In the 1820s two young chemists in Germany ran into a puzzle that should not have existed. Friedrich Wöhler was studying silver cyanate, a quiet, well-behaved salt. Justus von Liebig was studying silver fulminate, a compound so touchy that a small quantity could detonate on being struck. When each of them carefully analysed his compound, the results came out the same: the same elements, present in the same proportions. By the chemical thinking of the time this was impossible. If two substances contained identical amounts of identical elements, they had to be the same substance — yet one sat harmlessly in a bottle while the other exploded.
At first each chemist suspected the other of sloppy analysis. Careful repetition showed both were right. The Swedish chemist Jöns Jacob Berzelius drew the conclusion that changed organic chemistry: composition alone does not decide what a substance is. What also matters is how the atoms are joined and arranged. Berzelius gave the phenomenon a name — isomerism. The formula told chemists what was inside a molecule; only the structure could tell them what the molecule would do. Two compounds of identical composition and very different behaviour are compared in Fig. 14.0.
As you work through this chapter, keep these five questions in mind — each of them will be answered as the ideas behind them are developed.
14.2 Structural Representation of Organic Molecules
- Draw the structural formula of ethane.
- Draw the electron-dot structure of propane.
In earlier standards you learnt how to write structural formulae and electron dot structures of hydrocarbons. Both are ways of putting a three-dimensional molecule on flat paper, and each shows a different amount of detail.
Electron dot structures
In the electron dot structure of a molecule, the valence electrons of all the atoms are shown as dots placed around the atomic symbols. Two dots drawn in between two atoms indicate one covalent bond between them. The electron dot structure of methane is shown in Fig. 14.17.
To read such a structure, count what each atom brings. Carbon has four valence electrons and each hydrogen has one, so the four shared pairs around carbon add up to eight electrons — a complete octet — while each hydrogen reaches its duplet of two.
Electron dot structures are called Lewis structures, and the dash formula represents a simplified Lewis formula: it keeps the bonds but drops the individual dots. Chemists have developed some more ways of representing organic molecules, each fulfilling a specific requirement.
14.2.2 Bond Line Formula or Zig-Zag Formula
In this representation of a molecule, the symbols of carbon and hydrogen atoms are not written at all. The lines representing carbon–carbon bonds are drawn in a zig-zag manner, and the terminals of the zig-zag line represent methyl groups.
The intersection of two lines denotes a carbon atom bonded to the appropriate number of hydrogen atoms — that is, however many hydrogens are needed to satisfy the tetravalency of carbon. For example, propane is represented by a simple two-line zig-zag, and the bond line formula of \(\mathrm{C_2H_5OH}\) is a single zig-zag carrying \(\mathrm{OH}\) at one end. Both are drawn in Fig. 14.18.
However, the heteroatoms, or hydrogen atoms bonded to heteroatoms, are written clearly. This is why the \(\mathrm{OH}\) of ethanol appears in the drawing while the six hydrogens on the carbons of propane do not: only carbon-bound hydrogens can be worked out silently from tetravalency.
There is a reason the line is drawn zig-zag rather than straight. A saturated carbon is tetrahedral, with bond angles close to \(109.5^\circ\), so a real carbon chain is not a straight rod. The zig-zag is an honest, quick sketch of that geometry, and it also keeps each carbon at a clear corner so no atom is missed while counting.
To count hydrogens at any carbon in a bond line formula, count the lines meeting at that point and subtract from four. A chain end has one line, so it carries three hydrogens — a methyl group. An ordinary corner has two lines and carries two hydrogens.
dash formula vs condensed formula vs bond line formula | Feature | Dash (structural) formula | Condensed formula | Bond line (zig-zag) formula | |—|—|—|—| | Atoms shown | All C and H shown | All C and H shown | C and H on carbon not shown; heteroatoms shown | | Bonds shown | Every bond as a dash | Some or all bonds hidden | C–C bonds shown as lines | | Example (butane) | H and C written out with all bonds as dashes | \(\mathrm{CH_3(CH_2)_2CH_3}\) | Four-carbon zig-zag line | | Chief advantage | Complete, unambiguous connectivity | Compact, easy to type in one line | Fastest to draw; skeleton and functional groups stand out | | Chief limitation | Slow and crowded for large molecules | Branching can be hard to read | Needs practice; hydrogens must be inferred |
The choice of representation is therefore decided by what you need to show, as summarised in FC 14.1.
The condensed formula of a compound is \(\mathrm{CH_3(CH_2)_3CH_3}\). Write it with all its groups and dashes shown, give its molecular formula, and state how many carbon atoms its bond line formula would contain.
Solution: First unpack the bracket. The subscript 3 means three \(\mathrm{-CH_2-}\) units lie between the two terminal methyl groups, so the expanded form is
\[\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!CH_2\!-\!CH_3}\]
Now count. Carbon atoms: \(1 + 1 + 1 + 1 + 1 = 5\). Hydrogen atoms: \(3 + 2 + 2 + 2 + 3 = 12\). The molecular formula is therefore \(\mathrm{C_5H_{12}}\).
The bond line formula contains the same five carbon atoms, drawn as four zig-zag lines: the two free ends are the terminal methyl carbons and the three corners in between are the \(\mathrm{-CH_2-}\) carbons. Note that no carbon atom disappears when hydrogens and symbols are dropped — only the writing does.
Table 14.1 : Representation of structural formula
| 1. |
\(\mathrm{H\!-\!C\!-\!C\!-\!C\!-\!C\!-\!H}\) with all H shown |
\(\mathrm{CH_3(CH_2)_2CH_3}\) |
\(\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!CH_3}\) |
(four-carbon zig-zag, given) |
| 2. |
……………… |
……………… |
……………… |
(zig-zag formula given in the figure) |
| 3. |
……………… |
\(\mathrm{CH_3(CH_2)_2CHO}\) |
……………… |
……………… |
| 4. |
……………… |
\(\mathrm{CH_3CH_2CH(CH_3)OH}\) |
……………… |
……………… |
| 5. |
\(\mathrm{N}\equiv\mathrm{C\!-\!CH(OH)\!-\!C}\equiv\mathrm{N}\) (given with the H and OH drawn on the central carbon) |
……………… |
……………… |
……………… |
(question only — no solution) Write the expanded condensed formula and the molecular formula of \(\mathrm{CH_3(CH_2)_4CH_3}\), and state how many hydrogen atoms sit on the third carbon from either end.
14.2.3 Drawing the Molecules in Three Dimensions
Most organic molecules have three-dimensional shapes. Since paper is flat, four different methods are used to represent three-dimensional molecules on plane paper. Each of the four keeps the shape information that a flat dash formula throws away.
Molecular models
Drawings can only suggest depth. Molecular models, shown in Fig. 14.5, are used for an easier visualisation of the three-dimensional shapes of organic molecules. Three types of molecular models are common.
A framework model emphasizes the skeletal pattern of bonds, ignoring the size of the atoms. In a ball and stick model, a ball represents an atom and a stick represents a bond. A space filling model emphasizes the relative size of each atom; bonds are not shown, and the model conveys the volume actually occupied by each atom in the molecule. Nowadays computer graphics can be used for molecular modelling.
Each model answers a different question, which is why all three survive. If you want to trace connectivity, the framework model is clearest. If you want bond angles and geometry, the ball and stick model shows them. If you want to know whether a molecule can fit into a small space, only the space filling model will tell you.
The space filling idea is the working basis of modern drug design. A medicine acts by fitting into a pocket on a protein, and whether it fits depends on the volume its atoms occupy, not on how neatly its bonds can be drawn. Pharmaceutical chemists therefore build molecules on screen as space filling models and test the fit computationally before a single gram is synthesised — saving both chemicals and solvent waste. Before reading on, look again at Fig. 14.5 and predict which of the three models would be least useful for this purpose.
