---
title: "Biomolecules: Amino Acids, Sugars, Lipids and Nucleotides"
book: "University Chemistry — Year 2"
subject: chemistry
language: en
chapter: 30
exercises: 12
source: https://one-course.com/books/chemistry/3/en/chapter/30-biomolecules-amino-acids-sugars-lipids-and-nucleotides
license: CC-BY-NC-SA-4.0
credit: "One Chemistry Book, One Course (one-course.com)"
---

# Chapter 30 — Biomolecules: Amino Acids, Sugars, Lipids and Nucleotides

A common sweetener, far sweeter than sugar, is two amino acids joined by a single amide bond, one of them as its methyl ester: aspartic acid and phenylalanine, both with the configuration living cells use. The molecules of life are built from a few families of small units, amino acids, sugars, [fatty acids](#def-b2-biomolecules-lipid) and [nucleotides](#def-b2-biomolecules-nucleotide), joined by the reactions of the previous chapters: amides, acetals, esters. This chapter reads them as an organic chemist does: functional groups, stereochemistry, acid–base behaviour, and the reactions that join and cut them.

**You already know.**

The Year 1 volume: CIP rules, Fischer projections, chirality, hemiacetals and acetals, $\mathrm pK_a$ and distribution diagrams, amphiphilic molecules, Biot’s law. [Chapter 24](https://one-course.com/books/chemistry/3/en/chapter/24-carboxylic-acid-derivatives#ch-b2-acyl-substitution): amides, esters, [saponification](https://one-course.com/books/chemistry/3/en/chapter/24-carboxylic-acid-derivatives#def-b2-acyl-substitution-saponification), activated acids. [Chapter 23](https://one-course.com/books/chemistry/3/en/chapter/23-amines-and-nitrogen-compounds#ch-b2-amines): amines as bases. [Chapter 28](https://one-course.com/books/chemistry/3/en/chapter/28-retrosynthesis-and-multistep-strategy#ch-b2-retrosynthesis): protecting groups in a route.

## 30.1 Amino acids

**Definition 30.1 (Amino acids).**

An *$\alpha$-amino acid* $\ce{H2N-CHR-COOH}$ carries an amino group and a carboxylic acid group on the same carbon; $\ce{R}$ is its *side chain*. In water near neutral pH it exists as a *zwitterion*, $\ce{H3N+-CHR-COO-}$, carrying a positive and a negative charge. The *isoelectric point* pI is the pH at which its average net charge is zero.

Twenty amino acids build the proteins of living things. All but glycine ($\ce{R = H}$) are chiral, and the natural ones have the L configuration ([Definition 30.8](#def-b2-biomolecules-d-l)). Their [side chains](#def-b2-biomolecules-amino-acid) sort them into families: nonpolar (alanine, $\ce{R = CH3}$; phenylalanine, $\ce{R = CH2C6H5}$), polar, acidic (aspartic and glutamic acids, with a second carboxylic group) and basic (lysine, with a second amino group; histidine, with an imidazole ring).

**Proposition 30.2 (Isoelectric point).**

For an amino acid with a [side chain](#def-b2-biomolecules-amino-acid) that is neither acidic nor basic, $\mathrm{pI} = (\mathrm pK_{a1} +
\mathrm pK_{a2})/2$, where $\mathrm pK_{a1}$ belongs to the carboxylic group and $\mathrm pK_{a2}$ to the ammonium group. For an acidic or basic [side chain](#def-b2-biomolecules-amino-acid), pI is the mean of the two $\mathrm pK_a$ that frame the neutral form.

**Proof.** Write $\ce{H2A+}$ for the cation, $\ce{HA}$ for the [zwitterion](#def-b2-biomolecules-amino-acid) and $\ce{A-}$ for the anion. The net charge is zero when $[\ce{H2A+}] = [\ce{A-}]$. From $K_{a1} = [\ce{HA}]h/[\ce{H2A+}]$ and $K_{a2} =
[\ce{A-}]h/[\ce{HA}]$, with $h = [\ce{H3O+}]$, the ratio $[\ce{H2A+}]/[\ce{A-}] = h^2/(K_{a1}K_{a2})$ equals 1 for $h^2 = K_{a1}K_{a2}$, that is $\mathrm{pH} = (\mathrm pK_{a1} + \mathrm pK_{a2})/2$. With a third ionisable group the same argument applies to the two equilibria on either side of the neutral form, the third group being then entirely in one state. ∎

For glycine, $\mathrm pK_{a1} = 2.35$ and $\mathrm pK_{a2} = 9.78$: pI $\approx 6.1$. Aspartic acid (2.0, 3.9, 9.9) has pI $\approx 3.0$; lysine (2.2, 9.1, 10.7) has pI $\approx 9.9$.

