Organic Chemistry (Advanced)

1. Reaction Mechanisms

SN1 Mechanism (Unimolecular Nucleophilic Substitution)

Definition. SN1 proceeds via a two-step mechanism where the rate depends only on the concentration of the substrate.

Step 1 (slow, rate-determining): Heterolytic cleavage of the C--X bond to form a carbocation intermediate.

$$\mathrm{(CH_3)_3C\mathrm{-}Br} \to \mathrm{(CH_3)_3C^+} + \mathrm{Br}^-$$

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

$$\mathrm{(CH_3)_3C^+} + \mathrm{OH}^- \to \mathrm{(CH_3)_3COH}$$

Rate law: $\mathrm{rate} = k[\mathrm{substrate}]$ (first order)

Characteristics:

Feature	SN1
Kinetics	First order
Mechanism	Two steps, carbocation intermediate
Stereochemistry	Racemisation (attack from both sides)
Substrate preference	Tertiary $\gt$ secondary (primary never)
Nucleophile strength	Weak nucleophiles are sufficient
Leaving group	Needs a good leaving group
Rearrangements	Possible (hydride or alkyl shifts)

SN2 Mechanism (Bimolecular Nucleophilic Substitution)

Definition. SN2 proceeds via a single concerted step where the nucleophile attacks from the back side as the leaving group departs.

$$\mathrm{OH}^- + \mathrm{CH_3\mathrm{-}Br} \to \mathrm{TS} \to \mathrm{CH_3OH} + \mathrm{Br}^-$$

Rate law: $\mathrm{rate} = k[\mathrm{substrate}][\mathrm{nucleophile}]$ (second order)

Characteristics:

Feature	SN2
Kinetics	Second order
Mechanism	One concerted step, transition state
Stereochemistry	Walden inversion (back-side attack)
Substrate preference	Primary $\gt$ secondary (tertiary never)
Nucleophile strength	Strong nucleophiles required
Leaving group	Needs a good leaving group
Rearrangements	Not possible

E1 Mechanism (Unimolecular Elimination)

Two-step mechanism via a carbocation intermediate (same intermediate as SN1):

Step 1: Formation of carbocation (rate-determining).

Step 2: Base removes a proton from an adjacent carbon, forming a double bond.

$$\mathrm{rate} = k[\mathrm{substrate}]$$

E2 Mechanism (Bimolecular Elimination)

Concerted mechanism: the base removes a proton while the leaving group departs, forming a double bond in a single step.

$$\mathrm{rate} = k[\mathrm{substrate}][\mathrm{base}]$$

Comparing SN1, SN2, E1, E2

Factor	Favours SN1/E1	Favours SN2/E2
Substrate	Tertiary	Primary, methyl
Nucleophile/Base	Weak	Strong
Leaving group	Good (I$^-$, Br$^-$)	Good (I$^-$, Br$^-$)
Temperature	Lower (substitution)	Higher (elimination)
Solvent	Polar protic (stabilises carbocation)	Polar aprotic (does not stabilise carbocation)
Base concentration	Low	High

Zaitsev's Rule

When multiple alkenes can form, the more substituted (more stable) alkene is the major product:

$$\mathrm{CH_3CH_2CH(CH_3)CH_2Br} \xrightarrow{\mathrm{E2}} \mathrm{CH_3CH_2C(CH_3)=CH_2} \mathrm{ (minor, Hofmann)}$$ $$\mathrm{CH_3CH_2CH(CH_3)CH_2Br} \xrightarrow{\mathrm{E2}} \mathrm{CH_3CH=C(CH_3)CH_3} \mathrm{ (major, Zaitsev)}$$

Common Pitfalls

• Primary substrates do not undergo SN1 or E1 (they cannot form stable carbocations).
• Tertiary substrates do not undergo SN2 (steric hindrance blocks back-side attack).
• Strong bases favour E2 over SN2, especially at elevated temperatures.

---

2. Stereochemistry

Chirality

Definition. A molecule is chiral if it is not superimposable on its mirror image. A chiral centre (stereocentre) is a carbon atom bonded to four different groups.

Enantiomers

Definition. Enantiomers are non-superimposable mirror images. They have identical physical properties (melting point, solubility) except for their interaction with plane-polarised light and with other chiral molecules.

Optical Activity

• Dextrorotatory ($+$ or $d$): rotates plane-polarised light clockwise.
• Laevorotatory ($-$ or $l$): rotates plane-polarised light anticlockwise.
• A racemic mixture ($50:50$ mixture of enantiomers) shows no optical rotation.

