Organic Chemistry 202-SNU-RE • CEGEP, Montreal

Revision sheet: biomolecules (202-SNU-RE)

Biomolecules look like a chapter apart, reserved for biology. In reality, everything in it has already been seen: the glycosidic bond is an acetal, the triglyceride is a triester, and the zwitterion is the product of an internal acid-base reaction.

This sheet leaves the structures in the course notes and gathers the rules for READING: the Fischer convention, the move to Haworth, the criterion for a reducing sugar, and the calculation of the isoelectric point.

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The thread of the chapter

Biomolecules require no new reaction: they are acetals, esters and zwitterions, and the whole difficulty is to READ a projection correctly.

This chapter is part of 202-SNU-RE, Organic Chemistry

The essentials

Monosaccharides and counting stereoisomers

• An ALDOSE carries an aldehyde, a KETOSE a ketone. Carbons are counted to name them: triose, tetrose, pentose, hexose.
• An aldose with $n$ carbons has $n-2$ asymmetric carbons, hence $2^{n-2}$ stereoisomers.
• An aldohexose therefore has $2^{4}=16$ stereoisomers, of which $8$ are of the D series.
• In a FISCHER PROJECTION, the carbon chain is VERTICAL, the most oxidized carbon at the top; horizontal bonds point TOWARD the viewer.

The Fischer projection of D-glucose: it is the asymmetric carbon FARTHEST from the aldehyde, at the bottom, that decides the series.

The first and last carbons of an aldose are not asymmetric: one carries the aldehyde, the other two hydrogens. That is where the $-2$ in the exponent comes from.

D or L series, and epimers

• The series is decided by the asymmetric carbon FARTHEST from the carbonyl: $-OH$ on the RIGHT in Fischer means D series.
• Almost all natural sugars are of the D series, almost all natural amino acids of the L series.
• EPIMERS: two diastereomers that differ at ONE SINGLE asymmetric carbon. Glucose and mannose are epimers at $C_{2}$ , glucose and galactose at $C_{4}$ .
• The D or L series says NOTHING about the direction in which polarized light is rotated, which is written $(+)$ or $(-)$ and is measured.

Mannose differs from glucose only at C2, galactose only at C4 (purple lines): two epimers, hence two diastereomers.

Two epimers are diastereomers, not enantiomers: they have different physical properties and can be separated by ordinary means.

From Fischer to Haworth, and the glycosidic bond

• What is on the RIGHT in Fischer ends up BELOW the plane in Haworth.
• $\alpha$ anomer: anomeric $-OH$ on the side OPPOSITE the $CH_{2}OH$ . $\beta$ anomer: on the SAME side.
• The GLYCOSIDIC BOND is an ACETAL formed between the anomeric carbon of one sugar and a hydroxyl of another.
• A disaccharide is REDUCING if at least one anomeric carbon remains free, hence in hemiacetal form.

The two anomers differ only by one side: the anomeric hydroxyl group is opposite the

CH_{2}OH

in alpha, on the same side in beta.

STARCH: $\alpha(1\rightarrow 4)$ bonds with $\alpha(1\rightarrow 6)$ branches. CELLULOSE: $\beta(1\rightarrow 4)$ . This is the ONLY difference, and it decides digestibility.

Lipids and amino acids

• A TRIGLYCERIDE is the triester of glycerol with three fatty acids: basic hydrolysis gives soap.
• IODINE VALUE: mass of $I_{2}$ , in grams, taken up by $100$ g of fat. It measures UNSATURATION, with $M(I_{2})=253.81$ g/mol.
• In solution, an amino acid exists as a ZWITTERION $^{+}H_{3}N-CHR-COO^{-}$ : the neutral, non-ionized form is negligible.
• For a neutral amino acid, $pK_{a1}\approx 2.3$ for the $COOH$ and $pK_{a2}\approx 9.6$ for the $NH_{3}^{+}$ , and $pI=\dfrac{pK_{a1}+pK_{a2}}{2}$ .