![Left: net charge of three amino acids against pH, from their pK_a at 25 C; each curve crosses zero at the pI. Right: the three forms of glycine, H3N+CH2COOH, H3N+CH2COO- and H2NCH2COO-; the zwitterion dominates from about pH 3 to pH 9.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-biomolecules/fig-caf078162ee8.svg)

![Left: net charge of three amino acids against pH, from their pK_a at 25 C; each curve crosses zero at the pI. Right: the three forms of glycine, H3N+CH2COOH, H3N+CH2COO- and H2NCH2COO-; the zwitterion dominates from about pH 3 to pH 9.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-biomolecules/fig-c409f4bc5020.svg)

*Left: net charge of three amino acids against pH, from their $\mathrm pK_a$ at $25{}^{\circ}\mathrm{C}$; each curve crosses zero at the pI. Right: the three forms of glycine, $\ce{H3N+CH2COOH}$, $\ce{H3N+CH2COO-}$ and $\ce{H2NCH2COO-}$; the [zwitterion](#def-b2-biomolecules-amino-acid) dominates from about pH 3 to pH 9.*

**Method 30.3 (Net charge at a given pH).**

1. List every ionisable group with its $\mathrm pK_a$ : the terminal $\ce{NH3+}$ and $\ce{COOH}$ , and the ionisable [side chains](#def-b2-biomolecules-amino-acid) .
2. For each group, compare pH and $\mathrm pK_a$ : a group is mostly protonated below its $\mathrm pK_a$ , mostly deprotonated above (by more than 1 unit: over 90 %).
3. Add the charges: $\ce{NH3+}$ $+1$ , $\ce{NH2}$ 0, $\ce{COOH}$ 0, $\ce{COO-}$ $-1$ (imidazolium $+1$ ).
4. Near a $\mathrm pK_a$ , count that group as half charged; for an exact value, use the fractions $1/(1 + 10^{\mathrm pK_a - \mathrm{pH}})$ .

## 30.2 Peptides

**Definition 30.4 (Peptides).**

The amide bond formed between the carboxylic group of one amino acid and the amino group of another is a *peptide bond*; a chain of amino acids joined by peptide bonds is a *peptide* (a protein when it is long), written from its free amino end (N-terminus) to its free carboxylic end (C-terminus). A *peptide coupling* forms a peptide bond in the laboratory with a *coupling agent*, a reagent that converts the carboxylic group into an activated derivative under mild conditions (a carbodiimide, for instance).

![The two resonance structures of the peptide bond. The nitrogen lone pair is shared with the carbonyl: the C-N bond has a partial double-bond character, and the six atoms C, C, O, N, H, C lie in one plane.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-biomolecules/fig-435742d7b042.svg)

*The two resonance structures of the [peptide bond](#def-b2-biomolecules-peptide). The nitrogen lone pair is shared with the carbonyl: the $\ce{C-N}$ bond has a partial double-bond character, and the six atoms C, C, O, N, H, C lie in one plane.*

**Proposition 30.5 (A planar bond).**

The [peptide bond](#def-b2-biomolecules-peptide) is planar, rotation about the $\ce{C-N}$ bond is restricted, and the amide nitrogen is neither basic nor nucleophilic.

**Argument.** The second resonance structure gives the $\ce{C-N}$ bond a share of double-bond character: rotating it would break the overlap of the nitrogen lone pair with the $\ce{C=O}$ $\pi$ system, which costs energy, so the atoms around it stay in one plane, the trans arrangement (the two chain carbons on opposite sides of the $\ce{C-N}$ bond) being the usual one. The lone pair engaged in the resonance is not available for a proton or an electrophile, as for every amide ([Chapter 24](https://one-course.com/books/chemistry/3/en/chapter/24-carboxylic-acid-derivatives#ch-b2-acyl-substitution)). Proteins fold around chains of these rigid planes. ∎

**Proposition 30.6 (Why protect and activate).**

Making one given dipeptide from two amino acids needs both the protection of the groups that must not react and the activation of the carboxylic group that must.

