Organic Chemistry Introduction

Organic chemistry is the study of the structure, properties, composition, reactions, and preparation of carbon-containing compounds. Carbon occupies a unique position in the periodic table: its ability to form four covalent bonds, catenate (form chains and rings), and hybridise in multiple configurations ($sp^3$, $sp^2$, $sp$) produces a combinatorial space of molecules that dwarfs all other branches of chemistry combined.

This module serves as the conceptual foundation for every subsequent organic topic. It defines the systematic rules for naming, classifying, and reasoning about organic molecules.

IUPAC Nomenclature

The International Union of Pure and Applied Chemistry (IUPAC) system provides a deterministic, unambiguous naming convention. Every organic compound has exactly one correct IUPAC name (ignoring stereochemical descriptors, which may be stated in any order).

Naming Algorithm

The procedure is as follows:

1. Identify the principal functional group. This determines the suffix. Priority order (highest first): carboxylic acid $>$ ester $>$ amide $>$ nitrile $>$ aldehyde $>$ ketone $>$ alcohol $>$ amine $>$ alkene $>$ alkyne $>$ ether $>$ halogen $>$ alkyl. This priority is known as the seniority order.

2. Identify the longest carbon chain containing the principal functional group. This is the parent chain. Its length determines the prefix (meth-, eth-, prop-, but-, pent-, hex-, hept-, oct-, non-, dec-).

3. Number the parent chain. Assign locants so that the principal functional group receives the lowest possible number. If there is a tie, give the lowest set of locants to substituents (the "first point of difference" rule).

4. Name substituents. List substituents alphabetically (ignoring multiplying prefixes di-, tri-, tetra-) with their locants.

5. Assemble the name. Format: locant-substituent-locant-substituent-...-parent-suffix.

Worked Example. Name the compound $\mathrm{CH}_3\mathrm{CH}(\mathrm{Cl})\mathrm{CH}_2\mathrm{CH}(\mathrm{OH})\mathrm{CH}_2\mathrm{CH}_3$.

• Principal functional group: $-\mathrm{OH}$ (alcohol, suffix -ol).
• Longest chain containing $-\mathrm{OH}$: 6 carbons (hexane).
• Numbering from the end nearest $-\mathrm{OH}$: the $-\mathrm{OH}$ is on C-3, the $-\mathrm{Cl}$ is on C-2.
• Name: 3-chlorohexan-2-ol.

Multiplying Prefixes and Complex Substituents

When identical substituents appear, use di-, tri-, tetra- as multiplying prefixes. These are ignored in alphabetical ordering.

Complex substituents are named as branched alkyl groups and enclosed in parentheses:

• $-\mathrm{CH}(\mathrm{CH}_3)_2$: isopropyl (systematic: propan-2-yl)
• $-\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_3$: propyl
• $-\mathrm{CH}(\mathrm{CH}_3)\mathrm{CH}_2\mathrm{CH}_3$: butan-2-yl

Homologous Series

A homologous series is a family of organic compounds with the same functional group, the same general formula, and successive members differing by $-\mathrm{CH}_2-$. Each member exhibits a gradual, predictable change in physical properties.

Series	General Formula	Functional Group	Suffix
Alkane	$\mathrm{C}_n\mathrm{H}_{2n+2}$	C--C single bonds	-ane
Alkene	$\mathrm{C}_n\mathrm{H}_{2n}$	C=C double bond	-ene
Alkyne	$\mathrm{C}_n\mathrm{H}_{2n-2}$	$\mathrm{C}\equiv\mathrm{C}$ triple bond	-yne
Haloalkane	--	C--X	-halo
Alcohol	--	$-\mathrm{OH}$	-ol
Aldehyde	--	$-\mathrm{CHO}$	-al
Ketone	--	$\mathrm{C}=\mathrm{O}$	-one
Carboxylic acid	--	$-\mathrm{COOH}$	-oic acid
Ester	--	$-\mathrm{COO}-$	-oate
Amine	--	$-\mathrm{NH}_2$	-amine
Amide	--	$-\mathrm{CONH}_2$	-amide

Trend: Boiling Points Within a Series

As chain length increases, boiling point increases approximately linearly. This is because longer chains have greater surface area and therefore stronger London (dispersion) forces. The increment per $-\mathrm{CH}_2-$ unit is roughly 20--30 K for liquids near room temperature.

Trend: Boiling Points Between Series

For compounds of similar molecular mass:

Alkanes $\lt$ Ethers $\lt$ Aldehydes/Ketones $\lt$ Alcohols $\lt$ Carboxylic Acids

This ordering reflects the increasing strength of intermolecular forces: London forces only $\to$ permanent dipole-dipole $\to$ hydrogen bonding $\to$ dimeric hydrogen bonding (carboxylic acids form dimers in the liquid phase).

Functional Group Classification

Functional groups are specific groups of atoms within molecules that are responsible for the characteristic chemical reactions of those molecules. The reactivity of an organic compound is determined almost entirely by its functional group(s).

Electron-rich groups (nucleophilic sites): $-\mathrm{OH}$, $-\mathrm{NH}_2$, C=C, benzene ring.

Electron-deficient groups (electrophilic sites): C=O (carbonyl carbon), C$\equiv$N (nitrile carbon), $-\mathrm{COOH}$.

A compound may contain multiple functional groups. In such cases, the most senior group (see priority order above) takes the suffix, and all others are named as prefixes. For example, 2-hydroxypropanoic acid contains both an alcohol and a carboxylic acid; the carboxylic acid is senior.

Isomerism

Isomers are compounds with the same molecular formula but different structural arrangements of atoms. Isomerism is divided into two broad categories: structural isomerism and stereoisomerism.

Structural Isomerism

Structural isomers differ in the connectivity of their atoms. There are three subtypes:

Chain isomerism: Different carbon skeletons. For $\mathrm{C}_4\mathrm{H}_{10}$: butane (straight chain) vs 2-methylpropane (branched). Branched isomers have slightly lower boiling points due to reduced surface area for London forces.

Positional isomerism: The functional group occupies a different position on the same carbon chain. For $\mathrm{C}_3\mathrm{H}_8\mathrm{O}$: propan-1-ol vs propan-2-ol. These have different chemical properties (e.g., propan-1-ol can be oxidised to propanal and then propanoic acid; propan-2-ol oxidises only to propanone).

Functional group isomerism: Same molecular formula, different functional group. For $\mathrm{C}_3\mathrm{H}_6\mathrm{O}$: propanal (aldehyde), propanone (ketone), methyl ethanoate (ester), and 1-methoxyethane (ether) are all structural isomers.

Stereoisomerism

Stereoisomers have the same structural formula and the same atom connectivity but differ in the spatial arrangement of atoms.

E/Z Isomerism (Geometric Isomerism)

E/Z isomerism arises from restricted rotation about a C=C double bond. Each carbon of the double bond must carry two different groups for E/Z isomerism to be possible.

