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.

1. Identify the longest carbon chain (parent chain)
2. Number the chain to give the lowest possible locants to substituents
3. Name substituents alphabetically with their position numbers
4. 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.

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)

(c) Alcohols: butan-1-ol, butan-2-ol, 2-methylpropan-1-ol, 2-methylpropan-2-ol

(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)

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.

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.

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

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

Distinguishing tests:

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 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

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 $S_N1$.

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).

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.

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_3And \mathrm{COOH. This makes it a chiral centre. The two enantiomers are non-superimposable mirror Images that rotate plane-polarised light in opposite directions.

1. Markovnikov’s Rule: The hydrogen adds to the carbon with MORE hydrogens, not fewer.

2. Oxidation of alcohols: Primary alcohols can be oxidised to aldehydes (with distillation) or carboxylic acids (with reflux). Use the correct conditions.

3. IUPAC numbering: Always number from the end that gives the lowest locants, not from the left.

4. $S_N1$ vs. $S_N2$: Primary halogenoalkanes favour $S_N2$; tertiary favour $S_N1$.

5. Stereoisomerism: E/Z isomerism requires two different groups on each carbon of the double bond.

6. Ester naming: The alkyl part of the ester name comes from the alcohol (not the acid).

7. Polymer vs. Monomer: The repeating unit of a condensation polymer is NOT the same as the monomer. Be careful to show the bonds correctly.

1. 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.

2. Draw the structural isomers of \mathrm{C_4\mathrm{H_8\mathrm{O that are (a) aldehydes, (b) ketones, (c) alcohols, (d) ethers.

3. Write the mechanism for the addition of \mathrm{Br_2 to ethene, showing the bromonium ion intermediate.

4. Explain why 2-bromo-2-methylpropane undergoes $S_N1$ hydrolysis faster than 1-bromopropane.

5. Write balanced equations for the complete combustion of (a) propane and (b) cyclohexane.

6. A student oxidises propan-1-ol using acidified potassium dichromate under reflux. Write the equation and name the organic product.

7. Draw the repeating unit of the polyester formed from benzene-1,4-dicarboxylic acid and ethane-1,2-diol.

8. Explain why the compound \mathrm{CH_3\mathrm{CH(\mathrm{OH)\mathrm{COOH is optically active, and draw both enantiomers.

9. Describe the mechanism of the electrophilic addition of \mathrm{Br_2 to propene, explaining why 1,2-dibromopropane is the major product.

10. Draw and name all structural isomers of \mathrm{C_4\mathrm{H_9\mathrm{Cl.

11. Explain the difference between addition and condensation polymerisation, giving one example of each.

12. Explain why butan-2-ol is optically active but butan-1-ol is not.

13. 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).

14. Describe how you would distinguish experimentally between pentan-2-one and pentanal using simple chemical tests.

15. For the reaction of 2-methylpropan-2-ol with HCl, explain whether $S_N1$ or $S_N2$ is the dominant mechanism and why.

16. 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.

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.

• 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.

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.

Key absorptions for functional group identification:

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.

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^+)

• 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:

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

To assign E/Z to alkenes:

1. On each carbon of the double bond, assign priority to the two substituents using atomic number (higher atomic number = higher priority).
2. If the two higher-priority groups are on the same side of the double bond, the isomer is Z (zusammen, together).
3. 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.

E and Z isomers have different physical properties:

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

• 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.

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.

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:

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