Organic Chemistry | IB - Wyatt's Notes

Intuition

Organic chemistry is like the chemistry of life — carbon’s ability to form four bonds creates the backbone of all living things: Functional groups are the reactive centers of organic molecules — mastering them unlocks the chemistry of millions of compounds

Why it matters: Organic chemistry is essential for pharmaceuticals, polymers, food science, and understanding biological processes

The key insight: Functional groups are the reactive centers of organic molecules — mastering them unlocks the chemistry of millions of compounds

Homologous Series and Naming

IUPAC Nomenclature

Organic compounds are named systematically using IUPAC rules:

Identify the longest carbon chain (parent chain).
Number the chain to give substituents the lowest possible numbers.
Name and number substituents (alkyl groups, halogens, etc.).
Identify and name the principal functional group (gets the lowest number).

Alkyl Groups

Name	Formula
Methyl	—CH $_3$
Ethyl	—C $_2$ H $_5$
Propyl	—C $_3$ H $_7$
Butyl	—C $_4$ H $_9$

Carbon Chain Prefixes

Carbons	Prefix
1	Meth-
2	Eth-
3	Prop-
4	But-
5	Pent-
6	Hex-
7	Hept-
8	Oct-
9	Non-
10	Dec-

Suffixes for Functional Groups

Functional Group	Suffix	Example
Alkane	-ane	Ethane
Alkene	-ene	Ethene
Alkyne	-yne	Ethyne
Alcohol	-ol	Ethanol
Aldehyde	-al	Ethanal
Ketone	-one	Propanone
Carboxylic acid	-oic acid	Ethanoic acid
Ester	-oate	Methyl ethanoate
Amine	-amine	Ethanamine
Amide	-amide	Ethanamide

Example C $_4$ H $_8$ O has multiple isomers: butan-1-ol, butan-2-ol, 2-methylpropan-1-ol, butanal, butanone, Methyl propanoate, ethyl ethanoate, etc.

Alkanes

Structure

General formula: C $_n$ H $_{2n+2}$
Single covalent bonds only (C—C and C—H).
sp $^3$ hybridisation, tetrahedral geometry (bond angles $\approx 109.5\degree$ ).
Saturated hydrocarbons (maximum number of hydrogens).

Physical Properties

Property	Trend
Boiling point	Increases with chain length (more London forces)
Melting point	Increases with chain length
State at room temperature	C $_1$ —C $_4$ : gas; C $_5$ —C $_{17}$ : liquid; C $_{18}^+$ : solid
Solubility	Non-polar, insoluble in water

Reactions

Combustion

Complete combustion (excess oxygen):

\mathrm{C}_n\mathrm{H}_{2n+2} + \frac{3n+1}{2}\mathrm{O}_2 \to n\mathrm{CO}_2 + (n+1)\mathrm{H}_2\mathrm{O}

Incomplete combustion (limited oxygen):

\mathrm{C}_n\mathrm{H}_{2n+2} + \frac{2n+1}{2}\mathrm{O}_2 \to n\mathrm{CO} + (n+1)\mathrm{H}_2\mathrm{O}

Or: produces carbon (soot) and water.

Halogenation (Free Radical Substitution)

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

Mechanism (free radical substitution):

Initiation: Cl $_2 \xrightarrow{\mathrm{UV}} 2\mathrm{Cl}^\bullet$ (homolytic fission)
Propagation:

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

Termination: radicals combine in various ways:

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

Cracking

Breaking large hydrocarbons into smaller, more useful molecules.

Thermal cracking: high temperature, produces alkenes.

Catalytic cracking: uses a zeolite catalyst, lower temperature, produces branched alkanes and Cycloalkanes.

Alkenes

Structure

General formula: C $_n$ H $_{2n}$
Contain at least one C=C double bond.
sp $^2$ hybridisation, trigonal planar geometry around the double bond (bond angle $\approx 120\degree$ ).
Unsaturated hydrocarbons.
The double bond consists of one $\sigma$ bond and one $\pi$ bond.
Restricted rotation about the C=C bond leads to cis-trans (E/Z) isomerism.

Physical Properties

Similar to alkanes of comparable molecular mass but with slightly higher boiling points due to the $\pi$ electron cloud.

