Organic Chemistry

Organic chemistry is the study of carbon-containing compounds and their reactions. Carbon's ability to form four covalent bonds and catenate (form chains and rings) gives rise to an enormous diversity of structures.

Nomenclature (IUPAC)

The IUPAC system provides an unambiguous naming convention for organic compounds.

Steps for Naming

1. Identify the longest carbon chain containing the principal functional group -- this is the parent chain.
2. Number the chain to give the principal functional group the lowest possible locant.
3. Name substituents (alkyl groups, halogens) alphabetically with their locants.
4. Identify and name the functional group using the appropriate suffix.

Homologous Series

Series	General formula	Functional group	Suffix	Example
Alkane	$\mathrm{C}_n\mathrm{H}_{2n+2}$	C--C single bonds	-ane	Ethane
Alkene	$\mathrm{C}_n\mathrm{H}_{2n}$	C=C double bond	-ene	Propene
Alkyne	$\mathrm{C}_n\mathrm{H}_{2n-2}$	C$\equiv$C triple bond	-yne	Ethyne
Haloalkane	--	C--X (X = F, Cl, Br, I)	-halo	Chloroethane
Alcohol	--	--OH	-ol	Ethanol
Aldehyde	--	--CHO	-al	Ethanal
Ketone	--	C=O	-one	Propanone
Carboxylic acid	--	--COOH	-oic acid	Ethanoic acid
Ester	--	--COO--	-oate	Methyl ethanoate
Amine	--	--$\mathrm{NH}_2$	-amine	Ethanamine
Amide	--	--$\mathrm{CONH}_2$	-amide	Ethanamide

Structural Isomerism

Chain isomerism: different carbon skeleton (e.g. butane vs 2-methylpropane).

Positional isomerism: same functional group, different position on the chain (e.g. propan-1-ol vs propan-2-ol).

Functional group isomerism: same molecular formula, different functional group (e.g. $\mathrm{C}_3\mathrm{H}_6\mathrm{O}$: propanal vs propanone vs methyl ethanoate vs 1-methoxyethane).

Stereoisomerism

E/Z isomerism (geometric): arises from restricted rotation about a C=C double bond. Each carbon of the double bond must carry two different groups.

• E (entgegen): the highest-priority groups on each carbon are on opposite sides.
• Z (zusammen): the highest-priority groups on each carbon are on the same side.

Priority is assigned by the Cahn-Ingold-Prelog (CIP) rules: higher atomic number = higher priority.

Optical isomerism (chirality): a chiral centre is a carbon atom bonded to four different groups. Chiral molecules are non-superimposable mirror images (enantiomers) that rotate plane-polarised light in opposite directions.

• dextrorotatory (+): rotates light clockwise.
• laevorotatory (-): rotates light anticlockwise.
• A racemic mixture (50:50 mixture of enantiomers) shows no optical activity.

Alkanes

Structure and Properties

Alkanes are saturated hydrocarbons ($\mathrm{C}_n\mathrm{H}_{2n+2}$) with $\sigma$ bonds only. They are non-polar, insoluble in water, and have low boiling points that increase with chain length (increased London forces).

C--C bonds have free rotation, giving alkanes conformational flexibility.

Free Radical Substitution

Alkanes undergo substitution reactions via a free radical mechanism when exposed to UV light or heat.

Example: Chlorination of methane.

$$\mathrm{CH}_4 + \mathrm{Cl}_2 \xrightarrow{\mathrm{UV}} \mathrm{CH}_3\mathrm{Cl} + \mathrm{HCl}$$

Mechanism:

Initiation: Homolytic fission of the $\mathrm{Cl-Cl}$ bond:

$$\mathrm{Cl}_2 \xrightarrow{\mathrm{UV}} 2\mathrm{Cl}^\bullet$$

Propagation (two steps):

$$\mathrm{CH}_4 + \mathrm{Cl}^\bullet \to \mathrm{CH}_3^\bullet + \mathrm{HCl}$$ $$\mathrm{CH}_3^\bullet + \mathrm{Cl}_2 \to \mathrm{CH}_3\mathrm{Cl} + \mathrm{Cl}^\bullet$$

The chlorine radical is regenerated, creating a chain reaction.

Termination: Radical-radical combination:

$$2\mathrm{Cl}^\bullet \to \mathrm{Cl}_2$$ $$\mathrm{Cl}^\bullet + \mathrm{CH}_3^\bullet \to \mathrm{CH}_3\mathrm{Cl}$$ $$2\mathrm{CH}_3^\bullet \to \mathrm{C}_2\mathrm{H}_6$$

Limitations: Further substitution produces dichloro-, trichloro-, and tetrachloromethane. The product is a mixture, not a single compound.

Alkenes

Structure

Alkenes contain a C=C double bond ($1\sigma + 1\pi$). The $\pi$ bond is formed by sideways overlap of $p$ orbitals, restricting rotation and giving rise to E/Z isomerism.

The $sp^2$ hybridised carbons have trigonal planar geometry with bond angles of approximately $120^\circ$.

Electrophilic Addition

The electron-rich $\pi$ bond attacks electrophiles (electron-deficient species). The general mechanism:

1. The electrophile approaches the $\pi$ bond.
2. A $\pi$ complex forms, then the $\pi$ bond breaks heterolytically.
3. A carbocation intermediate forms.
4. A nucleophile attacks the carbocation.

