Back to homeChemistry · 6th Year HL · Change subject
Start for free
6th Year HL · Chemistry
Organic chemistry · Organic structure and reactivity

Main organic reactions

Imagine a chemist at a local co-operative making soap from fat, or a student in the school lab turning ethanol into a fruity-smelling ester. Neither is memorising hundreds of separate reactions. In Leaving Cert Chemistry, almost every organic reaction belongs to one of five types: addition, substitution, elimination, redox and acid-base. Once you can spot the type from the functional group and the conditions, you can predict the product.

The five reaction types

Summary of the five organic reaction types with a one-line example of each Addition CH₂=CH₂ + HCl → CH₃CH₂Cl Substitution CH₃Br + OH⁻ → CH₃OH + Br⁻ Elimination CH₃CH₂OH → CH₂=CH₂ + H₂O Redox CH₃CH₂OH → CH₃CHO (oxidation) Acid-base CH₃COOH + NaOH → CH₃COONa + H₂O

Addition to alkenes

Propene reacts with hydrogen bromide and almost all the product is 2-bromopropane, with only traces of 1-bromopropane. Why does the reagent choose one carbon of the double bond rather than splitting 50:50?

The π bond is a region of high electron density, so an alkene attracts electron-poor species (electrophiles). That is why alkenes undergo electrophilic addition.

Mechanism: ionic addition of HCl to ethene

You must be able to describe this mechanism using curved arrows. A curved arrow shows the movement of a pair of electrons.

Two-step mechanism of ionic addition of hydrogen chloride to ethene with curved arrows Step 1: H–Cl is polar (Hδ⁺–Clδ⁻) The π electrons attack the Hδ⁺: CH₂=CH₂ + H–Cl → CH₃–CH₂⁺ + Cl⁻ Step 2: Cl⁻ uses a lone pair to bond to the carbocation: CH₃–CH₂⁺ + Cl⁻ → CH₃–CH₂Cl
  1. The π bond breaks heterolytically: its electron pair forms a new C–H bond with the δ⁺ hydrogen.

  2. The H–Cl bond breaks and both electrons go to chlorine, giving Cl⁻ and a carbocation.

  3. The lone pair on Cl⁻ forms a bond to the positive carbon.

The same pattern works for Cl₂ with ethene (the first step makes a Cl⁺-like attack and Cl⁻), giving 1,2-dichloroethane.

Practice★☆☆Not started
Propene ($\mathrm{CH_3-CH=CH_2}$) undergoes an electrophilic addition reaction when treated with water in the presence of dilute sulfuric acid as catalyst, following…

Propene ($\mathrm{CH_3-CH=CH_2}$) undergoes an electrophilic addition reaction when treated with water in the presence of dilute sulfuric acid as catalyst, following Markovnikov's rule (the proton is added preferentially to the carbon with the greater number of hydrogens).

  1. a) Write the formula and name of the starting reagent and of the major organic product obtained in this hydration.
  2. b) Justify, with reasoning, why propan-1-ol is not obtained as the major product.
Solve
Practice★★★Not started
It is desired to synthesise propanone in the laboratory by a two-step sequence starting from propene. In step 1, propene is hydrated in the presence of aqueous sulfuric…

It is desired to synthesise propanone in the laboratory by a two-step sequence starting from propene. In step 1, propene is hydrated in the presence of aqueous sulfuric acid following Markovnikov's rule, giving a secondary alcohol as the major product. In step 2, this alcohol is purified and oxidised with potassium dichromate in acidic medium to obtain propanone.

  1. a) Write the equations for both steps, classifying each reaction, with justification, according to the type of organic process (addition, substitution, elimination, oxidation-reduction).
  2. b) If you start with $84.0\,\text{g}$ of propene and step 1 has a yield of $75.0\,\%$ while step 2 has a yield of $80.0\,\%$, calculate the final mass of pure propanone obtained. Atomic masses: $\mathrm{C}=12.0$; $\mathrm{H}=1.0$; $\mathrm{O}=16.0$.
Solve

Substitution and elimination in haloalkanes

Mechanism: free-radical substitution of methane

Alkanes have no π bond and are not polar, so they do not react with ionic reagents. With chlorine and UV light they undergo free-radical substitution. Here the bond breaks homolytically, one electron going to each atom, shown with a fishhook (half-headed) arrow.

Stage Equation
Initiation Cl₂ → 2Cl• (UV light splits the Cl–Cl bond)
Propagation 1 CH₄ + Cl• → CH₃• + HCl
Propagation 2 CH₃• + Cl₂ → CH₃Cl + Cl•
Termination Cl• + Cl• → Cl₂; CH₃• + Cl• → CH₃Cl; CH₃• + CH₃• → C₂H₆

The overall equation is CH₄ + Cl₂ → CH₃Cl + HCl. Ethane reacts in exactly the same way, and further substitution gives CH₂Cl₂ and so on.