**Argument.** A carboxylic acid and an amine mixed at room temperature give a salt, not an amide; heating to drive off water would also racemise the amino acids. Activation (an [acyl chloride](https://one-course.com/books/chemistry/3/en/chapter/24-carboxylic-acid-derivatives#def-b2-acyl-substitution-derivative), an anhydride, or a [coupling agent](#def-b2-biomolecules-peptide)) is therefore needed. But each amino acid has both groups: activating a mixture of glycine and alanine would give Gly–Ala, Ala–Gly, Gly–Gly, Ala–Ala and longer chains. Blocking the amine of the first and the acid of the second leaves only one possible bond. ∎

**Method 30.7 (Synthesis of a dipeptide).**

1. Protect the amino group of the N-terminal amino acid (as a carbamate, for instance a *tert* -butoxycarbonyl group).
2. Protect the carboxylic group of the C-terminal amino acid (as a methyl or benzyl ester), and any reactive [side chain](#def-b2-biomolecules-amino-acid) .
3. Activate the free carboxylic group with a [coupling agent](#def-b2-biomolecules-peptide) and add the protected partner: one [peptide bond](#def-b2-biomolecules-peptide) forms.
4. Remove the protecting groups, preferably an orthogonal set ( [Definition 28.7](https://one-course.com/books/chemistry/3/en/chapter/28-retrosynthesis-and-multistep-strategy#def-b2-retrosynthesis-orthogonal) ); to go on, remove only the amine protection and repeat.

Repeating the cycle many times in solution would need a purification at each step. In solid-phase synthesis, the growing chain is attached by its C-terminus to a [polymer](https://one-course.com/books/chemistry/3/en/chapter/29-polymers-synthesis-structure-and-properties#def-b2-polymer-synthesis-polymer) bead; excess reagents and by-products are washed away after each coupling, and the finished [peptide](#def-b2-biomolecules-peptide) is cut from the support at the end.

![Eggs frying. The proteins of the white unfold on heating and bond to one another: the translucent liquid turns into an opaque white solid. No peptide bond breaks; the change is in the way the chains are folded (illustration).](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-biomolecules/img-1204f7bcf2a1.jpg)

*Eggs frying. The proteins of the white unfold on heating and bond to one another: the translucent liquid turns into an opaque white solid. No [peptide bond](#def-b2-biomolecules-peptide) breaks; the change is in the way the chains are folded (illustration).*

## 30.3 Carbohydrates

**Definition 30.8 (D and L).**

A sugar or an amino acid is drawn in Fischer projection with its most oxidised carbon at the top. It is D if the $\ce{OH}$ on the stereocentre farthest from the top (or, for an $\alpha$-amino acid, the $\ce{NH2}$ on the $\alpha$ carbon) points to the right, and L if it points to the left. The descriptors compare configurations with that of glyceraldehyde; they are not the CIP descriptors.

**Definition 30.9 (Monosaccharides).**

A *monosaccharide* is a polyhydroxy aldehyde or ketone, usually $\ce{C_nH_{2n}O_n}$ with $n$ from 3 to 7, that cannot be hydrolysed into smaller sugars. An *aldose* has an aldehyde group at C1, a *ketose* a ketone group, usually at C2. Glucose is an aldohexose, fructose a ketohexose.

In water an aldohexose is almost entirely cyclic: the $\ce{OH}$ on C5 adds to the aldehyde, forming a six-membered hemiacetal ring (a pyranose). The cyclisation creates a new stereocentre at C1.

**Definition 30.10 (Anomers).**

In a *Haworth projection* the ring of a cyclic sugar is drawn flat, perpendicular to the page, with its oxygen at the back right and its substituents above or below the ring. The hemiacetal carbon (C1 of an [aldose](#def-b2-biomolecules-sugar)) is the *anomeric carbon*; the two configurations it can take give two diastereoisomers, the *anomers*: for a D sugar, $\alpha$ has the anomeric $\ce{OH}$ below the ring (opposite the $\ce{CH2OH}$), $\beta$ above. In water the anomers interconvert through the open chain, and the optical rotation of a freshly dissolved anomer drifts to an equilibrium value: *mutarotation*.