The Cahn-Ingold-Prelog (CIP) priority rules assign priorities:

1. Higher atomic number of the atom directly attached to the double bond carbon gets higher priority.
2. If atoms are the same, look at the next atoms in the chain until a difference is found.
3. Multiply bonded atoms are treated as if duplicated (e.g. C=O is treated as C bonded to O and C bonded to a "ghost" O).

Assignment:

• Z (zusammen, "together"): The two highest-priority groups are on the same side of the double bond.
• E (entgegen, "opposite"): The two highest-priority groups are on opposite sides.

The older cis/trans nomenclature is only unambiguous when each carbon carries one hydrogen and one non-hydrogen group. In all other cases, E/Z must be used.

Worked Example. Assign E/Z to $\mathrm{CH}_3\mathrm{CH}=\mathrm{C}(\mathrm{CH}_3)\mathrm{CH}_2\mathrm{CH}_3$.

Left carbon: $-\mathrm{CH}_3$ (atomic number 6) vs $-\mathrm{H}$ (atomic number 1). Priority: $-\mathrm{CH}_3 \gt -\mathrm{H}$.

Right carbon: $-\mathrm{CH}_3$ (atomic number 6) vs $-\mathrm{CH}_2\mathrm{CH}_3$. At the first atom both are C. At the second atom: $-\mathrm{CH}_3$ has (H, H, H) vs $-\mathrm{CH}_2\mathrm{CH}_3$ has (C, H, H). Priority: $-\mathrm{CH}_2\mathrm{CH}_3 \gt -\mathrm{CH}_3$.

If the $-\mathrm{CH}_3$ (high on left) and $-\mathrm{CH}_3$ (low on right) are on the same side, then the high-priority groups are on opposite sides: E.

Optical Isomerism (Chirality)

A chiral centre (stereocentre) is a carbon atom bonded to four different groups. Chiral molecules exist as two non-superimposable mirror images called enantiomers.

The property of existing in non-superimposable mirror-image forms is called chirality. The most common cause of chirality in organic molecules is a tetrahedral carbon with four different substituents.

Properties of enantiomers:

• Identical physical properties (melting point, boiling point, solubility, Rf value).
• Identical chemical properties with achiral reagents.
• Different interactions with plane-polarised light: one enantiomer rotates it clockwise (dextrorotatory, $+$ or $d$), the other anticlockwise (laevorotatory, $-$ or $l$).
• Different biological activity (e.g. thalidomide: one enantiomer is a sedative, the other is teratogenic).

Racemic mixture: A 50:50 mixture of both enantiomers. It shows no net optical rotation because the rotations cancel. Racemic mixtures have different physical properties from the pure enantiomers (e.g. different melting points).

CIP assignment for chirality (R/S system):

1. Assign priorities 1--4 to the four groups attached to the chiral centre using CIP rules.
2. Orient the molecule so the lowest-priority group (4) points away from you.
3. Trace a path from priority 1 to 2 to 3. If clockwise: R (rectus). If anticlockwise: S (sinister).

Summary of Isomerism Types

Type	Basis	Subtypes
Structural	Different connectivity	Chain, positional, functional group
Stereoisomerism	Same connectivity, different spatial arrangement	E/Z isomerism, optical isomerism

General Principles of Organic Reactions

Classification of Reagents

Nucleophiles ("nucleus-loving") are electron-rich species that donate a lone pair to form a new bond. They are attracted to electron-deficient centres. Examples: $\mathrm{OH}^-$, $\mathrm{CN}^-$, $\mathrm{NH}_3$, $\mathrm{H}_2\mathrm{O}$, $-\mathrm{OH}$.

Electrophiles ("electron-loving") are electron-deficient species that accept a lone pair. They are attracted to regions of high electron density. Examples: $\mathrm{H}^+$, $\mathrm{Br}^+$, $\mathrm{NO}_2^+$, $\mathrm{AlCl}_3$, $\delta^+$ carbons.

Radicals are species with an unpaired electron. They are highly reactive and are generated by homolytic fission. Examples: $\mathrm{Cl}^\bullet$, $\mathrm{CH}_3^\bullet$, $\mathrm{OH}^\bullet$.

Homolytic vs Heterolytic Fission

Homolytic fission: A covalent bond breaks symmetrically, with one electron going to each atom. Produces two radicals. Initiator: UV light or heat. Denoted with a half-arrow (fishhook).

$$\mathrm{Cl-Cl} \xrightarrow{\mathrm{UV}} \mathrm{Cl}^\bullet + \mathrm{Cl}^\bullet$$

Heterolytic fission: A covalent bond breaks asymmetrically, with both electrons going to the more electronegative atom. Produces a cation and an anion. Denoted with a full curly arrow.

$$\mathrm{H-Cl} \to \mathrm{H}^+ + \mathrm{Cl}^-$$

Mechanism Types

Nucleophilic substitution ($\mathrm{S_N}$): A nucleophile replaces a leaving group on a saturated carbon. Two mechanisms exist: $\mathrm{S_N}2$ (concerted, bimolecular) and $\mathrm{S_N}1$ (stepwise, unimolecular). Covered in detail in halogenoalkanes.

Electrophilic addition ($\mathrm{Ad}_E$): An electrophile adds across a C=C double bond. The $\pi$ bond acts as a nucleophile. The mechanism proceeds via a carbocation intermediate. Covered in alkenes.

Electrophilic aromatic substitution ($\mathrm{SE}_Ar$): An electrophile substitutes a hydrogen on an aromatic ring. The aromatic system is temporarily disrupted (arenium ion intermediate) but restored by loss of a proton. Covered in arenes.

Nucleophilic addition ($\mathrm{Ad}_N$): A nucleophile adds to the carbonyl carbon of an aldehyde or ketone. The C=O $\pi$ bond is broken. Covered in carbonyl compounds.

Elimination ($E$): A small molecule (e.g. $\mathrm{H}_2\mathrm{O}$, $\mathrm{HX}$) is removed from adjacent carbons to form a C=C double bond. Favoured by strong bases and high temperatures. Covered in halogenoalkanes.

Curly Arrow Conventions

Curly arrows are the formal notation for showing electron movement in organic mechanisms. The rules are strict:

1. A curly arrow starts from a lone pair, a bond, or a negative charge.
2. A curly arrow ends at an atom, a bond, or a positive charge.
3. A full arrow (two-barbed) represents the movement of an electron pair.
4. A half arrow (one-barbed, fishhook) represents the movement of a single electron (used in radical mechanisms).
5. Curly arrows show electron movement, not atom movement.
6. Every curly arrow must be accompanied by the corresponding change in bonding.

Reaction Coordinate Diagrams

A reaction coordinate diagram plots the potential energy of the system against the progress of the reaction. Key features:

• Reactants: Starting energy level.
• Transition state: The highest-energy point on the reaction pathway. Denoted with a double dagger ($\ddagger$). Bond breaking and forming are partially complete.
• Activation energy ($E_a$): The energy difference between reactants and the transition state. $E_a = E_{\mathrm{TS}} - E_{\mathrm{reactants}}$.
• Products: Final energy level.
• Enthalpy change ($\Delta H$): The energy difference between products and reactants. $\Delta H = E_{\mathrm{products}} - E_{\mathrm{reactants}}$.
• Intermediate: A local energy minimum (valley) between two transition states, present in multi-step reactions.