(question only — no solution) A student draws a Newman projection in which the three bonds of the front carbon and the three bonds of the rear carbon all radiate from the centre of the circle. Identify the mistake and state what the rear-carbon bonds should radiate from.
14.3 Classification of Organic Compounds
With so many different molecules to study, it is important to find ways of relating the different structures of molecules to their chemical properties. That link between structure and property provides the basis for classifying organic compounds.
Organic compounds are broadly classified in two ways: on the basis of (i) the carbon skeleton and (ii) the functional group. The complete scheme based on the carbon skeleton is drawn as a tree in Fig. 14.6.
Note that these two schemes are not rivals. Every organic compound can be placed in both at once: ethanol, for instance, is acyclic by skeleton and belongs to the alcohol family by functional group.
14.3.1 Classification Based on Carbon Skeleton
Construct a model of the propane (\(\mathrm{C_3H_8}\)) molecule. Which atoms are on the surface of the molecule? Which atoms lie in the core of the molecule?
The carbon atoms bonded to each other lie in the core of an organic molecule and are described as the carbon skeleton of the molecule. The hydrogen atoms sit on the outside, covering that core.
It is the carbon skeleton of a molecule that decides its shape and size. A variety of carbon skeletons is found in different organic compounds, and accordingly they are classified into various types, as shown in Fig. 14.6.
The carbon skeleton of a molecule is the framework formed by its mutually bonded carbon atoms; it fixes the shape and size of the molecule.
Acyclic or open chain compounds
These are also known as aliphatic compounds. Molecules of these compounds have chains of carbon atoms with two free ends — the chain does not close on itself.
Such chains may be straight chains, formed by carbon atoms bonded to one or two other carbons, or branched chains, having at least one carbon bonded to three or four other carbons. Table 14.2 shows examples of acyclic compounds.
Table 14.2 : Acyclic/Aliphatic compounds
| Straight chain |
\(\mathrm{CH_3\!-\!CH_2\!-\!CH_3}\) (propane); \(\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!OH}\) (n-propyl alcohol) |
| Branched chain |
\(\mathrm{CH_3\!-\!CH(CH_3)\!-\!CH_3}\) (isobutane) |
The test for a branch is simply how many carbons a given carbon touches. In propane the middle carbon touches two others, so the chain runs straight through it. In isobutane the middle carbon touches three others, and the third one has nowhere to go except off to the side — that is the branch.
Cyclic compounds
These are compounds in which the molecules are formed by joining atoms to form ring-like structures. Here the chain has closed on itself, so there are no free ends.
The compounds in which the ring is made of only carbon atoms are known as homocyclic or carbocyclic compounds.
Carbon is an essential atom in any organic molecule, and one or more hydrogen atoms are present in most of the organic molecules. All the other atoms that are found in organic molecules are called heteroatoms. The cyclic compounds in which the ring includes one or more heteroatoms (oxygen, nitrogen, sulphur etc.) are known as heterocyclic compounds.
A heteroatom is any atom in an organic molecule other than carbon and hydrogen — for example oxygen, nitrogen or sulphur.
Read the two terms carefully, because students often mix them up. “Heterocyclic” is about the ring only. A molecule can contain oxygen or nitrogen and still be homocyclic, provided the heteroatom hangs outside the ring rather than forming part of it — phenol is a good example.
Observe the compounds (a) to (e) shown in Fig. 14.21 — cyclohexene, piperidine, cyclopentane, benzene and pyrrole. Identify the compounds whose ring contains a ring of carbon atoms only. Then identify the compounds in which the ring contains at least one atom other than carbon.
Alicyclic compounds
These are cyclic compounds exhibiting properties similar to those of aliphatic compounds. In other words, closing the chain into a ring has not given the molecule any new kind of stability; it still behaves much like an open-chain compound. For example: cyclobutane and cyclohexene. These are drawn in Fig. 14.7.
Aromatic compounds
Aromatic compounds have special stability. This is the feature that separates them from alicyclic compounds, and it is why they resist the addition reactions that ordinary alkenes undergo readily.
Aromatic compounds containing a benzene ring are known as benzenoid aromatics, while those not containing a benzene ring are called non-benzenoid aromatics. For example: benzene and tropone. See Fig. 14.7.
Heterocyclic compounds
Heterocyclic compounds are divided in the same way, since a ring built around a heteroatom may or may not be aromatic. Figure 14.8 illustrates some heterocyclic compounds of both kinds.
Compare furan with tetrahydrofuran in Fig. 14.8 and the pattern becomes clear. Both are five-membered rings containing one oxygen; only furan carries the complete set of alternating double bonds, and only furan is aromatic. Piperidine stands in the same relation to pyridine.
aliphatic vs alicyclic vs aromatic vs heterocyclic compounds | Feature | Aliphatic (acyclic) | Alicyclic | Aromatic | Heterocyclic | |—|—|—|—|—| | Skeleton | Open chain, two free ends | Closed ring of carbons only | Closed conjugated ring system | Ring containing one or more heteroatoms | | Heteroatom in the ring | No ring at all | Absent | Absent in benzenoid hydrocarbons | Present | | Special stability | No | No | Yes | Only in the aromatic members | | Examples | Propane, n-propyl alcohol, isobutane | Cyclobutane, cyclohexene | Benzene, naphthalene, phenol, tropone | Furan, thiophene, pyridine, pyrrole, THF, piperidine, 4H-pyran |
To place an unfamiliar compound, work through the decisions in FC 14.2 in order.
Naphthalene, the benzenoid aromatic in Fig. 14.7, is the compound of traditional mothballs; it sublimes slowly, so a solid lump becomes vapour without melting. Benzene itself, however, is treated very differently in a modern laboratory. It is a confirmed human carcinogen, and where a similar solvent is needed it is now routinely replaced by toluene or by less hazardous alternatives. Aromatic stability makes these rings chemically durable — which is exactly why they also persist in the body and in the environment once released.
Classify each of the following on the basis of its carbon skeleton, giving a reason: (i) cyclopentane, (ii) pyrrole, (iii) naphthalene, (iv) propan-1-ol.
Solution: Work through the two questions of FC 14.2 for each compound — is there a ring, and if so, is it made only of carbon?
Cyclopentane is a closed five-membered ring of carbon atoms with no heteroatom and no conjugated double bonds, so it is homocyclic and alicyclic.
Pyrrole is a five-membered ring in which one ring position is occupied by nitrogen, a heteroatom. It is therefore heterocyclic, and since it carries a complete conjugated system it is heterocyclic aromatic.
Naphthalene consists of two fused rings made entirely of carbon, and it contains benzene rings. It is homocyclic, aromatic and specifically benzenoid aromatic.
Propan-1-ol, \(\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!OH}\), has no ring at all. It is acyclic, that is aliphatic, and its chain is a straight chain. The oxygen present does not make it heterocyclic, because there is no ring for the oxygen to be part of.
(question only — no solution) Classify each of the following by carbon skeleton and justify your choice in one line: (i) thiophene, (ii) cyclobutane, (iii) isobutane, (iv) tropone, (v) tetrahydrofuran.
14.3.2 Classification Based on Functional Group
Consider the following reaction:
\[2\,\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!OH} + 2\,\mathrm{Na} \longrightarrow 2\,\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!ONa} + \mathrm{H_2}\]
Compare the structure of the substrate propanol with that of the product sodium propoxide. Which part of the substrate — the carbon skeleton or the \(\mathrm{OH}\) group — has undergone a change during the reaction?
A part of an organic molecule which undergoes change as a result of a reaction is called a functional group. In the above illustration, \(\mathrm{OH}\) is the functional group present in propanol; the three-carbon skeleton emerges from the reaction untouched.
