Organic Chemistry

Higher Organic Chemistry

Hydrocarbons

Alkanes: Saturated hydrocarbons with the general formula $\mathrm{C_n\mathrm{H_{2n+2}$ . All C-C Bonds are single. Names follow the IUPAC system.

Alkenes: Unsaturated hydrocarbons with the general formula $\mathrm{C_n\mathrm{H_{2n}$ . Contain at Least one C=C double bond.

Alkynes: Unsaturated hydrocarbons with the general formula $\mathrm{C_n\mathrm{H_{2n-2}$ . Contain At least one C=C triple bond.

Comparison of Hydrocarbons

Property	Alkanes	Alkenes	Alkynes
General formula	$\mathrm{C_n\mathrm{H_{2n+2}$	$\mathrm{C_n\mathrm{H_{2n}$	$\mathrm{C_n\mathrm{H_{2n-2}$
Bonding	Single C-C only	At least one C=C	At least one C=C
Reactivity	Relatively unreactive	Reactive (electrophilic addition)	Very reactive
Test	Burns with clean flame	Decolourises bromine water	Decolourises bromine water
Flammability	Burns in air	Burns with smoky flame	Burns with very smoky flame

IUPAC Nomenclature

Identify the longest carbon chain (parent chain)
Number the chain to give the lowest possible locants to substituents
Name substituents alphabetically with their position numbers
For multiple identical substituents, use prefixes di-, tri-, tetra-

Worked Example 1: Name the compound $\mathrm{CH_3\mathrm{CH(\mathrm{CH_3)\mathrm{CH_2\mathrm{CH(\mathrm{C_2\mathrm{H_5)\mathrm{CH_2\mathrm{CH_3$ .

Longest chain: 6 carbons (hexane).

Numbering from the end nearest a branch:

$\mathrm{CH_3-\mathrm{CH(\mathrm{CH_3)-\mathrm{CH_2-\mathrm{CH(\mathrm{C_2\mathrm{H_5)-\mathrm{CH_2-\mathrm{CH_3$

Position 2: methyl. Position 4: ethyl.

Name: 4-ethyl-2-methylhexane.

Worked Example 2: Name $\mathrm{CH_3\mathrm{CH_2\mathrm{CH(\mathrm{CH_3)\mathrm{CH_2\mathrm{CH_2\mathrm{Cl$ .

Longest chain: 6 carbons. Chlorine on C-1, methyl on C-3.

Name: 1-chloro-3-methylpentane.

Worked Example 3: Name $\mathrm{CH_3\mathrm{COCH_2\mathrm{CH_3$ .

Longest chain: 4 carbons. Carbonyl on C-2.

Name: butan-2-one.

Isomerism

Structural isomers have the same molecular formula but different structural arrangements.

Types of structural isomerism:

Chain isomers: Different carbon skeleton (e.g., butane vs. 2-methylpropane)
Positional isomers: Same skeleton, different position of functional group (e.g., 1-propanol vs. 2-propanol)
Functional group isomers: Different functional groups (e.g., $\mathrm{C_3\mathrm{H_6\mathrm{O$ : propanal vs. Propanone)

Worked Example 4: Draw the structural isomers of $\mathrm{C_4\mathrm{H_8\mathrm{O$ that are (a) Aldehydes, (b) ketones, (c) alcohols, (d) ethers.

(a) Aldehydes: butanal ( $\mathrm{CH_3\mathrm{CH_2\mathrm{CH_2\mathrm{CHO$ ), 2-methylpropanal ( $(\mathrm{CH_3)_2\mathrm{CHCHO$ )

(b) Ketones: butan-2-one ( $\mathrm{CH_3\mathrm{COCH_2\mathrm{CH_3$ )

(d) Ethers: ethoxyethane ( $\mathrm{CH_3\mathrm{CH_2\mathrm{OCH_2\mathrm{CH_3$ ), 1-methoxypropane, 2-methoxypropane

Stereoisomers have the same structural formula but different spatial arrangement.

E/Z isomerism (geometric): Restricted rotation around C=C bond. Priority rules (Cahn-Ingold-Prelog) determine E (entgegen, opposite) or Z (zusammen, together).