For a one-step reaction (e.g. $\mathrm{S_N}2$), there is one transition state and no intermediate. For a two-step reaction (e.g. $\mathrm{S_N}1$), there is an intermediate separated by two transition states.

Leaving Group Ability

In substitution and elimination reactions, the quality of the leaving group is critical. A good leaving group is a weak base that can stabilise the negative charge after departure.

Leaving group ability: $\mathrm{I}^- \gt \mathrm{Br}^- \gt \mathrm{Cl}^- \gg \mathrm{F}^- \gg \mathrm{OH}^-$, $\mathrm{NH}_2^-$, $\mathrm{CH}_3^-$.

The trend follows basicity (weaker base = better leaving group) and the stability of the conjugate acid. Water ($\mathrm{H}_2\mathrm{O}$) is a much better leaving group than $-\mathrm{OH}$, which is why acid catalysis is often used to protonate $-\mathrm{OH}$ groups before substitution or elimination.

Common Pitfalls

1. Incorrect numbering of the parent chain. The principal functional group must receive the lowest locant. If there is a tie between the functional group and a substituent, the functional group wins. If two substituents tie, use the "first point of difference" rule (compare locants sequentially and prefer the lower at the first point of difference).

2. Confusing E/Z with cis/trans. Cis/trans is a subset of E/Z that only works when each carbon has one hydrogen. For general cases, always use E/Z with CIP priority rules.

3. Identifying chirality incorrectly. Having four different groups on a carbon is necessary but not sufficient for optical activity -- the molecule must not possess an internal plane of symmetry. For example, meso-tartaric acid has two chiral centres but is achiral overall because it has a plane of symmetry.

4. Drawing curly arrows from atoms instead of electrons. Arrows always originate from electron sources (lone pairs, bonds, charges), never from bare atoms.

5. Forgetting that the aromatic ring is nucleophilic. Students often confuse the reactivity of benzene with that of alkenes. Benzene undergoes substitution (preserving the aromatic system), not addition (which would destroy the delocalisation energy).

Practice Problems

Problem 1

Draw all structural isomers of $\mathrm{C}_5\mathrm{H}_{11}\mathrm{Br}$. Identify which isomers contain chiral centres.

Solution:

There are eight structural isomers of $\mathrm{C}_5\mathrm{H}_{11}\mathrm{Br}$:

1. 1-bromopentane (no chiral centre)
2. 2-bromopentane (chiral centre at C-2: bonded to $-\mathrm{Br}$, $-\mathrm{CH}_3$, $-\mathrm{CH}_2\mathrm{CH}_3$, $-\mathrm{H}$)
3. 3-bromopentane (no chiral centre: C-3 has two identical $-\mathrm{CH}_2\mathrm{CH}_3$ groups)
4. 1-bromo-2-methylbutane (chiral centre at C-2: bonded to $-\mathrm{Br}$, $-\mathrm{CH}_3$, $-\mathrm{CH}_2\mathrm{CH}_3$, $-\mathrm{CH}_2\mathrm{Br}$) -- wait, this is 2-methylbutane with bromine on C-1. C-2 is bonded to $-\mathrm{CH}_3$, $-\mathrm{CH}_3$, $-\mathrm{CH}_2\mathrm{Br}$, $-\mathrm{H}$ -- two methyl groups are the same, so not chiral.
5. 1-bromo-3-methylbutane (no chiral centre)
6. 2-bromo-2-methylbutane (no chiral centre: C-2 has three different groups but also has two $-\mathrm{CH}_3$? No: $-\mathrm{CH}_3$, $-\mathrm{CH}_3$, $-\mathrm{CH}_2\mathrm{CH}_3$, $-\mathrm{Br}$ -- two methyls are identical, not chiral)
7. 1-bromo-2,2-dimethylpropane (no chiral centre)
8. 2-bromo-3-methylbutane (chiral centre at C-2: bonded to $-\mathrm{Br}$, $-\mathrm{CH}_3$, $-\mathrm{CH}(\mathrm{CH}_3)_2$, $-\mathrm{H}$)

Isomers with chiral centres: 2-bromopentane and 2-bromo-3-methylbutane.

Problem 2

Assign R/S configuration to the chiral centre in (2R)-2-chlorobutane.

Solution:

The chiral centre is C-2, bonded to: $-\mathrm{Cl}$ (priority 1, atomic number 17), $-\mathrm{CH}_2\mathrm{CH}_3$ (priority 2, C bonded to C,H,H), $-\mathrm{CH}_3$ (priority 3, C bonded to H,H,H), $-\mathrm{H}$ (priority 4, atomic number 1).

Orient the molecule with $-\mathrm{H}$ (lowest priority) pointing away. Tracing from priority 1 ($-\mathrm{Cl}$) to priority 2 ($-\mathrm{CH}_2\mathrm{CH}_3$) to priority 3 ($-\mathrm{CH}_3$): this is clockwise, giving R configuration.

If the lowest-priority group is pointing towards you (as drawn in a Fischer projection or a dashed wedge), the observed direction is reversed: clockwise becomes S and anticlockwise becomes R.

Problem 3

Name the following compound using IUPAC nomenclature:

$\mathrm{CH}_3\mathrm{CH}(\mathrm{Cl})\mathrm{CH}(\mathrm{CH}_3)\mathrm{CH}_2\mathrm{CH}_3$

Solution:

Step 1: The longest chain containing the principal functional group (halogen) is 5 carbons (pentane).

Step 2: Number from the end that gives the lowest locants to the substituents. Numbering from the left: Cl on C-2, methyl on C-3. Numbering from the right: Cl on C-4, methyl on C-3. Left numbering gives lower first locant (2 < 4), so number from the left.

Step 3: Name substituents alphabetically: chloro- (before methyl-).

The name is 3-chloro-2-methylpentane.

Step 4: Check for chirality. C-2 is bonded to $-\mathrm{Cl}$, $-\mathrm{CH}_3$, $-\mathrm{CH}(\mathrm{CH}_3)\mathrm{CH}_2\mathrm{CH}_3$, and $-\mathrm{H}$. All four groups are different, so C-2 is a chiral centre. The compound exists as a pair of enantiomers: (2R)-3-chloro-2-methylpentane and (2S)-3-chloro-2-methylpentane.

Mechanisms: Curly Arrow Convention

Rules for Drawing Curly Arrows

1. Curly arrows show the movement of electron pairs. A full-headed arrow ($\curvearrowright$) shows movement of a pair of electrons. A half-headed arrow ($\hookrightarrow$, fishhook) shows movement of a single electron (used in radical mechanisms).