A functional group is that part of an organic molecule which undergoes change as a result of a reaction.
This single observation carries most of the working logic of organic chemistry. If the skeleton stays and the functional group changes, then compounds sharing a functional group should react alike, whatever the length of their chains. That is exactly what is found, and it is why the second classification is so useful.
Inspection of the structures of various organic compounds tells us that there is a large variety of functional groups present in them. The second method of classification of organic compounds is based on the nature of the functional group present.
The resulting individual class is called a family, named after the constituent functional group. For example, the family of alcohols and the family of halogen derivatives. Table 14.3 shows some common functional groups with their names, bond structures and examples.
Table 14.3 : Functional groups in organic compounds
| 1. |
Halide |
\(\mathrm{-X}\) |
\(\mathrm{CH_3Br}\) methyl bromide |
| 2. |
Cyanide (or Nitrile) |
\(\mathrm{-C}\equiv\mathrm{N}\) |
\(\mathrm{CH_3CN}\) methyl cyanide (or acetonitrile) |
| 3. |
Isocyanide |
\(\mathrm{-N^{\oplus}}\!\equiv\!\mathrm{C^{\ominus}}\) |
\(\mathrm{CH_3NC}\) methyl isocyanide |
| 4. |
Nitro compound/group |
\(\mathrm{-N^{\oplus}(=O)O^{\ominus}}\) |
\(\mathrm{CH_3NO_2}\) Nitromethane |
| 5. |
Alcohol |
\(\mathrm{-OH}\) |
\(\mathrm{CH_3OH}\) methyl alcohol |
| 6. |
Phenol |
\(\mathrm{Ar\!-\!OH}\) |
\(\mathrm{C_6H_5OH}\) Phenol |
| 7. |
Primary amine |
\(\mathrm{-NH_2}\) |
\(\mathrm{CH_3NH_2}\) methyl amine |
| 8. |
Secondary amine |
\(\mathrm{-NH\!-}\) |
\(\mathrm{CH_3\!-\!NH\!-\!CH_3}\) Dimethyl amine |
| 9. |
Tertiary amine |
\(\mathrm{-N\!-}\) (three bonds to N) |
\(\mathrm{(CH_3)_3N}\) Trimethyl amine |
| 10. |
Ether |
\(\mathrm{-C\!-\!O\!-\!C\!-}\) |
\(\mathrm{CH_3\!-\!O\!-\!CH_3}\) Dimethyl ether |
| 11. |
Aldehyde |
\(\mathrm{-C(=\!O)\!-\!H}\) |
\(\mathrm{CH_3CHO}\) Acetaldehyde |
| 12. |
Ketone |
\(\mathrm{-C(=\!O)\!-}\) |
\(\mathrm{CH_3\!-\!CO\!-\!CH_3}\) Acetone |
| 13. |
Carboxylic acid |
\(\mathrm{-C(=\!O)\!-\!OH}\) |
\(\mathrm{CH_3\!-\!COOH}\) Acetic acid |
| 14. |
Ester |
\(\mathrm{-C(=\!O)\!-\!O\!-}\) |
\(\mathrm{CH_3\!-\!COOC_2H_5}\) Ethyl acetate |
| 15. |
Amide |
\(\mathrm{-C(=\!O)\!-\!NH_2}\) |
\(\mathrm{CH_3\!-\!CONH_2}\) Acetamide |
| 16. |
Secondary amide |
\(\mathrm{-C(=\!O)\!-\!NH\!-}\) |
\(\mathrm{CH_3\!-\!CO\!-\!NH\!-\!CH_3}\) N-methyl acetamide |
| 17. |
Tertiary amide |
\(\mathrm{-C(=\!O)\!-\!N\!-}\) |
\(\mathrm{CH_3\!-\!CO\!-\!N(CH_3)_2}\) N, N - dimethyl acetamide |
| 18. |
Acid anhydride |
\(\mathrm{-C(=\!O)\!-\!O\!-\!C(=\!O)\!-}\) |
\(\mathrm{CH_3\!-\!CO\!-\!O\!-\!CO\!-\!CH_3}\) Acetic anhydride |
| 19. |
Acyl chloride |
\(\mathrm{-C(=\!O)\!-\!Cl}\) |
\(\mathrm{CH_3\!-\!COCl}\) Acetyl chloride |
| 20. |
Sulphonic acid |
\(\mathrm{-S(=\!O)_2\!-\!O\!-\!H}\) |
\(\mathrm{C_6H_5SO_3H}\) Benzene sulphonic acid |
| 21. |
Alkene/olefin |
\(\mathrm{C\!=\!C}\) |
\(\mathrm{CH_2\!=\!CH_2}\) Ethylene |
| 22. |
Alkyne |
\(\mathrm{-C}\equiv\mathrm{C}\!-\) |
\(\mathrm{CH}\equiv\mathrm{CH}\) Acetylene |
| 23. |
Alkane |
No functional group |
\(\mathrm{CH_3\!-\!CH_3}\) Ethane |
Table 14.3 repays slow reading rather than memorising in one go. Look for the repeating carbonyl unit, \(\mathrm{C\!=\!O}\): it appears in aldehydes, ketones, carboxylic acids, esters, amides, anhydrides and acyl chlorides. What distinguishes those families is only the atom or group sitting next to that carbonyl carbon.
Both classification words have older meanings hidden in them. Aliphatic compounds are named so because of their attraction towards fats. Aromatic compounds are named so because some of these compounds, discovered in the early nineteenth century, possessed an aroma. Neither name is a chemical definition today — plenty of aromatic compounds have no smell at all — but both have stayed in use.
Homologous series
A family of a functional group, in turn, comprises different homologous series. A series of compounds of the same family in which each member has the same type of carbon skeleton and functional group, and differs from the next member by a constant difference of one methylene group (\(\mathrm{-CH_2-}\)) in its molecular and structural formula, is called a homologous series.
An individual member of a homologous series is called a homologue. All the members, or homologues, of such a series are represented by the same general formula.
A homologous series is a series of compounds of the same family, each member having the same type of carbon skeleton and the same functional group, and differing from the next member by one \(\mathrm{-CH_2-}\) group; its members share one general formula.
The members of a particular homologous series possess similar chemical properties, and vary only very gradually in their physical properties, namely melting point, boiling point, density, solubility, etc. Table 14.4 illustrates this with the help of the homologous series of straight chain aldehydes, having the general formula \(\mathrm{C_nH_{2n}O}\).
Table 14.4 : Homologous series of straight chain aldehydes
| Formaldehyde |
n = 1 |
\(\mathrm{CH_2O}\) |
\(\mathrm{H\!-\!C(=\!O)\!-\!H}\) |
\(-21\ ^0\text{C}\) |
| Acetaldehyde |
n = 2 |
\(\mathrm{C_2H_4O}\) |
\(\mathrm{CH_3\!-\!C(=\!O)\!-\!H}\) |
\(21\ ^0\text{C}\) |
| Propionaldehyde |
n = 3 |
\(\mathrm{C_3H_6O}\) |
\(\mathrm{CH_3\!-\!CH_2\!-\!C(=\!O)\!-\!H}\) |
\(48\ ^0\text{C}\) |
| Butyraldehyde |
n = 4 |
\(\mathrm{C_4H_8O}\) |
\(\mathrm{CH_3\!-\!(CH_2)_2\!-\!C(=\!O)\!-\!H}\) |
\(75\ ^0\text{C}\) |
| Valeraldehyde |
n = 5 |
\(\mathrm{C_5H_{10}O}\) |
\(\mathrm{CH_3\!-\!(CH_2)_3\!-\!C(=\!O)\!-\!H}\) |
\(103\ ^0\text{C}\) |
The boiling points in Table 14.4 are plotted against the number of carbon atoms in Fig. 14.20.
the boiling point versus number of carbons plot for straight chain aldehydes in Fig. 14.20 (i) By roughly how many degrees does the boiling point rise for each \(\mathrm{-CH_2-}\) group added? (ii) Is that rise constant, or does it get smaller as the chain lengthens? (iii) Formaldehyde is a gas at room temperature while valeraldehyde is a liquid — read the graph and explain why, in terms of molecular size. (iv) Use the trend to predict, without any calculation, whether the next member of the series would boil above or below \(103\ ^0\text{C}\).