Example: For 1,2-dichloroethene:

Both Cl atoms on same side: Z (cis)
Cl atoms on opposite sides: E (trans)

Functional Groups

Class	Functional Group	Suffix	Example
Alcohol	$-\mathrm{OH$	-ol	Ethanol
Aldehyde	$-\mathrm{CHO$	-al	Ethanal
Ketone	$>\mathrm{C=\mathrm{O$	-one	Propanone
Carboxylic acid	$-\mathrm{COOH$	-oic acid	Ethanoic acid
Ester	$-\mathrm{COO-$	-oate	Methyl ethanoate
Amine	$-\mathrm{NH_2$	-amine	Ethanamine
Halogenoalkane	$-\mathrm{Cl, -\mathrm{Br$	halogeno-	Chloroethane

Reactions of Alkanes

Combustion:

Complete: $\mathrm{C_n\mathrm{H_{2n+2} + \dfrac{3n+1}{2}\mathrm{O_2 \to n\mathrm{CO_2 + (n+1)\mathrm{H_2\mathrm{O$

Worked Example 5: Write the balanced equation for the complete combustion of (a) propane and (b) Cyclohexane.

(a) $\mathrm{C_3\mathrm{H_8 + 5\mathrm{O_2 \to 3\mathrm{CO_2 + 4\mathrm{H_2\mathrm{O$

(b) $\mathrm{C_6\mathrm{H_{12} + 9\mathrm{O_2 \to 6\mathrm{CO_2 + 6\mathrm{H_2\mathrm{O$

Incomplete (limited oxygen): produces $\mathrm{CO$ and/or $\mathrm{C$ (soot).

Free radical substitution (with halogens):

Initiation: $\mathrm{Cl_2 \xrightarrow{\mathrm{UV} 2\mathrm{Cl^\bullet$

Propagation: $\mathrm{CH_4 + \mathrm{Cl^\bullet \to \mathrm{CH_3^\bullet + \mathrm{HCl$

$\mathrm{CH_3^\bullet + \mathrm{Cl_2 \to \mathrm{CH_3\mathrm{Cl + \mathrm{Cl^\bullet$

Termination: Various radical combinations.

Worked Example 6: Describe the free radical substitution mechanism for the reaction of methane With chlorine, including all three stages.

Initiation: UV light provides enough energy to break the Cl-Cl bond homolytically.

$\mathrm{Cl_2 \xrightarrow{\mathrm{UV} 2\mathrm{Cl^\bullet$

Propagation: A chlorine radical abstracts a hydrogen atom from methane, forming HCl and a methyl Radical. The methyl radical then reacts with another chlorine molecule.

$\mathrm{CH_4 + \mathrm{Cl^\bullet \to \mathrm{CH_3^\bullet + \mathrm{HCl$ $\mathrm{CH_3^\bullet + \mathrm{Cl_2 \to \mathrm{CH_3\mathrm{Cl + \mathrm{Cl^\bullet$

Termination: Any two radicals combine: $\mathrm{Cl^\bullet + \mathrm{Cl^\bullet \to \mathrm{Cl_2$ $\mathrm{CH_3^\bullet + \mathrm{Cl^\bullet \to \mathrm{CH_3\mathrm{Cl$ $\mathrm{CH_3^\bullet + \mathrm{CH_3^\bullet \to \mathrm{C_2\mathrm{H_6$

Further substitution can occur: $\mathrm{CH_3\mathrm{Cl \to \mathrm{CH_2\mathrm{Cl_2 \to \mathrm{CHCl_3 \to \mathrm{CCl_4$ .

Reactions of Alkenes

Electrophilic addition: The C=C double bond is an electron-rich site susceptible to attack by Electrophiles.

Addition of HBr:

$\mathrm{CH_2=\mathrm{CH_2 + \mathrm{HBr \to \mathrm{CH_3\mathrm{CH_2\mathrm{Br$

Markovnikov’s Rule: When HX adds to an unsymmetrical alkene, the hydrogen adds to the carbon With more hydrogens already attached.

$\mathrm{CH_3\mathrm{CH=\mathrm{CH_2 + \mathrm{HBr \to \mathrm{CH_3\mathrm{CHBrCH_3 \quad \mathrm{(major product)$

Addition of water (hydration):

$\mathrm{CH_2=\mathrm{CH_2 + \mathrm{H_2\mathrm{O \xrightarrow{\mathrm{H^+} \mathrm{CH_3\mathrm{CH_2\mathrm{OH$

Worked Example 7: Write the mechanism for the addition of $\mathrm{Br_2$ to ethene, showing the Bromonium ion intermediate.

Step 1: The $\pi$ electrons of the C=C bond attack a bromine molecule, forming a bromonium ion Intermediate and a bromide ion.

$\mathrm{CH_2=\mathrm{CH_2 + \mathrm{Br_2 \to \mathrm{CH_2\mathrm{CH_2\mathrm{Br^+ + \mathrm{Br^-$

The bromonium ion is a three-membered ring with a positive charge on bromine.