Cahn-Ingold-Prelog (CIP) Priority Rules

Assign priorities to the four groups on a stereocentre:

1. Higher atomic number = higher priority.
2. If tied, look at the next atoms outward.
3. Double bonds are treated as if each atom is duplicated.
4. Orient the molecule so the lowest priority group points away.
5. If the remaining three groups go clockwise: $R$ (rectus). Anticlockwise: $S$ (sinister).

Diastereomers

Definition. Diastereomers are stereoisomers that are not mirror images. They have different physical properties.

Geometric (cis/trans) Isomerism

Restricted rotation around a C=C double bond or a ring gives rise to geometric isomers.

For C=C: assign priorities using CIP rules. If the two higher-priority groups are on the same side: $Z$ (zusammen). Opposite sides: $E$ (entgegen).

Isomer	Description	Example
$cis$	Same side (similar groups)	$cis$-but-2-ene
$trans$	Opposite sides	$trans$-but-2-ene
$Z$	Higher priority groups same side	$(Z)$-1-bromo-1-chloropropene
$E$	Higher priority groups opposite	$(E)$-1-bromo-1-chloropropene

Optical vs Geometric Isomerism

Property	Optical Isomerism	Geometric Isomerism
Cause	Chiral centre	Restricted rotation (C=C, ring)
Mirror images	Non-superimposable	Not necessarily mirror images
Physical properties	Identical (except optical)	Different

Common Pitfalls

• A molecule with a plane of symmetry is never chiral, even if it has stereocentres (meso compounds).
• Not all stereocentres produce chirality — internal symmetry can make a molecule achiral.
• $R$/$S$ refers to absolute configuration at a single stereocentre; it does not predict the direction of optical rotation.

---

3. Addition Polymers

Mechanism

Addition polymers form from alkene monomers via free-radical addition. The double bond opens and monomers link together in a chain.

Initiation: A peroxide or other radical initiator generates free radicals.

$$\mathrm{ROOR} \to 2\mathrm{RO}^{\bullet}$$

Propagation: The radical adds to a monomer, and the new radical adds to another monomer.

$$\mathrm{RO}^{\bullet} + \mathrm{CH_2=CHR} \to \mathrm{ROCH_2CHR}^{\bullet} \to \mathrm{ROCH_2CHRCH_2CHR}^{\bullet} \to \cdots$$

Termination: Two radicals combine, ending chain growth.

Common Addition Polymers

Monomer	Polymer	Uses
Ethene	Polyethene (PE)	Bags, bottles, film
Propene	Polypropene (PP)	Containers, ropes, carpet fibre
Chloroethene (vinyl chloride)	PVC	Pipes, window frames, insulation
Phenylethene (styrene)	Polystyrene (PS)	Packaging, insulation, CD cases
Tetrafluoroethene	PTFE (Teflon)	Non-stick coatings, electrical insulation

Biodegradability

Most addition polymers are non-biodegradable because the strong C--C backbone resists chemical and biological breakdown. Disposal by landfill or incineration creates environmental problems.

Common Pitfalls

• The repeating unit of an addition polymer is not the same as the monomer (the double bond is gone).
• Condensation polymers and addition polymers are formed by different mechanisms — do not confuse them.

---

4. Condensation Polymers

Mechanism

Condensation polymers form when monomers join with the elimination of a small molecule (usually $\mathrm{H}_2\mathrm{O}$ or $\mathrm{HCl}$). Two types:

Polyesters

Formed from a diol and a dicarboxylic acid:

$$\mathrm{diol} + \mathrm{dicarboxylic acid} \to \mathrm{polyester} + \mathrm{H}_2\mathrm{O}$$

Example — PET (polyethylene terephthalate)

Monomers: ethane-1,2-diol and benzene-1,4-dicarboxylic acid.

$$n\mathrm{HOCH_2CH_2OH} + n\mathrm{HOOC\mathrm{-}C_6H_4\mathrm{-}COOH} \to \mathrm{PET} + 2n\mathrm{H}_2\mathrm{O}$$

Uses: fibres (clothing), bottles, food containers.

