Amino acids, amides and chirality

In a nutshell

An α-amino acid carries an amine group and a carboxylic acid group on the same carbon. Each group keeps its own chemistry: the -COOH reacts with alkalis and alcohols, and the -NH₂ reacts with acids.

This subtopic also covers primary and secondary amides, and optical isomerism: molecules that exist as two non-superimposable mirror images because they contain a chiral centre.

Assumed knowledge: Alkenes (stereoisomers), Carboxylic acids and esters, Amines.

Core content

Reactions of amino acids: the general formula of an α-amino acid

The general formula of an α-amino acid is RCH(NH₂)COOH.

• The -NH₂ (amine group) and the -COOH (carboxylic acid group) are both bonded to the same carbon atom. That carbon is the α-carbon.
• R is the side chain. It is what makes one amino acid different from another, and it can contain other functional groups (an extra -COOH, an extra -NH₂, an -OH).
• The simplest α-amino acid has R = H: H₂NCH₂COOH, aminoethanoic acid (glycine).

You do not need the common names of amino acids. You may be asked for systematic names, which follow the usual rules: the -COOH carbon is carbon 1, so CH₃CH(NH₂)COOH is 2-aminopropanoic acid.

Reactions of amino acids: the carboxylic acid group with alkalis

The -COOH group reacts with an alkali such as NaOH(aq) like any carboxylic acid: it is neutralised to a carboxylate salt and water.

H₂NCH(CH₃)COOH + NaOH → H₂NCH(CH₃)COO^-Na⁺ + H₂O

General: RCH(NH₂)COOH + NaOH → RCH(NH₂)COO^-Na⁺ + H₂O

• Only the -COOH changes. The -NH₂ is untouched, because it is basic and does not react with an alkali.
• The H lost from -COOH ends up in the water.

Reactions of amino acids: the carboxylic acid group in the formation of esters

The -COOH group reacts with an alcohol to form an ester and water, exactly as in Carboxylic acids and esters.

H₂NCH(CH₃)COOH + CH₃OH ⇌ H₂NCH(CH₃)COOCH₃ + H₂O

The ester is methyl 2-aminopropanoate. As with every ester, the alcohol part is named first (methyl) and the acid part second (2-aminopropanoate).

Because the mixture contains acid, the -NH₂ group can also be protonated to -NH₃⁺. Write whichever form the question asks for; if it just asks for the ester, the equation above is the one to give.

Reactions of amino acids: the amine group with acids

The -NH₂ group is basic. The lone pair on the nitrogen accepts a proton (H⁺) from an acid such as HCl(aq), forming a salt.

H₂NCH(CH₃)COOH + HCl → Cl^{- +}H₃NCH(CH₃)COOH

General: RCH(NH₂)COOH + HCl → Cl^{- +}H₃NCH(R)COOH

• -NH₂ becomes -NH₃⁺, and Cl^- is the counter-ion.
• Only the amine group changes; the -COOH does not react with an acid.
• There is no water formed: the whole HCl is taken up into the salt. This is the same reaction as an amine with an acid (see Amines).

Reactions of amino acids: side chains react too

If the R group contains an extra -COOH or -NH₂, that group reacts as well. Count the groups before balancing.

Worked equations for both cases are in Worked example 2.

Amides: structures of primary and secondary amides

An amide contains the group -CONH-: a C=O carbon bonded directly to a nitrogen.

Amides are classed by how many carbon atoms are bonded to the nitrogen:

• Primary amide, RCONH₂: the N is bonded to one carbon (the C=O carbon) and two H atoms. Example: ethanamide, CH₃CONH₂.
• Secondary amide, RCONHR': the N is bonded to two carbons (the C=O carbon and an alkyl or aryl group) and one H atom. Example: N-methylethanamide, CH₃CONHCH₃.

Naming. Take the stem from the acid part and add -amide. In a secondary amide, the group on the nitrogen is given the locant N-: CH₃CH₂CONHCH₃ is N-methylpropanamide.

Don't confuse amides with amines. In an amide the N is bonded to a C=O carbon; in an amine it is not. Ethanamide (CH₃CONH₂) is an amide; ethylamine (CH₃CH₂NH₂) is an amine.

Making amides (from Carboxylic acids and esters): an acyl chloride reacts with ammonia to give a primary amide, or with a primary amine to give a secondary amide. Two moles of the nitrogen compound are used, because the second one takes up the HCl formed.

CH₃COCl + 2NH₃ → CH₃CONH₂ + NH₄Cl

CH₃COCl + 2CH₃NH₂ → CH₃CONHCH₃ + CH₃NH₃⁺Cl^-

The -CONH- link that joins the units of a polyamide is a secondary amide group. That is the next subtopic, Polyesters and polyamides.

