Reactions of Organic Compounds

What is it?

Organic reactions look complicated, but almost all of them fall into a few types. Once you know the functional group in a molecule, you can predict which type of reaction it undergoes and what the product will be.

Type	What happens	Typical example
Addition	two molecules join to make one; a C=C becomes C–C	alkene + $\ce{Br2}$
Substitution	one atom or group swaps for another	alkane + $\ce{Cl2}$
Elimination	a small molecule (often $\ce{H2O}$ ) is removed, forming C=C	alcohol → alkene
Oxidation	gain of O or loss of H	alcohol → aldehyde → acid
Condensation	two molecules join and release a small molecule ( $\ce{H2O}$ )	acid + alcohol → ester

Key idea

The functional group decides the reaction. Alkenes add across the C=C; alkanes and haloalkanes substitute; alcohols can be oxidized, eliminated (dehydrated) or condensed with acids. Learn one example of each and you can apply it to the whole family.

Why does it matter?

Making things. Medicines, plastics, dyes and fuels are made by chains of these reactions, converting one functional group into another.
Everyday chemistry. Margarine is made by adding hydrogen to vegetable oils; wine turns to vinegar by oxidation; the breathalyser test once relied on dichromate turning green.
Analysis. Simple colour tests (bromine water, acidified dichromate) tell you which functional groups a compound contains.

How does it work?

1. Addition reactions of alkenes

The C=C double bond opens, and one atom or group adds to each carbon:

Hydrogen (with a nickel catalyst, about 150 °C): $\ce{CH2=CH2 + H2 -> CH3CH3}$
Bromine: $\ce{CH2=CH2 + Br2 -> CH2BrCH2Br}$ (1,2-dibromoethane)
Hydrogen bromide: $\ce{CH2=CH2 + HBr -> CH3CH2Br}$
Water (steam, acid catalyst): $\ce{CH2=CH2 + H2O -> CH3CH2OH}$ , the industrial route to ethanol.

Test for C=C: shake with orange-brown bromine water. An alkene decolourises it (the bromine adds across the double bond); an alkane does not.

2. Substitution reactions

Alkanes + halogens in ultraviolet light: $\ce{CH4 + Cl2 -> CH3Cl + HCl}$ . A hydrogen is swapped for a chlorine. Further substitution gives a mixture ( $\ce{CH2Cl2}$ , $\ce{CHCl3}$ , $\ce{CCl4}$ ).
Haloalkanes + hydroxide ions (warm, aqueous): $\ce{CH3CH2Br + OH- -> CH3CH2OH + Br-}$ . The halogen is replaced by –OH, giving an alcohol.

3. Elimination: dehydrating alcohols

Heating an alcohol with concentrated sulfuric acid (or passing its vapour over hot aluminium oxide) removes water and forms an alkene:

\ce{CH3CH2OH -> CH2=CH2 + H2O}

4. Oxidation of alcohols

Warming with acidified potassium dichromate oxidizes alcohols; the orange dichromate turns green as it is reduced to $\ce{Cr^3+}$ . Writing [O] for the oxidizing agent:

Primary alcohol (–OH on an end carbon) → aldehyde (if distilled off at once) → carboxylic acid (if heated under reflux):

\begin{aligned} &\small \ce{CH3CH2OH + [O] -> CH3CHO + H2O} \\[2pt] &\small \ce{CH3CHO + [O] -> CH3COOH} \end{aligned}

Secondary alcohol → ketone: $\ce{CH3CH(OH)CH3 + [O] -> CH3COCH3 + H2O}$
Tertiary alcohol (–OH on a carbon with no hydrogen): not oxidized, so the dichromate stays orange.

5. Condensation

A carboxylic acid and an alcohol, with an acid catalyst, condense to an ester and water (see Functional Groups). Repeating condensation many times builds polymers such as polyesters and nylon.

Think of it like this

Think of the reaction types as moves in a game of building blocks. Addition snaps a new piece onto an open slot (the C=C). Substitution swaps one piece for another. Elimination pulls two pieces off and leaves a new open slot. Condensation clicks two models together and drops a small spare piece (water).

More precisely

When an unsymmetrical reagent such as HBr adds to an unsymmetrical alkene such as propene, two products are possible. The major product follows Markovnikov’s rule: the hydrogen adds to the carbon that already has more hydrogens, giving mostly 2-bromopropane. Substitution of haloalkanes is a nucleophilic substitution: the electron-rich $\ce{OH-}$ attacks the carbon bonded to the electronegative halogen. Oxidation of alcohols is a redox reaction: carbon’s oxidation number rises while chromium’s falls from +6 to +3.