![D-Glucose. Left: the open chain in Fischer projection, OH on C5 to the right (D). Middle and right: the two pyranose anomers in Haworth projection; the groups on the right of the Fischer projection point down, those on the left point up, and C5 carries the CH2OH up. The anomers differ only at C1. Hydrogens on C2 to C5 are omitted.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-biomolecules/fig-cf6efa355cfd.svg)

![D-Glucose. Left: the open chain in Fischer projection, OH on C5 to the right (D). Middle and right: the two pyranose anomers in Haworth projection; the groups on the right of the Fischer projection point down, those on the left point up, and C5 carries the CH2OH up. The anomers differ only at C1. Hydrogens on C2 to C5 are omitted.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-biomolecules/fig-12576f6af98b.svg)

![D-Glucose. Left: the open chain in Fischer projection, OH on C5 to the right (D). Middle and right: the two pyranose anomers in Haworth projection; the groups on the right of the Fischer projection point down, those on the left point up, and C5 carries the CH2OH up. The anomers differ only at C1. Hydrogens on C2 to C5 are omitted.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-biomolecules/fig-9b09db362e7d.svg)

*D-Glucose. Left: the open chain in Fischer projection, $\ce{OH}$ on C5 to the right (D). Middle and right: the two pyranose [anomers](#def-b2-biomolecules-anomer) in [Haworth projection](#def-b2-biomolecules-anomer); the groups on the right of the Fischer projection point down, those on the left point up, and C5 carries the $\ce{CH2OH}$ up. The [anomers](#def-b2-biomolecules-anomer) differ only at C1. Hydrogens on C2 to C5 are omitted.*

**Method 30.11 (From Fischer to Haworth (D-aldopyranose)).**

1. Draw the ring with O at the back right, C1 at the right, then C2, C3, C4 clockwise along the front and left, C5 at the back left.
2. Groups on the right in the Fischer projection go down, groups on the left go up.
3. On C5 (D series), the $\ce{CH2OH}$ goes up.
4. On C1, the $\ce{OH}$ goes down for $\alpha$ , up for $\beta$ .

**Proposition 30.12 (Mutarotation equilibrium).**

If the specific rotations of the pure [anomers](#def-b2-biomolecules-anomer) are $[\alpha]_\alpha$ and $[\alpha]_\beta$ and that of the equilibrium mixture is $[\alpha]_{\mathrm{eq}}$, the mole fraction of the $\alpha$ anomer at equilibrium is

$$
x_\alpha = \frac{[\alpha]_{\mathrm{eq}} - [\alpha]_\beta}{[\alpha]_\alpha - [\alpha]_\beta}.
$$

**Proof.** The open chain is a tiny fraction of the mixture and is neglected. The rotations of the two [anomers](#def-b2-biomolecules-anomer) add in proportion to their concentrations (Biot’s law, Year 1 volume); since they have the same molar mass, $[\alpha]_{\mathrm{eq}} = x_\alpha[\alpha]_\alpha + (1 - x_\alpha)[\alpha]_\beta$, which is solved for $x_\alpha$. ∎

D-Glucose at equilibrium in water has $[\alpha] = +52.7$ (in ${}^{\circ}\,\mathrm{dm}^{-1}\,\mathrm{g}^{-1}\,\mathrm{mL}$). With anomer rotations of $+112$ and $+18.7$ (exercise data), $x_\alpha = (52.7 - 18.7)/(112 - 18.7)
\approx 0.36$: about 36 % $\alpha$ and 64 % $\beta$. The $\beta$ anomer, with all its substituents equatorial in the chair, is the more stable.

**Definition 30.13 (Glycosides).**

A *glycoside* is the acetal formed when the anomeric $\ce{OH}$ of a sugar is replaced by an $\ce{OR}$ group from an alcohol, often another sugar; the bond from the [anomeric carbon](#def-b2-biomolecules-anomer) to that oxygen is the *glycosidic bond*. A *reducing sugar* is one that reduces mild oxidants such as Fehling’s solution or Tollens’ reagent.

**Proposition 30.14 (Reducing or not).**

A sugar with a free hemiacetal (or hemiketal) group is reducing; a [glycoside](#def-b2-biomolecules-glycoside) whose [anomeric carbons](#def-b2-biomolecules-anomer) are all engaged in [glycosidic bonds](#def-b2-biomolecules-glycoside) is not.

**Argument.** A hemiacetal is in equilibrium with the open chain, whose aldehyde group is oxidised by the reagent; the equilibrium then reopens more rings until all the sugar has reacted. An acetal does not open in neutral or basic water: the [anomeric carbon](#def-b2-biomolecules-anomer) stays locked, with no aldehyde to oxidise. ∎

Maltose (two glucoses, C1 of one bonded to O4 of the other) keeps one free hemiacetal and is reducing; sucrose joins the [anomeric carbons](#def-b2-biomolecules-anomer) of glucose and fructose to each other and is not. Starch and cellulose are both chains of glucose joined C1–O4: $\alpha$ in starch, which coils and is digested; $\beta$ in cellulose, whose straight chains pack into fibres held by hydrogen bonds.

