thank you. So am I right in understanding that if the anomeric hydroxy group of α-glucose and the OH at position 4 of β-glucose are joined by an α1,4 bond, α-maltose and β-maltose would still be in similar chemical equilibrium? $\endgroup$ Commented May 16, 2023 at 14:44
  • $\begingroup$ I would be glad to have your answer to this question as well. chemistry.stackexchange.com/questions/173748/… $\endgroup$ Commented May 16, 2023 at 15:03

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