Polyamides (Nylons)

Formed from a diamine and a dicarboxylic acid:

$$\mathrm{diamine} + \mathrm{dicarboxylic acid} \to \mathrm{polyamide} + \mathrm{H}_2\mathrm{O}$$

Example — Nylon-6,6

Monomers: hexane-1,6-diamine and hexanedioic acid.

$$n\mathrm{H_2N(CH_2)_6NH_2} + n\mathrm{HOOC(CH_2)_4COOH} \to \mathrm{Nylon-6,6} + 2n\mathrm{H}_2\mathrm{O}$$

Uses: textiles, ropes, parachutes, engineering plastics.

Kevlar

An aromatic polyamide (aramid) with exceptional strength:

$$\mathrm{benzene-1,4-diamine} + \mathrm{benzene-1,4-dicarboxylic acid} \to \mathrm{Kevlar}$$

The rigid aromatic rings and strong hydrogen bonding between chains give Kevlar its high tensile strength. Used in body armour, tyres, and aerospace.

Comparison of Polymer Types

Property	Addition polymer	Condensation polymer
Monomers	Alkenes	Diols + diacids / diamines + diacids
By-product	None	$\mathrm{H}_2\mathrm{O}$ or $\mathrm{HCl}$
Biodegradability	Generally non-biodegradable	Some are biodegradable
Bond type	C--C backbone	Contains ester or amide bonds
Examples	PE, PP, PVC, PTFE	PET, Nylon, Kevlar

---

5. Spectroscopic Identification of Organic Compounds

Combined Strategy

1. Calculate the degree of unsaturation (double bond equivalents, DBE):

$$\mathrm{DBE} = C + 1 - \frac{H}{2} + \frac{N}{2}$$

(For each halogen, add 1 to H. For each oxygen, ignore.)

2. IR spectroscopy: identify functional groups from characteristic absorptions.

3. $\mathrm{^1H}$ NMR: determine proton environments, integration, and splitting.

4. Mass spectrometry: determine molecular mass and fragmentation pattern.

5. Assemble all fragments into a consistent structure.

Degree of Unsaturation Examples

Molecular formula	DBE	Possible features
$\mathrm{C}_4\mathrm{H}_{10}$	$0$	No double bonds, no rings (alkane)
$\mathrm{C}_4\mathrm{H}_8$	$1$	One double bond or one ring
$\mathrm{C}_4\mathrm{H}_6$	$2$	Two double bonds, one triple, or two rings
$\mathrm{C}_6\mathrm{H}_6$	$4$	Benzene ring (three double bonds + ring)

Characteristic IR Absorptions (Recap)

Bond	Range ($\mathrm{cm}^{-1}$)
O--H (acid)	$2500$--$3300$ (very broad)
O--H (alcohol)	$3200$--$3600$ (broad)
N--H	$3300$--$3500$
C--H (alkane)	$2850$--$3000$
C--H (alkene)	$3000$--$3100$
C$\equiv$N	$2200$--$2250$
C=O	$1700$--$1750$
C=C	$1600$--$1680$

Key NMR Chemical Shifts (Recap)

Proton type	$\delta$ (ppm)
Alkane	$0.7$--$1.5$
Adjacent to C=O	$2.0$--$2.7$
Alkene	$4.5$--$6.5$
Aromatic	$6.5$--$8.0$
Aldehyde	$9.0$--$10.0$
Carboxylic acid	$10.0$--$12.0$

Common Pitfalls

• DBE = 4 is a strong indicator of a benzene ring, but not proof by itself.
• In $\mathrm{^1H}$ NMR, the integration ratio must be multiplied by the total number of protons (determined from the molecular formula).
• $\mathrm{D}_2\mathrm{O}$ exchange removes OH and NH signals, confirming their presence.

---

Practice Problems

Problem 1

For each substrate, predict whether SN1, SN2, E1, or E2 will be the major pathway:

(a) $\mathrm{CH_3CH_2Br}$ with $\mathrm{NaOH}$ in $\mathrm{H}_2\mathrm{O}$ at $25\degree\mathrm{C}$

(b) $(\mathrm{CH_3})_3\mathrm{CBr}$ with $\mathrm{NaOH}$ in ethanol at $80\degree\mathrm{C}$

(c) $(\mathrm{CH_3})_3\mathrm{CBr}$ with $\mathrm{H}_2\mathrm{O}$ at $25\degree\mathrm{C}$

Solution:

(a) SN2: primary substrate, strong nucleophile/base, polar protic solvent, low temperature.

(b) E2: tertiary substrate, strong base, high temperature favours elimination over substitution.

(c) SN1: tertiary substrate, weak nucleophile (water), polar protic solvent, low temperature. The carbocation intermediate is stable.