## 30.4 Lipids

**Definition 30.15 (Lipids).**

A *fatty acid* is a carboxylic acid with a long unbranched chain, usually of 12 to 22 carbons, saturated or with (*Z*) double bonds. A *triglyceride* is the triester of propane-1,2,3-triol (glycerol) with three fatty acids. A *phospholipid* is a diester of glycerol with two fatty acids whose third position carries a phosphate group bonded to a small polar group.

The (*Z*) double bonds bend the chains, which then pack badly: stearic acid (18 carbons, saturated) melts at $69{}^{\circ}\mathrm{C}$, oleic acid (18 carbons, one *Z* double bond) at $13{}^{\circ}\mathrm{C}$. Fats rich in saturated chains are solid at room temperature; oils, rich in unsaturated chains, are liquid. [Triglycerides](#def-b2-biomolecules-lipid) are saponified into glycerol and soaps ([Chapter 24](https://one-course.com/books/chemistry/3/en/chapter/24-carboxylic-acid-derivatives#ch-b2-acyl-substitution)).

![A phospholipid bilayer, schematic: the polar heads face the water on both sides, the hydrocarbon tails meet inside. Cell membranes are built on this structure; soaps, with one tail per head, form instead small spheres with the tails inside, called micelles.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-biomolecules/fig-3149c5baec71.svg)

*A [phospholipid](#def-b2-biomolecules-lipid) bilayer, schematic: the polar heads face the water on both sides, the hydrocarbon tails meet inside. Cell membranes are built on this structure; soaps, with one tail per head, form instead small spheres with the tails inside, called micelles.*

## 30.5 Nucleotides

**Definition 30.16 (Nucleotides).**

A *nucleobase* is a nitrogen heterocycle: adenine (A) and guanine (G), purines; cytosine (C), thymine (T) and uracil (U), pyrimidines. A *nucleoside* is a nucleobase bonded by a nitrogen to C1 of ribose or 2-deoxyribose (an $N$-glycoside). A *nucleotide* is a phosphate ester of a nucleoside. In nucleic acids the nucleotides are joined by *phosphodiester bonds*: one phosphate links O3 of one sugar to O5 of the next.

DNA uses 2-deoxyribose and the bases A, G, C, T; RNA uses ribose and U in place of T. The chain has a direction, from its 5$'$ end to its 3$'$ end, and the phosphates, ionised at physiological pH, make it a polyanion. Two DNA strands run in opposite directions and are held together by hydrogen bonds between bases facing each other.

**Proposition 30.17 (Base pairing).**

In double-stranded DNA, adenine pairs with thymine through two hydrogen bonds, and guanine with cytosine through three.

**Argument.** The pairs are those whose edges match donors ($\ce{N-H}$) with acceptors ($\ce{C=O}$ oxygen, ring nitrogen) at the right distances, with a purine facing a pyrimidine so that every pair has the same width. A offers $\ce{N-H}$ (its amino group) and a ring $\ce{N}$; T offers $\ce{C=O}$ and $\ce{N-H}$: two bonds. G offers $\ce{C=O}$, $\ce{N-H}$ and $\ce{N-H}$ (amino); C offers $\ce{N-H}$ (amino), $\ce{N}$ and $\ce{C=O}$: three bonds. The figure shows them. ∎

![The two base pairs of DNA. Left: thymine and adenine, two hydrogen bonds (dotted). Right: cytosine and guanine, three. R is the deoxyribose of each strand; the two R groups of a pair point to the same side, towards the two backbones.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-biomolecules/fig-fcc2635bae7c.svg)

*The two base pairs of DNA. Left: thymine and adenine, two hydrogen bonds (dotted). Right: cytosine and guanine, three. R is the deoxyribose of each strand; the two R groups of a pair point to the same side, towards the two backbones.*

**History — The sugars, 1891.**

![](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-biomolecules/img-b204aaeb5672.jpg)

Emil Fischer worked out, by 1891, the configurations of glucose and of its isomers, before any physical method could see a molecule: by converting sugars into one another and counting the stereoisomers obtained. The projection he invented to write them still bears his name. He also made the first synthetic [peptides](#def-b2-biomolecules-peptide), and received the 1902 Nobel Prize in Chemistry. (Portrait: Atelier Victoria, Berlin, around 1895, public domain; Wikimedia Commons.)

## 30.6 Exercises

**Exercise 30.1 ★.**

Draw alanine as a [zwitterion](#def-b2-biomolecules-amino-acid), then the forms that dominate at pH 1 and at pH 12.