Reactions

Addition Reactions

Hydrogenation:

\mathrm{CH}_2=\mathrm{CH}_2 + \mathrm{H}_2 \xrightarrow{\mathrm{Ni catalyst}} \mathrm{CH}_3\mathrm{CH}_3

Halogenation:

\mathrm{CH}_2=\mathrm{CH}_2 + \mathrm{Br}_2 \to \mathrm{CH}_2\mathrm{BrCH}_2\mathrm{Br}

The bromine water is decolourised — this is the test for unsaturation.

Hydration (with acid catalyst):

\mathrm{CH}_2=\mathrm{CH}_2 + \mathrm{H}_2\mathrm{O} \xrightarrow{\mathrm{H}_3\mathrm{PO}_4} \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}

Hydrohalogenation:

\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 the greater number of Hydrogen atoms (the more substituted carbon gets the halogen).

\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 + \mathrm{HBr} \to \mathrm{CH}_3\mathrm{CHBrCH}_3 \mathrm{ (major product)}

Polymerisation

Alkenes undergo addition polymerisation:

N\mathrm{CH}_2=\mathrm{CHCl} \to -(\mathrm{CH}_2\mathrm{CHCl})_n-

Poly(chloroethene) — PVC.

Alkynes

Structure

General formula: C $_n$ H $_{2n-2}$
Contain a C $\equiv$ C triple bond.
sp hybridisation, linear geometry around the triple bond.
Terminal alkynes have an acidic hydrogen.

Reactions

Similar to alkenes but can undergo two successive addition reactions due to two $\pi$ bonds.

Benzene and Aromatic Chemistry

Structure of Benzene

Formula: C $_6$ H $_6$ .
Delocalised $\pi$ electron system (6 $\pi$ electrons above and below the ring).
Planar hexagonal structure.
All C—C bonds are equal length (intermediate between single and double).
Represented by a hexagon with a circle inside (or alternating double bonds).

Stability

Benzene is more stable than predicted by Kekule’s structure because of delocalisation energy (resonance energy).

Electrophilic Substitution

Benzene undergoes electrophilic substitution (not addition) to preserve the aromatic system.

Nitration

\mathrm{C}_6\mathrm{H}_6 + \mathrm{HNO}_3 \xrightarrow{\mathrm{H}_2\mathrm{SO}_4, 50\degree\mathrm{C}} \mathrm{C}_6\mathrm{H}_5\mathrm{NO}_2 + \mathrm{H}_2\mathrm{O}

Electrophile: $\mathrm{NO}_2^+$ (nitronium ion).

Halogenation

\mathrm{C}_6\mathrm{H}_6 + \mathrm{Br}_2 \xrightarrow{\mathrm{FeBr}_3} \mathrm{C}_6\mathrm{H}_5\mathrm{Br} + \mathrm{HBr}

Electrophile: $\mathrm{Br}^+$ (generated by FeBr $_3$ ).

Friedel-Crafts Alkylation

\mathrm{C}_6\mathrm{H}_6 + \mathrm{RCl} \xrightarrow{\mathrm{AlCl}_3} \mathrm{C}_6\mathrm{H}_5\mathrm{R} + \mathrm{HCl}

Electrophile: $\mathrm{R}^+$ (carbocation).

Functional Groups

Alcohols (R—OH)

Classification:

Primary ( $1\degree$ ): the —OH carbon is attached to one other carbon.
Secondary ( $2\degree$ ): the —OH carbon is attached to two other carbons.
Tertiary ( $3\degree$ ): the —OH carbon is attached to three other carbons.

Reactions:

Reaction	Conditions	Product
Combustion	Burns in air	CO $_2$ + H $_2$ O
Oxidation	PCC	Aldehyde (from $1\degree$ )
Oxidation	Acidified K $_2$ Cr $_2$ O $_7$	Carboxylic acid (from $1\degree$ ), Ketone (from $2\degree$ )
Dehydration	H $_2$ SO $_4$ Heat	Alkene
Substitution	PBr $_3$ or SOCl $_2$	Alkyl halide
Esterification	Carboxylic acid, H $^+$ catalyst	Ester

Oxidation of Alcohols:

Primary alcohol $\to$ Aldehyde $\to$ Carboxylic acid

Secondary alcohol $\to$ Ketone (stops here)

Tertiary alcohol $\to$ Not oxidised

Aldehydes and Ketones (C=O)

Aldehydes (terminal C=O): suffix -al. Can be oxidised to carboxylic acids (reducing agents).