2. Arrows start from electron-rich sites: lone pairs, pi bonds, sigma bonds, or negative charges.

3. Arrows point towards electron-deficient sites: atoms (to form new bonds) or bonds (to break them).

4. The tail of the arrow is at the source of electrons. The head of the arrow is where the electrons go.

Common Arrow-Pushing Patterns

Nucleophilic attack: Arrow from a lone pair on the nucleophile to the electrophilic atom.

$$\mathrm{Nu}: \curvearrowright \mathrm{C}^{\delta+}$$

Pi bond as nucleophile: Arrow from the middle of a pi bond to an electrophile (electrophilic addition to alkenes).

$$\mathrm{C}=\mathrm{C} \curvearrowright \mathrm{E}^+$$

Proton transfer: Arrow from a lone pair to $\mathrm{H}^+$, or from a bond to $\mathrm{H}$ (to break it).

$$\mathrm{B}: \curvearrowright \mathrm{H}-\mathrm{A}$$

Bond cleavage: Arrow from the middle of a bond to one of the bonded atoms (heterolytic fission).

$$\mathrm{A}-\mathrm{B} \curvearrowright \mathrm{B}$$

Worked Example: Drawing the SN2 Mechanism

The reaction of $\mathrm{CH}_3\mathrm{CH}_2\mathrm{Br}$ with $\mathrm{OH}^-$:

1. Arrow from the lone pair on $\mathrm{O}$ of $\mathrm{OH}^-$ to the $\delta^+$ carbon of the $\mathrm{C}-\mathrm{Br}$ bond (nucleophilic attack).
2. Arrow from the $\mathrm{C}-\mathrm{Br}$ bond to $\mathrm{Br}$ (bond cleavage, heterolytic).

Both arrows are drawn simultaneously, showing the concerted nature of SN2. The transition state is shown with a dotted line between C and both the incoming $\mathrm{OH}$ and outgoing $\mathrm{Br}$.

Stereoisomerism Summary

Types of Stereoisomerism

Type	Requirement	Example
Optical (enantiomers)	Chiral centre (no plane of symmetry)	$(R)$- and $(S)$-2-chlorobutane
E/Z (geometric)	Restricted rotation about C=C or C=N	$(E)$- and $(Z)$-but-2-ene
Optical (diastereomers)	Multiple chiral centres, not mirror images	$(2R,3R)$- and $(2R,3S)$-tartaric acid

Enantiomers vs Diastereomers

• Enantiomers: Non-superimposable mirror images. Same physical properties except for interaction with plane-polarised light (and other chiral environments).
• Diastereomers: Stereoisomers that are not mirror images. Different physical properties (melting point, boiling point, solubility, $R_f$ in TLC).

Optical Activity

A chiral compound rotates the plane of plane-polarised light. The observed rotation $\alpha$ depends on:

$$[\alpha] = \frac◆LB◆\alpha_\mathrm{obs}◆RB◆◆LB◆l \times c◆RB◆$$

where $[\alpha]$ is the specific rotation (constant for a given compound), $l$ is the path length (in dm), and $c$ is the concentration (in $\mathrm{g/cm}^3$).

A racemic mixture (50:50 enantiomers) has $[\alpha] = 0$ (no net rotation) because the two enantiomers rotate light by equal amounts in opposite directions.

Racemisation

Racemisation is the conversion of a single enantiomer into a racemic mixture. It occurs when a chiral centre is temporarily destroyed, for example:

• Via a planar carbocation intermediate (SN1 mechanism).
• Via enolisation (alpha carbon of a carbonyl compound).
• Under acidic or basic conditions for certain compounds.

The rate of racemisation depends on the mechanism and conditions. Compounds with stable chiral centres (no carbocation intermediates, no enolisable protons) are configurationally stable.

Worked Example: Identifying Chirality

Is 2,3-dibromobutane chiral?

Draw the structure: $\mathrm{CH}_3\mathrm{CHBrCHBrCH}_3$. C-2 is bonded to $-\mathrm{Br}$, $-\mathrm{CH}_3$, $-\mathrm{CHBrCH}_3$, $-\mathrm{H}$. C-3 is bonded to $-\mathrm{Br}$, $-\mathrm{CH}_3$, $-\mathrm{CHBrCH}_3$, $-\mathrm{H}$.

Both C-2 and C-3 are chiral centres. There are three stereoisomers:

1. $(2R,3R)$-2,3-dibromobutane (meso: has a plane of symmetry through C-2--C-3 bond, achiral)
2. $(2S,3S)$-2,3-dibromobutane (identical to $(2R,3R)$ by $180^\circ$ rotation)
3. $(2R,3S)$-2,3-dibromobutane (no plane of symmetry, chiral; this is the meso compound)

Wait: $(2R,3R)$ and $(2S,3S)$ are enantiomers of each other. $(2R,3S)$ is a diastereomer of both and is the meso form (has a plane of symmetry bisecting the molecule). So there are three stereoisomers: one pair of enantiomers $(2R,3R)/(2S,3S)$ and one meso compound $(2R,3S)$.

Functional Group Interconversion Summary

The following table summarises the key interconversions covered across the organic chemistry modules:

From	To	Reagent/conditions	Mechanism
Alkene	Alkane	$\mathrm{H}_2$ / Ni catalyst	Addition (hydrogenation)
Alkene	Halogenoalkane	HX ($\mathrm{HBr}$, $\mathrm{HCl}$)	Electrophilic addition
Alkene	Dihalogenoalkane	$\mathrm{Br}_2$ or $\mathrm{Cl}_2$	Electrophilic addition
Alkene	Alcohol	$\mathrm{H}_3\mathrm{PO}_4$ / steam, $300^\circ\mathrm{C}$	Electrophilic addition
Alkene	Diol	Cold dilute $\mathrm{KMnO}_4$	Oxidation
Alkane	Halogenoalkane	$\mathrm{Cl}_2$ / UV	Free radical substitution
Alcohol	Halogenoalkane	$\mathrm{PBr}_3$, $\mathrm{SOCl}_2$, or HX	Substitution
Alcohol	Aldehyde	Distillation with $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ / $\mathrm{H}^+$	Oxidation
Alcohol	Carboxylic acid	Reflux with $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ / $\mathrm{H}^+$	Oxidation
Alcohol	Alkene	$\mathrm{H}_2\mathrm{SO}_4$, $170^\circ\mathrm{C}$	Elimination (dehydration)
Alcohol	Ester	Carboxylic acid / $\mathrm{H}_2\mathrm{SO}_4$ reflux	Esterification
Halogenoalkane	Alcohol	$\mathrm{NaOH}(aq)$, heat	Nucleophilic substitution (SN2/SN1)
Halogenoalkane	Alkene	$\mathrm{NaOH}$ in ethanol, heat	Elimination (E2/E1)
Halogenoalkane	Nitrile	$\mathrm{KCN}$ in ethanol	Nucleophilic substitution
Halogenoalkane	Amine	Excess $\mathrm{NH}_3$	Nucleophilic substitution
Halogenoalkane	Amine (primary)	Phthalimide / $\mathrm{KOH}$ (Gabriel)	Nucleophilic substitution
Halogenoalkane	Ether	Alkoxide ($\mathrm{RO}^-\mathrm{Na}^+$)	Williamson synthesis (SN2)
Nitrile	Primary amine	$\mathrm{LiAlH}_4$ then $\mathrm{H}_2\mathrm{O}$	Reduction
Nitrile	Carboxylic acid	$\mathrm{H}_3\mathrm{O}^+$, reflux	Hydrolysis
Amide	Amine	$\mathrm{LiAlH}_4$ then $\mathrm{H}_2\mathrm{O}$	Reduction
Amide	Carboxylic acid	$\mathrm{HCl}$ reflux or $\mathrm{NaOH}$ reflux	Hydrolysis
Aldehyde	Primary alcohol	$\mathrm{NaBH}_4$ then $\mathrm{H}^+$	Nucleophilic addition
Aldehyde	Carboxylic acid	$\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ / $\mathrm{H}^+$ reflux	Oxidation
Ketone	Secondary alcohol	$\mathrm{NaBH}_4$ then $\mathrm{H}^+$	Nucleophilic addition
Aldehyde/Ketone	Cyanohydrin	$\mathrm{HCN}$ / $\mathrm{NaCN}$	Nucleophilic addition
Aldehyde/Ketone	2,4-DNPH derivative	2,4-dinitrophenylhydrazine	Nucleophilic addition-elimination
Carboxylic acid	Ester	Alcohol / $\mathrm{H}_2\mathrm{SO}_4$ reflux	Esterification
Carboxylic acid	Acyl chloride	$\mathrm{SOCl}_2$	Nucleophilic acyl substitution
Acyl chloride	Ester	Alcohol	Nucleophilic acyl substitution
Acyl chloride	Amide	Amine (excess)	Nucleophilic acyl substitution
Acyl chloride	Carboxylic acid	Water	Nucleophilic acyl substitution (hydrolysis)
Benzene	Nitrobenzene	$\mathrm{HNO}_3$ / $\mathrm{H}_2\mathrm{SO}_4$, $50$--$60^\circ\mathrm{C}$	Electrophilic substitution
Benzene	Bromobenzene	$\mathrm{Br}_2$ / $\mathrm{FeBr}_3$	Electrophilic substitution
Benzene	Methylbenzene	$\mathrm{CH}_3\mathrm{Cl}$ / $\mathrm{AlCl}_3$ (Friedel-Crafts)	Electrophilic substitution
Benzene	Phenyl ketone	$\mathrm{RCOCl}$ / $\mathrm{AlCl}_3$ (Friedel-Crafts)	Electrophilic substitution
Nitrobenzene	Phenylamine	$\mathrm{Sn}$ / $\mathrm{HCl}$ then $\mathrm{NaOH}$	Reduction
Phenylamine	Diazonium salt	$\mathrm{NaNO}_2$ / $\mathrm{HCl}$, $0$--$5^\circ\mathrm{C}$	Diazotisation

Determining the Structure of an Unknown Organic Compound

A systematic approach to identifying an unknown compound using spectroscopic data:

Step 1: Determine the Molecular Formula

From the molecular ion peak in mass spectrometry ($M^+$), determine the molecular mass. Use the rule of 13 or isotope patterns to suggest possible formulas. High-resolution MS gives the exact mass to distinguish between isobaric formulas.

Step 2: Calculate the Degree of Unsaturation (Index of Hydrogen Deficiency)

$$\text{DoU} = \frac{2C + 2 + N - H - X}{2}$$

where $C$ = number of carbons, $H$ = hydrogens, $N$ = nitrogens, $X$ = halogens. Oxygen does not affect the DoU.

DoU	Possible structural features
0	Acyclic, saturated (only single bonds)
1	One ring OR one double bond
2	Two rings, two double bonds, one triple bond, or one ring + one double bond
4	Often suggests a benzene ring (3 double bonds + 1 ring)

Step 3: IR Spectroscopy -- Identify Functional Groups

Check for characteristic absorptions: O--H, C=O, C=C, C$\equiv$N, C--O, N--H. The absence of a peak is as informative as its presence (e.g. no C=O rules out aldehydes, ketones, acids, esters).

Step 4: NMR Spectroscopy -- Piece Together the Structure

${}^1\mathrm{H}$ NMR tells you the number of hydrogen environments (signals), the ratio of protons in each (integration), the neighbouring environments (splitting), and the electronic environment (chemical shift).

${}^{13}\mathrm{C}$ NMR tells you the number of carbon environments. Combine this with the DoU and IR data.

Step 5: Mass Spectrometry -- Confirm and Find Fragments

The molecular ion confirms the molecular mass. Fragment peaks provide information about substructures (e.g. loss of 15 = $\mathrm{CH}_3$, loss of 17 = $\mathrm{OH}$, loss of 29 = $\mathrm{CHO}$ or $\mathrm{C}_2\mathrm{H}_5$, loss of 31 = $\mathrm{OCH}_3$, loss of 45 = $\mathrm{COOH}$).

Worked Example: Complete Structure Determination

An unknown compound has:

• Mass spec: $M^+ = 120$, M+1 peak approximately $7.7\%$ of $M^+$
• IR: $1705\,\mathrm{cm}^{-1}$ (strong), $2850$--$2950\,\mathrm{cm}^{-1}$ (strong), no O--H, no C--O
• ${}^1\mathrm{H}$ NMR: $\delta\, 1.2$ (t, 6H), $\delta\, 2.9$ (q, 4H), $\delta\, 7.5$ (t, 2H), $\delta\, 7.9$ (d, 2H)
• ${}^{13}\mathrm{C}$ NMR: 6 peaks

Step 1: $M = 120$. M+1 = $7.7\% \implies$ approximately 7 carbons ($1.1\%$ per carbon). $\mathrm{C}_7\mathrm{H}_8\mathrm{O}$: $7(12) + 8(1) + 16 = 84 + 8 + 16 = 108 \neq 120$. $\mathrm{C}_8\mathrm{H}_8\mathrm{O}$: $96 + 8 + 16 = 120$. Correct.

Step 2: $\text{DoU} = \frac{2(8) + 2 - 8}{2} = \frac{10}{2} = 5$. Likely a benzene ring (DoU = 4) plus one more unsaturation (C=O from IR).

Step 3: IR: $1705\,\mathrm{cm}^{-1}$ = C=O (ketone). No broad O--H, no strong C--O, so not a carboxylic acid or ester.

Step 4: ${}^1\mathrm{H}$ NMR: 4 signals.

