Sandbox 1A2
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Shown in the box to the right is an animation of the cyclization of D-glucose to form D-glucopyranose. The electron-withdrawing oxygen atom of the carbonyl group carries a permanent positive dipole while the alcohol group bound to the penultimate carbon acts as a nucleophile due to the two lone pairs of electrons around the oxygen atom. Listed below is an explanation of the animation broken down into two major steps.
(This button may be used at any time to pause animation for a clearer view of the process.)
Step 1 - Starting from its linear form, glucose begins to cyclize with the 90 degree rotation of the bond between C4 and C5. This causes the molecule to be "bent" in such a way that the C5 atom, or the penultimate carbon, is near the first carbon. The glucose molecule is now configured in a way that the "tail" of glucose (the penultimate carbon containing the alcohol) is close to the "head" of glucose (the carbon containing the aldehyde).
Step 2 - The nucleophilic group on the penultimate carbon is then able to undergo an intramolecular reaction with the carbonyl group of the aldehyde. This nucleophilic attack releases the hydrogen atom from the alcohol (in white) and attaches to the oxygen that was originally an aldehyde, now making it an alcohol.
Free rotation around the first carbon and second carbon allows for the formation of 2 anomers - alpha-D-glucopyranose & beta-D-glucopyranose. The specific anomer is decided by the position of the -OH group of the anomeric carbon; and the anomeric carbon can always be spotted by finding the carbon that is attached to two oxygens.