Step 2: The bromide ion attacks one carbon of the bromonium ion from the opposite side (anti-addition), opening the ring.

$\mathrm{CH_2\mathrm{CH_2\mathrm{Br^+ + \mathrm{Br^- \to \mathrm{CH_2\mathrm{BrCH_2\mathrm{Br$

Product: 1,2-dibromoethane.

Reactions of Alcohols

Oxidation:

Primary alcohol $\to$ aldehyde $\to$ carboxylic acid (with acidified $\mathrm{K_2\mathrm{Cr_2\mathrm{O_7$ ).

$\mathrm{CH_3\mathrm{CH_2\mathrm{OH \xrightarrow{[O]} \mathrm{CH_3\mathrm{CHO \xrightarrow{[O]} \mathrm{CH_3\mathrm{COOH$

Secondary alcohol $\to$ ketone (stops here).

$\mathrm{CH_3\mathrm{CH(\mathrm{OH)\mathrm{CH_3 \xrightarrow{[O]} \mathrm{CH_3\mathrm{COCH_3$

Tertiary alcohol: not oxidised.

Worked Example 8: A student oxidises propan-1-ol using acidified potassium dichromate under Reflux. Write the equation and name the organic product.

$\mathrm{CH_3\mathrm{CH_2\mathrm{CH_2\mathrm{OH + 2[\mathrm{O] \xrightarrow{\mathrm{reflux} \mathrm{CH_3\mathrm{CH_2\mathrm{COOH + \mathrm{H_2\mathrm{O$

Product: propanoic acid. Under reflux, the aldehyde intermediate is further oxidised to the Carboxylic acid. To stop at the aldehyde, distillation would be used instead of reflux.

Dehydration: Alcohols can be dehydrated to form alkenes using concentrated $\mathrm{H_2\mathrm{SO_4$ or $\mathrm{Al_2\mathrm{O_3$ at high temperature.

$\mathrm{CH_3\mathrm{CH_2\mathrm{OH \xrightarrow{\mathrm{conc. \mathrm{H_2\mathrm{SO_4, 170°C} \mathrm{CH_2=\mathrm{CH_2 + \mathrm{H_2\mathrm{O$

Carbonyl Compounds

Aldehydes and ketones both contain the carbonyl group $\mathrm{C=\mathrm{O$ .

Distinguishing tests:

Test	Aldehyde	Ketone
Tollens’ reagent	Silver mirror	No change
Fehling’s solution	Brick-red precipitate	No change
2,4-DNP test	Orange precipitate	Orange precipitate

Note: 2,4-DNP tests for both aldehydes and ketones (confirms carbonyl group), while Tollens’ and Fehling’s distinguish between them.

Aldehyde reactions:

Oxidation to carboxylic acids
Reduction to primary alcohols (using $\mathrm{NaBH_4$ )
Nucleophilic addition with $\mathrm{HCN$ to form hydroxynitriles

Ketone reactions:

Reduction to secondary alcohols (using $\mathrm{NaBH_4$ )
Do not oxidise under mild conditions

Carboxylic Acids and Esters

Carboxylic acid reactions:

With alcohols to form esters (esterification): $\mathrm{CH_3\mathrm{COOH + \mathrm{CH_3\mathrm{CH_2\mathrm{OH \rightleftharpoons \mathrm{CH_3\mathrm{COOCH_2\mathrm{CH_3 + \mathrm{H_2\mathrm{O$
As weak acids: react with bases and carbonates

Ester hydrolysis:

Acid hydrolysis: $\mathrm{CH_3\mathrm{COOCH_2\mathrm{CH_3 + \mathrm{H_2\mathrm{O \rightleftharpoons \mathrm{CH_3\mathrm{COOH + \mathrm{CH_3\mathrm{CH_2\mathrm{OH$

Base hydrolysis (saponification): $\mathrm{CH_3\mathrm{COOCH_2\mathrm{CH_3 + \mathrm{NaOH \to \mathrm{CH_3\mathrm{COONa + \mathrm{CH_3\mathrm{CH_2\mathrm{OH$

Advanced Higher Organic Chemistry

Reaction Mechanisms

Electrophilic Addition (alkenes):

The $\pi$ electrons in the C=C bond attack an electrophile, forming a carbocation intermediate.

Worked Example 9: Addition of HBr to propene.

Step 1: $\mathrm{CH_3\mathrm{CH=\mathrm{CH_2 + \mathrm{H^+ \to \mathrm{CH_3\mathrm{CH^+\mathrm{CH_3$ (secondary carbocation, more stable)

Step 2: $\mathrm{CH_3\mathrm{CH^+\mathrm{CH_3 + \mathrm{Br^- \to \mathrm{CH_3\mathrm{CHBrCH_3$

The minor product (1-bromopropane) forms from the less stable primary carbocation $\mathrm{CH_3\mathrm{CH_2\mathrm{CH_2^+$ .

