Biomolecules

Glucose is an aldose and fructose a ketose — an aldehyde and a ketone, as different as two sugars can be. Yet both give exactly the same osazone, indistinguishable in melting point and crystal form. How can a reagent fail to tell an aldehyde from a ketone?

Because osazone formation consumes both carbon one and carbon two.

Phenylhydrazine reacts first at the carbonyl carbon, then oxidises the adjacent carbon and reacts there too. Whatever distinguished those two carbons is destroyed in the process. Glucose and fructose differ only at carbons one and two; from carbon three onward they are identical. Once the first two carbons have been converted to the same bis-hydrazone unit, nothing remains to tell them apart.

Mannose joins them for the same reason: it is the carbon-two epimer of glucose, so it too differs only within the region the reaction consumes. Reading which carbons a reagent touches is the key to the whole of carbohydrate chemistry, and it also settles which sugars reduce Tollens' reagent and which do not.

1. Anomers, epimers and mutarotation

Glucose in solution is almost entirely cyclic, the carbonyl having been attacked by the hydroxyl on carbon five. That creates a new stereocentre at carbon one, the anomeric carbon, and the two forms produced are called $\alpha$ and $\beta$.

Glucose and mannose are carbon-two epimers; glucose and galactose are carbon-four epimers.

Mutarotation is the drift in optical rotation when either pure anomer is dissolved. Pure $\alpha$-D-glucose starts at $+112^{\circ }$ and pure $\beta$ at $+19^{\circ }$; both settle at $+52.7^{\circ }$, the rotation of the equilibrium mixture. The interconversion goes through the open chain, which is why the ring must be able to open for mutarotation to occur at all.

Illustration 1

Explain why $\alpha$-methyl-D-glucoside does not show mutarotation, although glucose itself does.

In glucose the anomeric hydroxyl is a free hydroxyl, so the ring can open to the aldehyde and reclose the other way.

In the methyl glucoside that hydroxyl has been converted to an acetal, with a methoxy group in its place.

An acetal is stable to neutral and basic conditions and cannot open to the carbonyl, so the anomeric configuration is locked and no interconversion occurs.

The same fact makes glucosides non-reducing. Whatever prevents ring opening prevents both mutarotation and reduction of Tollens' reagent, since both need the open chain.

2. Reducing and non-reducing sugars

A sugar reduces Tollens' or Fehling's reagent only if it can open to a free carbonyl, which requires a free anomeric hydroxyl. All monosaccharides qualify, including ketoses such as fructose, which isomerise to an aldose under the alkaline conditions of the test.

For disaccharides the question is whether the glycosidic link has consumed one anomeric centre or both:

Sucrose is the only common non-reducing disaccharide, and the reason is entirely structural: its glycosidic bond joins the anomeric carbon of glucose to the anomeric carbon of fructose, so neither ring can open.

Illustration 2

Explain the term invert sugar and calculate the change in rotation on hydrolysing sucrose.

Sucrose has a specific rotation of $+66.5^{\circ }$.

Hydrolysis gives an equimolar mixture of glucose at $+52.7^{\circ }$ and fructose at $-92^{\circ }$.

The mixture's rotation is the average, $\frac{52.7-92}{2}=-19.7^{\circ }$.

The sign has inverted from positive to negative, which is the origin of the name.

Fructose is the more strongly rotating of the two, and its large negative value outweighs glucose's positive one — which is why the product mixture is laevorotatory despite containing equal amounts of a dextrorotatory sugar.

3. Amino acids: the zwitterion

An amino acid has both an acidic and a basic group, so in the solid state and in water near neutrality it exists as a zwitterion, with the carboxyl deprotonated and the amino group protonated. That internal salt explains their high melting points, their solubility in water and their insolubility in organic solvents.

The isoelectric point is the pH at which the molecule carries no net charge and therefore does not migrate in an electric field. For an amino acid with a neutral side chain,

$pI=\frac{p{K}_{a1}+p{K}_{a2}}{2}$

All twenty standard amino acids are $\alpha$-amino acids and all except glycine are chiral, occurring naturally in the L configuration.

Illustration 3

Glycine has $p{K}_a$ values of $2.34$ and $9.60$. Find its isoelectric point and state the direction it migrates at pH $6$ and at pH $12$.

$pI=\frac{2.34+9.60}{2}=5.97$

At pH $6$, essentially the isoelectric point, the molecule is a zwitterion with no net charge and does not migrate.

At pH $12$, well above $pI$, the amino group is deprotonated and the molecule carries a net negative charge, so it migrates towards the anode.