Ketones (internal C=O): suffix -one. Cannot be further oxidised.

Tests:

Tollens’ reagent: aldehydes give a silver mirror; ketones do not.
Fehling’s solution: aldehydes give a brick-red precipitate; ketones do not.
2,4-DNPH: both aldehydes and ketones give an orange precipitate (test for carbonyl group).

Carboxylic Acids (R—COOH)

Properties:

Weak acids (partially dissociate in water).
Hydrogen bonding gives relatively high boiling points.
Form dimers in non-polar solvents.

Reactions:

Reaction	Product
With base	Salt + water
With alcohol (esterification)	Ester + water
With ammonia	Amide
Reduction (LiAlH $_4$ )	Primary alcohol

Esters (R—COO—R’)

Formed by condensation (esterification) of a carboxylic acid and an alcohol:

\mathrm{RCOOH} + \mathrm{R'OH} \rightleftharpoons \mathrm{RCOOR}' + \mathrm{H}_2\mathrm{O}

Catalysed by concentrated H $_2$ SO $_4$ .

Uses: flavourings, fragrances, plasticisers, solvents.

Amines (R—NH $_2$ )

Weak bases (the lone pair on nitrogen accepts a proton).
Form salts with acids.
Primary amines can be formed by nucleophilic substitution of halogenoalkanes with ammonia.

Reaction Mechanisms

Nucleophilic Substitution ( $S_N1$ and $S_N2$ )

$S_N2$ (Bimolecular)

One-step mechanism.
Concerted: bond breaking and forming happen simultaneously.
Inversion of configuration (Walden inversion).
Rate depends on both substrate and nucleophile concentration.

\mathrm{Rate} = k[\mathrm{R--X}][\mathrm{Nu}^-]

Favoured by: primary substrates, strong nucleophiles, polar aprotic solvents.
Sterically hindered substrates react slowly.

Mechanism:

Nucleophile attacks from the back of the C—X bond.
Transition state with partial bonds.
X $^-$ leaves.
Product has inverted configuration.

$S_N1$ (Unimolecular)

Two-step mechanism.
Rate depends only on substrate concentration.

\mathrm{Rate} = k[\mathrm{R--X}]

Step 1 (slow, rate-determining): R—X $\to$ R $^+$ + X $^-$ (carbocation formation).
Step 2 (fast): R $^+$ + Nu $^-$ $\to$ R—Nu.
Racemisation occurs (equal mixture of inverted and retained configuration).
Favoured by: tertiary substrates, weak nucleophiles, polar protic solvents.

Feature	$S_N1$	$S_N2$
Steps	Two (carbocation intermediate)	One (concerted)
Rate law	Rate $= k[\mathrm{RX}]$	Rate $= k[\mathrm{RX}][\mathrm{Nu}]$
Stereochemistry	Racemisation	Inversion
Substrate preference	Tertiary	Primary
Carbocation	Yes	No

Elimination ( $E1$ and $E2$ )

Elimination reactions remove HX to form an alkene.

$E2$ (Bimolecular)

Concerted one-step mechanism.
Strong base removes a proton while X $^-$ leaves.
Follows Zaitsev’s rule: the more substituted alkene is the major product.

\mathrm{Rate} = k[\mathrm{R--X}][\mathrm{base}]

$E1$ (Unimolecular)

Two-step mechanism.
Step 1: carbocation formation (rate-determining).
Step 2: base removes a proton.

\mathrm{Rate} = k[\mathrm{R--X}]

Factors Affecting $S_N$ vs $E$

Condition	Favours Substitution ( $S_N$ )	Favours Elimination ( $E$ )
Temperature	Lower	Higher
Base strength	Weak	Strong
Base concentration	Low	High
Substrate	Primary ( $S_N2$ )	Tertiary
Steric hindrance	Low	High
Solvent	Polar protic ( $S_N1$ )	-

Polymer Chemistry

Addition Polymers

Formed by addition polymerisation of alkenes (monomers with C=C bonds). No by-product.