Carbocation stability: tertiary > secondary > primary > methyl.

This is because alkyl groups donate electron density by induction (+I effect), stabilising the Positive charge.

Nucleophilic Substitution ( $S_N1$ and $S_N2$ ):

$S_N2$ (bimolecular):

Single step, concerted mechanism
Backside attack, inversion of configuration
Rate: $\mathrm{Rate = k[\mathrm{halogenoalkane][\mathrm{nucleophile]$
Favoured by primary halogenoalkanes, strong nucleophiles

$S_N1$ (unimolecular):

Two steps: formation of carbocation, then nucleophilic attack
Racemisation (mixture of configurations)
Rate: $\mathrm{Rate = k[\mathrm{halogenoalkane]$
Favoured by tertiary halogenoalkanes, weak nucleophiles

Worked Example 10: Explain why 2-bromo-2-methylpropane undergoes $S_N1$ hydrolysis faster than 1-bromopropane.

2-bromo-2-methylpropane is a tertiary halogenoalkane. In $S_N1$ The rate-determining step is Formation of the carbocation. The tertiary carbocation $(\mathrm{CH_3)_3\mathrm{C^+$ is stabilised by Three methyl groups (+I effect), making it relatively easy to form. 1-bromopropane would form a Primary carbocation ( $\mathrm{CH_3\mathrm{CH_2\mathrm{CH_2^+$ ), which is much less stable. Therefore, 2-bromo-2-methylpropane reacts faster via $S_N1$ .

Comparison of $S_N1$ and $S_N2$

Feature	$S_N1$	$S_N2$
Steps	Two (carbocation intermediate)	One (concerted)
Rate law	First order	Second order
Stereochemistry	Racemisation	Inversion
Favoured substrate	Tertiary	Primary
Favoured nucleophile	Weak (e.g., $\mathrm{H_2\mathrm{O$ )	Strong (e.g., $\mathrm{OH^-$ )
Carbocation rearrangement	Possible	Not applicable
Solvent effect	Favoured by polar protic	Favoured by polar aprotic

Elimination (E1 and E2):

Competes with substitution. Requires a strong base and heat. Produces alkenes.

Zaitsev’s Rule: In elimination, the more substituted alkene is the major product (more stable).

Condensation Polymers

Formed by joining monomers with the elimination of a small molecule (e.g., water).

Polyesters: Dicarboxylic acid + diol.

$\mathrm{HOOC-\mathrm{R-\mathrm{COOH + \mathrm{HO-\mathrm{R'-\mathrm{OH \to \mathrm{[OC-\mathrm{R-\mathrm{COO-\mathrm{R'\mathrm{]_n + n\mathrm{H_2\mathrm{O$

Worked Example 11: Draw the repeating unit of the polyester formed from benzene-1,4-dicarboxylic Acid and ethane-1,2-diol.

The repeating unit is:

$\mathrm{[-OC-C_6\mathrm{H_4\mathrm{-COO-CH_2\mathrm{CH_2\mathrm{O-]_n$

Polyamides: Dicarboxylic acid + diamine.

$\mathrm{HOOC-\mathrm{R-\mathrm{COOH + \mathrm{H_2\mathrm{N-\mathrm{R'-\mathrm{NH_2 \to \mathrm{[OC-\mathrm{R-\mathrm{CONH-\mathrm{R'-\mathrm{NH]_n + n\mathrm{H_2\mathrm{O$

Nylon and Kevlar are examples of polyamides.

Comparison: Addition vs. Condensation Polymers

Feature	Addition polymer	Condensation polymer
Monomers	Alkenes	Two different functional groups
Byproduct	None	Small molecule (e.g., $\mathrm{H_2\mathrm{O$ )
Bond formed	C-C	Ester or amide
Example	Polythene, PVC	Nylon, polyester
Biodegradability	Generally not biodegradable	Can be biodegradable

Optical Activity

A molecule is optically active if it is chiral (has a non-superimposable mirror image). This Requires a chiral centre (carbon with four different groups attached).

A racemic mixture is a 50:50 mixture of enantiomers (optically inactive)
Enantiomers have identical physical properties except for their effect on plane-polarised light

Worked Example 12: Explain why $\mathrm{CH_3\mathrm{CH(\mathrm{OH)\mathrm{COOH$ is optically active, And draw both enantiomers.