Practice★☆☆Not started
Alkyl halides can undergo selective transformations depending on the nature of the reagent and the solvent used. a) Write the balanced equation for the nucleophilic…

Alkyl halides can undergo selective transformations depending on the nature of the reagent and the solvent used.

  1. a) Write the balanced equation for the nucleophilic substitution reaction between 1-bromopropane ($\mathrm{CH_3-CH_2-CH_2Br}$) and a dilute aqueous solution of sodium hydroxide at moderate temperature, naming the organic product formed.
  2. b) State the semi-structural formula and the name of the product that would be obtained if the propene formed in a side reaction were subjected to catalytic hydrogenation with $\mathrm{H_2}$ in the presence of nickel.
Solve

Oxidation of alcohols and combustion

Oxidation routes: primary alcohol to aldehyde to carboxylic acid, secondary alcohol to ketone primary alcohol R–CH₂OH [O], distil R–CHO [O], excess, reflux R–COOH secondary alcohol R–CH(OH)–R' R–CO–R' (ketone) tertiary alcohol: no oxidation

Telling the products apart

Test Aldehyde Ketone Carboxylic acid
Fehling's solution, warm blue → red-brown precipitate (Cu₂O) no change no change
Sodium carbonate / hydrogencarbonate no reaction no reaction fizzes (CO₂)
Magnesium no reaction no reaction fizzes (H₂)

Carboxylic acids also react with NaOH in an acid-base reaction: CH₃COOH + NaOH → CH₃COONa + H₂O. Bromine water (or acidified KMnO₄) is decolourised by alkenes, which tests for unsaturation. Aldehydes and ketones can be reduced back to alcohols with H₂/Ni.

Combustion

Complete combustion of a hydrocarbon or alcohol gives CO₂ and H₂O and releases heat. Limited oxygen gives CO or soot (carbon).

Practice★☆☆Not started
Complete combustion of alcohols produces carbon dioxide and water vapour, releasing a considerable amount of heat energy. a) Write and balance, using whole-number…

Complete combustion of alcohols produces carbon dioxide and water vapour, releasing a considerable amount of heat energy.

  1. a) Write and balance, using whole-number stoichiometric coefficients, the chemical equation for the complete combustion of ethanol ($\mathrm{CH_3-CH_2OH}$).
  2. b) Calculate the volume of carbon dioxide, measured at $25\,^{\circ}\text{C}$ and $1.0\,\text{atm}$, released when $23.0\,\text{g}$ of ethanol is burned completely. Data: relative atomic masses: $\mathrm{C}=12.0$; $\mathrm{H}=1.0$; $\mathrm{O}=16.0$; gas constant $R = 0.082\,\text{atm}\times\text{L}\times\text{mol}^{-1}\times\text{K}^{-1}$.
Solve
Practice★★☆Not started
The chemical behaviour of a primary alcohol towards a strong oxidising agent such as potassium dichromate in acidic medium depends on the experimental conditions used.…

The chemical behaviour of a primary alcohol towards a strong oxidising agent such as potassium dichromate in acidic medium depends on the experimental conditions used. Observe the following transformation scheme

Ethanol + K₂Cr₂O₇ / H⁺ Heating (Δ) Immediate distillation (continuous removal of vapour) Product A ¿ ? Prolonged reflux (excess oxidant) Product B ¿ ?
  1. a) Identify, with justification, the semi-structural formulae and systematic names of the organic products $\mathrm{A}$ and $\mathrm{B}$.
  2. b) Justify why immediate distillation of the mixture allows compound $\mathrm{A}$ to be isolated selectively, whereas prolonged reflux with excess oxidant leads to $\mathrm{B}$.
Solve
Practice★★★Not started
A historical organic analysis report describes four unlabelled bottles ($\mathrm{A}$, $\mathrm{B}$, $\mathrm{C}$ and $\mathrm{D}$), each of which contains one of the…

A historical organic analysis report describes four unlabelled bottles ($\mathrm{A}$, $\mathrm{B}$, $\mathrm{C}$ and $\mathrm{D}$), each of which contains one of the following pure liquid substances: propanal, propanone, propanoic acid and hex-1-ene. To identify them, three independent chemical tests were recorded, which are analysed as data without reproducing them:

  1. Addition of a reddish solution of bromine in carbon tetrachloride: only substance $\mathrm{C}$ decolourises the solution quickly and immediately without evolution of gas.

  2. Addition of saturated aqueous sodium hydrogencarbonate solution ($\mathrm{NaHCO_3}$): only substance $\mathrm{D}$ produces clearly visible effervescence.

  3. Test with Tollens' reagent (ammoniacal solution of $\mathrm{Ag^+}$): substance $\mathrm{A}$ forms a silver mirror of metallic silver on the walls of the tube when gently heated, whereas $\mathrm{B}$ produces no visible change.