• $\delta\, 7.5$ (t, 2H) and $\delta\, 7.9$ (d, 2H) = 4 aromatic protons = monosubstituted benzene ring (but with 4 protons, it is para-disubstituted: two doublets would be expected for AA'BB' pattern; the triplet and doublet suggest a different arrangement). Actually, a triplet and doublet for 2H each suggests a 1,4-disubstituted benzene (AA'BB' pattern can appear as approximate triplet/doublet at low resolution).
• $\delta\, 1.2$ (t, 6H) = two equivalent $\mathrm{CH}_3$ groups.
• $\delta\, 2.9$ (q, 4H) = two equivalent $\mathrm{CH}_2$ groups adjacent to the methyls.

This pattern suggests an $\mathrm{N}(\mathrm{CH}_2\mathrm{CH}_3)_2$ group, but the formula has no nitrogen. Reconsidering: it suggests two $\mathrm{CH}_2\mathrm{CH}_3$ groups. With only C, H, O available: the 4 aliphatic protons at $\delta\, 2.9$ (adjacent to an aromatic or carbonyl) and 6 at $\delta\, 1.2$ suggest two ethyl groups.

Structure: 1,4-diacetylbenzene or 4-ethoxyacetophenone, etc. The correct structure consistent with all data is 1-phenylpropan-1-one (propiophenone) if we have 5 aromatic protons... but we only have 4 aromatic protons. With $\mathrm{C}_8\mathrm{H}_8\mathrm{O}$, DoU = 5, a benzene ring (4 unsaturations) plus C=O: para-ethylacetophenone ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{C}_6\mathrm{H}_4\mathrm{COCH}_3$) = $\mathrm{C}_9\mathrm{H}_{10}\mathrm{O}$ (wrong formula).

The correct answer is 4-methylpropiophenone: $\mathrm{CH}_3\mathrm{COC}_6\mathrm{H}_4\mathrm{CH}_3$ = $\mathrm{C}_9\mathrm{H}_{10}\mathrm{O}$ (still wrong).

Returning to $\mathrm{C}_8\mathrm{H}_8\mathrm{O}$: acetophenone ($\mathrm{C}_6\mathrm{H}_5\mathrm{COCH}_3$) = $\mathrm{C}_8\mathrm{H}_8\mathrm{O}$, $M = 120$. But acetophenone would have 5 aromatic protons and a $\mathrm{CH}_3$ singlet, not matching the triplet/quartet pattern.

Correct answer: The NMR pattern (t, 6H; q, 4H) is characteristic of a diethyl ketone fragment. With $M = 120$ and $\mathrm{C}_8\mathrm{H}_8\mathrm{O}$, the structure is phenacyl ethyl ether ($\mathrm{C}_6\mathrm{H}_5\mathrm{COCH}_2\mathrm{CH}_3$) = butyrophenone. But this is $\mathrm{C}_{10}\mathrm{H}_{12}\mathrm{O}$...

The key insight: with triplet (6H) and quartet (4H), there are two $-\mathrm{CH}_2\mathrm{CH}_3$ groups. With 4 aromatic protons (1,4-disubstituted), C=O, and formula $\mathrm{C}_8\mathrm{H}_8\mathrm{O}$, the molecule is too small. The formula must be reconsidered: $\mathrm{C}_9\mathrm{H}_{12}\mathrm{O}$ has $M = 136$. With $M = 120$, the only option with two ethyl groups is not possible with $\mathrm{C}_8\mathrm{H}_8\mathrm{O}$.

This example illustrates the importance of cross-checking all data. In an exam, if the data are internally consistent, proceed with the structure that fits best. Here, the triplet (6H)/quartet (4H) pattern requires two ethyl groups (4 aliphatic carbons + 10 aliphatic H), leaving only 4 carbons and no hydrogens for the aromatic portion -- impossible. The data likely represent a larger molecule or the molecular formula needs verification.

Additional Practice Problems

Problem 4

For each of the following molecules, state whether it is chiral and explain your reasoning:

(a) $\mathrm{CH}_3\mathrm{CHClCH}_3$ (2-chloropropane) (b) $\mathrm{CH}_3\mathrm{CHClCH}_2\mathrm{CH}_3$ (2-chlorobutane) (c) $\mathrm{CH}_2\mathrm{ClCH}_2\mathrm{CH}_2\mathrm{Cl}$ (1,3-dichloropropane) (d) $\mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{COOH}$ (2-hydroxypropanoic acid, lactic acid)

Solution:

(a) Not chiral. The central carbon (C-2) is bonded to $\mathrm{CH}_3$, $\mathrm{CH}_3$, $\mathrm{Cl}$, and $\mathrm{H}$. Two of the substituents are the same ($\mathrm{CH}_3$), so there is a plane of symmetry through C-2, Cl, and H. The molecule has a superimposable mirror image.

(b) Chiral. The central carbon (C-2) is bonded to $\mathrm{CH}_3$, $\mathrm{C}_2\mathrm{H}_5$, $\mathrm{Cl}$, and $\mathrm{H}$. All four substituents are different. The molecule has a non-superimposable mirror image (exists as $R$ and $S$ enantiomers).

(c) Not chiral. No carbon atom is bonded to four different groups. C-2 is bonded to two $\mathrm{CH}_2\mathrm{Cl}$ groups (identical).

(d) Chiral. C-2 is bonded to $\mathrm{CH}_3$, $\mathrm{OH}$, $\mathrm{COOH}$, and $\mathrm{H}$. All four substituents are different. Lactic acid exists as two enantiomers. L-(+)-lactic acid is the naturally occurring form found in muscle tissue.

Problem 5

Classify each of the following reactions as substitution, addition, elimination, or hydrolysis, and identify the reagent as a nucleophile, electrophile, or both:

(a) $\mathrm{CH}_3\mathrm{CH}_2\mathrm{Br} + \mathrm{OH}^- \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} + \mathrm{Br}^-$ (b) $\mathrm{CH}_2=\mathrm{CH}_2 + \mathrm{HBr} \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{Br}$ (c) $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{Br} + \mathrm{OH}^- \xrightarrow{\text{ethanol, }\Delta} \mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 + \mathrm{Br}^- + \mathrm{H}_2\mathrm{O}$ (d) $\mathrm{CH}_3\mathrm{COOCH}_3 + \mathrm{NaOH} \to \mathrm{CH}_3\mathrm{COO}^-\mathrm{Na}^+ + \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}$

Solution:

(a) Nucleophilic substitution (SN2). $\mathrm{OH}^-$ is the nucleophile (attacks the electrophilic carbon bearing Br). $\mathrm{Br}^-$ is the leaving group.

(b) Electrophilic addition. The $\pi$ electrons of the C=C act as a nucleophile, attacking the $\delta^+$ hydrogen of HBr (electrophile). The $\mathrm{Br}^-$ then acts as a nucleophile in the second step.

(c) Elimination (E2). $\mathrm{OH}^-$ acts as a base (not a nucleophile under these conditions), abstracting a $\beta$-hydrogen. The C--Br bond breaks simultaneously, forming the C=C double bond. The high temperature and ethanol solvent favour elimination over substitution.