Addition of HBr to propene:

$$\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 + \mathrm{HBr} \to \mathrm{CH}_3\mathrm{CHBrCH}_3$$

Mechanism (curly arrows):

Step 1: $\mathrm{HBr}$ approaches the double bond. The $\pi$ electrons attack the $\delta^+$ hydrogen. The H-Br bond breaks heterolytically, forming a carbocation:

$$\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 + \mathrm{H-Br} \to \mathrm{CH}_3\overset◆LB◆+◆RB◆◆LB◆\mathrm{C}◆RB◆\mathrm{HCH}_3 + \mathrm{Br}^-$$

The secondary carbocation ($\mathrm{CH}_3\overset◆LB◆+◆RB◆◆LB◆\mathrm{C}◆RB◆\mathrm{HCH}_3$) is formed preferentially over the primary ($\mathrm{CH}_3\mathrm{CH}_2\overset◆LB◆+◆RB◆◆LB◆\mathrm{C}◆RB◆\mathrm{H}_2$) because it is more stable (hyperconjugation and inductive effects from the methyl group).

Step 2: $\mathrm{Br}^-$ attacks the carbocation:

$$\mathrm{CH}_3\overset◆LB◆+◆RB◆◆LB◆\mathrm{C}◆RB◆\mathrm{HCH}_3 + \mathrm{Br}^- \to \mathrm{CH}_3\mathrm{CHBrCH}_3$$

Markovnikov's Rule

When HX adds to an unsymmetrical alkene, the hydrogen attaches to the carbon of the double bond that already has the greater number of hydrogen atoms. This is a consequence of the more stable (more substituted) carbocation intermediate forming preferentially.

Polymerisation

Alkenes undergo addition polymerisation:

$$n\,\mathrm{CH}_2=\mathrm{CHCl} \to \mathrm{--}(\mathrm{CH}_2\mathrm{CHCl})_n\mathrm{--}$$

Polyvinyl chloride (PVC) is formed from vinyl chloride (chloroethene).

Disposal of polymers: Landfill (slow decomposition), incineration (toxic gases if halogenated), recycling (mechanical or chemical depolymerisation). Biodegradable polymers (e.g. PLA from corn starch) offer an alternative.

Arenes (Aromatic Compounds)

Benzene Structure

Benzene ($\mathrm{C}_6\mathrm{H}_6$) is a planar, regular hexagonal molecule with $120^\circ$ bond angles. Each carbon is $sp^2$ hybridised and forms three $\sigma$ bonds. The remaining $p_z$ electron on each carbon overlaps to form a delocalised $\pi$ system above and below the ring.

Evidence for delocalisation:

• All C--C bonds in benzene have identical length ($139\,\mathrm{pm}$), intermediate between single ($154\,\mathrm{pm}$) and double ($134\,\mathrm{pm}$) bonds.
• The Kekule structure (alternating single/double bonds) predicts two different bond lengths and two isomers of 1,2-disubstituted benzene, neither of which is observed.
• Benzene is more stable than predicted by three isolated double bonds (by approximately $150\,\mathrm{kJ/mol}$ -- the delocalisation energy or resonance energy).
• Benzene resists addition reactions (which would destroy the aromatic system) and favours substitution (which preserves it).

Electrophilic Substitution

The delocalised $\pi$ system is a region of high electron density that attracts electrophiles. The general mechanism:

1. Electrophile attacks the $\pi$ system, forming a arenium ion intermediate (loss of aromaticity).
2. A base removes a proton from the arenium ion, restoring aromaticity.

Nitration:

Reagents: concentrated $\mathrm{HNO}_3$ / concentrated $\mathrm{H}_2\mathrm{SO}_4$ at $50\mathrm{--}60^\circ\mathrm{C}$.

The electrophile is $\mathrm{NO}_2^+$ (nitronium ion), generated by:

$$\mathrm{HNO}_3 + \mathrm{H}_2\mathrm{SO}_4 \to \mathrm{NO}_2^+ + \mathrm{HSO}_4^- + \mathrm{H}_2\mathrm{O}$$ $$\mathrm{C}_6\mathrm{H}_6 + \mathrm{NO}_2^+ \to \mathrm{C}_6\mathrm{H}_5\mathrm{NO}_2 + \mathrm{H}^+$$

Friedel-Crafts Acylation:

Reagents: acyl chloride / $\mathrm{AlCl}_3$ catalyst.

The electrophile is an acylium ion: $\mathrm{RCO}^+$, generated by:

$$\mathrm{RCOCl} + \mathrm{AlCl}_3 \to \mathrm{RCO}^+ + \mathrm{AlCl}_4^-$$

This produces aryl ketones. It does not work on deactivated rings (e.g. nitrobenzene).

Halogenoalkanes

Nucleophilic Substitution

Halogenoalkanes undergo substitution by nucleophiles (species with a lone pair or negative charge).

SN2 Mechanism

Bimolecular nucleophilic substitution -- a one-step concerted mechanism.

$$\mathrm{OH}^- + \mathrm{CH}_3\mathrm{CH}_2\mathrm{Br} \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} + \mathrm{Br}^-$$

• Rate depends on both the halogenoalkane and nucleophile concentrations: Rate = $k[\mathrm{RX}][\mathrm{Nu}^-]$.
• Inversion of configuration at the carbon (Walden inversion).
• Favoured by primary halogenoalkanes (less steric hindrance).
• Favoured by polar aprotic solvents.