Problem 2

An unknown compound has molecular formula $\mathrm{C}_3\mathrm{H}_6\mathrm{O}_2$. Its IR spectrum shows a broad peak at $3000\mathrm{ cm}^{-1}$ and a strong peak at $1710\mathrm{ cm}^{-1}$. Its $\mathrm{^1H}$ NMR spectrum shows: $\delta\ 1.2\ (d,\ 3\mathrm{H})$, $\delta\ 4.1\ (q,\ 1\mathrm{H})$, $\delta\ 11.0\ (s,\ 1\mathrm{H})$, and a singlet at $\delta\ 2.0$ that integrates to $1\mathrm{H}$.

Wait — the formula only has 6 H. Let me correct: $\delta\ 1.2\ (d,\ 3\mathrm{H})$, $\delta\ 2.5\ (q,\ 2\mathrm{H})$, $\delta\ 11.5\ (s,\ 1\mathrm{H})$. Identify the compound.

Solution:

• $\mathrm{C}_3\mathrm{H}_6\mathrm{O}_2$: $\mathrm{DBE} = 3 + 1 - 6/2 = 1$ (one C=O).
• IR: broad $3000\mathrm{ cm}^{-1}$ (O--H) + $1710\mathrm{ cm}^{-1}$ (C=O) = carboxylic acid.
• NMR: $d, 3\mathrm{H}$ + $q, 2\mathrm{H}$ = ethyl group ($\mathrm{CH_3CH_2}$--). $s, 1\mathrm{H}$ = COOH proton.
• Structure: $\mathrm{CH_3CH_2COOH}$ (propanoic acid).

Check: M = $3(12) + 6(1) + 2(16) = 74$. But $\mathrm{C}_3\mathrm{H}_6\mathrm{O}_2$ has M = 74. Confirmed.

Problem 3

Draw the repeating unit of the polyester formed from propane-1,3-diol and butanedioic acid. Write the equation for its formation.

Solution:

$$n\mathrm{HOCH_2CH_2CH_2OH} + n\mathrm{HOOCCH_2CH_2COOH} \to \left[-\mathrm{OCH_2CH_2CH_2OOCCH_2CH_2CO}-\right]_n + 2n\mathrm{H}_2\mathrm{O}$$

Repeating unit: $-\mathrm{OCH_2CH_2CH_2OOCCH_2CH_2CO}-$

Problem 4

A compound $\mathrm{C}_5\mathrm{H}_{10}\mathrm{O}$ shows the following spectra. IR: strong peak at $1700\mathrm{ cm}^{-1}$, no broad O--H. $\mathrm{^1H}$ NMR: $\delta\ 1.0\ (t,\ 3\mathrm{H})$, $\delta\ 1.6\ (m,\ 2\mathrm{H})$, $\delta\ 2.1\ (s,\ 3\mathrm{H})$, $\delta\ 2.3\ (t,\ 2\mathrm{H})$. Identify the compound.

Solution:

• $\mathrm{C}_5\mathrm{H}_{10}\mathrm{O}$: $\mathrm{DBE} = 5 + 1 - 10/2 = 1$ (one C=O).
• IR: $1700\mathrm{ cm}^{-1}$ (C=O), no O--H = ketone or aldehyde. No aldehyde peak around $\delta\ 9$--$10$, so it is a ketone.
• NMR: triplet + multiplet + triplet = propyl chain ($\mathrm{CH_3CH_2CH_2}$--). Singlet at $\delta\ 2.1$ ($3\mathrm{H}$) = $\mathrm{CH_3CO}$--.
• Structure: \mathrm{CH_3COCH_2CH_2CH_3 (pentan-2-one).

Total protons: $3 + 2 + 3 + 2 = 10 = \mathrm{C}_5\mathrm{H}_{10}\mathrm{O}$. Confirmed.

---

Worked Examples

Worked Example: Predicting the major organic product of a substitution reaction

Predict the major product when $(CH_3)_3CBr$ is heated with $\mathrm{KOH}$ in ethanol. State the mechanism, draw the transition state or intermediate, and explain the regiochemistry.

Solution

Substrate analysis: $(CH_3)_3CBr$ is a tertiary alkyl halide.

Conditions: $\mathrm{KOH}$ (strong base) in ethanol (polar protic solvent) at elevated temperature.

Mechanism determination: Tertiary substrate + strong base + heat $\to$ E2 is favoured over SN2. SN1 is also possible but elevated temperature shifts the product distribution toward elimination.

Major product: E2 elimination.