Polymer	Monomer	Uses
Polyethene	Ethene	Bags, bottles
Polypropene	Propene	Ropes, containers
PVC	Chloroethene	Pipes, insulation
Polystyrene	Phenylethene	Packaging, insulation
PTFE	Tetrafluoroethene	Non-stick coatings

Condensation Polymers

Formed when monomers join with the elimination of a small molecule ( water).

Polyesters

Monomer: dicarboxylic acid + diol.

N\mathrm{HOOC--R--COOH} + n\mathrm{HO--R'--OH} \to -(\mathrm{OC--R--COO--R'O})_n- + 2n\mathrm{H}_2\mathrm{O}

Example: PET (polyethylene terephthalate) — used in fibres and bottles.

Polyamides (Nylons)

Monomer: dicarboxylic acid + diamine.

N\mathrm{HOOC--R--COOH} + n\mathrm{H}_2\mathrm{N--R'--NH}_2 \to -(\mathrm{OC--R--CONH--R'--NH})_n- + 2n\mathrm{H}_2\mathrm{O}

Proteins

Proteins are natural polyamides formed from amino acid monomers via peptide bonds.

Biodegradability

Polymer	Biodegradable?	Reason
Polyethene	No	C—C backbone resists hydrolysis
Polyesters	Yes (some)	Ester bonds can be hydrolysed
Polyamides	Partially	Amide bonds can be hydrolysed (slowly)
Polylactic acid (PLA)	Yes	Ester linkages, derived from renewable sources
Cellulose	Yes	Natural polymer, readily broken down

Identifying Monomers

To find the monomer from an addition polymer, break single C—C bonds alternately to recover the C=C Double bond.

For condensation polymers, identify the repeating unit and add back the eliminated molecule (H $_2$ O).

IB Exam-Style Questions

Question 1 (Paper 1 style)

Name the following compound: CH $_3$ CH(Cl)CH(CH $_3$ )CH $_2$ CH $_3$ .

Longest chain: 5 carbons (pentane).
Number from the end nearest the substituent with the lowest number: Cl at C2, CH $_3$ at C3.
Name: 2-chloro-3-methylpentane.

Question 2 (Paper 2 style)

Compare the mechanisms of $S_N1$ and $S_N2$ reactions.

$S_N2$ : One-step bimolecular mechanism. The nucleophile attacks the carbon bearing the leaving group From the opposite side, leading to inversion of configuration. The rate depends on both [substrate] And [nucleophile]. Favoured for primary substrates.

$S_N1$ : Two-step unimolecular mechanism. The leaving group departs first to form a carbocation Intermediate, which is then attacked by the nucleophile. This leads to racemisation. The rate Depends only on [substrate]. Favoured for tertiary substrates.

Question 3 (Paper 2 style)

Ethanol can be oxidised to ethanal and then to ethanoic acid.

(a) Describe the conditions for each oxidation.

Ethanol $\to$ Ethanal: Use PCC (pyridinium chlorochromate) in CH $_2$ Cl $_2$ at room temperature (mild Oxidation).

Ethanol $\to$ Ethanoic acid: Use acidified K $_2$ Cr $_2$ O $_7$ (potassium dichromate) under reflux (strong oxidation).

(b) How would you distinguish between ethanol, ethanal, and ethanoic acid?

Tollens’ reagent: silver mirror with ethanal, no reaction with ethanol or ethanoic acid.
Acidified K $_2$ Cr $_2$ O $_7$ : orange to green with ethanol and ethanal, no change with ethanoic acid.
NaHCO $_3$ : bubbles of CO $_2$ with ethanoic acid, no reaction with ethanol or ethanal.

Question 4 (Paper 1 style)

Which compound is the major product when 2-methylpropene reacts with HBr?

By Markovnikov’s rule, H adds to the less substituted carbon (C1) and Br adds to the more Substituted carbon (C2):

\mathrm{CH}_2=\mathrm{C}(\mathrm{CH}_3)_2 + \mathrm{HBr} \to \mathrm{CH}_3\mathrm{CBr}(\mathrm{CH}_3)_2

Product: 2-bromo-2-methylpropane.

Question 5 (Paper 2 style)

Describe the electrophilic substitution mechanism for the nitration of benzene.