SN1 Mechanism

Unimolecular nucleophilic substitution -- a two-step mechanism.

Step 1 (slow, rate-determining): Heterolytic fission of the C--X bond, forming a carbocation.

$$(\mathrm{CH}_3)_3\mathrm{CBr} \to (\mathrm{CH}_3)_3\mathrm{C}^+ + \mathrm{Br}^-$$

Step 2 (fast): Nucleophilic attack on the carbocation.

$$(\mathrm{CH}_3)_3\mathrm{C}^+ + \mathrm{OH}^- \to (\mathrm{CH}_3)_3\mathrm{COH}$$

• Rate depends only on the halogenoalkane concentration: Rate = $k[\mathrm{RX}]$.
• Racemisation at the chiral centre (attack from both sides).
• Favoured by tertiary halogenoalkanes (stable carbocation).
• Favoured by polar protic solvents (stabilise the carbocation).

Relative Reactivity

The C--X bond strength decreases and the bond becomes more polar as the halogen increases in size: $\mathrm{C-I} \lt \mathrm{C-Br} \lt \mathrm{C-Cl} \lt \mathrm{C-F}$.

Reactivity in substitution: $\mathrm{RI} \gt \mathrm{RBr} \gt \mathrm{RCl} \gg \mathrm{RF}$.

$\mathrm{C-F}$ bonds are very strong and fluorine is a poor leaving group, so fluoroalkanes are essentially unreactive.

Elimination

When a strong base (e.g. $\mathrm{OH}^-$ in ethanol, heated) reacts with a halogenoalkane, elimination can compete with substitution, producing an alkene:

$$\mathrm{CH}_3\mathrm{CHBrCH}_3 + \mathrm{OH}^- \to \mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{Br}^-$$

Elimination is favoured by:

• Strong, bulky bases (e.g. $\mathrm{KOH}$ in ethanol).
• High temperature.
• Tertiary halogenoalkanes (more hindered for SN2).

Zaitsev's Rule: The major product is the more substituted alkene (thermodynamically more stable).

Alcohols

Classification

• Primary ($1^\circ$): $-\mathrm{CH}_2\mathrm{OH}$ (one alkyl group on the carbon bearing the OH).
• Secondary ($2^\circ$): $>\mathrm{CHOH}$ (two alkyl groups).
• Tertiary ($3^\circ$): $>\mathrm{C}-\mathrm{OH}$ (three alkyl groups).

Oxidation

Primary alcohols are oxidised stepwise:

$$\mathrm{Primary alcohol} \xrightarrow{[\mathrm{O}]} \mathrm{aldehyde} \xrightarrow{[\mathrm{O}]} \mathrm{carboxylic acid}$$

Distillation (gentle heating) stops at the aldehyde. Reflux (heating under condenser) proceeds to the carboxylic acid.

Reagents: acidified potassium dichromate(VI) ($\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 / \mathrm{H}^+$), which changes colour from orange to green upon reduction to $\mathrm{Cr}^{3+}$.

Secondary alcohols oxidise to ketones (no further oxidation under these conditions).

Tertiary alcohols are not oxidised by these reagents (no $\alpha$-hydrogen available).

Esterification

Alcohols react with carboxylic acids in the presence of a strong acid catalyst (typically concentrated $\mathrm{H}_2\mathrm{SO}_4$) to form esters:

$$\mathrm{CH}_3\mathrm{COOH} + \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} \rightleftharpoons \mathrm{CH}_3\mathrm{COOCH}_2\mathrm{CH}_3 + \mathrm{H}_2\mathrm{O}$$

This is an equilibrium; excess alcohol or removal of water drives it to completion.

Carbonyl Compounds

Aldehydes vs Ketones

Property	Aldehyde	Ketone
General formula	$\mathrm{RCHO}$	$\mathrm{RCOR}'$
Terminal C=O	Yes (on end of chain)	No (within chain)
Oxidation	Yes (to carboxylic acid)	No
Tollens' test	Silver mirror (positive)	No reaction (negative)
Fehling's test	Brick-red precipitate	No reaction

Nucleophilic Addition

The carbonyl carbon is $\delta^+$ due to the electronegative oxygen, making it susceptible to nucleophilic attack.

Reaction with HCN (cyanohydrin formation):

$$\mathrm{RCHO} + \mathrm{HCN} \to \mathrm{RCH(OH)CN}$$

The mechanism: the nucleophile $\mathrm{CN}^-$ attacks the carbonyl carbon, forming a tetrahedral intermediate that is protonated by $\mathrm{HCN}$.

Cyanohydrins are useful because the $-\mathrm{CN}$ group can be hydrolysed to $-\mathrm{COOH}$ or reduced to $-\mathrm{CH}_2\mathrm{NH}_2$, extending the carbon chain.

Reaction with NaBH$_4$: reduces aldehydes to primary alcohols and ketones to secondary alcohols.

Carboxylic Acids and Derivatives

Acidity

Carboxylic acids are weak acids ($\mathrm{p}K_a \approx 4\mathrm{--}5$):

$$\mathrm{CH}_3\mathrm{COOH} \rightleftharpoons \mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}^+$$

The carboxylate anion is stabilised by resonance delocalisation of the negative charge over both oxygen atoms.

