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Subject:
From:
Dr W Meier-Augenstein <[log in to unmask]>
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Date:
Wed, 20 Jul 2005 12:34:08 +0000
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Hi Tim,


From a carbohydrate chemistry point of view, I'm afraid the answer is yes.

If one dissolves sucrose (= saccharose, a disaccharide) in water, sucrose
will start to become hydrolysed into D-glucose (C6H12O6) and D-fructose
(C6H2O6). This reaction consumes one mol equivalent of water per mol
equivalent sucrose (C12H22O11), so the 18O composition of the product will
change and the degree of this change will depend how much of the sucrose
was hydrolysed before the reaction was stopped.

Before anybody says it, I know this reaction is pH dependent, i.e. the
reaction rate is directly proportional to the H+ concentration; but it will
take place even in 'pure' water albeit slow. One can follow the reaction
quite easily with a polarimeter as the polarisation angle of the (pure
sucrose) solution (+65°) becomes smaller and smaller and eventually (at
equilibrium) ends up being -20°.

While on this subject, again for reasons of sugar chemistry, I would be
cautious even when it comes to monosaccharides such as D-glucose since
glucose dissolved in water undergoes a change from alpha- to beta-glucose
involving an opening of the ring at the acetal-O between C5 and C1. On
paper, this step does require a "helper" molecule of water but I don't
think anybody has ever looked at this by e.g. dissolving D-glucose in
18O-labelled water.


Best regards,

Wolfram



On Jul 20 2005, Heaton, Timothy HE wrote:

> Can sucrose undergo O-isotope exchange with water? In preliminary tests
> we are finding slight differences in d18O between IAEA CH-6 weighed out
> into capsules as a solid powder, and the same IAEA CH-6 dissolved in
> water and pipetted into capsules. All cases used well-homogenised powder
> and thorough freeze-drying.
>
> Tim H.E. Heaton
>
> NERC Isotope Geosciences Laboratory
> British Geological Survey
> Keyworth, Nottingham NG12 5GG, England
> (www.bgs.ac.uk/nigl/index.htm)
>
> Tel. +44(0)115 936 3401
> Email: [log in to unmask]
>
>
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