(d) Hydrolysis (base-catalysed). $\mathrm{OH}^-$ is the nucleophile that attacks the carbonyl carbon of the ester, forming a tetrahedral intermediate that collapses to give the carboxylate and alcohol.

Advanced Organic Chemistry Introduction

Functional Group Priority Table (IUPAC Nomenclature)

When naming organic compounds, the principal functional group gets the lowest possible locant. Priority order (highest to lowest):

Priority	Functional Group	Suffix	Prefix
1	Carboxylic acid	-oic acid	--
2	Ester	-oate	alkoxycarbonyl-
3	Amide	-amide	amido-
4	Nitrile	-nitrile	cyano-
5	Aldehyde	-al	oxo-
6	Ketone	-one	oxo-
7	Alcohol	-ol	hydroxy-
8	Amine	-amine	amino-
9	Alkene	-ene	--
10	Alkyne	-yne	--
11	Halogenoalkane	--	halo- (chloro-, bromo-, iodo-)
12	Ether	--	alkoxy-

Worked Example: Name $\mathrm{HOOC}-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NH}_2$.

The carboxylic acid has highest priority (suffix). Number from the carboxylic acid end:

$\mathrm{HOOC}$ (C-1), $\mathrm{CH}(\mathrm{OH})$ (C-2), $\mathrm{CH}_2$ (C-3), $\mathrm{CH}_2(\mathrm{NH}_2)$ (C-4).

Name: 4-amino-2-hydroxybutanoic acid

CIP Priority Rules for $R$/$S$ Configuration

The Cahn-Ingold-Prelog (CIP) rules assign priority to substituents on a chiral centre:

1. Higher atomic number = higher priority. $\mathrm{I} > \mathrm{Br} > \mathrm{Cl} > \mathrm{F} > \mathrm{O} > \mathrm{N} > \mathrm{C} > \mathrm{H}$

2. If the first atoms are the same, look at the next atoms in the chain until a difference is found. Treat multiple bonds as if the atom is bonded to an equivalent number of the same atom by single bonds.

3. Orient the molecule so the lowest priority group (usually H) points away from you. Trace 1 $\to$ 2 $\to$ 3:

   • Clockwise = $R$ (Rectus)
   • Anticlockwise = $S$ (Sinister)

Worked Example: Assign $R$/$S$ to lactic acid: $\mathrm{CH}_3-\mathrm{CH}(\mathrm{OH})-\mathrm{COOH}$.

The chiral centre is C-2, bonded to: $\mathrm{CH}_3$, $\mathrm{OH}$, $\mathrm{COOH}$, $\mathrm{H}$.

Priority order: $\mathrm{OH}$ (O bonded to H) > $\mathrm{COOH}$ (C bonded to O,O) > $\mathrm{CH}_3$ (C bonded to H,H,H) > $\mathrm{H}$.

With H pointing away: 1($\mathrm{OH}$) $\to$ 2($\mathrm{COOH}$) $\to$ 3($\mathrm{CH}_3$) is clockwise.

Lactic acid is $(S)$-lactic acid (L-(+)-lactic acid).

Reaction Classification: Complete Reference

Reaction type	Bond change	Typical reagents	Example
Nucleophilic substitution (SN1/SN2)	One group replaced by a nucleophile	$\mathrm{OH}^-$, $\mathrm{CN}^-$, $\mathrm{NH}_3$	Halogenoalkane $\to$ alcohol
Electrophilic addition	Two groups add across C=C or C$\equiv$C	$\mathrm{HBr}$, $\mathrm{Br}_2$, $\mathrm{H}_2\mathrm{SO}_4$	Alkene $\to$ dibromoalkane
Electrophilic substitution	Electrophile replaces H on aromatic ring	$\mathrm{Br}^+$, $\mathrm{NO}_2^+$, $\mathrm{CH}_3\mathrm{CO}^+$	Benzene $\to$ bromobenzene
Elimination (E1/E2)	Loss of small molecule to form C=C	Base in ethanol, $\Delta$	Halogenoalkane $\to$ alkene
Nucleophilic addition	Nucleophile adds to C=O	$\mathrm{NaBH}_4$, $\mathrm{LiAlH}_4$, $\mathrm{HCN}$	Aldehyde $\to$ alcohol
Nucleophilic addition-elimination	Addition to C=O followed by loss	$\mathrm{NH}_3$, $\mathrm{H}_2\mathrm{O}$, alcohols	Acyl chloride $\to$ amide
Oxidation	Increase in O:C ratio or loss of H	$\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7/\mathrm{H}^+$, acidified $\mathrm{KMnO}_4$	Primary alcohol $\to$ aldehyde $\to$ acid
Reduction	Decrease in O:C ratio or gain of H	$\mathrm{NaBH}_4$, $\mathrm{LiAlH}_4$, $\mathrm{H}_2$/Ni	Aldehyde $\to$ primary alcohol
Hydrolysis	Reaction with water (often base/acid-catalysed)	$\mathrm{NaOH}(\mathrm{aq})$, $\mathrm{H}^+(\mathrm{aq})$	Ester $\to$ acid + alcohol
Condensation	Two molecules join with loss of small molecule	Concentrated $\mathrm{H}_2\mathrm{SO}_4$, $\Delta$	Alcohol + acid $\to$ ester + $\mathrm{H}_2\mathrm{O}$
Free radical substitution	Atom replaced via radical chain	$\mathrm{Cl}_2/h\nu$, $\mathrm{Br}_2/h\nu$	Alkane $\to$ haloalkane
Free radical addition	Radicals add across C=C	$\mathrm{HF}$, peroxides, $\Delta$	Anti-Markovnikov alkene addition

Structural Isomerism: Complete Classification

Chain isomerism: Different carbon skeleton.

• $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_3$ (butane) vs $\mathrm{CH}(\mathrm{CH}_3)_3$ (2-methylpropane)

Positional isomerism: Same functional group, different position on the chain.

• $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH}$ (propan-1-ol) vs $\mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{CH}_3$ (propan-2-ol)

Functional group isomerism: Different functional groups, same molecular formula.

• $\mathrm{C}_2\mathrm{H}_6\mathrm{O}$: ethanol ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}$) vs methoxymethane ($\mathrm{CH}_3\mathrm{OCH}_3$)
• $\mathrm{C}_3\mathrm{H}_6\mathrm{O}$: propanal vs propanone (acetone) vs methyl ethanoate vs prop-2-en-1-ol

Metamerism (less common): Same functional group but different alkyl groups on either side.

• $\mathrm{CH}_3\mathrm{CH}_2\mathrm{OCH}_2\mathrm{CH}_3$ (diethyl ether) vs $\mathrm{CH}_3\mathrm{OCH}_2\mathrm{CH}_2\mathrm{CH}_3$ (methyl propyl ether)

Stereoisomerism: Complete Classification

Geometric (cis-trans / E-Z) isomerism:

• Requires restricted rotation (C=C double bond or cyclic structure)
• Each carbon of the double bond must have two different groups
• E (entgegen): highest priority groups on opposite sides
• Z (zusammen): highest priority groups on same side

Optical isomerism:

• Requires a chiral centre (carbon bonded to four different groups)
• Non-superimposable mirror images (enantiomers)
• Same physical properties except for rotation of plane-polarised light and reactions with other chiral molecules

Curly Arrow Conventions: Full Reference

1. Curly arrows show the movement of electron pairs. A full-headed arrow ($\curvearrowright$) shows movement of a pair of electrons. A half-headed arrow ($\hookrightarrow$) shows movement of a single electron (used in radical mechanisms).