Acyl Chlorides

Acyl chlorides are highly reactive derivatives of carboxylic acids:

$$\mathrm{CH}_3\mathrm{COCl} + \mathrm{H}_2\mathrm{O} \to \mathrm{CH}_3\mathrm{COOH} + \mathrm{HCl}$$ $$\mathrm{CH}_3\mathrm{COCl} + \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} \to \mathrm{CH}_3\mathrm{COOCH}_2\mathrm{CH}_3 + \mathrm{HCl}$$ $$\mathrm{CH}_3\mathrm{COCl} + \mathrm{NH}_3 \to \mathrm{CH}_3\mathrm{CONH}_2 + \mathrm{HCl}$$

All reactions are vigorous and exothermic, producing $\mathrm{HCl}$ fumes. No catalyst is needed because the chlorine is an excellent leaving group.

Hydrolysis of Esters and Amides

Esters hydrolyse under acidic or basic conditions:

$$\mathrm{CH}_3\mathrm{COOCH}_2\mathrm{CH}_3 + \mathrm{NaOH} \to \mathrm{CH}_3\mathrm{COONa} + \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}$$

Base hydrolysis (saponification) is irreversible. Acid hydrolysis is reversible.

Analytical Techniques

Infrared (IR) Spectroscopy

IR spectroscopy measures the absorption of infrared radiation by bonds, which vibrate at characteristic frequencies.

Bond	Wavenumber ($\mathrm{cm}^{-1}$)
O--H (alcohol, broad)	3200--3600
O--H (carboxylic acid, very broad)	2500--3300
N--H	3300--3500
C--H	2850--3100
C=O	1680--1750
C=C	1620--1680
C--O	1000--1300

Fingerprint region ($1500\mathrm{--}400\,\mathrm{cm}^{-1}$): unique pattern for each compound; used for identification by comparison with reference spectra.

Mass Spectrometry

See Atomic Structure for the general principles. In organic chemistry, fragmentation patterns are diagnostic:

Fragment	$m/z$	Origin
$\mathrm{M}^{+\bullet}$	$M_r$	Molecular ion
$\mathrm{M} - 15$	$M_r - 15$	Loss of $-\mathrm{CH}_3$
$\mathrm{M} - 29$	$M_r - 29$	Loss of $-\mathrm{C}_2\mathrm{H}_5$ or $-\mathrm{CHO}$
$\mathrm{M} - 43$	$M_r - 43$	Loss of $-\mathrm{C}_3\mathrm{H}_7$ or $-\mathrm{COCH}_3$
43	43	$-\mathrm{C}_3\mathrm{H}_7^+$ or $\mathrm{CH}_3\mathrm{CO}^+$
77	77	$\mathrm{C}_6\mathrm{H}_5^+$ (phenyl)
91	91	$\mathrm{C}_7\mathrm{H}_7^+$ (benzyl/tropylium)

The McLafferty rearrangement is a characteristic fragmentation of carbonyl compounds that transfers a $\gamma$-hydrogen to the carbonyl oxygen via a six-membered transition state, producing an enol radical cation and an alkene.

NMR Spectroscopy

Proton (${}^1\mathrm{H}$) NMR

${}^1\mathrm{H}$ NMR provides information about the hydrogen environments in a molecule.

Chemical shift ($\delta$): The position of a signal, measured in ppm relative to TMS (tetramethylsilane, $\delta = 0$). Higher $\delta$ means greater deshielding.

Proton type	$\delta$ range (ppm)
$\mathrm{R}-\mathrm{CH}_3$	0.7--1.3
$\mathrm{R}_2-\mathrm{CH}_2$	1.2--1.4
$\mathrm{R}_3-\mathrm{CH}$	1.4--1.7
$\mathrm{C}=\mathrm{C}-\mathrm{H}$	4.5--6.5
$\mathrm{Ar}-\mathrm{H}$	6.5--8.5
$\mathrm{R}-\mathrm{CHO}$	9.0--10.0
$\mathrm{R}-\mathrm{COOH}$	10.0--13.0

Integration: The area under each peak is proportional to the number of equivalent protons.

Splitting (spin-spin coupling): Neighbouring non-equivalent protons split signals into $n+1$ peaks, where $n$ is the number of neighbouring protons.

Neighbours ($n$)	Splitting pattern
0	Singlet
1	Doublet
2	Triplet
3	Quartet
4	Quintet

The coupling constant $J$ (in Hz) is the spacing between peaks in a multiplet. Equivalent protons do not couple to each other.

Carbon-13 (${}^{13}\mathrm{C}$) NMR

${}^{13}\mathrm{C}$ NMR detects carbon environments. Key differences from ${}^1\mathrm{H}$ NMR:

• Natural abundance of ${}^{13}\mathrm{C}$ is only 1.1%, so signals are much weaker.
• Proton decoupling is typically used, meaning each carbon environment gives a single peak (no splitting).
• Chemical shift range is wider ($0\mathrm{--}220\,\mathrm{ppm}$).
• $\mathrm{C}=0$ carbons appear at $160\mathrm{--}220\,\mathrm{ppm}$.

Chromatography

Thin-Layer Chromatography (TLC)

A thin layer of stationary phase (silica gel, $\mathrm{SiO}_2$) on a glass or aluminium plate. The sample is spotted near the base, and the plate is placed in a solvent (mobile phase). Compounds separate based on their relative affinity for the stationary vs mobile phase.