The three hydroxyls of glycerol are esterified by three fatty acids

R-COOH

: three ester bonds, which basic hydrolysis cleaves into soaps.

At the $pI$ , the overall charge is zero and the molecule does not migrate in an electric field: this is the principle of electrophoresis, and the question comes up almost every session.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Counting two to the power n stereoisomers

2 marks, and a complete count to redo

What not to write

“An aldohexose has six carbons, so $2^{6}=64$ stereoisomers.”

What to write

“Only $n-2$ carbons are asymmetric: $2^{4}=16$ stereoisomers, of which $8$ are of the D series.”

Only C2 to C5 carry four different groups; C1 (the aldehyde) and C6 (

CH_{2}OH

) are not asymmetric:

2^{4}=16

Why: The carbonyl carbon and the terminal $CH_{2}OH$ carbon are not asymmetric: the first carries a double bond, the second two hydrogens.

2. Reading the series on the first asymmetric carbon

2 marks, and half the sugars will be misclassified

What not to write

“The $-OH$ of carbon $2$ is on the right, so it is the D series.”

What to write

“The series is read on the asymmetric carbon FARTHEST from the carbonyl, hence the lowest one in the projection.”

D-mannose has its

-OH

at C2 on the LEFT and remains of the D series: only the

-OH

of C5, the lowest, on the right, counts.

Why: The convention fixes a single reference carbon, the one farthest from the functional group. On a hexose, it is carbon $5$ , not carbon $2$ .

3. Confusing the series and the direction of rotation

1.5 marks, and a confusion that recurs in biochemistry

What not to write

“It is a D-series sugar, so it rotates light to the right.”

What to write

“The D series is a drawing CONVENTION; the direction of rotation is measured and written $(+)$ or $(-)$ , independently.”

Natural glucose and fructose are both of the D series (

-OH

of C5 on the right), but one is dextrorotatory and the other levorotatory.

Why: Natural fructose is of the D series and levorotatory: the two notations do not coincide. One describes a configuration, the other a physical property.

4. Going to Haworth by swapping sides

3 marks, the whole ring is drawn upside down

What not to write

“The $-OH$ is on the right in Fischer, so it will be above the plane in Haworth.”

What to write

“What is on the RIGHT ends up BELOW: that is the rule, and it holds for every carbon of the ring.”

The

-OH

of C2, on the right in Fischer, goes under the ring; that of C3, on the left, goes above: same colors in both drawings.

Why: The rule comes from the vertical Fischer chain folding into a horizontal ring. A single inversion is enough to turn the sugar into another one, sometimes a nonexistent one.

5. Confusing the two anomers

2 marks, and the nature of the polysaccharide changes

What not to write

“The alpha anomer has its $-OH$ above the plane.”

What to write

“Alpha: anomeric $-OH$ on the side OPPOSITE the $CH_{2}OH$ , hence below when the $CH_{2}OH$ is above. Beta: on the SAME side.”

Why: The definition is always given RELATIVE TO the $CH_{2}OH$ , never in absolute terms. It is this relationship that distinguishes starch from cellulose.

6. Calling a sugar reducing when both anomeric carbons are engaged

2 marks, and it is the test question of the chapter

What not to write

“Sucrose is a disaccharide, so it reduces Fehling's solution.”

What to write

“Its TWO anomeric carbons are engaged in the glycosidic bond: no open form is possible, it is NOT reducing.”

Maltose keeps a free anomeric carbon and remains reducing; sucrose engages both and is not.

Why: Reducing power comes from the open form, which presents a free aldehyde. A full acetal does not open, whereas a hemiacetal does so constantly.

7. Thinking starch and cellulose have different skeletons

2 marks on the applied biochemistry question

What not to write

“Cellulose is made of different sugars than starch, which is why we do not digest it.”

What to write

“Both are polymers of glucose: only the geometry of the bond changes, $\alpha(1\rightarrow 4)$ against $\beta(1\rightarrow 4)$ .”