The base abstracts a proton from a $\beta$-carbon while the leaving group departs. By Zaitsev's rule, the more substituted (more stable) alkene is the major product:

$\mathrm{(CH_3)_3CBr + KOH \to (CH_3)_2C=CH_2 + KBr + H_2O}$

This is the Hofmann product (less substituted). The Zaitsev product would require removing a proton from a methyl group:

$\mathrm{(CH_3)_3CBr + KOH \to CH_3CH=C(CH_3)_2 + KBr + H_2O}$

Actually, for $(CH_3)_3CBr$, there are no $\beta$-hydrogens on the carbon bearing two methyl groups that are distinct from the terminal methyl groups. The only elimination products are $(CH_3)_2C=CH_2$ (the only possible alkene). Since there is only one type of $\beta$-hydrogen, Zaitsev's rule does not apply here --- there is only one elimination product.

SN1 product (minor): $(CH_3)_3COH$ (tert-butanol), formed via a carbocation intermediate with water as the nucleophile.

Worked Example: CIP priority rules and R/S assignment

Assign the absolute configuration (R or S) to the stereocentre in $\mathrm{CH_3CHClCH_2CH_3}$ (2-chlorobutane).

Solution

Step 1: Identify the stereocentre.

Carbon-2 is bonded to four different groups: $-\mathrm{H}$, $-\mathrm{Cl}$, $-\mathrm{CH_3}$, $-\mathrm{CH_2CH_3}$.

Step 2: Assign CIP priorities.

1. $-\mathrm{Cl}$: atomic number 17 (highest priority)
2. $-\mathrm{CH_2CH_3}$: the first atom is C, and the next atoms are C, H, H (by expansion: C is bonded to C, H, H)
3. $-\mathrm{CH_3}$: the first atom is C, and the next atoms are H, H, H
4. $-\mathrm{H}$: atomic number 1 (lowest priority)

Priority order: $\mathrm{Cl} \gt \mathrm{CH_2CH_3} \gt \mathrm{CH_3} \gt \mathrm{H}$

Step 3: Orient the molecule.

Place the lowest priority group ($-\mathrm{H}$) pointing away (dashed wedge). If $-\mathrm{H}$ is on a wedge in the given structure, invert the assignment.

Step 4: Determine R or S.

Looking at the remaining three groups (Cl, ethyl, methyl) in order of decreasing priority: $1 \to 2 \to 3$. If the sequence is clockwise, the configuration is R. If anticlockwise, it is S.

For the standard representation where Cl is on a wedge and H is on a dash: the sequence \mathrm{Cl} \to \mathrm{CH_2CH_3} \to \mathrm{CH_3 goes anticlockwise, so the configuration is S.

If H were on a wedge (pointing toward you), the apparent direction would be reversed, and the true configuration would be R. Always orient H away before assigning.

Worked Example: Drawing the repeating unit of a condensation polymer

Draw the repeating unit of the polyester formed from propane-1,3-diol and butanedioic acid. Write the balanced equation for its formation and identify the by-product.

Solution

Step 1: Write the structures of the monomers.

Propane-1,3-diol: $\mathrm{HOCH_2CH_2CH_2OH}$

Butanedioic acid (succinic acid): $\mathrm{HOOCCH_2CH_2COOH}$

Step 2: Write the condensation equation.

$n\mathrm{HOCH_2CH_2CH_2OH} + n\mathrm{HOOCCH_2CH_2COOH} \to \left[-\mathrm{OCH_2CH_2CH_2OOCCH_2CH_2CO}-\right]_n + 2n\mathrm{H_2O}$

Step 3: Identify the repeating unit.

The repeating unit is: $-\mathrm{OCH_2CH_2CH_2OOCCH_2CH_2CO}-$

Step 4: Verify atom conservation.

Left side per repeat: $\mathrm{C}_7\mathrm{H}_{12}\mathrm{O}_4$

Right side per repeat: repeating unit $\mathrm{C}_7\mathrm{H}_{10}\mathrm{O}_4$ + $2\mathrm{H_2O}$ = $\mathrm{C}_7\mathrm{H}_{10}\mathrm{O}_4 + \mathrm{H_4}\mathrm{O}_2$ = $\mathrm{C}_7\mathrm{H}_{14}\mathrm{O}_6$

Wait --- let me recount. Each repeat consumes one diol ($\mathrm{C}_3\mathrm{H}_8\mathrm{O}_2$) and one diacid ($\mathrm{C}_4\mathrm{H}_6\mathrm{O}_4$):