Chirality: optical isomerism

Optical isomerism is a type of stereoisomerism. Recall that stereoisomers have the same structural formula but a different arrangement (of atoms) in space.

Optical isomers are non-superimposable mirror images of each other, about a chiral centre. (They are also called enantiomers.)

A chiral centre is a carbon atom attached to four different atoms or groups of atoms.

• The four groups sit tetrahedrally round the carbon.
• Because all four are different, the molecule and its mirror image are two different molecules: however you rotate one, it cannot be laid exactly on top of the other.
• So any molecule with one chiral centre exists as a pair of optical isomers.

Every α-amino acid except aminoethanoic acid (glycine) has a chiral centre at the α-carbon: H, NH₂, COOH and R are four different groups. In glycine R = H, so the α-carbon has two H atoms and is not chiral.

Chirality: drawing 3-D diagrams of optical isomers

OCR expects you to draw the two optical isomers in 3-D. Use the tetrahedral convention:

1. Draw the chiral carbon in the centre.
2. Draw two bonds in the plane of the paper (plain lines).
3. Draw one bond coming towards you (a solid wedge) and one going away (a dashed wedge).
4. Draw a vertical mirror line, then draw the second isomer as the exact reflection of the first: groups on the mirror side swap sides, groups pointing up or down stay where they are.
5. Check every bond joins to the correct atom: the bond goes to the N of NH₂, the C of COOH, the C of CH₃ and the O of OH.

Still don't get it? · non-superimposable mirror images

Think about a pair of gloves. The left glove is the mirror image of the right one: hold the left glove up to a mirror and the reflection looks exactly like the right glove. Yet you cannot turn a left glove into a right glove. Flip it, spin it, turn it over: the thumb always ends up on the wrong side. Mirror images, but not identical.

Now rebuild it for a molecule, one step at a time:

1. A carbon with four single bonds points them to the corners of a tetrahedron, like a camera tripod with one leg sticking straight up.
2. Put four different groups on those corners. That carbon is now a "glove": it has a definite handedness.
3. Reflect the molecule in a mirror. The reflection has the same four groups bonded to the same carbon, so its structural formula is identical.
4. Try to place the reflection on top of the original so all four groups match. Line up any two groups and the other two land the wrong way round. It never fits. So the two are different molecules: optical isomers.
5. Now swap one group for a copy of another, so two groups are the same (like glycine's two H atoms). Reflect it, and you can always rotate the reflection to sit exactly on the original. No handedness, no optical isomers.

Exam wording: a carbon atom attached to four different atoms or groups is a chiral centre, and the molecule exists as two optical isomers that are non-superimposable mirror images. Draw them in 3-D, as reflections of each other.

Chirality: identifying chiral centres in any organic molecule

You may be given any organic molecule (often a drug or natural product, as a skeletal formula) and asked to mark its chiral centres, usually with an asterisk (*).

Check each carbon in turn:

1. Skip carbons that cannot be chiral: any C in a CH₃ or CH₂ group (two or more identical H atoms), and any C in a C=C or C=O (only three groups attached, including the carbons of a benzene ring).
2. For each carbon left, list its four attached groups. Do not forget an H that a skeletal formula does not draw.
3. Compare the groups as whole groups, not just the first atom. CH₃ and CH₂CH₃ both start with C, but they are different groups.
4. If all four are different, mark the carbon *.

Rings need care. A ring carbon can be a chiral centre. Its two ring bonds count as two groups, and they are different if the ring "looks different" going round each way.

Still don't get it? · chiral centres in a ring

Imagine standing at one house on a circular street. You can walk to the left or to the right, and both routes eventually bring you back home. If the houses you pass are in exactly the same order whichever way you go, the two routes are the same. If one route passes the post office first and the other passes it last, the routes are different.

Now in chemistry, step by step:

1. Pick a ring carbon. Two of its four bonds go round the ring, one each way. The other two are its substituents (often an H and something else).
2. If the two substituents are the same (two H atoms, a CH₂ in the ring), stop: it is not chiral.
3. If they are different, walk round the ring both ways from that carbon, atom by atom.
4. The first atom you meet may be the same both ways (a CH₂ each side). Keep walking until the two routes differ, for example one route meets a carbon carrying a CH₃ sooner than the other.
5. If the two routes ever differ, the two ring bonds lead to different groups, so the carbon has four different groups.

Exam wording: the carbon is a chiral centre because it is attached to four different groups. In cyclohexanol the C-OH carbon is not chiral (both routes round the ring are identical), but in 2-methylcyclohexan-1-ol it is (Worked example 3).