Same glucoses, same carbons 1 and 4: the bridge leaves from under the ring in starch (

\alpha

), from above in cellulose (

\beta

Why: Our enzymes recognize only the alpha geometry. A single stereochemical difference separates a food from an indigestible fiber: it is the most striking example in the course.

8. Mistaking two epimers for enantiomers

1.5 marks, and the physical properties announced are wrong

What not to write

“Glucose and mannose differ by one carbon, so they are enantiomers.”

What to write

“Enantiomers differ at ALL their asymmetric carbons: glucose and mannose change only one, they are DIASTEREOMERS.”

L-glucose inverts ALL the asymmetric carbons of D-glucose (enantiomer); D-mannose inverts only one (epimer, hence diastereomer).

Why: Two enantiomers have the same melting points and solubilities; two diastereomers do not, and this is what allows them to be separated by crystallization.

9. Having an amino acid exist in neutral form in solution

2 marks, and the whole discussion of the

pI

loses its meaning

What not to write

“In solution, the amino acid carries a $COOH$ and an $NH_{2}$ .”

What to write

“It exists as a ZWITTERION: the carboxyl proton has moved to the nitrogen, and the molecule carries two opposite charges.”

The neutral form we usually draw is negligible in solution: the carboxyl proton has moved to the nitrogen.

Why: The $pK_{a}$ gap between $2.3$ and $9.6$ makes the internal transfer completely favorable. The non-ionized form accounts for less than one molecule in a million.

Which method to choose

Reading a projection, in order

First look at which projection it is.

If vertical chain, most oxidized carbon at the top → FISCHER projection: horizontal bonds come toward the viewer

Example: D-glucose and its four asymmetric carbons
If looking for the D or L series → look at the LOWEST asymmetric carbon: $-OH$ on the right gives D

Example: on a hexose, it is carbon $5$
If moving to Haworth → what is on the RIGHT goes BELOW, and the $CH_{2}OH$ is placed above for a D-series sugar

Example: the $-OH$ of carbon $3$ , on the left, ends up above
If naming the anomer → compare the position of the anomeric $-OH$ with that of the $CH_{2}OH$ : opposite gives alpha, same side gives beta

Example: it is a comparison, never an absolute position
If checking whether the sugar is reducing → look for a FREE anomeric carbon, that is, a hemiacetal

Example: maltose yes, sucrose no

The whole reading rests on comparisons, never on absolute positions: right or left of the skeleton, above or below the plane, same side or not as the $CH_{2}OH$ .

What is the charge of an amino acid at a given pH

Compare the $pH$ of the medium with the $pI$ of the amino acid.

Below

pK_{a1}

the cationic form, between the two

pK_{a}

values the zwitterion, above

pK_{a2}

the anionic form.

If $pH$ clearly below $pK_{a1}$ → CATIONIC form: the carboxyl is protonated and so is the nitrogen

Example: at $pH$ $1$ , the overall charge is $+1$
If $pH$ equal to the $pI$ → ZWITTERION, overall charge ZERO, no migration in electrophoresis

Example: $pI=\dfrac{2.3+9.6}{2}=5.95$
If $pH$ above the $pI$ but below $pK_{a2}$ → the overall charge becomes NEGATIVE: migration toward the anode

Example: at $pH$ $7$ , the molecule migrates toward the positive pole
If $pH$ clearly above $pK_{a2}$ → ANIONIC form, charge $-1$

Example: at $pH$ $12$

For an amino acid with an ionizable side chain, the $pI$ is calculated with the two $pK_{a}$ values that BRACKET the neutral form, not systematically the first two.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Going from a Fischer projection to a Haworth projection

When to use it: The problem gives a sugar in Fischer projection and asks for its cyclic form.

1 Number the carbons and locate the anomeric carbon, the one of the carbonyl.
2 Apply the rule: what is on the RIGHT in Fischer goes BELOW the plane.
3 Place the $CH_{2}OH$ above the plane for a D-series sugar.
4 Name the anomer by comparing the anomeric $-OH$ with the $CH_{2}OH$ .

Concluding sentence

“The anomeric $-OH$ is on the side opposite the $CH_{2}OH$ : this is the alpha anomer of D-glucopyranose.”