Alkanes constitute a homologous series of straight chain saturated hydrocarbons. Write down the structural formulae of the first five homologues of this series. Write their molecular formulae and deduce the general formula of such a homologous series.
Solution: The first five homologues are generated by adding one \(\mathrm{-CH_2-}\) at a time, starting with the first homologue \(\mathrm{CH_4}\).
\[\mathrm{CH_4}\ (\text{1st homologue}) \xrightarrow{\ \text{add}\ \mathrm{-CH_2-}\ } \mathrm{CH_3\!-\!CH_3}\ (\text{2nd homologue})\]
\[\xrightarrow{\ \text{add}\ \mathrm{-CH_2-}\ } \mathrm{CH_3\!-\!CH_2\!-\!CH_3}\ (\text{3rd homologue})\]
\[\xrightarrow{\ \text{add}\ \mathrm{-CH_2-}\ } \mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!CH_3}\ (\text{4th homologue})\]
\[\xrightarrow{\ \text{add}\ \mathrm{-CH_2-}\ } \mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!CH_2\!-\!CH_3}\ (\text{5th homologue})\]
By counting carbon and hydrogen atoms in the five homologues we get their molecular formulae as \(\mathrm{CH_4}\), \(\mathrm{C_2H_6}\), \(\mathrm{C_3H_8}\), \(\mathrm{C_4H_{10}}\) and \(\mathrm{C_5H_{12}}\).
Comparing these molecular formulae, the following general formula is deduced: \(\mathrm{C_nH_{2n+2}}\).
The pattern behind that general formula is worth seeing rather than memorising. Each added carbon brings two extra hydrogens, and the two chain ends together account for the extra “+2”.
Write down structural formulae of (a) the third higher and (b) the second lower homologue of \(\mathrm{CH_3CH_2COOH}\).
Solution:
- The structural formula of the third higher homologue is obtained by adding three \(\mathrm{-CH_2-}\) units to the carbon chain of the given structure.
\[\mathrm{CH_3\!-\!CH_2\!-\!COOH} \xrightarrow{\ \text{add three}\ \mathrm{-CH_2-}\ \text{units}\ } \mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!CH_2\!-\!CH_2\!-\!COOH}\]
This product is the third higher homologue.
- The structural formula of the second lower homologue is obtained by removing two \(\mathrm{-CH_2-}\) units from the carbon chain of the given structure.
\[\mathrm{CH_3-CH_2-COOH} \xrightarrow{\ \text{remove two}\ \mathrm{-CH_2-}\ \text{units}\ } \mathrm{H-COOH}\]
Notice that the \(\mathrm{-COOH}\) group survives both operations untouched. Moving up or down a homologous series changes only the skeleton; the functional group, and therefore the family, must stay the same.
(question only — no solution) The general formula of the straight chain aldehyde series is \(\mathrm{C_nH_{2n}O}\). Write the molecular formula and the condensed structural formula of (i) the second higher homologue of acetaldehyde and (ii) the first lower homologue of butyraldehyde.
Primary, secondary, tertiary and quaternary carbons
One more piece of vocabulary belongs here, because it labels positions on the carbon skeleton and will be needed constantly in later chapters. The saturated (\(sp^3\)) carbons in a molecule are labelled as primary, secondary, tertiary and quaternary in accordance with the number of other carbons bonded to them by single bonds.
A primary carbon (\(1^0\)) is bonded to only one other carbon. For example, both the carbons in ethane, \(\mathrm{CH_3\!-\!CH_3}\).
A secondary carbon (\(2^0\)) is bonded to two other carbons. For example, the middle carbon in propane, \(\mathrm{CH_3\!-\!CH_2\!-\!CH_3}\).
A tertiary carbon (\(3^0\)) is bonded to three other carbons. For example, the middle carbon in isobutane, \(\mathrm{CH_3CH(CH_3)_2}\).
A quaternary carbon (\(4^0\)) is bonded to four other carbons. For example, the middle carbon in neo-pentane, \(\mathrm{C(CH_3)_4}\).
Count carbons only — never hydrogens. A primary carbon carries three hydrogens and a quaternary carbon carries none, but it is the carbon count that fixes the label.
primary vs secondary vs tertiary vs quaternary carbon | Feature | Primary (\(1^0\)) | Secondary (\(2^0\)) | Tertiary (\(3^0\)) | Quaternary (\(4^0\)) | |—|—|—|—|—| | Other carbons attached | 1 | 2 | 3 | 4 | | Hydrogens on that carbon | 3 | 2 | 1 | 0 | | Example | Either carbon of \(\mathrm{CH_3\!-\!CH_3}\) | Middle carbon of \(\mathrm{CH_3\!-\!CH_2\!-\!CH_3}\) | Middle carbon of \(\mathrm{CH_3CH(CH_3)_2}\) | Middle carbon of \(\mathrm{C(CH_3)_4}\) |
This labelling is the reason examiners can ask a question as short as “how many primary carbons does this molecule contain?” and still test whether you can read a branched skeleton correctly.
(question only — no solution) In \(\mathrm{CH_3\!-\!CH(CH_3)\!-\!CH_2\!-\!C(CH_3)_3}\), state the number of primary, secondary, tertiary and quaternary carbon atoms present.
14.4 Nomenclature of Organic Compounds
14.4.1 Common/Trivial Names
As organic chemistry developed, attempts were made to devise a name for every organic compound. These old names are now referred to as common or trivial names.
The common name of a compound usually has some history behind it, and is usually accepted on account of its long usage. Though a systematic method of nomenclature was developed later, common names are still useful and in many cases cannot be avoided, particularly for commonly used commercial organic compounds.
Stem names of certain common organic compounds are given in Table 14.5.
Table 14.5 : Trivial names of some organic compounds
| \(\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!CH_3}\) |
n - Butane |
| \(\mathrm{CH_3\!-\!CH(CH_3)\!-\!CH_3}\) |
Isobutane |
| \(\mathrm{C_6H_5NH_2}\) |
Aniline |
| \(\mathrm{CH_3\!-\!CH(OH)\!-\!COOH}\) |
Lactic acid |
| \(\mathrm{CHCl_3}\) |
Chloroform |
| \(\mathrm{CH}\equiv\mathrm{CH}\) |
Acetylene |
| \(\mathrm{C_2H_5\!-\!O\!-\!C_2H_5}\) |
Diethyl ether |
| \(\mathrm{CH_3COOH}\) |
Acetic acid |
| \(\mathrm{H\!-\!C(=\!O)\!-\!H}\) |
Formaldehyde |
| \(\mathrm{CH_3\!-\!CH(NH_2)\!-\!COOH}\) |
Glycine |
| \(\mathrm{CH_2(OH)\!-\!CH(OH)\!-\!CH_2(OH)}\) |
Glycerol |
Notice that a trivial name tells you nothing about the structure. “Lactic acid” does not reveal three carbons, an \(\mathrm{-OH}\) and a \(\mathrm{-COOH}\); you have to already know the compound. That limitation is exactly what the systematic system was built to remove.
Several trivial names in Table 14.5 record where the compound was first found. Lactic acid takes its name from milk, in which it forms as milk sours (Latin lac, milk). Acetic acid is named from vinegar (Latin acetum). Glycine and glycerol both begin with a Greek root meaning sweet — glycine has a sweetish taste and glycerol a distinctly sweet one. Formic acid, the first member of the acid series, is named from the Latin word for ant, the insect from which it was originally obtained.