Carbon-2 has four different groups attached: $\mathrm{H$$\mathrm{OH$$\mathrm{CH_3$ And $\mathrm{COOH$ . This makes it a chiral centre. The two enantiomers are non-superimposable mirror Images that rotate plane-polarised light in opposite directions.

Common Pitfalls

Markovnikov’s Rule: The hydrogen adds to the carbon with MORE hydrogens, not fewer.
Oxidation of alcohols: Primary alcohols can be oxidised to aldehydes (with distillation) or carboxylic acids (with reflux). Use the correct conditions.
IUPAC numbering: Always number from the end that gives the lowest locants, not from the left.
$S_N1$ vs. $S_N2$ : Primary halogenoalkanes favour $S_N2$ ; tertiary favour $S_N1$ .
Stereoisomerism: E/Z isomerism requires two different groups on each carbon of the double bond.
Ester naming: The alkyl part of the ester name comes from the alcohol (not the acid).
Polymer vs. Monomer: The repeating unit of a condensation polymer is NOT the same as the monomer. Be careful to show the bonds correctly.

Practice Questions

Name the following compounds: (a) $\mathrm{CH_3\mathrm{CH_2\mathrm{CH(\mathrm{CH_3)\mathrm{CH_2\mathrm{CH_2\mathrm{Cl$ (b) $\mathrm{CH_3\mathrm{COCH_2\mathrm{CH_3$ .
Draw the structural isomers of $\mathrm{C_4\mathrm{H_8\mathrm{O$ that are (a) aldehydes, (b) ketones, (c) alcohols, (d) ethers.
Write the mechanism for the addition of $\mathrm{Br_2$ to ethene, showing the bromonium ion intermediate.
Explain why 2-bromo-2-methylpropane undergoes $S_N1$ hydrolysis faster than 1-bromopropane.
Write balanced equations for the complete combustion of (a) propane and (b) cyclohexane.
A student oxidises propan-1-ol using acidified potassium dichromate under reflux. Write the equation and name the organic product.
Draw the repeating unit of the polyester formed from benzene-1,4-dicarboxylic acid and ethane-1,2-diol.
Explain why the compound $\mathrm{CH_3\mathrm{CH(\mathrm{OH)\mathrm{COOH$ is optically active, and draw both enantiomers.
Describe the mechanism of the electrophilic addition of $\mathrm{Br_2$ to propene, explaining why 1,2-dibromopropane is the major product.
Draw and name all structural isomers of $\mathrm{C_4\mathrm{H_9\mathrm{Cl$ .
Explain the difference between addition and condensation polymerisation, giving one example of each.
Explain why butan-2-ol is optically active but butan-1-ol is not.
Write equations for the reaction of ethanoic acid with (a) sodium, (b) sodium hydroxide, (c) sodium carbonate, and (d) ethanol (in the presence of concentrated sulfuric acid).
Describe how you would distinguish experimentally between pentan-2-one and pentanal using simple chemical tests.
For the reaction of 2-methylpropan-2-ol with HCl, explain whether $S_N1$ or $S_N2$ is the dominant mechanism and why.
A compound with molecular formula $\mathrm{C_5\mathrm{H_{10}\mathrm{O$ shows the following properties: (a) it decolourises bromine water, (b) it gives a positive Tollens’ test, (c) it has a chiral centre. Suggest a structure and explain.

Organic Synthesis Strategies

Two-Step Synthesis Plans

Worked Example 17: Describe how to synthesise propanoic acid from propene.

Step 1: Add HBr to propene (electrophilic addition):

$\mathrm{CH_3\mathrm{CH=\mathrm{CH_2 + \mathrm{HBr \to \mathrm{CH_3\mathrm{CHBrCH_3$

Step 2: Hydrolyse using $\mathrm{NaOH$ (aq) to form the alcohol, then oxidise:

$\mathrm{CH_3\mathrm{CHBrCH_3 + \mathrm{NaOH \to \mathrm{CH_3\mathrm{CH(OH)CH_3 + \mathrm{NaBr$

$\mathrm{CH_3\mathrm{CH(OH)CH_3 + 2[\mathrm{O] \to \mathrm{CH_3\mathrm{COOH + \mathrm{H_2\mathrm{O$

Wait — oxidation of propan-2-ol gives propanone, not propanoic acid. Let us use a different route.

Alternative: Use anti-Markovnikov addition (not available without peroxides in the Higher course). Better route:

Step 1: Hydration of propene to propan-1-ol (indirect, via addition then hydrolysis): $\mathrm{CH_3\mathrm{CH=\mathrm{CH_2 + \mathrm{H_2\mathrm{O \xrightarrow{\mathrm{H^+} \mathrm{CH_3\mathrm{CH(OH)CH_3$

This gives propan-2-ol (Markovnikov). To get propanoic acid, we need propan-1-ol.