  1. a) Assign, with justification, each substance to the letters $\mathrm{A}$, $\mathrm{B}$, $\mathrm{C}$ and $\mathrm{D}$.
  2. b) Write the chemical equations corresponding to the positive tests described for $\mathrm{C}$ and $\mathrm{D}$.
Solve

Esterification and base hydrolysis

Warm ethanoic acid (sharp, vinegary smell) with ethanol and a few drops of concentrated sulfuric acid, and the smell changes to something fruity. Which reaction joins two families of compounds to make a new one and release water?

Interconversion between carboxylic acid plus alcohol and ester plus water by esterification and acid hydrolysis R–COOH + R'–OH H⁺, reflux H⁺, H₂O (hydrolysis) R–COO–R' + H₂O

Esters from fats and oils (triesters of propane-1,2,3-triol, glycerol) are hydrolysed by NaOH to give glycerol and the sodium salts of fatty acids. These salts are soap.

Practice★★☆Not started
In a closed reactor of $1.00\,\text{L}$ at a constant temperature of $25\,^{\circ}\text{C}$, $2.00\,\text{mol}$ of acetic acid (ethanoic acid) and $2.00\,\text{mol}$ of…

In a closed reactor of $1.00\,\text{L}$ at a constant temperature of $25\,^{\circ}\text{C}$, $2.00\,\text{mol}$ of acetic acid (ethanoic acid) and $2.00\,\text{mol}$ of ethanol are mixed in the presence of a catalytic amount of concentrated sulfuric acid.

  1. a) Write the esterification reaction, giving the names of the ester and of the inorganic by-product formed.
  2. b) Given that the dimensionless equilibrium constant for this esterification in solution is $K_c = 4.00$ at this temperature, calculate the amount of ester (in moles) present at equilibrium and the percentage yield of the process. Initially there is no ester or water. Assume an ideal medium of constant volume prepared with anhydrous reagents and, if necessary, an inert solvent; include the water formed in the equilibrium expression.
Solve
Practice★★☆Not started
Basic hydrolysis, or saponification, of ethyl acetate, $\mathrm{CH_3-COO-CH_2-CH_3}$, consumes a strong base. a) Write the molecular equation for its reaction with…

Basic hydrolysis, or saponification, of ethyl acetate, $\mathrm{CH_3-COO-CH_2-CH_3}$, consumes a strong base. a) Write the molecular equation for its reaction with aqueous sodium hydroxide solution and name the two organic products obtained, one of them in the form of a salt. b) $17.6\,\mathrm{g}$ of the ester is treated with $250\,\mathrm{mL}$ of $\mathrm{NaOH}$ $1.00\,\mathrm{mol}\times\mathrm{L}^{-1}$. Identify the limiting reagent and calculate the mass of sodium acetate isolated if the yield with respect to the stoichiometric maximum is $85.0\,\%$. Data: relative atomic masses, C = 12.0; H = 1.0; O = 16.0; Na = 23.0.

Solve
Practice★★★Not started
A route of three consecutive steps is designed to synthesise ethyl acetate from ethene ($\mathrm{CH_2=CH_2}$): Step 1: Acid hydration of ethene to obtain ethanol with a…

A route of three consecutive steps is designed to synthesise ethyl acetate from ethene ($\mathrm{CH_2=CH_2}$):

Step 1: Acid hydration of ethene to obtain ethanol with a yield of $80.0\,\%$.

Step 2: Vigorous oxidation under reflux of half of the moles of ethanol obtained in step 1 using excess $\mathrm{KMnO_4/H^+}$ to acetic acid, with a yield of $90.0\,\%$.

Step 3: Condensation (Fischer esterification) between the acetic acid obtained in step 2 and the other half of the ethanol set aside from step 1.

  1. a) Write the equations for the three steps, stating the usual conditions and catalysts.
  2. b) If the process is started with $112\,\text{g}$ of ethene and the yield of the final esterification (step 3) with respect to the limiting reagent is $70.0\,\%$, calculate the mass of ethyl acetate obtained. Atomic masses: $\mathrm{C}=12.0$; $\mathrm{H}=1.0$; $\mathrm{O}=16.0$.
Solve

Reduction and dehydration

Reaction schemes link everything together. For each arrow, give the reagent or conditions, name the product and check that the atoms balance.

Map linking carbonyl compounds, alcohols, alkenes and alkanes by reduction, dehydration and hydrogenation Aldehyde → primary alcohol (reduction) Ketone → secondary alcohol (reduction) Alcohol → alkene + H₂O (heat, conc. H₂SO₄) Alkene + H₂ → alkane (Ni catalyst) Check conditions, formulae and atom balance

Back to the start: to chain alkene → alcohol → acid → ester, hydrate ethene to ethanol (addition), oxidise it with acidified potassium manganate(VII) under reflux to ethanoic acid (redox), then heat the acid with more ethanol and concentrated H₂SO₄ to form ethyl ethanoate (esterification).

Summary

See all the exercises in this section