Above the isoelectric point an amino acid is anionic and below it cationic. That single rule predicts every electrophoresis result.

4. Proteins: four levels of structure

The peptide bond is planar, because the nitrogen lone pair is delocalised onto the carbonyl oxygen and the bond acquires partial double-bond character. That rigidity is what makes regular secondary structures possible at all.

An $\alpha$-helix is held by hydrogen bonds within one chain; a $\beta$-pleated sheet by hydrogen bonds between chains or between distant parts of one chain.

Denaturation destroys secondary and higher structure while leaving the primary sequence intact, which is why a boiled egg cannot be unboiled but its protein is still the same sequence.

Illustration 4

Explain why heating, extremes of pH and heavy metal ions all denature proteins, and why the primary structure survives.

Secondary, tertiary and quaternary structures are held almost entirely by non-covalent interactions: hydrogen bonds, ionic attractions and hydrophobic effects.

Heating supplies enough thermal energy to break these; extreme pH changes the charges on ionisable side chains and destroys the ionic interactions; heavy metal ions bind to thiol groups and disrupt disulphide bridges.

The primary structure is held by peptide bonds, which are covalent amides and far stronger than any of these.

Denaturation is a loss of shape, not of sequence, which is precisely why it is usually irreversible in practice but not in principle.

5. Polysaccharides: one linkage decides everything

Starch and cellulose are both polymers of glucose alone. They differ in one stereochemical detail at the glycosidic bond, and that single difference produces two materials with nothing in common.

The $\alpha$ linkage forces a bend at every unit, so the chain coils into a helix that packs loosely and is easily reached by water and enzymes. The $\beta$ linkage lets each unit sit rotated by half a turn from the last, so the chain runs straight. Straight chains lie alongside one another and hydrogen bond extensively into rigid fibres, which is why cellulose is the structural material of plants and starch is not.

Human digestive enzymes hydrolyse $\alpha$-$1,4$ bonds only. We can therefore digest starch and glycogen but not cellulose, despite all three being made of the same sugar.

Illustration 5

Explain why cellulose is fibrous and insoluble while starch swells and disperses in hot water.

Cellulose's $\beta$-$1,4$ linkage gives straight chains that lie parallel and form extensive hydrogen bonds between neighbouring strands.

Those inter-chain bonds must all be broken for water to penetrate, which is energetically prohibitive, so cellulose remains as insoluble fibres.

Starch's $\alpha$-$1,4$ linkage produces a helix, and helices cannot pack closely or hydrogen bond to one another as effectively.

Water reaches the individual chains far more easily, so starch granules swell and disperse on heating, forming the familiar paste.

Illustration 6

Ruminants digest cellulose but humans cannot, although both eat the same plant material. Explain.

Human enzymes are specific to the $\alpha$-$1,4$ glycosidic bond, so they hydrolyse starch and glycogen readily and cellulose not at all.

Enzyme specificity arises from the precise three-dimensional fit between substrate and active site, and the $\beta$ linkage presents a different geometry entirely.

Ruminants do not produce a cellulase either. They host bacteria in the rumen that do, and absorb the products of that bacterial digestion.

The difference is microbiological rather than chemical. The cellulose is identical in both cases; only the available enzymes differ.

6. Nucleic acids

Base pairing is specific: adenine with thymine through two hydrogen bonds, guanine with cytosine through three. A purine always pairs with a pyrimidine, which keeps the width of the helix constant along its length. Chargaff's rules follow directly: in any DNA the amount of adenine equals that of thymine and guanine equals cytosine.

Because the guanine-cytosine pair has an extra hydrogen bond, DNA rich in those bases requires a higher temperature to separate the strands.

Illustration 7

A sample of DNA contains $22\%$ adenine. Deduce the percentage of each of the other three bases.

By Chargaff's rules, thymine equals adenine, so thymine is also $22\%$.

Together they account for $44\%$, leaving $56\%$ for guanine and cytosine combined.

Since those two are equal, each is $28\%$.

The rules follow from pairing, not from any chemical accident. Each adenine on one strand is opposite a thymine on the other, so the totals must match exactly.

Illustration 8

Two DNA samples melt at $85^{\circ }$C and $95^{\circ }$C. Which has the higher guanine content, and why?

The sample melting at $95^{\circ }$C.

Melting means separating the two strands, which requires breaking every hydrogen bond between them.

A guanine-cytosine pair is held by three hydrogen bonds against two for adenine-thymine, so a strand richer in guanine and cytosine needs more energy to separate.

Melting temperature is a direct measure of base composition, and it is used routinely to estimate it without any sequencing.

Illustration 9

Explain why sucrose does not reduce Fehling's solution while both maltose and lactose do.