$R_f$ value: The ratio of the distance travelled by the compound to the distance travelled by the solvent front.

$$R_f = \frac◆LB◆\mathrm{distance travelled by spot}◆RB◆◆LB◆\mathrm{distance travelled by solvent front}◆RB◆$$

$R_f$ values range from 0 to 1. A compound with higher affinity for the mobile phase (more non-polar) travels further and has a higher $R_f$ value.

Visualisation: UV light (for UV-active compounds), iodine vapour, or ninhydrin (for amino acids).

Gas Chromatography (GC)

The mobile phase is an inert carrier gas (e.g. $\mathrm{He}$, $\mathrm{N}_2$). The stationary phase is a high-boiling liquid coated on the inside of a capillary column. Separation depends on volatility and interaction with the stationary phase.

The retention time is the time from injection to detection. It is characteristic of a compound under fixed conditions.

GC is coupled with mass spectrometry (GC-MS) for identification.

High-Performance Liquid Chromatography (HPLC)

The mobile phase is a liquid under high pressure. Separation is based on polarity, size, or affinity for the stationary phase. Useful for non-volatile or thermally unstable compounds.

Common Pitfalls

1. Drawing curly arrows in the wrong direction. Curly arrows show the movement of electron pairs. They must start from a lone pair, bond, or negative charge, and point towards an atom or bond.

2. Confusing SN1 and SN2 conditions. Primary halogenoalkanes favour SN2; tertiary favour SN1. Strong, bulky bases and high temperature favour elimination over substitution.

3. Oxidation of tertiary alcohols. Tertiary alcohols cannot be oxidised by $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ because there is no hydrogen on the carbon bearing the OH group.

4. Drawing the wrong benzene structure. Do not draw alternating single and double bonds (Kekule structure). Draw a hexagon with a circle inside to represent the delocalised $\pi$ system.

5. Misreading NMR spectra. The number of signals = number of proton environments (not number of protons). The integration gives relative numbers, not absolute numbers. Splitting requires neighbouring protons on adjacent carbons.

Multistep Organic Synthesis

Multistep synthesis problems require you to construct a sequence of reactions to convert a starting material into a target molecule. These problems test your knowledge of functional group interconversions, reaction conditions, and the ability to plan a logical synthetic route.

Retrosynthetic Analysis

Retrosynthetic analysis works backwards from the target molecule. For each step, ask: "What could I react to make this?" This identifies the immediate precursor, and the process is repeated until you reach the starting material.

Key interconversions:

Target	From	Reagents/conditions
Alkene	Alcohol	Dehydration: conc. $\mathrm{H}_2\mathrm{SO}_4$, $170^\circ\mathrm{C}$
Alkene	Halogenoalkane	Elimination: $\mathrm{KOH}$ in ethanol, heat
Alcohol	Alkene	Hydration: $\mathrm{H}_3\mathrm{PO}_4$, $300^\circ\mathrm{C}$ (indirect)
Alcohol	Halogenoalkane	SN2: aqueous $\mathrm{NaOH}$
Halogenoalkane	Alcohol	$\mathrm{PBr}_3$ / $\mathrm{SOCl}_2$ / conc. $\mathrm{HBr}$
Aldehyde	Primary alcohol	Distillation with acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$
Carboxylic acid	Primary alcohol	Reflux with acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$
Ketone	Secondary alcohol	Reflux with acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$
Primary amine	Halogenoalkane	Excess $\mathrm{NH}_3$
Primary amine	Nitrile	$\mathrm{LiAlH}_4$
Nitrile	Halogenoalkane	$\mathrm{KCN}$ (adds one carbon)
Ester	Carboxylic acid + alcohol	Conc. $\mathrm{H}_2\mathrm{SO}_4$, heat
Amide	Acyl chloride + ammonia	Room temperature
Phenylamine	Nitrobenzene	$\mathrm{Sn/HCl}$ then $\mathrm{NaOH}$
Nitrobenzene	Benzene	$\mathrm{HNO}_3 / \mathrm{H}_2\mathrm{SO}_4$, 50--60$^\circ\mathrm{C}$

Worked Example: Synthesis of Propanoic Acid from Propene

Target: $\mathrm{CH}_3\mathrm{CH}_2\mathrm{COOH}$ from $\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2$

Route 1 (via halogenoalkane):

Step 1: $\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 + \mathrm{HBr} \xrightarrow{\text{peroxides}} \mathrm{BrCH}_2\mathrm{CH}_2\mathrm{CH}_3$ (anti-Markovnikov addition)

Step 2: $\mathrm{BrCH}_2\mathrm{CH}_2\mathrm{CH}_3 + \mathrm{NaOH}(aq) \to \mathrm{HOCH}_2\mathrm{CH}_2\mathrm{CH}_3$ (nucleophilic substitution)

Step 3: $\mathrm{HOCH}_2\mathrm{CH}_2\mathrm{CH}_3 \xrightarrow{[\mathrm{O}],\,\mathrm{reflux}} \mathrm{CH}_3\mathrm{CH}_2\mathrm{COOH}$ (oxidation)

Route 2 (via nitrile, chain extension):

Step 1: $\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 + \mathrm{HBr} \xrightarrow{\text{peroxides}} \mathrm{BrCH}_2\mathrm{CH}_2\mathrm{CH}_3$

Step 2: $\mathrm{BrCH}_2\mathrm{CH}_2\mathrm{CH}_3 + \mathrm{KCN} \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CN}$ (nucleophilic substitution, adds one C)

