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

Organic functional groups and formulation

Propanal and propanone are both C₃H₆O. The atoms are the same; the difference is how they are joined and which functional group they carry: one is an aldehyde, the other a ketone. Spotting these groups and naming them without ambiguity is the first step in all of organic chemistry.

Functional groups

Table of general formulae of common organic functional groups and their decreasing IUPAC order of priority. haloalkane R–X alcohol R–OH ether (ext.) R–O–R' amine (ext.) R–NH₂ aldehyde R–CHO ketone R–CO–R' carboxylic acid R–COOH ester R–COO–R' amide R–CO–NH₂ nitrile: R–C≡N decreasing IUPAC priority (highest first) acid ester amide nitrile aldehyde ketone alcohol amine ether
Practice★☆☆Not started
Consider the following two oxygenated organic compounds with three carbon atoms: propanal and propanone. a) Write the semi-structural formula of both compounds and state…

Consider the following two oxygenated organic compounds with three carbon atoms: propanal and propanone.

  1. a) Write the semi-structural formula of both compounds and state the functional group to which each belongs.
  2. b) Explain what type of structural isomerism exists between them.
Solve

Hydrocarbons and haloalkanes

Skeletal formula of 3-chloropentane showing the numbering of the five carbon atoms and the position of the chlorine substituent. C1 C2 C3 C4 C5 Cl
Practice★☆☆Not started
The organic compound with semi-structural formula $\mathrm{CH_3-CH_2-CH(Cl)-CH_2-CH_3}$ is an open-chain halogenated derivative. a) Write the systematic name of this…

The organic compound with semi-structural formula $\mathrm{CH_3-CH_2-CH(Cl)-CH_2-CH_3}$ is an open-chain halogenated derivative.

  1. a) Write the systematic name of this compound according to IUPAC rules.
  2. b) Write the formula and name of a positional isomer belonging to the same functional group that keeps the same linear carbon chain.
Solve

Oxygen-containing compounds: alcohols to esters

An ester R–COO–R' splits on hydrolysis into a carboxylic acid R–COOH and an alcohol R'–OH. ester R–COO–R' hydrolysis acid R–COOH + alcohol R'–OH
Practice★★☆Not started
The following figure shows the molecular structure of an open-chain oxygenated organic compound with explicit heteroatomic bonds: a) Identify, with justification, the…

The following figure shows the molecular structure of an open-chain oxygenated organic compound with explicit heteroatomic bonds:

H₃C OH OH O
  1. a) Identify, with justification, the two oxygen-containing functional groups present in the molecule and state which of them has higher priority as the principal group according to IUPAC.
  2. b) Assign the correct numbering to the main chain and write the systematic name of the compound according to the official IUPAC rules.
Solve
Practice★★☆Not started
An aliphatic ester of molecular formula $\mathrm{C_4H_8O_2}$ undergoes complete acid hydrolysis, giving a carboxylic acid with two carbon atoms and an alcohol. a) Write…

An aliphatic ester of molecular formula $\mathrm{C_4H_8O_2}$ undergoes complete acid hydrolysis, giving a carboxylic acid with two carbon atoms and an alcohol.

  1. a) Write the semi-structural formula and the systematic IUPAC name of the starting ester, as well as of the two hydrolysis products.
  2. b) Write the semi-structural formula and the name of a straight-chain primary amide with four carbon atoms. Compare its molecular formula with that of the ester and justify why they are not isomers.
Solve

From percentage composition to formula

Analysing a pure compound gives its percentage composition by mass. This fixes the ratio of atoms but leaves many possible structures. Tests can help to choose between them: an aldehyde is easily oxidised to a carboxylic acid (it gives a red precipitate with Fehling's solution), whereas a ketone is not easily oxidised and gives no reaction. So a carbonyl compound that is not easily oxidised to an acid is a ketone. How do we get the molecular formula, and how many double bonds or rings it has, before drawing anything?

Flow from percentage composition and molar mass to the empirical formula and then the molecular formula. % by mass + molar mass empirical formula × k molecular formula
Practice★★★Not started
An organic compound $A$ made up exclusively of carbon, hydrogen and oxygen has a mass percentage of $66.63\,\%$ of $\mathrm{C}$ and of $11.18\,\%$ of $\mathrm{H}$. Its…

An organic compound $A$ made up exclusively of carbon, hydrogen and oxygen has a mass percentage of $66.63\,\%$ of $\mathrm{C}$ and of $11.18\,\%$ of $\mathrm{H}$. Its molar mass is $72.10\,\text{g}\times\text{mol}^{-1}$. The compound is known to contain a carbonyl group that is not easily oxidised to a carboxylic acid and to have no additional $\mathrm{C=C}$ bonds.