Reduction requires the sugar to open to a free carbonyl, which needs a free anomeric hydroxyl.

In maltose and lactose the glycosidic bond uses the anomeric carbon of only one of the two rings, leaving the other free to open.

In sucrose the bond joins the anomeric carbon of glucose to the anomeric carbon of fructose, so both are consumed and neither ring can open.

Sucrose is the standard example of a non-reducing disaccharide, and hydrolysing it restores the reducing behaviour of both components at once.

Illustration 10

Glucose does not give the Schiff test and does not react with sodium bisulphite, although it contains an aldehyde group. Explain.

Both tests require a free aldehyde group in appreciable concentration.

In aqueous solution glucose exists almost entirely in its cyclic hemiacetal forms, with less than one per cent present as the open chain at any instant.

That concentration is too low for these tests, which are not driven forward by consumption of the product.

Tollens' and Fehling's tests still work, because they consume the open chain irreversibly and pull the ring-opening equilibrium across — which is why glucose is a reducing sugar despite failing the aldehyde tests.

Illustration 11

Classify each pair as anomers, epimers or neither: $\alpha$-D-glucose and $\beta$-D-glucose; D-glucose and D-mannose; D-glucose and L-glucose.

$\alpha$- and $\beta$-D-glucose differ only at the anomeric carbon: anomers.

D-Glucose and D-mannose differ at carbon two only: epimers.

D-Glucose and L-glucose differ at every stereocentre: they are enantiomers, and are neither anomers nor epimers.

The distinction is simply how many centres differ: one anomeric, one other, or all of them.

Illustration 12

Lysine has $p{K}_a$ values of $2.2$, $9.0$ and $10.5$, the last belonging to its side chain. Predict whether its isoelectric point lies above or below $7$.

The side chain is basic and carries a positive charge when protonated.

To reach zero net charge, that positive charge must be removed, which requires a fairly high pH.

The isoelectric point is therefore well above $7$, at the average of the two values that bracket the neutral species: $\frac{9.0+10.5}{2}=9.75$.

Basic amino acids have high isoelectric points and acidic ones have low ones. Averaging the two $p{K}_a$ values on either side of the neutral form gives the right answer in every case.

Illustration 13

Explain why an enzyme is far more effective than an ordinary catalyst and why it works only within a narrow range of temperature and pH.

An enzyme binds its substrate in an active site shaped to fit the transition state rather than the reactant, which lowers the activation energy enormously and can accelerate a reaction by many millions of times.

That active site is held in shape by hydrogen bonds, ionic interactions and hydrophobic effects, all of them non-covalent.

Raising the temperature or changing the pH disrupts those interactions, so the site loses its shape and the enzyme is denatured.

Specificity and fragility have the same cause. The precise fit that makes an enzyme so effective is exactly what makes it so easily destroyed.

Summary

• Osazone formation consumes carbons one and two, which is why glucose, fructose and mannose all give the same product.
• Anomers differ only at the anomeric carbon, epimers at exactly one other, enantiomers at all of them.
• Mutarotation goes through the open chain: $\alpha$ at $+112^{\circ }$ and $\beta$ at $+19^{\circ }$ both settle at $+52.7^{\circ }$.
• A glycoside cannot mutarotate or reduce Tollens', because its acetal cannot open.
• A sugar is reducing only if it has a free anomeric hydroxyl. All monosaccharides qualify, including ketoses.
• Sucrose is non-reducing because its link joins both anomeric carbons; maltose and lactose leave one free.
• Invert sugar: $+66.5^{\circ }$ becomes $-19.7^{\circ }$, because fructose's $-92^{\circ }$ outweighs glucose's $+52.7^{\circ }$.
• Amino acids exist as zwitterions, hence high melting points and water solubility.
• $pI=\frac{p{K}_{a1}+p{K}_{a2}}{2}$ for neutral side chains; above $pI$ the molecule is anionic, below it cationic.
• The peptide bond is planar, from delocalisation of the nitrogen lone pair, which is what permits regular secondary structure.
• $\alpha$-Helix uses hydrogen bonds within a chain, $\beta$-sheet between chains.
• Denaturation destroys shape, not sequence, because only non-covalent forces are broken.
• Starch is $\alpha$-$1,4$ and coils; cellulose is $\beta$-$1,4$ and runs straight into hydrogen-bonded fibres.
• Human enzymes hydrolyse $\alpha$-$1,4$ only, which is why we digest starch and not cellulose.
• A-T has two hydrogen bonds and G-C three, so guanine-rich DNA melts higher; Chargaff's rules follow from pairing.