Generation of electrophile:

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

Mechanism:

The electron-rich benzene ring attacks the nitronium ion ( $\mathrm{NO}_2^+$ ), forming a delocalised carbocation intermediate.
The intermediate loses a proton to HSO $_4^-$ Regenerating the aromatic system and forming nitrobenzene.

flowchart TD
    A[1_Organic Chemistry] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Summary

Reaction Type	Description
Substitution	Atom/group replaced (alkanes, benzene)
Addition	Atoms added across double/triple bond (alkenes, alkynes)
Elimination	Small molecule removed to form double bond
Condensation	Monomers join, small molecule eliminated
Oxidation	Gain of oxygen or loss of hydrogen
Reduction	Loss of oxygen or gain of hydrogen
Polymerisation	Monomers join to form long chains

Mechanism	Characteristics
$S_N2$	One step, inversion, primary substrates
$S_N1$	Two steps, racemisation, tertiary substrates
$E2$	One step, strong base, Zaitsev product
$E1$	Two steps, carbocation, weak base

Exam Strategy For naming questions, always identify the longest chain first. For mechanism questions, draw clear Curly arrow mechanisms showing electron movement. For comparison questions ( $S_N1$ vs $S_N2$ ), use a Table for clarity. For polymer questions, be able to identify the monomer from the polymer and vice Versa.

Organic Chemistry: Extended Topics

Optical Isomerism

Optical isomers (enantiomers) are non-superimposable mirror images. They occur when a carbon atom Has four different groups attached (a chiral centre).

Properties of enantiomers:

Same physical properties (except interaction with plane-polarised light).
Rotate plane-polarised light in opposite directions.
Same chemical properties except with other chiral molecules.
A 50:50 mixture is called a racemic mixture (optically inactive).

Racemisation

If a reaction creates a chiral centre but has no stereochemical control, a racemic mixture is Produced.

$R/S$ Nomenclature

The Cahn-Ingold-Prelog priority rules assign $R$ or $S$ to chiral centres:

Assign priority (1—4) to groups around the chiral centre (atomic number).
Orient the molecule so the lowest priority group points away.
Trace 1 $\to$ 2 $\to$ 3: clockwise $= R$ Counterclockwise $= S$ .

Conformational Isomerism

Different spatial arrangements due to rotation about single bonds.

Ethane: staggered (most stable) and eclipsed (least stable) conformations.

Butane: anti (most stable), gauche, and eclipsed conformations.

Stereoisomerism in Cyclic Compounds

Cycloalkanes can have cis-trans isomerism when substituents are on different sides of the ring.

Spectroscopy in Organic Chemistry

Infrared (IR) Spectroscopy

Identifies functional groups by absorption of IR radiation.

Absorption (cm $^{-1}$ )	Bond
2500—3300	O—H (broad), N—H
2850—3100	C—H
1680—1750	C=O
1620—1680	C=C
2100—2260	C $\equiv$ C, C $\equiv$ N

Mass Spectrometry

Molecular ion peak: gives the molecular mass.
Fragmentation pattern: helps identify the structure.
Isotope pattern: Cl has $^{35}$ Cl: $^{37}$ Cl $\approx$ 3:1 ratio; Br has $^{79}$ Br: $^{81}$ Br $\approx$ 1:1.

Nuclear Magnetic Resonance (NMR) Spectroscopy

Chemical shift ( $\delta$ ): indicates the electronic environment.
Number of signals: indicates the number of different types of proton.
Integration: indicates the relative number of protons.
Splitting: indicates the number of neighbouring protons ( $n+1$ rule).

Additional Reactions

Oxidation of Primary Alcohols: Detailed

Primary alcohol $\xrightarrow{\mathrm{[O]} }$ Aldehyde $\xrightarrow{\mathrm{[O]} }$ Carboxylic acid

The aldehyde can be protected from over-oxidation by:

Using PCC (mild oxidising agent).
Distilling off the aldehyde as it forms (lower boiling point).

Reduction of Carboxylic Acids

\mathrm{RCOOH} \xrightarrow{\mathrm{LiAlH}_4} \mathrm{RCH}_2\mathrm{OH}

LiAlH $_4$ is more powerful than NaBH $_4$ (which does not reduce carboxylic acids).

Nucleophilic Addition to Carbonyls

With HCN

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

Forms a hydroxynitrile (cyanohydrin). Creates a new chiral centre.

