به انگلیسی

Rings, anomers and sweet links

Science · Chemistry · ages 21-22

Name ______________________ Date ____________

A D-glucose Fischer projection has C1 at the top and C3 OH on the left. Use a Haworth view with ring oxygen at upper right and C1 at right. Where does C3 OH point?
- Down, by keeping Fischer-left groups below the ring.
- Up, by mapping Fischer-left groups above the ring.
- Inward, by placing the C3 oxygen within the ring.
Which bond forms glucose's common glucopyranose ring, and which carbon becomes anomeric?
- C5 hydroxyl oxygen bonds to C1; C1 becomes anomeric.
- The C4 hydroxyl oxygen bonds to C1; C1 becomes anomeric.
- C5 hydroxyl oxygen bonds to C2; C2 becomes anomeric.
What does mutarotation describe when a pure glucose anomer dissolves in water?
- Changing optical rotation as the ring adopts another chair.
- Changing optical rotation as sugars form new O-glycosidic links.
- Changing optical rotation as free anomers approach equilibrium.
Why does an ordinary glucose O-glycoside usually resist base but hydrolyse under suitable aqueous-acid conditions?
- An acetal centre: acid-assisted cleavage permits hydrolysis with water.
- An ester centre: acid-assisted cleavage permits hydrolysis with water.
- A hemiacetal centre: base-assisted cleavage releases the linked alcohol.
In ordinary aqueous solution, which D-glucopyranose anomer is favored, and what contributes to that preference in its usual 4C1 chair?
- Alpha, with an equatorial anomeric OH that reduces axial crowding.
- Beta, with an equatorial anomeric OH that reduces axial crowding.
- Beta, with an axial anomeric OH that reduces axial crowding.
Under suitable Benedict's test conditions, maltose is reducing but intact sucrose is non-reducing.
Circle one: True False
Methyl glucoside can undergo ordinary mutarotation in water while its glycosidic link remains intact.
Circle one: True False
A soluble disaccharide gives a negative Benedict's test at suitable concentration with working controls. An analyst says this proves it is sucrose. What is justified?
- The result identifies sucrose because it establishes that both anomeric centres are linked.
- The result supports a reducing sugar whose free anomeric centre opens in water.
- The result supports non-reducing behavior; the disaccharide's identity remains open.

Answer key

For grown-ups. Fold this page away before handing over the rest.

Rings, anomers and sweet links W1-mt_XpSBU8b0HN-s1

Up, by mapping Fischer-left groups above the ring. · In the stated Haworth view, a Fischer-left OH at C2 to C4 maps to up. The C3 OH therefore points above the ring; its oxygen does not become the ring oxygen.
C5 hydroxyl oxygen bonds to C1; C1 becomes anomeric. · The C5 hydroxyl oxygen bonds to aldehyde carbon C1. Proton transfer completes the six-membered cyclic hemiacetal, with C1 as its anomeric centre.
Changing optical rotation as free anomers approach equilibrium. · Ring opening and closing change the anomer ratio. The observed optical rotation changes until equilibrium is reached.
An acetal centre: acid-assisted cleavage permits hydrolysis with water. · Acetal chemistry accounts for usual base stability. Acid-assisted cleavage allows hydrolysis and restores a free anomeric centre.
Beta, with an equatorial anomeric OH that reduces axial crowding. · Beta is favored under these conditions. Its equatorial anomeric OH avoids axial steric interactions present in the alpha chair.
True · Maltose keeps one free anomeric centre. Sucrose links glucose C1 to fructose C2, leaving neither free.
False · Its anomeric centre is an acetal, not a free hemiacetal. The intact glycoside cannot open through the ordinary mutarotation pathway.
The result supports non-reducing behavior; the disaccharide's identity remains open. · The controlled result supports non-reducing behavior under those conditions. It does not uniquely identify sucrose.