Worked examples

Example 1: writing the three reactions of an amino acid

2-Aminopropanoic acid, CH₃CH(NH₂)COOH, reacts with (a) NaOH(aq), (b) HCl(aq), (c) ethanol with concentrated H₂SO₄. Write an equation for each.

(a) The -COOH is neutralised by the alkali:

CH₃CH(NH₂)COOH + NaOH → CH₃CH(NH₂)COO^-Na⁺ + H₂O

(b) The -NH₂ accepts a proton from the acid:

CH₃CH(NH₂)COOH + HCl → CH₃CH(NH₃⁺)COOH Cl^-

(c) The -COOH is esterified (acid catalyst, heat):

CH₃CH(NH₂)COOH + CH₃CH₂OH ⇌ CH₃CH(NH₂)COOCH₂CH₃ + H₂O

The ester is ethyl 2-aminopropanoate.

Check (a) by counting atoms: left C₃H₇NO₂ + NaOH; right C₃H₆NO₂Na + H₂O. Balanced, and the charges on the salt cancel.

Where marks go: putting the change on the wrong group (for example NaOH reacting with NH₂), missing the H₂O in (a) and (c), adding a water to (b), or leaving the + charge off NH₃⁺.

Example 2: amino acids with a reactive side chain

(a) 2-Aminobutanedioic acid, HOOCCH₂CH(NH₂)COOH, reacts with excess NaOH(aq). (b) 2,6-Diaminohexanoic acid, H₂N(CH₂)₄CH(NH₂)COOH, reacts with excess HCl(aq). Write an equation for each.

(a) Count the acid groups: two -COOH, so 2NaOH and 2H₂O.

HOOCCH₂CH(NH₂)COOH + 2NaOH → Na^{+ -}OOCCH₂CH(NH₂)COO^- Na⁺ + 2H₂O

(b) Count the amine groups: two -NH₂, so 2HCl, both become -NH₃⁺.

H₂N(CH₂)₄CH(NH₂)COOH + 2HCl → Cl^{- +}H₃N(CH₂)₄CH(NH₃⁺)COOH Cl^-

Check (b): left C₆H₁₄N₂O₂ + 2HCl; right C₆H₁₆N₂O₂Cl₂, with charges +2 and -2 cancelling. Balanced.

Where marks go: treating the side chain as unreactive and using 1 mol of NaOH or HCl.

Example 3: identifying chiral centres

How many chiral centres are in each molecule? (a) 2-amino-3-hydroxybutanoic acid, CH₃CH(OH)CH(NH₂)COOH; (b) 1-phenylethanol, C₆H₅CH(OH)CH₃; (c) 2-methylcyclohexan-1-ol.

(a) Go carbon by carbon (C1 is the COOH carbon):

2 chiral centres.

(b) The six benzene-ring carbons are each bonded to only three atoms, so none can be chiral. The CH(OH) carbon has H, OH, CH₃ and C₆H₅: four different groups. The CH₃ carbon has three H. 1 chiral centre.

(c) Number the ring from the C-OH carbon (C1) towards the methyl-bearing carbon (C2).

• C1: H, OH, the route to C2 (which carries CH₃) and the route to C6 (a CH₂). The two ring routes differ, so chiral.
• C2: H, CH₃, the route to C1 (which carries OH) and the route to C3 (a CH₂). Again different, so chiral.
• C3 to C6: all CH₂ (two H atoms each), so not chiral. The CH₃ carbon is not chiral.

2 chiral centres.

Where marks go: starring a CH₂ or a benzene-ring carbon, or missing the ring chiral centres in (c) because the first atom each way round the ring is "just a carbon".

Example 4: model answer for drawing optical isomers

Draw 3-D diagrams to show the optical isomers of 2-aminopropanoic acid. Each point is something an examiner looks for. Compare your drawing with Figure 3.

1. Two structures, each with the chiral carbon bonded to H, NH₂, COOH and CH₃.
2. Each drawn in 3-D: two bonds in the plane, one solid wedge, one dashed wedge.
3. The second structure is the mirror image of the first (draw a mirror line to help you).
4. Bonds join the correct atoms: to the N of NH₂ and the C of COOH and CH₃.

Where marks go: two structures that are just the same molecule rotated (not mirror images), or a 2-D drawing with no wedges.

Example 5: classifying and naming amides

Classify each as a primary or secondary amide and name it: (a) CH₃CH₂CONH₂; (b) CH₃CONHCH₂CH₃.

(a) The N carries two H atoms and is bonded only to the C=O carbon: primary amide. Three carbons in the acid part: propanamide.

(b) The N carries one H and is bonded to two carbons (the C=O carbon and an ethyl group): secondary amide. Two carbons in the acid part, ethyl on the N: N-ethylethanamide.