With Grignard Reagents

\mathrm{RCHO} + \mathrm{R'MgBr} \to \mathrm{RCH(OMgBr)R'} \xrightarrow{\mathrm{H}_3\mathrm{O}^+} \mathrm{RCH(OH)R'}

Forms secondary alcohols from aldehydes, tertiary alcohols from ketones.

Hydrolysis of Esters

\mathrm{RCOOR}' + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{RCOOH} + \mathrm{R'OH}

Acid-catalysed (reversible) or base-catalysed (irreversible, saponification).

Saponification

Base-catalysed hydrolysis of fats produces soap (sodium salts of fatty acids):

\mathrm{Fat} + \mathrm{NaOH} \to \mathrm{glycerol} + \mathrm{soap}

Additional IB Exam-Style Questions

Question 6 (Paper 2 style)

Describe the mechanism for the nucleophilic substitution reaction between 1-bromopropane and Aqueous sodium hydroxide.

Mechanism:

The hydroxide ion (Nu $^-$ ) attacks the carbon bonded to bromine from the opposite side of the C—Br bond.
A transition state forms where the C—O bond is partially formed and the C—Br bond is partially broken.
The C—Br bond breaks completely, and Br $^-$ leaves as a leaving group.
The product is propan-1-ol, and the configuration at the carbon is inverted.

This is an $S_N2$ mechanism because:

It is a primary halogenoalkane (little steric hindrance).
The rate depends on both [halogenoalkane] and [Nu $^-$ ].
The reaction proceeds in a single concerted step.
Inversion of configuration occurs.

Question 7 (Paper 1 style)

Which compound is a product of the complete combustion of pentane?

A. C $_5$ H $_{12}$ O $_2$ B. C $_5$ H $_{10}$ C. CO $_2$ and H $_2$ O D. CO and H $_2$ O

Answer: C. Complete combustion of any hydrocarbon produces only CO $_2$ and H $_2$ O.

Question 8 (Paper 2 style)

Compare addition and condensation polymers, giving one example of each and their corresponding Monomers.

Addition polymer: Polyethene. Monomer: ethene (CH $_2$ =CH $_2$ ).

Formed from unsaturated monomers with C=C bonds.
No by-product.
The polymer contains the same atoms as the monomers.
Often non-biodegradable.

Condensation polymer: Nylon-6,6. Monomers: hexanedioic acid and 1,6-diaminohexane.

Formed from monomers with two functional groups.
A small molecule ( water) is eliminated.
The polymer contains fewer atoms than the monomers combined.
Can be more biodegradable due to hydrolysable bonds.

Question 9 (Paper 1 style)

Which compound can exhibit both geometrical (cis-trans) and optical isomerism?

A. CH $_3$ CH=CHCH $_3$ B. CH $_3$ CHClCOOH C. CH $_3$ CH(OH)CH=CH $_2$ D. BrCH $_2$ CH(NH $_2$ )COOH

Answer: D. It has a C=C bond (for cis-trans isomerism) and a chiral carbon (for optical Isomerism).

Question 10 (Paper 2 style)

Explain Markovnikov’s rule in terms of the mechanism of electrophilic addition to propene.

When HBr adds to propene (CH $_3$ CH=CH $_2$ ):

The H $^+$ electrophile attacks the double bond.
It can form either a secondary carbocation (CH $_3$ CH $^+$ CH $_3$ ) or a primary carbocation (CH $_3$ CH $_2$ CH $_2^+$ ).
The secondary carbocation is more stable (hyperconjugation and inductive effects).
The secondary carbocation forms preferentially.
Br $^-$ then attacks the carbocation, giving 2-bromopropane as the major product.

Markovnikov’s rule states that the hydrogen adds to the carbon with more hydrogen atoms because this Pathway proceeds through the more stable carbocation intermediate.

Practice Problems

Question 1: IUPAC Nomenclature

Name the following compound: $\mathrm{CH}_3\mathrm{CH}(\mathrm{CH}_3)\mathrm{CH}_2\mathrm{CH}(\mathrm{Cl})\mathrm{CH}_2\mathrm{CH}_3$ .

Answer

Longest carbon chain: 6 carbons (hexane).
Number from the end nearest the substituent with the lowest number.
Substituents: methyl at C2, chloro at C4.

Name: 4-chloro-2-methylhexane.