Best route: Use the Wacker process or go via the bromoalkane with anti-Markovnikov addition (requires HBr + peroxides, not in Higher syllabus). An acceptable Higher answer:

Propene $\xrightarrow{\mathrm{HBr}$ 2-bromopropane $\xrightarrow{\mathrm{NaOH (aq)}$ propan-2-ol $\xrightarrow{[O]}$ propanone. This gives a ketone, not an acid.

For propanoic acid: Propan-1-ol $\xrightarrow{[O]}$ propanal $\xrightarrow{[O]}$ propanoic acid.

Starting from propene, the most direct Higher-level route to propanoic acid involves: Propene $\xrightarrow{\mathrm{Br_2}$ 1,2-dibromopropane $\xrightarrow{\mathrm{NaOH (aq, excess)}$ Propane-1,2,3-triol (not ideal).

This example illustrates the limitations of certain synthetic routes at Higher level.

Functional Group Interconversion Summary

From	To	Reagent/Condition
Alkene	Alkane	$\mathrm{H_2$ Ni catalyst
Alkene	Haloalkane	$\mathrm{HX$ or $\mathrm{X_2$
Alkene	Alcohol	$\mathrm{H_2\mathrm{O$ , $\mathrm{H^+$
Alcohol	Aldehyde	$[\mathrm{O]$ Distillation
Alcohol	Carboxylic acid	$[\mathrm{O]$ Reflux
Alcohol	Alkene	$\mathrm{conc. H_2\mathrm{SO_4$ Heat
Haloalkane	Alcohol	$\mathrm{NaOH (aq)$
Haloalkane	Amine	$\mathrm{NH_3$ (excess)
Carboxylic acid	Ester	Alcohol, $\mathrm{conc. H_2\mathrm{SO_4$
Carboxylic acid	Salt	Base / carbonate
Aldehyde	Primary alcohol	$\mathrm{NaBH_4$
Ketone	Secondary alcohol	$\mathrm{NaBH_4$

Alcohols in Detail

Classification

Primary (1 degree): The carbon bearing the $-\mathrm{OH$ group is attached to at most one other carbon. Example: ethanol, propan-1-ol.
Secondary (2 degree): Attached to two other carbons. Example: propan-2-ol.
Tertiary (3 degree): Attached to three other carbons. Example: 2-methylpropan-2-ol.

Reactivity of Alcohols

Reaction type	Primary	Secondary	Tertiary
Oxidation	Aldehyde $\to$ acid	Ketone	No reaction
Dehydration	Multiple products possible	One major product	One product
$S_N1/S_N2$	$S_N2$ favoured	Both possible	$S_N1$ favoured
Reaction with HX	Slow	Moderate	Fast

Environmental and Industrial Chemistry

Polymers and the Environment

Polymer	Source	Environmental issue	Alternative
Polythene	Ethene	Non-biodegradable, landfill	Biodegradable polymers
PVC	Chloroethene	Releases HCl when burned	Use alternative plastics
PET	Condensation polymer	Recycling possible	Biodegradable polyester
Polylactic acid (PLA)	Corn starch	Biodegradable	—

Biofuels

Bioethanol: Produced by fermentation of sugars using yeast. Can be blended with petrol.

$\mathrm{C_6\mathrm{H_{12}\mathrm{O_6 \xrightarrow{\mathrm{yeast} 2\mathrm{C_2\mathrm{H_5\mathrm{OH + 2\mathrm{CO_2$

Biodiesel: Produced by transesterification of vegetable oils with methanol.

Advantages: renewable, carbon-neutral. Disadvantages: land use competition, lower energy density.

Spectroscopy in Organic Chemistry (Introduction)

Infrared Spectroscopy

Key absorptions for functional group identification:

Bond	Wavenumber (cm $^{-1}$ )
O-H (alcohol)	3200-3600 (broad)
O-H (acid)	2500-3300 (very broad)
C=O	1680-1750
C=C	1620-1680
C-O	1000-1300

Worked Example 18: An unknown compound $\mathrm{C_3\mathrm{H_6\mathrm{O$ shows a strong absorption At $1715 \mathrm{ cm^{-1}$ but no broad O-H absorption. Identify the compound.

The absorption at $1715 \mathrm{ cm^{-1}$ indicates a C=O group. The absence of a broad O-H peak Rules out a carboxylic acid. The molecular formula $\mathrm{C_3\mathrm{H_6\mathrm{O$ could be propanal Or propanone. Both are ketones/aldehydes with a C=O group. To distinguish, use Tollens’ test: Propanal gives a silver mirror, propanone does not.