$\mathrm{C}_3\mathrm{H}_8\mathrm{O}_2 + \mathrm{C}_4\mathrm{H}_6\mathrm{O}_4 \to \mathrm{C}_7\mathrm{H}_{12}\mathrm{O}_4 \mathrm{(repeating unit)} + 2\mathrm{H_2O}$

Check atoms: LHS = $\mathrm{C}_7\mathrm{H}_{14}\mathrm{O}_6$. RHS = $\mathrm{C}_7\mathrm{H}_{12}\mathrm{O}_4 + \mathrm{H_4}\mathrm{O}_2 = \mathrm{C}_7\mathrm{H}_{16}\mathrm{O}_6$.

That does not balance. The correct stoichiometry is:

$\mathrm{C}_3\mathrm{H}_8\mathrm{O}_2 + \mathrm{C}_4\mathrm{H}_6\mathrm{O}_4 \to \mathrm{C}_7\mathrm{H}_{10}\mathrm{O}_4 + 2\mathrm{H_2O}$

Check: LHS = $\mathrm{C}_7\mathrm{H}_{14}\mathrm{O}_6$. RHS = $\mathrm{C}_7\mathrm{H}_{10}\mathrm{O}_4 + \mathrm{H_4}\mathrm{O}_2 = \mathrm{C}_7\mathrm{H}_{14}\mathrm{O}_6$. Balanced.

The repeating unit has lost 4 H and 2 O relative to the monomers (two ester linkages formed).

Worked Example: Degree of unsaturation and structural elucidation

A compound $\mathrm{C}_8\mathrm{H}_8\mathrm{O}$ has the following spectra. IR: $3060\mathrm{ cm}^{-1}$ (medium), $1690\mathrm{ cm}^{-1}$ (strong), $1600\mathrm{ cm}^{-1}$ (medium), $1580\mathrm{ cm}^{-1}$ (medium), $750\mathrm{ cm}^{-1}$ (strong). $^{1}\mathrm{H}$ NMR: $\delta\ 7.5$--$7.9\ (m,\ 5\mathrm{H})$, $\delta\ 3.9\ (s,\ 2\mathrm{H})$. MS: $\mathrm{M}^+ = 120$. Identify the compound.

Solution

Step 1: Calculate the degree of unsaturation.

$\mathrm{DBE} = 8 + 1 - \frac{8}{2} = 5$

DBE = 5 is consistent with a benzene ring (DBE = 4) plus one additional unsaturation (likely C=O).

Step 2: Analyse IR data.

• $3060\mathrm{ cm}^{-1}$: aromatic C--H stretch (above $3000\mathrm{ cm}^{-1}$ confirms sp2 C--H).
• $1690\mathrm{ cm}^{-1}$: C=O stretch (slightly below $1700$, suggesting conjugation with the aromatic ring).
• $1600$, $1580\mathrm{ cm}^{-1}$: aromatic C=C stretches.
• $750\mathrm{ cm}^{-1}$: mono-substituted benzene (ortho-disubstituted typically shows near $750\mathrm{ cm}^{-1}$, but combined with other evidence, this suggests a single substituent on the benzene ring).

Step 3: Analyse NMR data.

• $\delta\ 7.5$--$7.9\ (m,\ 5\mathrm{H})$: 5 aromatic protons, consistent with a mono-substituted benzene ring ($\mathrm{C}_6\mathrm{H}_5$--).
• $\delta\ 3.9\ (s,\ 2\mathrm{H})$: isolated $\mathrm{CH}_2$ group, singlet (no adjacent protons). The chemical shift ($\delta\ 3.9$) suggests the $\mathrm{CH}_2$ is adjacent to an electron-withdrawing group (C=O).

Step 4: Assemble the structure.

Mono-substituted benzene ring: $\mathrm{C}_6\mathrm{H}_5$--. Remaining atoms: $\mathrm{C}_2\mathrm{H}_3\mathrm{O}$. With C=O at $1690\mathrm{ cm}^{-1}$ and a $\mathrm{CH}_2$ singlet at $\delta\ 3.9$:

Structure: $\mathrm{C}_6\mathrm{H}_5\mathrm{CH_2CHO}$ (phenylethanal, also called phenylacetaldehyde).

Step 5: Verify.