Question 2: Alkene Addition Reactions

Describe the products formed when propene ( $\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2$ ) reacts with Each of the following:

(a) $\mathrm{HBr}$

(b) $\mathrm{Br}_2$ in $\mathrm{CCl}_4$

Answer

(a) By Markovnikov’s rule, H adds to the less substituted carbon and Br adds to the more substituted Carbon:

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

Product: 2-bromopropane.

(b) Halogenation (anti addition):

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

Product: 1,2-dibromopropane.

$\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 + \mathrm{H}_2\mathrm{O} \xrightarrow{\mathrm{H}^+} \mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{CH}_3$

Product: propan-2-ol.

Question 3: Comparing $S_N1$ and $S_N2$ Mechanisms

For the reaction between 1-bromobutane and aqueous $\mathrm{NaOH}$ :

(a) Identify the mechanism and explain why.

(b) State the rate equation.

Answer

(a) This is an $S_N2$ mechanism because 1-bromobutane is a primary halogenoalkane with minimal Steric hindrance, and $\mathrm{OH}^-$ is a strong nucleophile. $S_N2$ is favoured for primary Substrates with strong nucleophiles.

(b) $\mathrm{Rate} = k[\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{Br}][\mathrm{OH}^-]$

(c) Inversion of configuration (Walden inversion) occurs at the carbon bearing the leaving group. The nucleophile attacks from the back of the C—Br bond, resulting in the product having the Opposite configuration.

Question 4: Polymer Identification

A polymer has the repeating unit: $-[\mathrm{NH}(\mathrm{CH}_2)_6\mathrm{NHCO}(\mathrm{CH}_2)_4\mathrm{CO}]-$ .

(a) Identify the type of polymer.

(b) Identify the two monomers.

Answer

(a) This is a condensation polymer (specifically a polyamide), identifiable by the amide Linkages ( $-\mathrm{NHCO}-$ ) in the repeating unit.

(b) The monomers are:

1,6-diaminohexane ( $\mathrm{H}_2\mathrm{N}(\mathrm{CH}_2)_6\mathrm{NH}_2$ )
hexanedioic acid (adipic acid, $\mathrm{HOOC}(\mathrm{CH}_2)_4\mathrm{COOH}$ )

This polymer is nylon-6,6.

Question 5: Alcohol Oxidation Pathway

Describe the products formed when each of the following alcohols is oxidised with acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ :

(a) Propan-1-ol ( $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH}$ )

(b) Propan-2-ol ( $\mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{CH}_3$ )

Answer

(a) Primary alcohol: oxidised first to propanal ( $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CHO}$ ), then Further oxidised to propanoic acid ( $\mathrm{CH}_3\mathrm{CH}_2\mathrm{COOH}$ ). With acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ under reflux, the final product is propanoic acid.

(b) Secondary alcohol: oxidised to propanone (acetone, $\mathrm{CH}_3\mathrm{COCH}_3$ ). The reaction Stops at the ketone because ketones cannot be further oxidised.

(c) Tertiary alcohol: tertiary alcohols cannot be oxidised by $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ because there is no hydrogen atom on the carbon bearing the $-\mathrm{OH}$ group. No reaction occurs (the orange solution remains orange).

A-Level Organic Chemistry: Organic Chemistry
DSE Carbon Chemistry: Carbon Chemistry

Common Pitfalls

Confusing structural isomers with stereoisomers. Structural isomers have different bonding arrangements; stereoisomers have the same connectivity.
Forgetting that primary alcohols can be oxidised to aldehydes then carboxylic acids, but secondary alcohols only to ketones.
Forgetting to balance equations before performing calculations. Always check that atoms and charges balance on both sides.
Drawing structural formulae incorrectly. Check the number of bonds each atom can form and the overall charge.
Confusing enthalpy of formation with enthalpy of combustion, or using the wrong sign convention.
Assuming that a strong acid always has a lower pH than a weak acid without considering concentration.

Cross-References

Topic	Site	Link
[Organic Chemistry]	A-Level	View
[Organic Chemistry]	IB	View
[Organic Chemistry]	DSE	View

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Organic Chemistry | IB - Wyatt's Notes

Intuition

Homologous Series and Naming

IUPAC Nomenclature

Alkyl Groups

Carbon Chain Prefixes

Suffixes for Functional Groups

See Also