**Solution of Exercise 30.1.**

- Between pH 3 and 9, the [zwitterion](#def-b2-biomolecules-amino-acid) $\ce{H3N+-CH(CH3)-COO-}$ .
- At pH 1, the cation $\ce{H3N+-CH(CH3)-COOH}$ .
- At pH 12, the anion $\ce{H2N-CH(CH3)-COO-}$ .

**Exercise 30.2 ★.**

Compute the [isoelectric point](#def-b2-biomolecules-amino-acid) of glycine, and that of alanine ($\mathrm pK_a$ 2.35 and 9.87).

**Solution of Exercise 30.2.**

Glycine: $(2.35 + 9.78)/2 \approx 6.1$. Alanine: $(2.35 + 9.87)/2 \approx 6.1$.

**Exercise 30.3 ★.**

Draw the dipeptide Ala–Gly and circle its [peptide bond](#def-b2-biomolecules-peptide). How does it differ from Gly–Ala?

**Solution of Exercise 30.3.**

$\ce{H2N-CH(CH3)-CO-NH-CH2-COOH}$: alanine provides the N-terminus, glycine the C-terminus. Gly–Ala, $\ce{H2N-CH2-CO-NH-CH(CH3)-COOH}$, is a constitutional isomer: the free amino group is on glycine.

**Exercise 30.4 ★.**

In a Fischer projection with $\ce{COOH}$ at the top and the [side chain](#def-b2-biomolecules-amino-acid) $\ce{CH2OH}$ at the bottom, the $\ce{NH2}$ of serine points to the left. Is it D or L? Give its CIP descriptor.

**Solution of Exercise 30.4.**

$\ce{NH2}$ on the left: L. Priorities $\ce{NH2}$ > $\ce{COOH}$ > $\ce{CH2OH}$ > $\ce{H}$: L-serine is (*S*).

**Exercise 30.5 ★★.**

Give the net charge of the tripeptide Lys–Gly–Asp at pH 1, 7 and 12.

**Solution of Exercise 30.5.**

Groups: N-terminal ammonium (about 9), lysine [side chain](#def-b2-biomolecules-amino-acid) (10.7), aspartate [side chain](#def-b2-biomolecules-amino-acid) (3.9), C-terminal carboxylic group (about 2). pH 1: $+1 +1 + 0 + 0 = +2$. pH 7: $+1 +1 -1 -1 = 0$. pH 12: $0 + 0 - 1 - 1 = -2$ (the lysine [side chain](#def-b2-biomolecules-amino-acid), 1.3 units below pH 12, is mostly neutral).

**Exercise 30.6 ★★.**

Why is a [coupling agent](#def-b2-biomolecules-peptide) used to make a [peptide bond](#def-b2-biomolecules-peptide), rather than an [acyl chloride](https://one-course.com/books/chemistry/3/en/chapter/24-carboxylic-acid-derivatives#def-b2-acyl-substitution-derivative) of the protected amino acid made with thionyl chloride and heated?

**Solution of Exercise 30.6.**

Thionyl chloride and heat would also attack other groups and, above all, the very reactive [acyl chloride](https://one-course.com/books/chemistry/3/en/chapter/24-carboxylic-acid-derivatives#def-b2-acyl-substitution-derivative) of an amino acid loses its configuration easily (the $\alpha$ hydrogen becomes acidic): the [peptide](#def-b2-biomolecules-peptide) would be partly racemised. A [coupling agent](#def-b2-biomolecules-peptide) activates the acid at room temperature, under mild conditions.

**Exercise 30.7 ★★.**

D-Fructose cyclises through its C5 $\ce{OH}$ onto the C2 ketone. Give the size of the ring and draw $\beta$-D-fructofuranose in [Haworth projection](#def-b2-biomolecules-anomer).

**Solution of Exercise 30.7.**

O5 bonded to C2: a ring of C2, C3, C4, C5 and O, five atoms (a furanose). In [Haworth projection](#def-b2-biomolecules-anomer), ring O at the back, C2 at the right with $\ce{CH2OH}$ (C1) down and $\ce{OH}$ up ($\beta$); C3 $\ce{OH}$ up (left in Fischer), C4 $\ce{OH}$ down, C5 carrying $\ce{CH2OH}$ (C6) up.

**Exercise 30.8 ★★.**

The pure [anomers](#def-b2-biomolecules-anomer) of D-galactose have specific rotations of $+150.7$ ($\alpha$) and $+52.8$ ($\beta$) and the equilibrium mixture $+80.2$ (exercise data). Compute the equilibrium composition.