Mass Spectrometry

The molecular ion peak ( $\mathrm{M^+$ ) gives the molecular mass. Fragmentation patterns help identify The structure.

For $\mathrm{C_3\mathrm{H_6\mathrm{O$ ( $M_r = 58$ ):

Propanone: major fragment at $m/z = 43$ ( $\mathrm{CH_3\mathrm{CO^+$ )
Propanal: major fragment at $m/z = 29$ ( $\mathrm{C_2\mathrm{H_5^+$ ) and $m/z = 28$ ( $\mathrm{CO^+$ )

NMR in Organic Chemistry (Introduction)

$^1\mathrm{H$ NMR

Number of signals: different proton environments
Integration: relative number of protons
Splitting: $n + 1$ rule

Worked Example 19: Predict the $^1\mathrm{H$ NMR spectrum of propanone ( $\mathrm{CH_3\mathrm{COCH_3$ ).

There is only one type of proton environment (the six protons are in two equivalent $\mathrm{CH_3$ Groups). The spectrum shows a single peak (singlet) at approximately $\delta = 2.1$ ppm with Integration 6.

Worked Example 20: Predict the $^1\mathrm{H$ NMR spectrum of ethanol ( $\mathrm{CH_3\mathrm{CH_2\mathrm{OH$ ).

Three proton environments:

$\mathrm{CH_3$ (3H): triplet at $\delta \approx 1.2$ (split by 2 neighbouring H)
$\mathrm{CH_2$ (2H): quartet at $\delta \approx 3.7$ (split by 3 neighbouring H)
$\mathrm{OH$ (1H): singlet at $\delta \approx 2.5$ (exchanges with $\mathrm{D_2\mathrm{O$ )

Alkene Stereochemistry in Detail

Cahn-Ingold-Prelog Priority Rules

To assign E/Z to alkenes:

On each carbon of the double bond, assign priority to the two substituents using atomic number (higher atomic number = higher priority).
If the two higher-priority groups are on the same side of the double bond, the isomer is Z (zusammen, together).
If on opposite sides, the isomer is E (entgegen, opposite).

Worked Example 21: Assign E/Z to 1-bromo-2-chloropropene.

$\mathrm{CH_3\mathrm{C(\mathrm{Br)=\mathrm{C(\mathrm{Cl)\mathrm{H$

On the left carbon: $\mathrm{Br$ (priority 1) vs. $\mathrm{CH_3$ (priority 2). On the right carbon: $\mathrm{Cl$ (priority 1) vs. $\mathrm{H$ (priority 2).

If $\mathrm{Br$ and $\mathrm{Cl$ are on the same side: Z isomer. If on opposite sides: E isomer.

Impact of Stereochemistry on Properties

E and Z isomers have different physical properties:

Property	Z-1,2-dichloroethene	E-1,2-dichloroethene
Boiling point	60°C	48°C
Dipole moment	Non-zero (polar)	Zero (non-polar)
Melting point	-80°C	-50°C

The Z isomer has a dipole moment (both C-Cl bonds on the same side), while the E isomer has Cancelling dipoles.

Halogenoalkanes in Detail

Classification and Reactivity

Type	Structure	$S_N1$ or $S_N2$	Reactivity with NaOH (aq)
Primary	R- $\mathrm{CH_2\mathrm{X$	$S_N2$	Slow
Secondary	R $_2$ - $\mathrm{CHX$	Both	Moderate
Tertiary	R $_3$ - $\mathrm{CX$	$S_N1$	Fast

Environmental Impact of Halogenoalkanes

CFCs (chlorofluorocarbons): Were used as refrigerants and propellants. They deplete the ozone layer because the C-Cl bond is weak enough to be broken by UV radiation in the stratosphere, releasing chlorine radicals that catalyse ozone destruction.

$\mathrm{Cl^\bullet + \mathrm{O_3 \to \mathrm{ClO^\bullet + \mathrm{O_2$ $\mathrm{ClO^\bullet + \mathrm{O \to \mathrm{Cl^\bullet + \mathrm{O_2$

HCFCs and HFCs: Replacement compounds that are less damaging to the ozone layer.

Nucleophilic Substitution with Different Nucleophiles

Nucleophile	Product	Notes
$\mathrm{OH^-$ (aq)	Alcohol	Hydrolysis
$\mathrm{CN^-$	Nitrile	Extends carbon chain by one
$\mathrm{NH_3$ (excess)	Primary amine	Produces $\mathrm{NH_4^+$ as byproduct
$\mathrm{H_2\mathrm{O$	Alcohol (slow)	Acidic conditions needed

Worked Example 22: Describe the reaction of 1-bromopropane with potassium cyanide and explain The importance of this reaction in organic synthesis.