14.4.2 IUPAC Nomenclature
With the rapidly growing number of organic compounds, having increasingly complicated structures, it becomes difficult to name them. Some confusion can be created when the same compound is named differently by different chemists.
In due course of time, the International Union of Pure and Applied Chemistry (IUPAC) was founded (1919), and a systematic method of nomenclature for organic compounds was developed under its banner. This system of nomenclature is accepted and widely used all over the world today. It gives a unique name to each organic compound.
To arrive at the IUPAC name of an organic compound, its structure is considered to be made of three main parts: the parent hydrocarbon, the branches and the functional groups.
The IUPAC name of a compound is obtained by modifying the name of its parent hydrocarbon, further incorporating names of the branches and functional groups as prefix and suffix. For doing this, certain rules were formulated by IUPAC.
In the following subsections we consider the IUPAC rules for naming saturated and unsaturated hydrocarbons, simple monocyclic hydrocarbons, simple mono- and polyfunctional compounds, and substituted benzenes.
trivial names vs IUPAC names | Feature | Trivial (common) name | IUPAC name | |—|—|—| | Origin | Source, history or long usage | A set of agreed rules | | Information about structure | None, or almost none | Chain length, branches, functional groups and their positions | | Uniqueness | One compound may carry several names | One compound, one name | | Example | Acetic acid, glycerol, chloroform | Ethanoic acid, propane-1,2,3-triol, trichloromethane | | Still in use? | Yes, especially for commercial compounds | Yes, universally, for all reporting |
Q3. Chemists already had short, comfortable names such as acetic acid and chloroform. Why was a long systematic naming system such as IUPAC nomenclature still felt to be necessary?
A. Because a trivial name is only a label agreed on by habit. It carries no structural information, so it cannot be worked out for a new compound, and nothing stops two laboratories from giving the same substance different names — or, worse, the same name to different substances. As the number of known compounds grew into the millions, this became unworkable. An IUPAC name is built from the structure itself, so it can be constructed for any compound, read back into a structure by anyone, and belongs to that compound alone.
14.4.3 IUPAC Names of Straight Chain Alkanes
The homologous series of straight chain alkanes forms the parent hydrocarbon part of the IUPAC names of aliphatic compounds. This is why these names have to be learnt first — everything else in the system is built on them.
The IUPAC name of a straight chain alkane is derived from the number of carbon atoms it contains. Table 14.6 gives the list of the IUPAC names of the first twenty straight chain alkanes. It may be noted that the stem names of the first four members are accepted as their IUPAC names.
Table 14.6 : IUPAC names of the first twenty alkanes
| 1 |
\(\mathrm{CH_4}\) |
\(\mathrm{CH_4}\) |
Methane |
| 2 |
\(\mathrm{C_2H_6}\) |
\(\mathrm{CH_3\!-\!CH_3}\) |
Ethane |
| 3 |
\(\mathrm{C_3H_8}\) |
\(\mathrm{CH_3\!-\!CH_2\!-\!CH_3}\) |
Propane |
| 4 |
\(\mathrm{C_4H_{10}}\) |
\(\mathrm{CH_3\!-\!(CH_2)_2\!-\!CH_3}\) |
Butane |
| 5 |
\(\mathrm{C_5H_{12}}\) |
\(\mathrm{CH_3\!-\!(CH_2)_3\!-\!CH_3}\) |
Pentane |
| 6 |
\(\mathrm{C_6H_{14}}\) |
\(\mathrm{CH_3\!-\!(CH_2)_4\!-\!CH_3}\) |
Hexane |
| 7 |
\(\mathrm{C_7H_{16}}\) |
\(\mathrm{CH_3\!-\!(CH_2)_5\!-\!CH_3}\) |
Heptane |
| 8 |
\(\mathrm{C_8H_{18}}\) |
\(\mathrm{CH_3\!-\!(CH_2)_6\!-\!CH_3}\) |
Octane |
| 9 |
\(\mathrm{C_9H_{20}}\) |
\(\mathrm{CH_3\!-\!(CH_2)_7\!-\!CH_3}\) |
Nonane |
| 10 |
\(\mathrm{C_{10}H_{22}}\) |
\(\mathrm{CH_3\!-\!(CH_2)_8\!-\!CH_3}\) |
Decane |
| 11 |
\(\mathrm{C_{11}H_{24}}\) |
\(\mathrm{CH_3\!-\!(CH_2)_9\!-\!CH_3}\) |
Undecane |
| 12 |
\(\mathrm{C_{12}H_{26}}\) |
\(\mathrm{CH_3\!-\!(CH_2)_{10}\!-\!CH_3}\) |
Dodecane |
| 13 |
\(\mathrm{C_{13}H_{28}}\) |
\(\mathrm{CH_3\!-\!(CH_2)_{11}\!-\!CH_3}\) |
Tridecane |
| 14 |
\(\mathrm{C_{14}H_{30}}\) |
\(\mathrm{CH_3\!-\!(CH_2)_{12}\!-\!CH_3}\) |
Tetradecane |
| 15 |
\(\mathrm{C_{15}H_{32}}\) |
\(\mathrm{CH_3\!-\!(CH_2)_{13}\!-\!CH_3}\) |
Pentadecane |
| 16 |
\(\mathrm{C_{16}H_{34}}\) |
\(\mathrm{CH_3\!-\!(CH_2)_{14}\!-\!CH_3}\) |
Hexadecane |
| 17 |
\(\mathrm{C_{17}H_{36}}\) |
\(\mathrm{CH_3\!-\!(CH_2)_{15}\!-\!CH_3}\) |
Heptadecane |
| 18 |
\(\mathrm{C_{18}H_{38}}\) |
\(\mathrm{CH_3\!-\!(CH_2)_{16}\!-\!CH_3}\) |
Octadecane |
| 19 |
\(\mathrm{C_{19}H_{40}}\) |
\(\mathrm{CH_3\!-\!(CH_2)_{17}\!-\!CH_3}\) |
Nonadecane |
| 20 |
\(\mathrm{C_{20}H_{42}}\) |
\(\mathrm{CH_3\!-\!(CH_2)_{18}\!-\!CH_3}\) |
Icosane |
From the fifth member onwards the names are simply Greek or Latin number words carrying the ending “ane”: pent- for five, hex- for six, hept- for seven and so on. Only methane, ethane, propane and butane have to be memorised as special cases.
14.4.4 IUPAC Names of Branched Saturated Hydrocarbons
The branches, or side chains, in saturated hydrocarbons are alkyl groups or alkyl substituents. Before looking at the IUPAC nomenclature of branched saturated hydrocarbons, we look at the naming of alkyl groups themselves.
Alkyl groups
An alkyl group may be a straight or a branched chain. A straight chain alkyl group is generated by removing one hydrogen atom from the terminal carbon of an alkane molecule, and is named by replacing the ending “ane” of the alkane by “yl”. Some straight chain alkyl groups are listed in Table 14.7.
Table 14.7 : Straight chain alkyl groups
| \(\mathrm{CH_4}\) |
methane |
\(\mathrm{-CH_3}\) |
methyl |
Me |
| \(\mathrm{C_2H_6}\) |
ethane |
\(\mathrm{-CH_2\!-\!CH_3}\) |
ethyl |
Et |
| \(\mathrm{C_3H_8}\) |
propane |
\(\mathrm{-(CH_2)_2\!-\!CH_3}\) |
propyl |
Pr |
| \(\mathrm{C_4H_{10}}\) |
butane |
\(\mathrm{-(CH_2)_3\!-\!CH_3}\) |
butyl |
Bu |
| \(\mathrm{C_{10}H_{22}}\) |
decane |
\(\mathrm{-(CH_2)_9\!-\!CH_3}\) |
decyl |
- |
An alkyl group is never a compound on its own. The dash written in front of \(\mathrm{-CH_3}\) marks the position where the group is bonded to something else, so it always has one bond left unused.