• $\mathrm{C}_8\mathrm{H}_8\mathrm{O}$: $8(12) + 8(1) + 16 = 120 = \mathrm{M}^+$. Confirmed.
• DBE = 5: benzene ring (4) + aldehyde C=O (1). Confirmed.
• NMR: 5 aromatic H + 2 aldehydic CH2 = 7 H, plus 1 aldehyde H = 8 H total. Confirmed.

The compound is phenylethanal ($\mathrm{C}_6\mathrm{H}_5\mathrm{CH_2CHO}$).

Worked Example: Reaction mechanism comparison

For each substrate, predict the major product and mechanism when treated with $\mathrm{NaOH}$ in $\mathrm{H}_2\mathrm{O}$ at $25\degree\mathrm{C}$:

(a) $\mathrm{CH_3CH_2Br}$

(b) $(\mathrm{CH_3})_3\mathrm{CBr}$

(c) $\mathrm{CH_3CHBrCH_3}$

Solution

(a) $\mathrm{CH_3CH_2Br}$ (primary substrate, strong nucleophile, polar protic solvent):

Mechanism: SN2 (primary substrates do not form stable carbocations, so SN1/E1 are not possible).

$\mathrm{CH_3CH_2Br + OH^- \to CH_3CH_2OH + Br^-}$

Product: ethanol ($\mathrm{CH_3CH_2OH}$). Stereochemistry: Walden inversion at carbon.

(b) $(\mathrm{CH_3})_3\mathrm{CBr}$ (tertiary substrate, strong nucleophile/base, low temperature):

Mechanism: SN1 (tertiary substrates form stable carbocations; back-side attack is blocked for SN2).

Step 1 (slow): $(\mathrm{CH_3})_3\mathrm{CBr} \to (\mathrm{CH_3})_3\mathrm{C}^+ + \mathrm{Br}^-$

Step 2 (fast): $(\mathrm{CH_3})_3\mathrm{C}^+ + \mathrm{H_2O} \to (\mathrm{CH_3})_3\mathrm{COH_2}^+$

Step 3 (fast): $(\mathrm{CH_3})_3\mathrm{COH_2}^+ \to (\mathrm{CH_3})_3\mathrm{COH} + \mathrm{H}^+$

Product: 2-methylpropan-2-ol (tert-butanol). Minor E1 product: 2-methylpropene.

(c) $\mathrm{CH_3CHBrCH_3}$ (secondary substrate):

Mechanism: competition between SN1 and SN2. Secondary substrates can proceed via either pathway depending on exact conditions. With $\mathrm{NaOH}$ in water at $25\degree\mathrm{C}$, both SN2 and SN1 are possible, but SN2 is slightly favoured because $\mathrm{OH}^-$ is a strong nucleophile.

Product: propan-2-ol ($\mathrm{CH_3CH(OH)CH_3}$). Minor elimination product: propene.

---

Common Pitfalls

• Assuming primary substrates can undergo SN1: Primary carbocations are too unstable to form. A primary alkyl halide with a weak nucleophile and polar protic solvent will still proceed via SN2 (just slowly), not SN1.

• Confusing stereocentres with chirality: A molecule can have stereocentres but be achiral if it has an internal plane of symmetry (meso compounds). For example, meso-tartaric acid has two stereocentres but is not chiral overall.

• Writing the monomer instead of the repeating unit: The repeating unit of an addition polymer is not the same as the monomer --- the double bond has opened. For polyethene, the monomer is $\mathrm{CH_2=CH_2}$ but the repeating unit is $-\mathrm{CH_2CH_2}-$.

• Using the wrong $\mathrm{p}K_a$ when working with bases: In the Henderson-Hasselbalch equation, always use the $\mathrm{p}K_a$ of the conjugate acid, not the $\mathrm{p}K_b$ of the base. $\mathrm{p}K_a + \mathrm{p}K_b = 14.00$.

• Ignoring conjugation effects on IR frequencies: A C=O conjugated with a C=C bond absorbs at a lower wavenumber ($\sim 1680\mathrm{ cm}^{-1}$) than an unconjugated C=O ($\sim 1715\mathrm{ cm}^{-1}$). This shift is diagnostic in structure determination.

• Misassigning NMR splitting patterns: The $n + 1$ rule only applies when the neighbouring protons are equivalent. Non-equivalent neighbouring protons produce complex multiplets, not simple doublets or triplets.

• Forgetting that E2 requires an anti-periplanar arrangement: The proton being removed and the leaving group must be in the same plane on opposite sides. This geometric requirement can make certain E2 eliminations stereospecific and can explain why some theoretical elimination products are not observed.