$\mathrm{CH_3\mathrm{CH_2\mathrm{CH_2\mathrm{Br + \mathrm{KCN \to \mathrm{CH_3\mathrm{CH_2\mathrm{CH_2\mathrm{CN + \mathrm{KBr$

This is an $S_N2$ reaction. The product is butanenitrile. The importance is that the carbon chain Has been extended by one carbon atom (from 3 to 4). The nitrile can subsequently be hydrolysed to a Carboxylic acid, providing a route to chain-elongated compounds.

Summary Table: Key Organic Reactions

Reactant	Reagent	Product	Reaction type
Alkene	$\mathrm{H_2$ /Ni	Alkane	Addition
Alkene	$\mathrm{HX$	Haloalkane	Electrophilic addition
Alkene	$\mathrm{X_2$	Dihaloalkane	Electrophilic addition
Alkene	$\mathrm{H_2\mathrm{O/\mathrm{H^+$	Alcohol	Electrophilic addition
Primary alcohol	$[\mathrm{O]$ Distil	Aldehyde	Oxidation
Primary alcohol	$[\mathrm{O]$ Reflux	Carboxylic acid	Oxidation
Secondary alcohol	$[\mathrm{O]$	Ketone	Oxidation
Alcohol	$\mathrm{conc. H_2\mathrm{SO_4$ Heat	Alkene	Elimination (dehydration)
Alcohol + acid	$\mathrm{conc. H_2\mathrm{SO_4$	Ester	Esterification
Haloalkane	$\mathrm{NaOH (aq)$	Alcohol	Nucleophilic substitution
Haloalkane	$\mathrm{KCN$	Nitrile	Nucleophilic substitution
Carboxylic acid	$\mathrm{NaOH$	Carboxylate salt	Acid-base
Ester	$\mathrm{NaOH (aq)$	Carboxylate + alcohol	Base hydrolysis
Aldehyde	$\mathrm{NaBH_4$	Primary alcohol	Reduction
Ketone	$\mathrm{NaBH_4$	Secondary alcohol	Reduction

Worked Examples

Example 1: Rate equation

The reaction $\text{A} + \text{B} \rightarrow \text{C}$ has rate equation $\text{rate} = k[\text{A}][\text{B}]^2$ . When $[\text{A}] = 0.10\,\text{mol\,dm}^{-3}$ and $[\text{B}] = 0.20\,\text{mol\,dm}^{-3}$ , the rate is $2.0 \times 10^{-3}\,\text{mol\,dm}^{-3}\text{s}^{-1}$ . Calculate $k$ .

Solution:

$k = \frac{\text{rate}}{[\text{A}][\text{B}]^2} = \frac{2.0 \times 10^{-3}}{(0.10)(0.20)^2} = \frac{2.0 \times 10^{-3}}{4.0 \times 10^{-3}} = 0.50\,\text{mol}^{-2}\,\text{dm}^6\text{s}^{-1}$

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Organic Chemistry

Organic Chemistry

Higher Organic Chemistry

Hydrocarbons

Comparison of Hydrocarbons

IUPAC Nomenclature

Isomerism

Functional Groups

Reactions of Alkanes

Reactions of Alkenes

Reactions of Alcohols

Carbonyl Compounds

Carboxylic Acids and Esters

Advanced Higher Organic Chemistry

Reaction Mechanisms

Comparison of SN1S_N1SN​1 and SN2S_N2SN​2

Condensation Polymers

Comparison: Addition vs. Condensation Polymers

Optical Activity

Common Pitfalls

Practice Questions

Organic Synthesis Strategies

Two-Step Synthesis Plans

Functional Group Interconversion Summary

Alcohols in Detail

Classification

Reactivity of Alcohols

Environmental and Industrial Chemistry

Polymers and the Environment

Biofuels

Spectroscopy in Organic Chemistry (Introduction)

Infrared Spectroscopy

Mass Spectrometry

NMR in Organic Chemistry (Introduction)

^1\mathrm{H NMR

Alkene Stereochemistry in Detail

Cahn-Ingold-Prelog Priority Rules

Impact of Stereochemistry on Properties

Halogenoalkanes in Detail

Classification and Reactivity

Environmental Impact of Halogenoalkanes

Nucleophilic Substitution with Different Nucleophiles

Summary Table: Key Organic Reactions

Worked Examples

Comparison of $S_N1$ and $S_N2$

$^1\mathrm{H$ NMR