U4AOS1 Topic 4: Organic reactions

Organic reactions

Substitution reaction

- All substitution reactions occur with saturated molecules (alkanes)

Halogenation

- Substitute H for a halogen using pure halogen. Eg Cl₂(g), Br₂(l) using UV light as catalyst

- For each reaction that occurs, one hydrogen is replaced with a halogen and hydrogen halide byproduct is formed (HX where X is halogen), therefore reactions can be chained and all hydrogens can be replaced by the halogen

Eg. Cl₂ + C₂H₆ –(UV light)-> C₂H₅Cl + HCl

Amine substitution

- Substitute halogen position from haloalkane with ammonia to form amine

- ammonia breaks apart into NH₂ and H where NH₂ substitutes with halogen on molecule whilst H forms a hydrogen halide byproduct with the newly substituted halogen

Eg. CH₃Cl + NH₃ àCH₃NH₂+ HCl

Alcohol substitution

- use either KOH and NaOH (as Na⁺ and K⁺ are weak oxidants from electrochemical series) where OH^- attacks the halogen position and substitutes to form hydroxyl group

Note: states not required for writing out substitution reactions

Author’s tip: With substitution reactions, think that a haloalkane is required for most substitutions because halogen to carbon covalent bonds are weak, a weakening of the intramolecular bonds are required for the substitution to occur as hydrogen to carbon covalent bonds are too strong.

The only substitution reaction that occurs with hydrogen is halogenation which requires UV light whilst the other substitutions with the halogen in the haloalkane require no catalysts due to the ‘weaker’ bond.

Addition reaction

- All addition reactions occur with unsaturated molecules (alkenes)

- All addition reactions involve the splitting of the added reactant to form bonds at each adjacent carbon at C=C to remove the double carbon to carbon covalent bonds

- for non-symmetrical molecules added eg. HCl, there is an assumed 50:50 chance of H and Cl landing on each C=C carbon

- for symmetrical molecules eg. Pure halogen such as Cl₂, there is only one possible product produced due to product symmetry

For example:

CH₂CHCH₃ + HCl àCH₃CHClCH₃ or CH₂ClCH₂CH₃

As you can see, the H and the Cl can go in either C=C positions, it never goes to the same carbon, always two different carbons

Amine addition

- Addition reaction with ammonia and alkene where ammonia splits to produce H on one carbon at the C=C and NH₂ on the other carbon at C=C bond, enabling the formation of an amine

Eg. CH₂CHCH₃ + NH₃ àCH₂NH₂CH₂CH₃ or CH₃CHNH₂CH₃

Hydrogenation

- Add H₂ to alkene to form alkane with the presence of Ni(s) catalyst

Eg. H₂(g) + C₂H₄(g) –(Ni(s))->C₂H₆(g) with Ni(s) catalyst

Hydration

- Add water and phosphoric acid (H₃PO₄) at 300°C to immediately form an alcohol

- Water is a gas (over 100°C) and phosphoric acid is a solid

- H₂O splits off into H and OH to attach to each carbon at C=C

Eg. CH₂CHCH₃ –(H₂O(g) and H₃PO₄(s))->CH₂OHCH2CH₃ or CH₃CHOHCH₃

Oxidation reaction

Oxidation reaction is called oxidation as linked to redox reactions from unit 3, the carbon is oxidized thereby losing electrons and increasing in oxidation number in the presence of an oxidant.

The oxidant used in this case is acidified dichromate (H⁺(aq)/Cr₂O₇^2-(aq)) or acidified permanganate (H⁺(aq)/MnO₄^-(aq)). Either one is perfectly acceptable therefore you only need to stick with one of them.

For the following examples, assume acidified dichromate and acidified permanganate is written on top of the reaction arrow.

Key thing to note is that acidified conditions are required therefore don’t be surprised if an acid is added for an oxidation reaction to occur.

The oxidation reaction for both acidified dichromate and acidified permanganate are on databook page 2 in the electrochemical series therefore doesn’t need to be memorized.

- primary alcohol àaldehyde àcarboxylic acid

Here is an example half reaction for subsequent reactions using ethanol as starting molecule:

Primary alcohol to aldehyde – partial oxidation

CH₃CH₂OH –(H⁺(aq)/Cr₂O₇^2-(aq))->2H⁺ + 2e^- + CH₃CHO

Aldehyde to carboxylic acid

CH₃CHO + H₂O –(H⁺(aq)/Cr₂O₇^2-(aq))->CH₃COOH + 2H⁺ + 2e^-

Now using propan-2-ol as the starting molecule (secondary alcohol)

Primary alcohol straight to carboxylic acid (full oxidation)

CH₃CH₂OH + H₂O –(H⁺(aq)/Cr₂O₇^2-(aq))-> CH₃COOH + 4H⁺ + 4e^-

Secondary alcohol to ketone

CH₃CHOHCH₃ –(H⁺(aq)/Cr₂O₇^2-(aq))->CH₃COCH₃ + 2H⁺ + 2e^-

Note: Tertiary alcohols such as methylpropan-2-ol do not oxidise (act inert)

Easy way to remember is that the functional group is being substituted out, must be primary alcohol as hydroxyl group is on the end of the molecule which is required position for carbonyl group for aldehyde and carboxyl group for carboxylic acid.

Note: when not specified, always assume oxidation of primary alcohol is a complete oxidation, meaning that aldehyde intermediary product is skipped (alcohol straight to carboxylic acid)

Authors note: When asked what reactants are used to form carboxylic acid or if question asks what product(s) are formed from oxidation of primary alcohol, reference the aldehyde

Secondary alcohol àketone

Secondary alcohols have hydroxyl group in the middle of the chain therefore position substituted out for a carbonyl group in the middle of the chain (ketone)