A branched chain alkyl group is obtained by removing a hydrogen atom from any one of the nonterminal carbons of an alkane, or any hydrogen atom from a branched alkane. Stem names of branched alkyl groups containing up to four carbons are used in IUPAC names (see Table 14.8).
Table 14.8 : Trivial names of small branched alkyl groups
| \(\mathrm{CH_3\!-\!CH(-)\!-\!CH_3}\) |
Isopropyl |
| \(\mathrm{CH_3\!-\!CH(CH_3)\!-\!CH_2\!-}\) |
Isobutyl |
| \(\mathrm{CH_3\!-\!CH(-)\!-\!CH_2\!-\!CH_3}\) |
sec-Butyl |
| \(\mathrm{CH_3\!-\!C(CH_3)_2\!-}\) |
tert-Butyl |
Read the free valence carefully in Table 14.8. In isopropyl and sec-butyl the attachment is at a middle carbon, which is why the free bond is drawn on the carbon in the middle of the formula. In isobutyl the attachment is at a terminal \(\mathrm{-CH_2-}\), and in tert-butyl at the carbon that carries three methyl groups.
One aryl group is important enough to have its own name. The group \(\mathrm{C_6H_5\!-}\), a benzene ring with one hydrogen removed, is named the phenyl group and abbreviated as Ph. The word comes from phene, an early name for benzene, built on a Greek root meaning “shining” — benzene was obtained from the gas once used for lighting streets and homes.
Rules for IUPAC nomenclature of branched saturated hydrocarbons
Rule 1. Select the longest continuous chain of carbon atoms to be called the parent chain. All other carbon atoms not included in this chain constitute side chains, or branches, or alkyl substituents. Two examples are drawn in Fig. 14.22: in the first the parent chain has five carbon atoms with a \(\mathrm{-CH_3}\) group as the alkyl substituent, and in the second the parent chain has six carbon atoms with a methyl group as the alkyl substituent.
The word “continuous” matters more than the way the formula happens to be printed. A chain may turn a corner on the page and still be continuous, so always count carbons along every possible path before deciding.
If two chains of equal length are located, the one with the maximum number of substituents is selected as the parent chain. Figure 14.22 also shows this case: the same molecule can be read as a hexane parent chain carrying two alkyl substituents, which is correct, or as a hexane parent chain carrying only one alkyl substituent, which is the incorrect choice.
Rule 2. The parent chain is numbered from one end to the other to locate the position, called the locant number, of the alkyl substituent. The numbering is done in that direction which will result in the lowest possible locant numbers.
For the molecule drawn in Fig. 14.23, numbering from the left gives the substituent a locant number of 2, which is the proper numbering, while numbering from the right gives it 3, which is improper.
Rule 3. Names of the alkyl substituents are added as a prefix to the name of the parent alkane. Different alkyl substituents are listed in alphabetical order, with each substituent name preceded by its appropriate locant number. The name of the substituent is separated from the locant number by a hyphen.
Thus for \(\mathrm{CH_3\!-\!CH_2\!-\!CH(CH_3)\!-\!CH(C_2H_5)\!-\!CH_2\!-\!CH_2\!-\!CH_3}\) the name is 4-ethyl-3-methylheptane and not 3-methyl-4-ethylheptane, because “ethyl” comes before “methyl” in the alphabet.
Rule 4. When both the numberings give the same set of locants, that numbering is chosen which gives the smaller locant to the substituent having alphabetical priority. Thus the IUPAC name of \(\mathrm{CH_3\!-\!CH_2\!-\!CH(CH_3)\!-\!CH(C_2H_5)\!-\!CH_2\!-\!CH_3}\) is 3-ethyl-4-methylhexane and not 3-methyl-4-ethylhexane.
Compare Rules 3 and 4 carefully, because they do different jobs. Rule 3 fixes the order in which substituent names are written; Rule 4 is used only to break a tie in numbering, and it hands the smaller number to whichever substituent comes first alphabetically.
Rule 5. If two or more identical substituents are present, the prefixes di (for 2), tri (for 3), tetra (for 4) and so on are used before the name of the substituent to indicate how many identical substituents there are. The locants of identical substituents are listed together, separated by commas.
There must be as many numbers in the name as there are substituents, even when two substituents sit on the same carbon. A digit and an alphabet are separated by a hyphen.
The prefixes di, tri, tetra, sec and tert are ignored in alphabetizing the substituent names. Substituent and parent hydrocarbon names are joined into one word — and note that isopropyl comes before methyl in the alphabetical order.
Rule 6. A branched alkyl group having no accepted trivial name is named with the longest continuous chain beginning at the point of attachment as the base name. The carbon atom of this group attached to the parent chain is numbered 1, and the name of such a substituent is enclosed in a bracket.
Three examples are drawn in Fig. 14.24: the 1,2-dimethylpropyl group, the 1,1,3-trimethylbutyl group, and the compound 2,3-dimethyl-4-(1methylpropyl)heptane, in which such a substituent appears inside a bracket in the full name.
The whole naming procedure for a branched alkane is summarised in FC 14.3.
Give the IUPAC name of \(\mathrm{CH_3\!-\!CH(CH_3)\!-\!CH_2\!-\!CH(C_2H_5)\!-\!CH_2\!-\!CH_3}\).
Solution: Begin with the longest continuous chain. Counting along the written chain gives six carbons, and no alternative path through the branches gives more than six, so the parent chain is hexane.
The substituents left outside that chain are one methyl group and one ethyl group. Now number the chain from both ends and compare. Numbering from the left places the methyl at C-2 and the ethyl at C-4, giving the locant set \(\{2, 4\}\). Numbering from the right places the ethyl at C-3 and the methyl at C-5, giving \(\{3, 5\}\).
The set \(\{2, 4\}\) is lower, so numbering from the left is correct. Rule 4 is not needed here, because the two directions did not give the same set of locants.
Finally, arrange the substituent names alphabetically — ethyl before methyl — each with its own locant, and join everything into one word. The IUPAC name is 4-ethyl-2-methylhexane.
(question only — no solution) Write the IUPAC name of \(\mathrm{CH_3\!-\!CH(CH_3)\!-\!CH(CH_3)\!-\!CH_2\!-\!CH_2\!-\!CH_3}\), and state how many primary carbon atoms it contains.
Complete the following table.
| \(\mathrm{CH_3\!-\!CH(CH_3)\!-\!CH_2\!-\!CH_3}\) |
\(\overset{1}{\mathrm{C}}-\overset{2}{\mathrm{C}}-\overset{3}{\mathrm{C}}-\overset{4}{\mathrm{C}}\) with one C branch at 2 |
2 - methyl |
2 - methylbutane |
| \(\mathrm{CH_3\!-\!C(CH_3)_2\!-\!CH_3}\) |
\(\overset{1}{\mathrm{C}}-\overset{2}{\mathrm{C}}-\overset{3}{\mathrm{C}}\) with two C branches at 2 |
2, 2 - dimethyl |
2, 2-dimethylpropane |
| \(\mathrm{CH_3\!-\!CH(CH_3)\!-\!CH(CH_3)\!-\!CH_2\!-\!CH_3}\) |
………. |
………. |
………. |
| \(\mathrm{CH_3\!-\!CH(CH_3)\!-\!CH(C_2H_5)\!-\!CH_2\!-\!CH_2\!-\!CH_3}\) |
………. |
………. |
………. |
| \(\mathrm{CH_3\!-\!CH(C_2H_5)\!-\!CH(C_2H_5)\!-\!CH_2\!-\!CH_2\!-\!CH_3}\) |
………. |
………. |
………. |
| \(\mathrm{C(C_2H_5)_4}\) |
………. |
………. |
………. |
Branched-chain names are printed, indirectly, on every petrol pump. The octane number of a fuel is fixed against a reference compound whose IUPAC name is 2,2,4-trimethylpentane — the substance older texts call isooctane — which is assigned a rating of 100 because it resists knocking in an engine, while straight-chain heptane is assigned 0. So the difference between a fuel that knocks and one that does not is, in part, a difference in branching. Nomenclature here is not book-keeping: it is the only unambiguous way to specify which of the many \(\mathrm{C_8H_{18}}\) isomers is meant.
