A brief note on carbohydrates: classification, the structure of glucose, anomers and mutarotation, the reactions of glucose, disaccharides and polysaccharides. With a worked example, practice, a program and questions to check yourself.
A brief note on Carbohydrates. Read it before a test. Try each question at the end before you open its answer.
Classification
A carbohydrate is a polyhydroxy aldehyde or ketone, or gives one on hydrolysis.
Monosaccharides (glucose, fructose), oligosaccharides (2 to 10 units, such as sucrose) and polysaccharides (starch, cellulose, glycogen).
Aldose: C=O at C-1. Ketose: C=O at C-2. Glucose is an aldohexose and fructose a ketohexose.
Reducing sugars reduce Benedict's, Fehling's and Tollens' reagents. They have a free anomeric carbon.
Stereochemistry
n stereocentres give 2n stereoisomers. An aldohexose has 4 centres and 16 isomers. A ketohexose has 3 and 8.
D or L depends on the OH on the last stereocentre (right is D). It is not the sign of the rotation.
Epimers differ at one stereocentre. Mannose is the C-2 epimer of glucose. Galactose is the C-4 epimer.
Glucose
Evidence: C6H12O6, a straight chain (n-hexane with HI), five OH (pentaacetate), an aldehyde (bromine water gives gluconic acid) and a primary OH (nitric acid gives saccharic acid).
The C-5 OH closes the ring: a six-membered pyranose with a hemiacetal at C-1.
Anomers: α has the C-1 OH down and β has it up (D sugars, Haworth ring). Right in Fischer is down in Haworth.
Mutarotation:+112.2 (α) and +18.7 (β) degrees go to +52.7 degrees, about 36% α and 64% β.
Glucose and fructose both give Tollens', Fehling's and the same osazone.
Disaccharides and polysaccharides
Sucrose: α(1→2)β, non-reducing, gives glucose and fructose (invert sugar, +66.5 to −19.9 degrees).
Maltose: α(1→4), reducing. Lactose: β(1→4), reducing, gives galactose and glucose.
Starch: amylose (about 20%, unbranched, α(1→4)) and amylopectin (about 80%, α(1→4) with α(1→6) branches).
Cellulose: β(1→4), straight fibres, not digested by humans. Glycogen: highly branched animal store.
n units joined in a chain release n−1 molecules of water.
A worked example
Find the molar mass of a chain of 100 glucose units.
Links. One hundred units have 99 links.
Mass.100×180.16−99×18.02=18016−1784=16232g/mol.
Practice with new numbers
A program to try
Mistakes to avoid
Counting units as links. A chain of n units has n−1 links.
Calling sucrose reducing. Both anomeric carbons are in the link.
Mixing D with the sign (+). They are different ideas.
Calling mannose an anomer of glucose. It is an epimer.
Check yourself
How many stereoisomers has an aldopentose?
Which anomer of D-glucopyranose has the C-1 OH on the same face as the CH2OH group?
Why does sucrose not reduce Fehling's solution?
How many molecules of water are released when 30 glucose units join in a chain?
Sources
Nelson and Cox, Lehninger Principles of Biochemistry, 8th edition, Chapter 7.
NCERT Chemistry for Class 12, Part II, the chapter Biomolecules.