Visualise it

A map of organic reactions linking eight families. Alkane goes to haloalkane with Cl2 and UV light (substitution). Alkene goes to alkane with H2 and Ni, to haloalkane with HBr, and to alcohol with H2O and H+ (all addition). Haloalkane goes to alcohol with OH− (substitution). Alcohol goes back to alkene with concentrated H2SO4 and heat (elimination). Alcohol is oxidized: a primary alcohol to aldehyde and then to carboxylic acid, and a secondary alcohol to ketone. Carboxylic acid plus alcohol with H+ gives an ester (condensation). — Functional groups are linked by a handful of reaction types; colours show the type.

Worked example

Worked example: Predicting products

Question: Give the product, its name and the type of reaction: (a) propene + bromine (b) ethanol heated with concentrated sulfuric acid (c) propan-2-ol warmed with acidified potassium dichromate.

(a) Bromine adds across the C=C: $\ce{CH2=CHCH3 + Br2 -> CH2BrCHBrCH3}$ , 1,2-dibromopropane; addition.
(b) Water is removed: $\ce{CH3CH2OH -> CH2=CH2 + H2O}$ , ethene; elimination (dehydration).
(c) A secondary alcohol is oxidized to a ketone: propanone, $\ce{CH3COCH3}$ ; oxidation (orange to green).

Worked example: Telling an alkane from an alkene

Question: Two colourless liquids are hexane and hex-1-ene. How can you tell them apart?

Add a few drops of orange-brown bromine water to each and shake.
Hex-1-ene decolourises it: $\ce{C6H12 + Br2 -> C6H12Br2}$ (addition across the C=C, giving 1,2-dibromohexane).
Hexane has no C=C, so the orange colour stays (substitution needs UV light).

Worked example: Mass of bromine that reacts

Question: What mass of bromine reacts completely with 5.00 g of hex-1-ene? ( $\ce{C6H12}$ = 84.16 g/mol; $\ce{Br2}$ = 159.80 g/mol)

Equation: $\ce{C6H12 + Br2 -> C6H12Br2}$ (1 : 1)
$n(\ce{C6H12}) = \dfrac{5.00\ \text{g}}{84.16\ \text{g/mol}} = 0.05941\ \text{mol}$
$n(\ce{Br2}) = 0.05941\ \text{mol}\ \ce{C6H12} \times \dfrac{1\ \text{mol}\ \ce{Br2}}{1\ \text{mol}\ \ce{C6H12}} = 0.05941\ \text{mol}$
$m(\ce{Br2}) = 0.05941\ \text{mol} \times 159.80\ \text{g/mol} =$ 9.49 g

Common mistake

Common mistake: Expecting alkanes to react with bromine water

Alkanes have no C=C, so they cannot undergo addition. They react with halogens only by substitution, and only in ultraviolet light. A colour change with bromine water in the dark means a C=C is present.

Common mistake: Oxidizing a tertiary alcohol

A tertiary alcohol has no hydrogen on the carbon bearing the –OH, so acidified dichromate cannot oxidize it: the solution stays orange. Only primary alcohols give aldehydes and acids; secondary alcohols give ketones.

Common mistake: Confusing addition and substitution products

Addition puts both parts of the reagent into one product ( $\ce{CH2BrCH2Br}$ ). Substitution gives two products, because the atom that is replaced leaves ( $\ce{CH3Cl + HCl}$ ).

Notation note

[O] means “oxygen from an oxidizing agent”; [H] is sometimes used for a reducing agent.
Conditions are written over the arrow in many books, for example UV light, Ni catalyst or reflux.
Primary, secondary and tertiary (1°, 2°, 3°) describe how many carbons are bonded to the carbon carrying the –OH: one, two or three.

Remember this

Addition (alkenes): + H₂ (Ni), + Br₂, + HBr, + H₂O (H⁺). Bromine water decolourised = C=C present.
Substitution: alkane + Cl₂ (UV); haloalkane + OH⁻ → alcohol.
Elimination: alcohol + conc. H₂SO₄, heat → alkene + water.
Oxidation (acidified dichromate, orange → green): 1° alcohol → aldehyde → acid; 2° → ketone; 3° → no reaction.
Condensation: acid + alcohol ⇌ ester + water; repeated, it makes polymers.

Practise this topic with flashcards and a quiz at chemistryclarity.com/chemistry/organic-reactions/