14.4.5 IUPAC Nomenclature of Unsaturated Hydrocarbons (Alkenes and Alkynes)
While writing IUPAC names of alkenes and alkynes, the following rules are to be followed in addition to the rules already discussed. The reason a fresh set is needed is that the multiple bond, not the chain length alone, now controls both the choice of chain and the direction of numbering.
Rule 1. The longest continuous chain must include the carbon–carbon multiple bond. Thus the longest continuous chains in structures I and II of Fig. 14.25 contain four and six carbons respectively, even though a longer chain could be traced if the multiple bond were abandoned.
Rule 2. Numbering of this chain must be done in such a way that the carbon–carbon multiple bond gets the lowest possible locant number.
| \(\overset{4}{\mathrm{C}}\mathrm{H_3}\!-\!\overset{3}{\mathrm{C}}\mathrm{H_2}\!-\!\overset{2}{\mathrm{C}}\mathrm{H}\!=\!\overset{1}{\mathrm{C}}\mathrm{H_2}\) |
\(\overset{1}{\mathrm{C}}\mathrm{H_3}\!-\!\overset{2}{\mathrm{C}}\mathrm{H_2}\!-\!\overset{3}{\mathrm{C}}\mathrm{H}\!=\!\overset{4}{\mathrm{C}}\mathrm{H_2}\) |
| \(\overset{1}{\mathrm{H}}\mathrm{C}\equiv\overset{2}{\mathrm{C}}\!-\!\overset{3}{\mathrm{C}}\mathrm{H_2}\!-\!\overset{4}{\mathrm{C}}\mathrm{H_3}\) |
\(\overset{4}{\mathrm{H}}\mathrm{C}\equiv\overset{3}{\mathrm{C}}\!-\!\overset{2}{\mathrm{C}}\mathrm{H_2}\!-\!\overset{1}{\mathrm{C}}\mathrm{H_3}\) |
Rule 3. The ending “ane” of the alkane is replaced by “ene” for an alkene and by “yne” for an alkyne.
Rule 4. The position of the carbon atom from which the multiple bond starts is indicated by the smaller locant number of the two multiply bonded carbons, written before the ending “ene” or “yne”. For example:
| \(\mathrm{CH_3\!-\!CH_2\!-\!CH\!=\!CH_2}\) |
but-1-ene |
| \(\mathrm{HC}\equiv\mathrm{C\!-\!CH_2\!-\!CH_3}\) |
but-1-yne |
Only the smaller of the two numbers is written because the bond joins consecutive carbons; once you know it begins at C-1, it must end at C-2.
Rule 5. If the multiple bond is equidistant from both the ends of a selected chain, then the carbon atoms are numbered from that end which is nearer to the first branching. For example, in \(\mathrm{CH_3\!-\!CH(CH_3)\!-\!CH\!=\!C(CH_3)\!-\!CH_2\!-\!CH_3}\) the double bond lies between C-3 and C-4 whichever end you start from, so numbering begins at the end nearer the branch — reading the chain as 2,4-dimethylhex-3-ene. The same reasoning applies to \(\mathrm{CH_3\!-\!CH(CH_3)\!-\!C}\equiv\mathrm{C\!-\!CH_2\!-\!CH_3}\), read as 2-methylhex-3-yne.
Rule 6. If the parent chain contains two double bonds or two triple bonds, then it is named as a diene or a diyne. In all these cases the “a” of “ane” (alkane) is retained.
\[\mathrm{CH_3\!-\!CH\!=\!CH\!-\!CH\!=\!CH_2} \qquad \text{penta-1, 3-diene}\]
\[\mathrm{CH_3\!-\!C}\equiv\mathrm{C\!-\!CH_2\!-\!C}\equiv\mathrm{C\!-\!H} \qquad \text{hexa - 1, 4 -diyne}\]
That retained “a” is a small but examinable detail. We write penta-1,3-diene, not pent-1,3-diene, because the vowel is kept before a consonant-starting suffix such as “diene”.
Rule 7. If the parent chain contains both a double and a triple bond, then the carbon atoms are numbered from that end where the multiple bond is nearer. Such systems are named by putting the “en” ending first, followed by “yne”. The number indicating the location of each multiple bond is placed before that ending.
\[\mathrm{CH_2\!=\!CH\!-\!C}\equiv\mathrm{C\!-\!CH_3} \qquad \text{pent-1-en-3-yne}\]
\[\mathrm{H\!-\!C}\equiv\mathrm{C\!-\!CH_2\!-\!CH\!=\!CH\!-\!CH_3} \qquad \text{hex-4-en-1-yne}\]
Rule 8. If there is a tie between a double bond and a triple bond — that is, if both would get the same locant from opposite ends — the double bond gets the lower number.
\[\mathrm{H\!-\!C}\equiv\mathrm{C\!-\!CH_2\!-\!CH\!=\!CH_2} \qquad \text{pent-1-en-4-yne}\]
(question only — no solution) Write the IUPAC names of (i) \(\mathrm{CH_3\!-\!CH\!=\!CH\!-\!CH_2\!-\!CH_3}\), (ii) \(\mathrm{CH_3\!-\!CH_2\!-\!C}\equiv\mathrm{C\!-\!CH_3}\) and (iii) \(\mathrm{CH_2\!=\!CH\!-\!CH_2\!-\!CH\!=\!CH_2}\).
14.4.6 IUPAC Names of Simple Monocyclic Hydrocarbons
A saturated monocyclic hydrocarbon is named by attaching the prefix “cyclo” to the name of the corresponding open chain alkane. For example, cyclopropane and cyclopentane are drawn in Fig. 14.26.
An unsaturated monocyclic hydrocarbon is named by substituting “ene”, “yne” etc. for “ane” in the name of the corresponding cycloalkane. For example, cyclohexene and cyclopentene, also shown in Fig. 14.26.
If side chains are present, then the rules discussed above are applied. The numbering of the ring carbon is started from a side chain. For example, 1-ethyl-3-methylcyclohexane and 3-ethyl-1,1-dimethylcyclohexane are drawn in Fig. 14.27.
If the alkyl groups contain a greater number of carbon atoms than the ring, the compound is named as a derivative of the alkane, and the ring is treated as the substituent. For example, in \(\mathrm{CH_3\!-\!CH_2\!-\!CH(C_3H_5)\!-\!CH_2\!-\!CH_2\!-\!CH_3}\) the six-carbon chain is longer than the three-membered ring, so it is named 3-cyclopropylhexane. In the other example of Fig. 14.28 the ring is the larger part, so it becomes the parent and the compound is isopropylcyclohexane.
The rule is therefore a simple size comparison. Whichever part — ring or chain — carries more carbon atoms becomes the parent, and the smaller part becomes the substituent.
(question only — no solution) Name the compound formed by attaching a methyl group and a chlorine atom to adjacent carbons of a cyclopentane ring, choosing locants by the rules of this subsection. Then state which part would become the parent if the methyl group were replaced by a hexyl group.