The trap: Forgetting that the anomeric carbon did not exist as an asymmetric center in the open form: it is cyclization that creates it, hence the two anomers.

Marking: 1 mark for the right-below rule applied to all carbons, 1 mark for the $CH_{2}OH$ , 1 mark for the name of the anomer.

Calculating an isoelectric point and concluding

When to use it: The problem gives the $pK_{a}$ values of an amino acid and a working $pH$ .

1 Write the two $pK_{a}$ values and identify the neutral form, the zwitterion.
2 Calculate the $pI$ as the half-sum of the two $pK_{a}$ values that bracket this form.
3 Compare the $pH$ of the medium with the $pI$ and deduce the sign of the overall charge.
4 Conclude on the direction of migration in electrophoresis.

Concluding sentence

“ $pI=\dfrac{2.3+9.6}{2}=5.95$ ; at $pH$ $7$ , above the $pI$ , the overall charge is negative and the amino acid migrates toward the anode.”

The trap: Taking the average of all the $pK_{a}$ values when the side chain is ionizable: only those that bracket the neutral form count.

Marking: 1 mark for the calculation of the $pI$ , 1 mark for the comparison with the $pH$ , 1 mark for the direction of migration.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Counting the asymmetric carbons

Recount the carbons bearing four different substituents: an aldose with $n$ carbons has $n-2$ .

Six carbons and six centers announced: the terminal carbons were counted by mistake.
The direction of the right-below rule

Check on a known carbon, for example carbon $5$ of a D-series sugar, whose $CH_{2}OH$ must end up above.

A $CH_{2}OH$ drawn below signals a general inversion.
Is the anomeric carbon free?

Look for a hemiacetal, that is, a carbon bearing both an $-OH$ and an $-OR$ : it is what makes the sugar reducing.

Two full acetals: the disaccharide is not reducing.
The charge at a given pH

Compare the $pH$ with the $pI$ : above it the charge is negative, below it it is positive.

A migration toward the cathode announced at $pH$ $7$ for a neutral amino acid is wrong.

The typical problem, taken apart

Charge and migration of a neutral amino acid

A neutral amino acid has a $pK_{a1}$ of $2.3$ for its carboxyl group and a $pK_{a2}$ of $9.6$ for its ammonium ion.

Calculate its isoelectric point, give the major form at $pH$ $7$ , and say toward which electrode it migrates.

The two

pK_{a}

values of the amino acid and the pH 7 of the medium: the place of this pH relative to the

pI

, still to be calculated, decides the migration.

Step 1

Neutral form: the ZWITTERION $^{+}H_{3}N-CHR-COO^{-}$ , with overall charge zero.

Why

The gap of more than seven units between the two $pK_{a}$ values makes the internal proton transfer complete. The non-ionized form is negligible in solution.

Step 2

$pI=\dfrac{2.3+9.6}{2}=5.95$ .

Why

The isoelectric point is the half-sum of the two $pK_{a}$ values that bracket the neutral form. For a neutral amino acid, these are simply the only two.

Step 3

At $pH$ $7$ : $7>5.95$ , so the overall charge is NEGATIVE.

Why

Above the $pI$ , the medium is too basic to keep the nitrogen fully protonated: the balance tips toward the carboxylate.

Step 4

Major form at $pH$ $7$ : the carboxylate is deprotonated, the nitrogen still mostly protonated, but the net charge is negative.

Why

We reason on the average of the species, not on a single one. It is the sign of the NET charge that decides the migration.

Step 5

Migration: toward the ANODE, the positive pole.

Why

A negative charge migrates toward the electrode of opposite sign. This is the principle of protein separation by electrophoresis.

Step 6

Check: at exactly $pH=5.95$ , the charge is zero and the amino acid does not migrate.

Why

This limiting case validates the reasoning, and it corresponds to the technique of isoelectric focusing, where each protein is immobilized at its own $pI$ .