• Confusing addition and condensation polymerisation mechanisms: Addition polymers form from alkene monomers with no by-product. Condensation polymers form from monomers with two functional groups and release a small molecule (usually $\mathrm{H}_2\mathrm{O}$ or $\mathrm{HCl}$).

• Over-relying on DBE alone for structural identification: DBE = 4 suggests an aromatic ring but does not prove it. A compound could have two C=C bonds and two rings. Always confirm with IR and NMR data.

---

Exam-Style Problems

1. Predict the major organic product(s) and state the mechanism for each reaction: (a) $\mathrm{CH_3CH_2CH_2Br}$ with $\mathrm{NaOH}$ in $\mathrm{H}_2\mathrm{O}$ at $25\degree\mathrm{C}$ (b) $\mathrm{(CH_3)_3CBr}$ with $\mathrm{NaOH}$ in ethanol at $80\degree\mathrm{C}$ (c) $\mathrm{CH_3CHBrCH_3}$ with $\mathrm{NaOH}$ in ethanol at $25\degree\mathrm{C}$ [Medium]

2. A compound $\mathrm{C}_4\mathrm{H}_8\mathrm{O}$ shows IR absorptions at $1720\mathrm{ cm}^{-1}$ (strong) and $2720\mathrm{ cm}^{-1}$ (weak). $^{1}\mathrm{H}$ NMR: $\delta\ 1.2\ (d,\ 3\mathrm{H})$, $\delta\ 2.5\ (q,\ 1\mathrm{H})$, $\delta\ 9.7\ (d,\ 1\mathrm{H})$. Identify the compound and explain the splitting pattern of the signal at $\delta\ 9.7$. [Medium]

3. Draw the repeating unit of the polyamide formed from hexane-1,6-diamine and pentanedioic acid. Write the balanced equation. Calculate the mass of polymer produced from $10.0\mathrm{ g}$ of each monomer, assuming 100% yield. [Medium]

4. Explain why 2-bromobutane reacts with $\mathrm{NaOH}$ in ethanol to produce a mixture of butan-2-ol and but-2-ene, but 2-bromo-2-methylpropane reacts under the same conditions to produce predominantly 2-methylpropene. Reference the relevant mechanisms in your answer. [Hard]

5. A compound $\mathrm{C}_9\mathrm{H}_{10}\mathrm{O}$ has the following spectra. IR: $3050\mathrm{ cm}^{-1}$, $1700\mathrm{ cm}^{-1}$, $1600$, $1580$, $1470\mathrm{ cm}^{-1}$, $690\mathrm{ cm}^{-1}$. $^{1}\mathrm{H}$ NMR: $\delta\ 2.6\ (s,\ 3\mathrm{H})$, $\delta\ 7.5$--$8.0\ (m,\ 5\mathrm{H})$, $\delta\ 9.9\ (s,\ 2\mathrm{H})$. MS: $\mathrm{M}^+ = 134$. Identify the compound. [Hard]

6. (a) Assign the CIP priority to each group on the stereocentre of 3-bromopentan-2-ol. (b) Determine the absolute configuration. (c) Draw both enantiomers and label them R and S. (d) Would a racemic mixture of this compound be optically active? [Medium]

7. Compare and contrast the structures and properties of low-density polyethene (LDPE) and high-density polyethene (HDPE). Reference the branching of polymer chains, intermolecular forces, density, melting point, and typical applications. [Medium]

8. Deduce the structure of a compound $\mathrm{C}_7\mathrm{H}_{14}\mathrm{O}$ given: IR: $3400\mathrm{ cm}^{-1}$ (broad), $2950\mathrm{ cm}^{-1}$ (sharp), no C=O absorption. $^{1}\mathrm{H}$ NMR: $\delta\ 0.9\ (t,\ 6\mathrm{H})$, $\delta\ 1.4\ (m,\ 4\mathrm{H})$, $\delta\ 1.6\ (s,\ 2\mathrm{H})$, $\delta\ 2.4\ (t,\ 2\mathrm{H})$. The signal at $\delta\ 1.6$ disappears upon $\mathrm{D}_2\mathrm{O}$ addition. [Hard]

---

If You Get These Wrong, Revise:

• VSEPR geometry and molecular shapes → Review ./chemical-bonding-advanced
• IR, MS, and NMR spectroscopy → Review ./measurement-and-data-processing
• Electron configurations and orbital theory → Review ./atomic-theory
• Stoichiometry of polymer reactions → Review ./organic-chemistry/organic-chemistry