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
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.
- a) Write the semi-structural formula of both compounds and state the functional group to which each belongs.
- b) Explain what type of structural isomerism exists between them.
Hydrocarbons and haloalkanes
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.
- a) Write the systematic name of this compound according to IUPAC rules.
- b) Write the formula and name of a positional isomer belonging to the same functional group that keeps the same linear carbon chain.
Oxygen-containing compounds: alcohols to esters
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:
- 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.
- b) Assign the correct numbering to the main chain and write the systematic name of the compound according to the official IUPAC rules.
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.
- a) Write the semi-structural formula and the systematic IUPAC name of the starting ester, as well as of the two hydrolysis products.
- 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.
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?
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 $\mathrm{C}=12.01$; $\mathrm{H}=1.008$; $\mathrm{O}=16.00$).
- a) Determine the empirical formula and the molecular formula of substance $A$.
- 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.
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.
- a) Calculate the number of unsaturations (double-bond equivalents or rings) present in the molecule.
- 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.
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 $Y$ ($\mathrm{C_7H_6O_2}$).
- a) Determine the degree of unsaturation of $X$, deduce its structure and write its systematic IUPAC name.
- 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.
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.
| 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) |
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.
- 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.
- 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.
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.