Ch25 : Carbohydrates

Chapter 25: Carbohydrates

Carbohydrate Answers

Qu1:

This is a Fischer projection.

Qu 2:

In a Fischer projection the horizontal bonds are towards you, the vertical away as shown below:

To assign the R / S configuration, need to assign the priorities of the groups at the chirality center.
HO > CH=O > CH2OH > H. Now with the lowest priority group away, the sense of rotation of highest to lowest priority is counter-clockwise, so this is the S stereoisomer. Review this ?

To assign the D / L configuration, need to remember that in the Fischer projection the -OH group is on the left in the L stereoisomer.

Qu 3:

This is the mirror image of that in qu2, so this is the R- or D-glyceraldehyde.

Note how important it is to remember that the Fischer diagram of sugars needs to be drawn with the carbonyl (oxidised) end uppermost.

Qu 4:

Both are Fischer projections, drawn the same way up, so a straight comparison can be made. The systems are isomers, differing only in the arrangement of the groups at one of the two chirality centers. So they are stereoisomers, specifically diastereomers.

Qu 5:

In each case, the anomeric center has been indicated by * . It is important that you can recognise this center... you need it for part (ii)

(a)		A hemi-acetal, pyranose The -OH group at the anomeric center indicates the hemi-. The -H group (not shown) at the anomeric carbon implies the structure is based on an aldehyde, so we have an acetal. The cyclic unit is 6 atoms so it is a pyranose.
(b))		A hemi-ketal, furanose The -OH group at the anomeric center indicates the hemi-. The anomeric carbon is attached to 2 other C atoms so the structure is based on a ketone so we have an ketal. The cyclic unit is 5 atoms so it is a furanose.
(c)		A hemi-acetal, pyranose The -OH group at the anomeric center indicates the hemi-. The -H group (not shown) at the anomeric center implies the structure is based on an aldehyde so we have an acetal. The cyclic unit is 6 atoms so it is a pyranose.

Qu 6:

In each case the anomeric center / carbonyl group has been indicated by * . It is important that you can recognise this center... you need it for part (i)

(a)		The anomeric carbon is only attached to one other C atom, so it corresponds to an aldehyde. Hence, the carbohydrate is an aldose. The carbohydrate contains 6 C atoms, so it is a hexose.
(b)		The carbonyl carbon is attached to two other C atoms, so it is a ketone. Hence, the carbohydrate is a ketose. The carbohydrate contains 6 C atoms, so it is a hexose.
(c)		The carbonyl carbon is attached to only one other C atom, so it is a aldehyde. Hence, the carbohydrate is an aldose. The carbohydrate contains 4 C atoms, so it is a tetrose.

Qu 7:

A reducing sugar needs to have an aldehyde, a hemi-acetal group or be a ketose that can tautomerise to a aldose.

(a) is not a reducing sugar as there is not an -OH at the anomeric center.
(b) Sucrose is not a reducing sugar as both anomeric centers are glycosides (i.e. acetals) not hemi-acetals.
(c) Glucose is a reducing sugar, here the aldehyde group is obvious.