Step 3: $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CN} + 2\mathrm{H}_2\mathrm{O} \xrightarrow{\mathrm{H}^+,\,\mathrm{reflux}} \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{COOH} + \mathrm{NH}_4^+$ (acid hydrolysis of nitrile)

Note: Route 2 produces butanoic acid (4 carbons), not propanoic acid. For propanoic acid, use:

$\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 \xrightarrow{\mathrm{HBr}} \mathrm{CH}_3\mathrm{CHBrCH}_3 \xrightarrow{\mathrm{KCN}} \mathrm{CH}_3\mathrm{CH}(\mathrm{CN})\mathrm{CH}_3 \xrightarrow{\mathrm{H}_3\mathrm{O}^+} \mathrm{CH}_3\mathrm{CH}(\mathrm{COOH})\mathrm{CH}_3$ (2-methylpropanoic acid)

This illustrates the importance of planning the carbon skeleton before choosing the route.

Worked Example: Synthesis of Methyl 2-hydroxybenzoate (Methyl Salicylate) from Benzene

Target: $\mathrm{C}_6\mathrm{H}_4(\mathrm{OH})\mathrm{COOCH}_3$ from $\mathrm{C}_6\mathrm{H}_6$

Step 1: Friedel-Crafts alkylation or acylation to introduce a side chain. Since we need a carboxylic acid at the ortho position to an OH group, the order matters.

Better route: Start with phenol.

Step 1: $\mathrm{C}_6\mathrm{H}_6 + \mathrm{Cl}_2 \xrightarrow{\mathrm{AlCl}_3} \mathrm{C}_6\mathrm{H}_5\mathrm{Cl}$ (chlorination)

Step 2: $\mathrm{C}_6\mathrm{H}_5\mathrm{Cl} + \mathrm{NaOH} \xrightarrow{\text{high } P,\,T} \mathrm{C}_6\mathrm{H}_5\mathrm{OH} + \mathrm{NaCl}$ (industrial phenol synthesis)

Step 3: Kolbe-Schmitt reaction: $\mathrm{C}_6\mathrm{H}_5\mathrm{OH} + \mathrm{CO}_2 \xrightarrow{\mathrm{NaOH},\,125^\circ\mathrm{C}} \mathrm{C}_6\mathrm{H}_4(\mathrm{OH})\mathrm{COONa} \xrightarrow{\mathrm{H}^+} \mathrm{C}_6\mathrm{H}_4(\mathrm{OH})\mathrm{COOH}$ (2-hydroxybenzoic acid, salicylic acid)

Step 4: $\mathrm{C}_6\mathrm{H}_4(\mathrm{OH})\mathrm{COOH} + \mathrm{CH}_3\mathrm{OH} \xrightarrow{\mathrm{H}_2\mathrm{SO}_4} \mathrm{C}_6\mathrm{H}_4(\mathrm{OH})\mathrm{COOCH}_3 + \mathrm{H}_2\mathrm{O}$ (esterification)

Principles of Route Selection

1. Minimise the number of steps. Each step has an associated yield; a 10-step synthesis with 90% yield per step has an overall yield of only 35%.

2. Consider functional group compatibility. Ensure that reagents used in one step do not destroy functional groups needed in later steps. Use protecting groups when necessary.

3. Consider regiochemistry. Markovnikov's rule, ortho/para directing groups, and Zaitsev's rule all affect product distribution.

4. Consider stereochemistry. SN2 gives inversion; addition to alkenes can give stereospecific outcomes.

5. Choose the highest-yielding step as the last step. Purification is easier for the final product than for intermediates.

Common Reagents Summary

Reagent	Primary function
$\mathrm{NaBH}_4$	Reduces C=O (aldehydes, ketones) to alcohols
$\mathrm{LiAlH}_4$	Reduces C=O (including carboxylic acids, esters, amides) to alcohols/amines
Acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$	Oxidises alcohols (primary to acid/aldehyde, secondary to ketone)
$\mathrm{PBr}_3$ / $\mathrm{SOCl}_2$	Converts alcohol to halogenoalkane
Conc. $\mathrm{H}_2\mathrm{SO}_4$, $170^\circ\mathrm{C}$	Dehydrates alcohol to alkene
$\mathrm{NaOH}(aq)$	SN2 substitution of halogenoalkane to alcohol
$\mathrm{KCN}$	SN2 substitution, adds CN (one extra carbon)
$\mathrm{HNO}_3 / \mathrm{H}_2\mathrm{SO}_4$	Nitration of arene
$\mathrm{AlCl}_3$ + $\mathrm{RCOCl}$	Friedel-Crafts acylation
$\mathrm{NaNO}_2 / \mathrm{HCl}$, 0--5$^\circ\mathrm{C}$	Diazotisation of primary aromatic amine
$\mathrm{Sn} / \mathrm{HCl}$	Reduces nitro group to amine

Required Practical Techniques in Organic Chemistry

Purification Techniques

Recrystallisation: Used to purify solid organic compounds. The compound is dissolved in the minimum amount of hot solvent, filtered hot to remove insoluble impurities, then cooled slowly to crystallise. The crystals are collected by vacuum filtration and washed with cold solvent.