(Data: relative atomic masses $\mathrm{C}=12.01$; $\mathrm{H}=1.008$; $\mathrm{O}=16.00$).

  1. a) Determine the empirical formula and the molecular formula of substance $A$.
  2. b) Deduce, with justification, the degree of unsaturation (number of equivalent unsaturations) and propose the semi-structural formula and the systematic IUPAC name for $A$. To calculate the unsaturations you may use $I=(2C+2+N-H-X)/2$, where $X$ is the number of halogens; oxygen does not change this expression.
Solve
Practice★★★Not started
Consider a polyfunctional compound of molecular formula $\mathrm{C_6H_9NO_3}$ whose structure has an open carbon chain with the following characteristics: It contains a…

Consider a polyfunctional compound of molecular formula $\mathrm{C_6H_9NO_3}$ whose structure has an open carbon chain with the following characteristics:

  • It contains a carboxyl group as the principal functional group.

  • It has an amino group and a carbon-carbon double bond.

  • It has a ketone function in a non-terminal position.

  1. a) Calculate the number of unsaturations (double-bond equivalents or rings) present in the molecule.
  2. b) Propose a semi-structural formula consistent with all the requirements, in which the main chain has 5 carbons with one methyl substituent, and assign the complete systematic IUPAC name, stating the priority of the functional groups. To calculate the unsaturations you may use $I=(2C+2+N-H-X)/2$, where $X$ is the number of halogens; oxygen does not change this expression.
Solve
Practice★★★Not started
A hydrocarbon $X$ of molecular formula $\mathrm{C_7H_8}$ contains a single benzene ring and undergoes vigorous oxidation with $\mathrm{KMnO_4}$ in acidic medium,…

A hydrocarbon $X$ of molecular formula $\mathrm{C_7H_8}$ contains a single benzene ring and undergoes vigorous oxidation with $\mathrm{KMnO_4}$ in acidic medium, yielding an aromatic monocarboxylic acid $Y$ ($\mathrm{C_7H_6O_2}$).

  1. a) Determine the degree of unsaturation of $X$, deduce its structure and write its systematic IUPAC name.
  2. b) Write the semi-structural formula and the IUPAC name of two derivatives of $Y$: an acid chloride and an aromatic primary amide. To calculate the unsaturations you may use $I=(2C+2+N-H-X)/2$, where $X$ is the number of halogens; oxygen does not change this expression.
Solve

Nitrogen compounds: amines, amides, nitriles

Do not confuse a nitrogen joined only to saturated carbons with one joined to a carbonyl carbon. Look at which atom the N is bonded to.

Comparison of an amine, an amide and a nitrile with one example of each. AmineCH₃–NH₂ AmideCH₃–C(=O)–NH₂ NitrileCH₃–C≡N
Family Group Example and name
Amine $\mathrm{RNH_2}$, $\mathrm{R_2NH}$ or $\mathrm{R_3N}$ $\mathrm{CH_3NH_2}$: methanamine
Amide $\mathrm{RCONH_2}$ $\mathrm{CH_3CONH_2}$: ethanamide
Nitrile $\mathrm{RC\equiv N}$ $\mathrm{CH_3CN}$: ethanenitrile (the nitrile carbon counts in the chain)
Practice★★☆Not started
Amines and aromatic compounds are widely used in chemical synthesis. a) Write the formulae and names of two different aromatic amines that contain a benzene ring and…

Amines and aromatic compounds are widely used in chemical synthesis.

  1. a) Write the formulae and names of two different aromatic amines that contain a benzene ring and have molecular formula $\mathrm{C_7H_9N}$, one being primary and the other secondary.
  2. b) State, with justification, the order of priority between the amine group and the alcohol group according to IUPAC rules for naming a compound that contains both.
Solve

Back to the start: propanal, $\mathrm{CH_3CH_2CHO}$, has the C=O group on the end carbon (an aldehyde), while propanone, $\mathrm{CH_3COCH_3}$, has it on the middle carbon (a ketone). Same formula, different functional group: they are functional group isomers.

Summary

See all the exercises in this section