The conclusion, written out

“The isoelectric point is $5.95$ ; at $pH$ $7$ , above the $pI$ , the amino acid carries a negative overall charge and migrates toward the anode.”

The classic mistake on this problem: Announcing a positive charge at $pH$ $7$ because this $pH$ is below $pK_{a2}$ : it is with the $pI$ that you must compare, not with one of the two $pK_{a}$ values.

Learn by heart

• An aldose with $n$ carbons has $n-2$ asymmetric carbons, hence $2^{n-2}$ stereoisomers.
• In Fischer: VERTICAL chain, most oxidized carbon at the top, horizontal bonds toward the viewer.
• The D or L series is read on the asymmetric carbon FARTHEST from the carbonyl.
• Toward Haworth: what is on the RIGHT goes BELOW.
• Alpha anomer: $-OH$ OPPOSITE the $CH_{2}OH$ . Beta: on the SAME side.
• REDUCING sugar: at least one free anomeric carbon, hence a hemiacetal.
• Starch $\alpha(1\rightarrow 4)$ , cellulose $\beta(1\rightarrow 4)$ : the only difference, and it decides digestibility.
• In solution, an amino acid is a ZWITTERION, and $pI=\dfrac{pK_{a1}+pK_{a2}}{2}$ .

Frequently asked questions

How many stereoisomers does an aldohexose have?

Sixteen, that is two to the power four. An aldose with six carbons has only four asymmetric carbons: the aldehyde carbon carries a double bond, and the last one carries two hydrogens. Of these sixteen stereoisomers, eight belong to the D series and eight to the L series.

Which carbon decides the D or L series?

The asymmetric carbon farthest from the carbonyl group, that is, the lowest one in the Fischer projection. If its hydroxyl group is on the right, the sugar belongs to the D series. The other asymmetric carbons distinguish diastereomers from one another, but do not change the series.

How do you go from Fischer to Haworth?

What is written on the right in the Fischer projection ends up below the plane in the Haworth projection, and vice versa. For a D-series sugar, the hydroxymethyl group is placed above the plane. It then remains to name the anomer by comparing the anomeric hydroxyl group with that hydroxymethyl group.

What makes a sugar reducing?

The presence of at least one free anomeric carbon, that is, one engaged in a hemiacetal and not in a full acetal. This carbon allows the ring to open and present a free aldehyde, which reduces Fehling's solution. Sucrose engages both its anomeric carbons and is therefore not reducing.

Why do we digest starch but not cellulose?

Because both are polymers of the same sugar, glucose, but the geometry of the bond differs: alpha in starch, beta in cellulose. Our digestive enzymes recognize only the alpha geometry. A single difference in configuration thus separates an energy food from an indigestible fiber.

Practise it

Corrected exercises: Biomolecules, 202-SNU-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

15 corrected exercises
150 points
225 minutes

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-snu-biomolecules. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: biomolecules (202-SNU-RE)

The essentials

Monosaccharides and counting stereoisomers

D or L series, and epimers

From Fischer to Haworth, and the glycosidic bond

Lipids and amino acids

The mistakes that cost marks

1. Counting two to the power n stereoisomers

2. Reading the series on the first asymmetric carbon

3. Confusing the series and the direction of rotation

4. Going to Haworth by swapping sides

5. Confusing the two anomers

6. Calling a sugar reducing when both anomeric carbons are engaged

7. Thinking starch and cellulose have different skeletons

8. Mistaking two epimers for enantiomers

9. Having an amino acid exist in neutral form in solution

Which method to choose

Reading a projection, in order

What is the charge of an amino acid at a given pH

How the answer is expected to be written

Going from a Fischer projection to a Haworth projection

Calculating an isoelectric point and concluding

Check before you hand in

The typical problem, taken apart

Charge and migration of a neutral amino acid

Learn by heart

Frequently asked questions

How many stereoisomers does an aldohexose have?

Which carbon decides the D or L series?

How do you go from Fischer to Haworth?

What makes a sugar reducing?

Why do we digest starch but not cellulose?

Corrected exercises: Biomolecules, 202-SNU-RE

See also

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