2. Arrows always start from a source of electrons: a lone pair, a $\pi$ bond, or a $\sigma$ bond.

3. Arrows always point to where the electrons are going: to an atom (forming a new bond or lone pair) or to the middle of a bond (breaking it).

4. Nucleophilic attack: Arrow starts from the nucleophile's lone pair and points to the electrophilic atom.

5. Electrophilic attack: Arrow starts from the $\pi$ bond and points to the electrophile.

6. Bond breaking (heterolytic): Arrow from the middle of the bond to the more electronegative atom (which takes both electrons).

7. Bond breaking (homolytic): Two half-headed arrows, one from the bond to each atom.

Common Pitfalls

1. Confusing nucleophiles and electrophiles: A nucleophile is electron-rich (lone pair or negative charge) and attacks electron-deficient centres. An electrophile is electron-deficient (positive charge, partial positive, or empty orbital) and accepts electrons.

2. Drawing curly arrows incorrectly: Arrows must start from electron sources and point to electron sinks. Common errors include drawing arrows from atoms without lone pairs, or arrows that point to nothing.

3. Identifying chiral centres: A carbon bonded to four different groups is chiral. "Different groups" means structurally different, not just different positions. $\mathrm{CH}_2\mathrm{ClCH}_2\mathrm{CH}_2\mathrm{Cl}$ is not chiral because the two $\mathrm{CH}_2\mathrm{Cl}$ groups are identical.

4. Nomenclature: functional group vs substituent: The principal functional group determines the suffix. All other functional groups are named as prefixes. Students often use the wrong suffix or forget to include all substituents.

5. E/Z vs cis/trans: Use E/Z for all cases. Use cis/trans only for simple cases with two identical substituents (e.g. cis-but-2-ene, trans-but-2-ene). When substituents are different, only E/Z is unambiguous.

Exam-Style Questions with Full Mark Schemes

Q1 (5 marks)

Identify the chiral centres in the following molecules and state the total number of stereoisomers for each:

(a) 2,3-dibromobutane (b) 2,3-dibromopentane

Mark Scheme:

(a) 2,3-dibromobutane: $\mathrm{CH}_3-\mathrm{CH}(\mathrm{Br})-\mathrm{CH}(\mathrm{Br})-\mathrm{CH}_3$. Both C-2 and C-3 are chiral centres (each bonded to $\mathrm{CH}_3$, $\mathrm{H}$, $\mathrm{Br}$, and $\mathrm{CH}(\mathrm{Br})\mathrm{CH}_3$) (1 mark for each chiral centre, 2 marks total).

Maximum number of stereoisomers $= 2^n = 2^2 = 4$. However, there is a meso compound (the $RS$ isomer has a plane of symmetry), so there are 3 stereoisomers ($RR$, $SS$, and $RS$ meso) (1 mark).

(b) 2,3-dibromopentane: $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}(\mathrm{Br})-\mathrm{CH}(\mathrm{Br})-\mathrm{CH}_3$. Both C-2 and C-3 are chiral centres (each bonded to four different groups: C-2 is bonded to $\mathrm{CH}_3$, $\mathrm{H}$, $\mathrm{Br}$, $\mathrm{CH}(\mathrm{Br})\mathrm{CH}_2\mathrm{CH}_3$) (1 mark).

No meso compound is possible (the two halves of the molecule are different: C-1 is $\mathrm{CH}_3$ but C-5 is $\mathrm{CH}_3$ on one side vs $\mathrm{CH}_2\mathrm{CH}_3$ on the other). Total stereoisomers $= 2^2 = 4$ ($RR$, $SS$, $RS$, $SR$) (1 mark).

Q2 (6 marks)

For the reaction $\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}\mathrm{CH}_3 + \mathrm{Br}_2 \to \mathrm{CH}_3\mathrm{CHBrCHBrCH}_3$:

(a) Draw the mechanism using curly arrows. (3 marks) (b) State the stereochemistry of the products. (3 marks)

Mark Scheme:

(a) Mechanism (3 marks):

• The $\pi$ electrons of the C=C attack one Br atom in $\mathrm{Br}_2$, forming a C--Br bond and heterolytically cleaving the Br--Br bond (the arrow goes from the $\pi$ bond to one Br, and another arrow goes from the Br--Br bond to the other Br, forming $\mathrm{Br}^-$) (2 marks).
• The $\mathrm{Br}^-$ ion then attacks the carbocation from the opposite side (back-side attack), forming the second C--Br bond (1 mark).

(b) The product has two new chiral centres (C-2 and C-3), each of which can be $R$ or $S$ (1 mark). The anti addition (bromine adds from opposite faces of the double bond) gives the $RS$ and $SR$ diastereomers as the products (2 marks). These are enantiomers of each other, forming a racemic mixture. (The $RR$ and $SS$ diastereomers are NOT formed because they would require syn addition.)

Q3 (4 marks)

Classify and explain the type of isomerism shown by each pair:

(a) $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CHO}$ and $\mathrm{CH}_3\mathrm{COCH}_3$ (b) $(R)$-2-chlorobutane and $(S)$-2-chlorobutane (c) $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{Cl}$ and $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CHClCH}_3$ (d) $\mathrm{CH}_3\mathrm{CH}=\mathrm{CHCH}_3$ (cis) and $\mathrm{CH}_3\mathrm{CH}=\mathrm{CHCH}_3$ (trans)

Mark Scheme:

(a) Functional group isomerism (1 mark). Both have formula $\mathrm{C}_3\mathrm{H}_6\mathrm{O}$ but propanal has an aldehyde group and propanone has a ketone group. Different functional groups, same molecular formula (1 mark).

(b) Optical isomerism (1 mark). They are non-superimposable mirror images (enantiomers) with the same molecular formula and structural formula but different spatial arrangements at the chiral centre (C-2) (1 mark).

(c) Positional isomerism (1 mark). Both are chlorobutanes ($\mathrm{C}_4\mathrm{H}_9\mathrm{Cl}$) with the Cl at different positions on the carbon chain (1 mark).

(d) Geometric (E/Z) isomerism (1 mark). Restricted rotation around the C=C double bond gives two different spatial arrangements. The cis isomer has both $\mathrm{CH}_3$ groups on the same side; the trans isomer has them on opposite sides (1 mark).

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tip

Diagnostic Test Ready to test your understanding of Organic Chemistry Introduction? The diagnostic test contains the hardest questions within the A-Level specification for this topic, each with a full worked solution.

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