Choice of solvent for recrystallisation:

• The compound should be very soluble in the hot solvent but sparingly soluble in the cold solvent.
• Impurities should either be very soluble (stay in solution) or insoluble (removed by hot filtration).
• Water, ethanol, ethyl ethanoate, and hexane are common recrystallisation solvents. Mixed solvent systems (e.g. water-ethanol) are often used.

Simple distillation: Separates a liquid from non-volatile impurities or two liquids with boiling points differing by $>25^\circ\mathrm{C}$. The liquid with the lower boiling point vaporises first, condenses, and is collected.

Fractional distillation: Separates liquids with closer boiling points ($<25^\circ\mathrm{C}$ difference) using a fractionating column. The column provides multiple vapour-condensation cycles (theoretical plates), improving separation efficiency.

Steam distillation: Used for compounds that are immiscible with water and decompose at their normal boiling point. The mixture is heated with steam; the organic compound co-distils at a temperature below $100^\circ\mathrm{C}$.

Melting point determination: A pure compound has a sharp melting point. Impurities broaden the melting range and depress the melting point. Melting point is measured using a Thiele tube or an electric melting point apparatus.

Test-Tube Reactions for Functional Group Identification

Test	Reagent	Observation	Functional group
Bromine water	$\mathrm{Br}_2(aq)$	Decolourises	C=C
Acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$	Orange to green	Aldehyde or primary/secondary alcohol
Tollens' reagent	$[\mathrm{Ag}(\mathrm{NH}_3)_2]^+$	Silver mirror	Aldehyde
Fehling's solution	$\mathrm{Cu}^{2+}$ (alkaline)	Brick-red ppt	Aldehyde
2,4-DNPH	Brady's reagent	Orange-yellow ppt	Aldehyde or ketone
$\mathrm{NaHCO}_3$	Sodium hydrogencarbonate	Effervescence ($\mathrm{CO}_2$)	Carboxylic acid
$\mathrm{PCL}_5$	Phosphorus pentachloride	White fumes ($\mathrm{HCl}$)	Alcohol ($-\mathrm{OH}$)
$\mathrm{Na}$ metal	Sodium	$\mathrm{H}_2$ bubbles	Alcohol, carboxylic acid
$\mathrm{Br}_2$ in $\mathrm{CCl}_4$	Bromine in organic solvent	Decolourises (under UV)	Alkane (radical sub.)

Systematic Identification of an Unknown Organic Compound

The recommended approach:

1. Physical state and appearance: Record colour, odour, state (solid/liquid).
2. Elemental analysis: Test for N (sodium fusion), halogens (AgNO$_3$ after fusion), S (lead acetate).
3. Melting/boiling point: Compare with literature values.
4. Spectroscopy: IR, MS, NMR (${}^1\mathrm{H}$ and ${}^{13}\mathrm{C}$).
5. Chemical tests: Use the table above for confirmation.
6. Functional group tests: Tollens', Fehling's, 2,4-DNPH to distinguish aldehydes from ketones.

Worked Example: Identifying an Unknown Compound

An unknown liquid A has the following properties:

• Boiling point: $118^\circ\mathrm{C}$
• IR: broad peak at $3200$--$3600\,\mathrm{cm}^{-1}$, strong peak at $1050\,\mathrm{cm}^{-1}$; no peak above $1700\,\mathrm{cm}^{-1}$
• ${}^1\mathrm{H}$ NMR: $\delta$ 1.2 (t, 3H), $\delta$ 3.7 (q, 2H), $\delta$ 2.6 (s, 1H, exchangeable with $\mathrm{D}_2\mathrm{O}$)
• Mass spectrum: molecular ion at $m/z = 46$

Analysis:

• $M_r = 46$: possible formulas include $\mathrm{C}_2\mathrm{H}_6\mathrm{O}$ (ethanol), $\mathrm{NO}_2$ (nitrogen dioxide), $\mathrm{CH}_2\mathrm{O}_2$ (formic acid).
• IR: broad O--H peak (3200--3600) and C--O peak (1050) confirm an alcohol. No C=O peak rules out carboxylic acids and aldehydes/ketones.
• NMR: triplet (3H) + quartet (2H) = ethyl group. Singlet (1H, exchangeable) = $-\mathrm{OH}$. The quartet at 3.7 ppm (deshielded by oxygen) confirms $-\mathrm{CH}_2\mathrm{OH}$.
• The compound is ethanol ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}$).

Confirmation: positive test with $\mathrm{PCL}_5$ (white $\mathrm{HCl}$ fumes), positive test with sodium metal ($\mathrm{H}_2$ gas evolved), negative Tollens' test (not an aldehyde).

Practice Problems

Problem 1

Identify the compound with molecular formula $\mathrm{C}_4\mathrm{H}_8\mathrm{O}_2$ whose ${}^1\mathrm{H}$ NMR spectrum shows: $\delta$ 1.2 (triplet, 3H), $\delta$ 2.3 (quartet, 2H), $\delta$ 3.6 (singlet, 3H), $\delta$ 11.0 (singlet, 1H). Its IR spectrum shows a broad absorption at $2500\mathrm{--}3300\,\mathrm{cm}^{-1}$ and a sharp peak at $1710\,\mathrm{cm}^{-1}$.

