Organic functional groups and formulation
Propanal and propanone are both C₃H₆O. You can show a molecule as a molecular formula (C₃H₆O), a condensed structural formula (CH₃CH₂CHO), an expanded (displayed) structure showing every bond, or a 3D model. 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.
Unsaturated hydrocarbons have double or triple bonds that determine their reactivity and IUPAC nomenclature. a) Write the semi-structural formula of the hydrocarbon…
Unsaturated hydrocarbons have double or triple bonds that determine their reactivity and IUPAC nomenclature.
- a) Write the semi-structural formula of the hydrocarbon named 4-methylpent-2-yne.
- b) Justify, with reasoning, whether this compound can show geometric isomerism (cis-trans).
Structural isomers
Structural isomers have the same molecular formula but different arrangements of atoms. There are three common types: chain isomers (different carbon skeleton, e.g. butane and 2-methylpropane), position isomers (same group in a different place, e.g. 1-chloropropane and 2-chloropropane) and functional group isomers (different groups, e.g. propanal and propanone). Cis-trans isomers, as in but-2-ene, have the same bonding but a different arrangement around a C=C.
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.