**Solution of Exercise 30.8.**

$x_\alpha = (80.2 - 52.8)/(150.7 - 52.8) = 27.4/97.9 \approx 0.28$: 28 % $\alpha$, 72 % $\beta$.

**Exercise 30.9 ★★.**

Explain why maltose reduces Fehling’s solution and sucrose does not, and why sucrose does after heating with dilute acid.

**Solution of Exercise 30.9.**

Maltose keeps a free hemiacetal on its second glucose, which opens to an aldehyde: reducing. In sucrose both [anomeric carbons](#def-b2-biomolecules-anomer) are engaged in the [glycosidic bond](#def-b2-biomolecules-glycoside): no hemiacetal, not reducing. Dilute acid hydrolyses the [glycosidic bond](#def-b2-biomolecules-glycoside), an acetal: glucose and fructose are released, and both reduce.

**Exercise 30.10 ★★★.**

The [saponification](https://one-course.com/books/chemistry/3/en/chapter/24-carboxylic-acid-derivatives#def-b2-acyl-substitution-saponification) value of a fat is the mass of potassium hydroxide, in mg, that saponifies 1 g of it. Compute it for triolein, $\ce{C57H104O6}$, and explain why a fat with shorter chains has a higher value.

**Solution of Exercise 30.10.**

Triolein, $885.45\,\mathrm{g}/\mathrm{mol}$, consumes three $\ce{KOH}$ ($56.105\,\mathrm{g}/\mathrm{mol}$): $3 \times 56.105/885.45 =
0.190$ g per g, a [saponification](https://one-course.com/books/chemistry/3/en/chapter/24-carboxylic-acid-derivatives#def-b2-acyl-substitution-saponification) value of 190. Shorter chains mean a smaller molar mass for the same three ester groups: more moles of fat per gram, more KOH.

**Exercise 30.11 ★★★.**

Plan the synthesis of Gly–Ala from glycine and alanine: protecting groups, activation, coupling, deprotection, with an orthogonal pair.

**Solution of Exercise 30.11.**

Protect the amine of glycine as a *tert*-butoxycarbonyl carbamate, and the acid of alanine as a methyl ester. Couple with a carbodiimide. Remove the carbamate with acid and the ester with dilute base (orthogonal conditions, so either end can be freed alone): Gly–Ala.

**Exercise 30.12 ★★★.**

Write the strand complementary to 5$'$-ATGCGT-3$'$, from its 5$'$ end, and count the hydrogen bonds that hold the double strand together.

**Solution of Exercise 30.12.**

5$'$-ACGCAT-3$'$ (read antiparallel: A pairs with T, G with C). Pairs: A–T, T–A, G–C, C–G, G–C, T–A: $3 \times 2 + 3 \times 3 = 15$ hydrogen bonds.

## 30.7 Problem: Aspartame

**Problem 30.1.**

Weekend problem — the two amino acids and their configurations, the charge of the dipeptide against pH, its synthesis and the $\alpha$/$\beta$ problem, and its hydrolysis in a soft drink

Aspartame is the methyl ester of the dipeptide formed by the $\alpha$-carboxylic group of L-aspartic acid with the amino group of L-phenylalanine. Data: $\mathrm pK_a$ of aspartic acid 2.0, 3.9 and 9.9 (model values for the groups of the dipeptide). Molar masses ($\mathrm{g}/\mathrm{mol}$): C 12.011, H 1.008, N 14.007, O 15.999.

**Part I — The two amino acids.**

1. Name the three compounds released by the complete hydrolysis of aspartame.
2. Draw L-aspartic acid and L-phenylalanine in Fischer projection.
3. Give the CIP descriptor of each stereocentre.
4. How many stereoisomers does aspartame have?
5. Draw aspartame and mark the [peptide bond](#def-b2-biomolecules-peptide) and the ester.
6. Why is the configuration of each amino acid important for the sweet taste?

**Part II — Charge against pH.**

7. List the ionisable groups of aspartame. Why is the C-terminus not one of them?
8. With the model values, give the net charge at pH 2.
9. The same at pH 7.
10. The same at pH 12.
11. Estimate the [isoelectric point](#def-b2-biomolecules-amino-acid) in this model.
12. Why would the true $\mathrm pK_a$ values of the dipeptide differ from those of free aspartic acid?