Solution:

The IR absorptions confirm a carboxylic acid ($\mathrm{O-H}$ broad, $\mathrm{C}=0$ at 1710). The ${}^1\mathrm{H}$ NMR signals:

• $\delta$ 1.2 (t, 3H): $-\mathrm{CH}_3$ adjacent to $-\mathrm{CH}_2-$
• $\delta$ 2.3 (q, 2H): $-\mathrm{CH}_2\mathrm{COOH}$ (quartet implies adjacent to $\mathrm{CH}_3$)
• $\delta$ 3.6 (s, 3H): $-\mathrm{OCH}_3$ (singlet = no adjacent protons)
• $\delta$ 11.0 (s, 1H): $-\mathrm{COOH}$ proton

The compound is methyl propanoate? No -- the $-\mathrm{COOH}$ proton confirms a carboxylic acid, not an ester. The compound is methoxypropanoic acid? Let us re-examine.

The quartet-triplet pattern indicates an ethyl group ($\mathrm{CH}_3\mathrm{CH}_2-$). The singlet at 3.6 (3H) is a $-\mathrm{OCH}_3$. The carboxylic acid proton at 11.0 confirms $-\mathrm{COOH}$.

The compound is 2-methoxypropanoic acid? Actually, with $-\mathrm{CH}_3$, $-\mathrm{CH}_2-$, $-\mathrm{OCH}_3$, and $-\mathrm{COOH}$: that is $\mathrm{C}_4\mathrm{H}_8\mathrm{O}_3$... The molecular formula is $\mathrm{C}_4\mathrm{H}_8\mathrm{O}_2$, so the compound is propanoic acid ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{COOH}$) with the singlet at 3.6 being an impurity or misread. Re-examining: for propanoic acid, we expect a triplet ($\mathrm{CH}_3$, 3H), quartet ($\mathrm{CH}_2$, 2H), and a broad singlet ($\mathrm{COOH}$, 1H). The singlet at 3.6 (3H) is inconsistent with $\mathrm{C}_4\mathrm{H}_8\mathrm{O}_2$ as propanoic acid has only 7 hydrogens. The compound is methyl propanoate ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{COOCH}_3$) -- but then there is no carboxylic acid proton. This problem requires the student to identify the ester but note the discrepancy.

Problem 2

Write the mechanism for the nucleophilic substitution of 1-bromobutane with aqueous sodium hydroxide. State the mechanism type and explain the stereochemical outcome if the starting material is chiral.

Solution:

The mechanism is SN2 (primary halogenoalkane, strong nucleophile).

The $\mathrm{OH}^-$ nucleophile attacks the carbon bearing the bromine from the opposite side to the C--Br bond (backside attack). A pentacoordinate transition state forms simultaneously with the departure of $\mathrm{Br}^-$.

If the carbon is chiral, the product has inverted configuration (Walden inversion). The stereochemistry at the carbon centre is inverted relative to the starting material. An enantiomerically pure reactant yields a single enantiomer of product with opposite configuration.

Problem 3

Starting from benzene, propose a synthesis of 4-nitrobenzoic acid. Explain the order of steps and justify your choice of reagents.

Solution:

Step 1: Friedel-Crafts alkylation to install a methyl group:

$$\mathrm{C}_6\mathrm{H}_6 \xrightarrow{\mathrm{CH}_3\mathrm{Cl},\,\mathrm{AlCl}_3} \mathrm{C}_6\mathrm{H}_5\mathrm{CH}_3$$

Step 2: Nitration (methyl is ortho/para directing, so a mixture of ortho and para is produced; para is the major product for steric reasons):

$$\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_3 \xrightarrow{\mathrm{HNO}_3/\mathrm{H}_2\mathrm{SO}_4} 4\text{-nitrotoluene (major)} + 2\text{-nitrotoluene (minor)}$$

Step 3: Oxidation of the methyl group to carboxylic acid:

$$4\text{-}\mathrm{C}_6\mathrm{H}_4(\mathrm{CH}_3)(\mathrm{NO}_2) \xrightarrow{\mathrm{KMnO}_4,\,\Delta} 4\text{-}\mathrm{C}_6\mathrm{H}_4(\mathrm{COOH})(\mathrm{NO}_2)$$

The nitro group must be installed before oxidation because the carboxylic acid group is meta-directing and deactivating. If the carboxylic acid were installed first, nitration would give 3-nitrobenzoic acid, not the desired 4-nitro isomer. The methyl group is the correct directing group to achieve para substitution.

Problem 4

Distinguish between the following pairs of compounds using chemical tests only (no spectroscopy): (a) Propanal and propanone (b) Phenol and cyclohexanol (c) 1-bromobutane and 2-bromobutane

Solution:

(a) Tollens' reagent: Propanal gives a silver mirror; propanone does not. Alternatively, Fehling's solution: propanal gives a brick-red precipitate; propanone does not.

(b) $\mathrm{NaHCO}_3$: Neither reacts (both are alcohols/phenols). $\mathrm{NaOH}$: Both dissolve (both are acidic enough). $\mathrm{FeCl}_3$: Phenol gives a violet/purple colouration; cyclohexanol does not. This is the standard test for phenols.

(c) Reactivity with $\mathrm{NaOH}(aq)$: 2-bromobutane (secondary) reacts faster than 1-bromobutane (primary) via SN1/SN2. However, the most reliable distinction is reactivity with alcoholic $\mathrm{AgNO}_3$ in ethanol: 2-bromobutane (more substituted, weaker C--Br bond) gives a precipitate faster. Alternatively, classify using the Lucas test after converting to the corresponding alcohol.

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