Where marks go: counting the carbons on the N as part of the main chain (calling (b) "butanamide").

Common exam mistakes

• Chiral centre defined loosely. "A carbon with four groups" or "an asymmetric carbon" is not enough. The mark needs four different atoms or groups attached to the carbon.
• Optical isomers described as "mirror images" only. The key word is non-superimposable. Mirror images that can be superimposed are the same molecule.
• Stereoisomers defined with "molecular formula". Stereoisomers have the same structural formula; "same molecular formula" describes structural isomers too and does not earn the mark.
• Flat drawings. Optical isomers must be drawn in 3-D (wedge and dashed wedge), and the two must be mirror images, not the same molecule turned round.
• Bonds to the wrong atom. A bond drawn to the H of "HO" or "H₂N", or to the O of "COOH", is a different molecule. Bond to the O of OH, the N of NH₂, the C of COOH and CH₃.
• Starring carbons that cannot be chiral. CH₂ and CH₃ carbons, C=O carbons and benzene-ring carbons are never chiral centres.
• Forgetting the hidden H. In a skeletal formula, a carbon with three drawn bonds has an H as its fourth group. It still counts.
• Missing chiral centres in rings. Compare the two routes round the ring as whole groups, not just the first atom.
• Saying glycine (aminoethanoic acid) is chiral. Its α-carbon carries two H atoms. Every other α-amino acid is chiral.
• Reacting the wrong group. NaOH reacts with the -COOH; HCl reacts with the -NH₂. Putting a COO^- in the HCl product, or NH₃⁺ in the NaOH product, loses the mark.
• Missing charges or water. The HCl product needs -NH₃⁺ with Cl^- and no water. The NaOH product needs -COO^-Na⁺ and H₂O.
• Ignoring a reactive side chain. An extra -COOH or -NH₂ in R reacts too, so the equation needs 2 mol of NaOH or HCl.
• Esterification without the catalyst. State the acid catalyst (concentrated H₂SO₄), and include water as a product.
• Ester names back to front. The alkyl group from the alcohol comes first: the ester from 2-aminopropanoic acid and methanol is methyl 2-aminopropanoate, not "2-aminopropyl methanoate".
• Confusing amides and amines, or primary and secondary. An amide has the N on a C=O carbon. Primary amide RCONH₂ (N bonded to one C); secondary amide RCONHR' (N bonded to two C).

Key definitions

• α-amino acid: a compound with general formula RCH(NH₂)COOH, in which the amine group and the carboxylic acid group are attached to the same carbon atom.
• Stereoisomers: compounds with the same structural formula but a different arrangement (of atoms) in space.
• Optical isomerism: a type of stereoisomerism in which the isomers are non-superimposable mirror images about a chiral centre.
• Optical isomers (enantiomers): non-superimposable mirror images.
• Chiral centre: a carbon atom attached to four different atoms or groups of atoms.
• Primary amide: an amide, RCONH₂, in which the nitrogen atom is bonded to one carbon atom.
• Secondary amide: an amide, RCONHR', in which the nitrogen atom is bonded to two carbon atoms.

Specification

• I can write the general formula of an α-amino acid as RCH(NH₂)COOH. (6.2.2 a)
• I can write equations for the reaction of the carboxylic acid group of an amino acid with an alkali. (6.2.2 a i)
• I can write equations for the formation of an ester from an amino acid and an alcohol with an acid catalyst. (6.2.2 a i, see also 6.1.3 c)
• I can write equations for the reaction of the amine group of an amino acid with an acid. (6.2.2 a ii)
• I can apply these reactions to amino acids whose R group contains an extra -COOH or -NH₂. (6.2.2 a)
• I can draw and recognise the structures of primary and secondary amides. (6.2.2 b, see also 6.1.3 f)
• I can explain optical isomerism in terms of non-superimposable mirror images about a chiral centre. (6.2.2 c)
• I can draw 3-D diagrams of a pair of optical isomers. (6.2.2 c, M4.2, M4.3)
• I can identify every chiral centre in a molecule of any organic compound, including in rings and skeletal formulae. (6.2.2 d, M4.2, M4.3)

Related notes

• Alkenes: stereoisomers and E/Z isomerism, the other type of stereoisomerism
• Basic concepts of organic chemistry: naming, structural and skeletal formulae
• Carboxylic acids and esters: reactions of -COOH with alkalis, esterification, and making amides from acyl chlorides
• Amines: basicity of the nitrogen lone pair and salt formation
• Polyesters and polyamides: the amide link in condensation polymers and its hydrolysis
• Transition elements: optical isomerism in octahedral complexes
• Organic synthesis: functional groups in multi-step routes