Question 19M.3.hl.TZ2.12
| Date | May 2019 | Marks available | [Maximum mark: 4] | Reference code | 19M.3.hl.TZ2.12 |
| Level | hl | Paper | 3 | Time zone | TZ2 |
| Command term | Classify, Sketch, State | Question number | 12 | Adapted from | N/A |
Sucrose is a disaccharide.
State the name of the functional group forming part of the ring structure of each monosaccharide unit.
[1]
acetal
OR
ether [✔]
Note: Accept “glycosidic bond/linkage” but not “glucosidic”.
Do not accept “hemiacetal”.
This was reasonably answered although there were some candidates who stated ester or hemiacetal, which is incorrect.
Classify, giving your reason, the hexose (six-membered) ring of sucrose as an α or β isomer.
[1]
α-isomer AND hydroxyl group on carbon 1 and –CH2OH are trans
OR
α-isomer AND hydroxyl group on carbon 1 is below plane of ring
OR
α-isomer AND glycosidic linkage between rings is below plane of ring [✔]
Note: Accept “ether linkage” for M3.
This part was very poorly answered. Majority of the candidates had no idea about the reason whether the six-membered ring was an alpha or beta isomer.
Sketch the cyclic structures of the two monosaccharides which combine to form sucrose.
[2]
[✔]
[✔]
This question was poorly answered. Many candidates lost marks due to sloppy drawing and incorrect bond linkages. Some candidates did not separate the two monosaccharides as instructed.
