comed

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When D-glucose undergoes hemiacetyl formation, it shows the hydroxyl group on C5 acting as the nucleophile to make the chair formation.

Why doesn't the hydroxyl group on C6 act as the nucleophile also (from the CH2OH), since ring structures are stable in both 5/6 member forms?

Thanks in advance!
 

theonlytycrane

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It would actually be the hydroxyl on C4 acting as a nucleophile since the oxygen would be included as one member of the 5 membered ring. It's possible actually. The 6 membered ring is just more commonly found due to stability of the cyclohexane chair (most notably the beta anomer).
 
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comed

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ah, i just realized my misunderstanding after i read your post, @theonlytycrane
in the kaplan book, it has C5 acting as a nucleophile to form a 6 membered ring, including the O.
if the hydroxyl in the CH2OH acted as a nucleophile, that would lead to a 7 membered ring (including the O), which wouldn't be stable d/t ring strain.

haha.... thanks for making me realize my mistake, @theonlytycrane ! glad i have that cleared up now.
 
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comed

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Feb 20, 2011
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@betterfuture
yeah, you're right. i just got confused because i forgot to count Oxygen as a member in the ring structure.

so, my initial questions were: why does C5 act as nucleophile when that will just make a 5 member ring structure (WRONG because C5 as nucleophile = 6 member ring)
why doesn't the CH2OH act as a nucleophile to make a 6 member structure also (WRONG because that will lead to 7 member ring, which won't happen because of steric hinderance)

and as @theonlytycrane pointed out, C4 can act as nucleophile to make a 5 membered ring.