**Part III — Synthesis.**

13. Why can aspartic acid and the methyl ester of phenylalanine not simply be heated together?
14. Which groups of aspartic acid must be protected?
15. What does a [coupling agent](#def-b2-biomolecules-peptide) do to the free carboxylic group?
16. Aspartic acid has two carboxylic groups. What is formed if the wrong one is coupled?
17. Propose an orthogonal protection that leaves only the $\alpha$ -carboxylic group free, and its removal.
18. Why must the activated amino acid not be left too long in basic conditions?
19. Write the sequence of steps.

**Part IV — Hydrolysis in a soft drink.**

20. Which bonds of aspartame can be hydrolysed in an acidic drink? Which goes first?
21. Give the products of the hydrolysis of the ester alone.
22. Why does an old drink taste less sweet?
23. Compute the molar mass of aspartame, $\ce{C14H18N2O5}$ .
24. Compute the amount of aspartame in a can containing $180\,\mathrm{mg}$ .
25. State the mass of methanol released by the complete hydrolysis of $180\,\mathrm{mg}$ of aspartame.

**Solution of Problem 30.1.**

**1.** L-Aspartic acid, L-phenylalanine and methanol. **2.** In both, $\ce{COOH}$ at the top, [side chain](#def-b2-biomolecules-amino-acid) at the bottom ($\ce{CH2COOH}$, $\ce{CH2C6H5}$), $\ce{NH2}$ on the left. **3.** (*S*) for both. **4.** Two stereocentres: four stereoisomers. **5.** $\ce{H2N-CH(CH2COOH)-CO-NH-CH(CH2C6H5)-COOCH3}$: the [peptide bond](#def-b2-biomolecules-peptide) joins the two residues; the ester is the $\ce{COOCH3}$ at the C-terminus. **6.** Taste receptors are chiral: they bind one stereoisomer and not its mirror image or diastereoisomers, as the opening said. **7.** The N-terminal $\ce{NH3+}$ and the side-chain $\ce{COOH}$ of aspartic acid; the C-terminal acid is a methyl ester, with no acidic hydrogen. **8.** At pH 2: $\ce{NH3+}$ ($+1$), side-chain $\ce{COOH}$ (0): $+1$. **9.** At pH 7: $+1 - 1 = 0$. **10.** At pH 12: $0 - 1 = -1$. **11.** Between the two $\mathrm pK_a$ of the groups that change: $(3.9 + 9.9)/2 \approx 6.9$. **12.** In the dipeptide the amine’s carbon carries an amide instead of a carboxylate: the neighbouring groups and charges change, and so do the $\mathrm pK_a$ (the N-terminal ammonium of a [peptide](#def-b2-biomolecules-peptide) is markedly more acidic than that of a free amino acid). The model gives the shape, not the exact values. **13.** Acid and amine first form a salt; heating would give mixtures (Asp–Asp, longer chains, the wrong carboxylic group) and racemise the stereocentres. **14.** Its amino group and its side-chain carboxylic group. **15.** It turns the $\ce{OH}$ of the acid into a good leaving group, an activated derivative that the amine attacks at room temperature (acyl substitution). **16.** The $\beta$-dipeptide, linked through the [side chain](#def-b2-biomolecules-amino-acid): an isomer of aspartame, not aspartame. **17.** The amine as a benzyloxycarbonyl carbamate and the [side chain](#def-b2-biomolecules-amino-acid) as a benzyl ester: both are removed together by hydrogenolysis over palladium, which leaves the methyl ester untouched (orthogonal to it). **18.** An activated acid has an acidic $\alpha$ hydrogen; base removes it and the stereocentre racemises. **19.** Protect aspartic acid (carbamate on N, benzyl ester on the [side chain](#def-b2-biomolecules-amino-acid)); activate the free $\alpha$-$\ce{COOH}$ with the [coupling agent](#def-b2-biomolecules-peptide); add the methyl ester of L-phenylalanine; hydrogenolyse. **20.** The methyl ester and the [peptide bond](#def-b2-biomolecules-peptide); the ester, an amide being far less reactive. **21.** The dipeptide Asp–Phe (free acid) and methanol. **22.** The sweet molecule is consumed: its hydrolysis products do not replace it. **23.** $14 \times 12.011 + 18 \times 1.008 + 2 \times 14.007 + 5 \times 15.999 =
294.307\,\mathrm{g}/\mathrm{mol}$. **24.** $0.180/294.307 = 6.116 \times 10^{-4}\,\mathrm{mol}$, $0.612\,\mathrm{mmol}$. **25.** One methanol per aspartame: $0.6116 \times 32.042 = \boldsymbol{\approx 19.6\,\mathrm{mg}}$ of methanol.