6
$\begingroup$

In converting glucose from its acyclic (open-chain) form to either cyclic form (alpha or beta) why is the C5 hydroxyl group used instead of the bottom-most hydroxyl group?

Also, are these structures considered isomers? All sources I've seen only say "form" or "representation."

$\endgroup$
5
$\begingroup$

Due to the internal ring strain the best possible conformation is a six-membered heterocycle in this case called pyranose . A five-membered heterocycle may in rare cases also be formed, called furanose. (From left to right: α-D-Glucopyranose, β-D-Glucopyranose, α-D-Glucofuranose, β-D-Glucofuranose; courtesy of wikipedia)
α-D-Glucopyranose β-D-Glucopyranose α-D-Glucofuranose β-D-Glucofuranose
Including the linear chain form they are all isomers (wikipedia), as they have the same chemical formula. More specifically the relationship between $\ce{linear <-> pyranose <-> furanose}$ is called constitutional or structural isomer.
The $\ce{\alpha-D <-> \beta-D}$ forms are diastereomers, or more specific diastereomeric conformers.

$\endgroup$
3
$\begingroup$

The cyclic form of glucose uses the C5 oxygen for bonding since it forms a 6 member ring (more stable then either 5 or 7 member ring).

They are isomers since they have the same chemical formula.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.