John McMurry, Professor Emeritus

Problem 25-1

(a)

Aldotetrose

(b)

Ketopentose

(c)

Ketohexose

(d)

Aldopentose

Problem 25-2

(a)

S

(b)

R

(c)

S

Problem 25-3

A, B, and C are the same.

Problem 25-4

The central carbon is bonded to C H 3 on right, C H 2 O H on left, hydrogen on the top and chlorine on the bottom.

Problem 25-5

Fischer projection of aldotetrose. It comprises four carbon where C 2 and C 3 have an R configuration. C H O is positioned on the top.

Problem 25-6

(a)

L-Erythrose; 2S,3S

(b)

D-Xylose; 2R,3S,4R

(c)

D-Xylulose; 3S,4R

Problem 25-7

The structure of L-arabinose which is an aldopentose containing five carbon atoms and including an aldehyde functional group.

Problem 25-8

(a)

(b)

(c)

Problem 25-9

16 D and 16 L aldoheptoses

α-D-Allopyranose

D-Galactitol has a plane of symmetry and is a meso compound, whereas D-glucitol is chiral.

Problem 25-18

The −CHO end of L-gulose corresponds to the −CH₂OH end of D-glucose after reduction.

Problem 25-19

D-Allaric acid has a symmetry plane and is a meso compound, but D-glucaric acid is chiral.

Problem 25-20

D-Allose and D-galactose yield meso aldaric acids; the other six D-hexoses yield optically active aldaric acids.

Problem 25-21

D-Allose + D-altrose

Problem 25-22

L-Xylose

Problem 25-23

D-Xylose and D-lyxose

Problem 25-24

Fischer projection of N-acetylmannosamine. The reaction of N-acetylmannosamine with pyruvate ion involving a base to form neuraminic acid. The reaction has multiple steps.

Problem 25-25

(a)

The hemiacetal ring is reduced.

(b)

The hemiacetal ring is oxidized.

(c)

All hydroxyl groups are acetylated.

Chapter 25