14.4.7 IUPAC Nomenclature of Compounds Containing One or More Functional Groups
The IUPAC names of compounds containing one or more functional groups include the names of the functional groups as prefix and/or suffix. Table 14.9 shows how the names of functional groups are modified when included as prefix or suffix in the IUPAC name. Some functional groups appear only as prefixes, and these are shown in Table 14.9 (b).
Table 14.9 : Modified names of functional groups as in IUPAC nomenclature
a. Functional groups appearing prefix and suffix
| \(\mathrm{-COOH}\) |
carboxy |
\(-\) oic acid |
| \(\mathrm{-COOR}\) |
alkoxycarbonyl |
\(-\) oate |
| \(\mathrm{-COCl}\) |
chlorocarbonyl |
\(-\) oyl chloride |
| \(\mathrm{-CONH_2}\) |
carbamoyl |
\(-\) amide |
| \(\mathrm{-CN}\) |
cyano |
\(-\) nitrile |
| \(\mathrm{-CHO}\) |
formyl |
\(-\) al |
| \(\mathrm{-CO}-\) |
oxo |
\(-\) one |
| \(\mathrm{-OH}\) |
hydroxy |
\(-\) ol |
| \(\mathrm{-NH_2}\) |
amino |
\(-\) amine |
b. Functional groups appearing only as prefix
| \(\mathrm{-NO_2}\) |
nitro |
| \(\mathrm{-X}\) (\(\mathrm{-F}\), \(\mathrm{-Cl}\), \(\mathrm{-Br}\), \(\mathrm{-I}\)) |
halo (fluro, chloro, bromo, iodo) |
| \(\mathrm{-OR}\) (\(\mathrm{-OCH_3}\), \(\mathrm{-OC_2H_5}\), etc.) |
alkoxy (methoxy, ethoxy, etc.) |
The two columns of Table 14.9 (a) are two different jobs for the same group. A suffix says “this is the group the compound is named after”; a prefix says “this group is only a substituent here”. The groups in Table 14.9 (b) can never take the first role.
Naming monofunctional compounds
When a molecule contains only one functional group, the longest carbon chain containing that functional group is identified as the parent chain, and is numbered so as to give the smallest locant number to the carbon bearing the functional group.
The parent name is modified by applying the appropriate suffix. The location of the functional group is indicated where necessary — that is, when it is NOT ‘1’. For example:
\[\mathrm{CH_3-CH_2-CH_2-COOH} \qquad \text{butanoic acid (No need to mention -1-)}\]
\[\mathrm{CH_3-CH_2-C}\equiv\mathrm{N} \qquad \text{propanenitrile (No need to mention -1-)}\]
\[\mathrm{CH_3-CO-CH_2-CH_2-CH_3} \qquad \text{pentan-2-one (Necessary to mention -2-)}\]
\[\mathrm{CH_3-CO-CH_2-CH_3} \qquad \text{butanone (No need to mention -2-)}\]
Butanone needs no number for a different reason from butanoic acid. In a four-carbon ketone the carbonyl group can only be at C-2, since a carbonyl at C-1 would be an aldehyde — so the number carries no extra information and is dropped.
When the functional group cannot be used as a suffix, and can be only the prefix, the molecule is named as the parent alkane carrying the functional group as a substituent at a specified carbon. For example:
\[\mathrm{H_3C-CH_2-CH_2-NO_2} \qquad \text{1 - Nitropropane}\]
\[\mathrm{CH_3-CH(OCH_3)-CH_3} \qquad \text{2- Methoxypropane}\]
Polyfunctional compounds
The organic compounds possessing two or more functional groups (same or different) in their molecules are called polyfunctional compounds.
When there are two or more different functional groups, one of them is selected as the principal functional group and the others are considered as substituents. The principal functional group is used as the suffix of the IUPAC name, while the other substituents are written with appropriate prefixes.
The principal functional group is decided on the basis of the following order of priority:
\[\mathrm{-COOH} > \mathrm{-SO_3H} > \mathrm{-COOR} > \mathrm{-COCl} > \mathrm{-CONH_2} > \mathrm{-CN} > \mathrm{-CHO} > \mathrm{C\!=\!O} > \mathrm{-OH} > \mathrm{-NH_2} > \mathrm{C\!=\!C} > \mathrm{-C}\equiv\mathrm{C}-\]
This order is worth learning as a sequence rather than as isolated facts, since almost every polyfunctional naming question begins by applying it. Broadly, the acid and its derivatives rank highest, the carbonyl groups come next, then \(\mathrm{-OH}\) and \(\mathrm{-NH_2}\), and the multiple bonds rank lowest of all.
IUPAC Rules for naming mono or polyfunctional compounds
Rule i. The longest carbon chain containing or bearing the single or the principal functional group is identified as the parent chain. Ether is named as an alkoxyalkane; while naming it, the larger alkyl group is chosen as the parent chain. The numbering of the parent chain is done so as to give the lowest possible locant number to the carbon atom of this functional group.
Rule ii. The name of the parent hydrocarbon is modified adequately with the appropriate suffix, in accordance with the single or principal functional group.
Rule iii. Names of the other functional groups (if any) are attached to this modified name as prefixes. The locant numbers of all the functional groups are indicated before the corresponding suffix or prefix. The carbon atom in \(\mathrm{-COOR}\), \(\mathrm{-COCl}\), \(\mathrm{-CONH_2}\), \(\mathrm{-CN}\) and \(\mathrm{-CHO}\) is C-1 by rule and therefore is not mentioned in the IUPAC name.
The complete procedure is set out in FC 14.4.
Write IUPAC names for the following structures:
i. \(\mathrm{CH_3-CH(OH)-CO-CH_2-CH_3}\)
ii. \(\mathrm{CH_3-CH(NH_2)-CH_2-COOH}\)
iii. \(\mathrm{CH_3-CH(OH)-CH_2-OH}\)
iv. \(\mathrm{CH_2\!=\!CH-CH\!=\!CH_2}\)
Solution:
- \(\mathrm{CH_3-CH(OH)-CO-CH_2-CH_3}\)
Here the principal functional group, ketone, is located at the C-3 on the five carbon chain. The \(\mathrm{-OH}\) group, the hydroxyl substituent, is at C-2. Therefore the IUPAC name is 2-hydroxypentan-3-one.
- \(\mathrm{CH_3-CH(NH_2)-CH_2-COOH}\)
Here the principal functional group is carboxylic acid. The amino substituent is located at C-3 on the four carbon chain. Therefore the IUPAC name is 3-aminobutanoic acid.
- \(\mathrm{CH_3-CH(OH)-CH_2-OH}\)
Both functional groups are alcohols, so there is no question of priority between them; the parent chain is the three-carbon chain and the suffix “ol” is used twice with the counting prefix “di”. Numbering from the end that gives the lower set of locants puts the two \(\mathrm{-OH}\) groups at C-1 and C-2. The name is propane-1,2-diol.
- \(\mathrm{CH_2\!=\!CH-CH\!=\!CH_2}\)
There is no functional group other than the two double bonds. The parent chain has four carbons and carries two double bonds, so by Rule 6 of subsection 14.4.5 it is a diene with the “a” retained. The double bonds begin at C-1 and C-3, giving buta-1,3-diene.
Note how the numbering in (i) and (ii) was decided in each case. It is the principal functional group, not the substituent, that must receive the lowest locant — which is why the \(\mathrm{-OH}\) in (i) ends up at C-2 rather than at C-4.
Give the IUPAC name of \(\mathrm{CH_3\!-\!CH(OH)\!-\!CH_2\!-\!CHO}\).
Solution: First list the functional groups present: one \(\mathrm{-CHO}\) and one \(\mathrm{-OH}\).
Next apply the priority order. In the sequence given above, \(\mathrm{-CHO}\) ranks higher than \(\mathrm{-OH}\), so the aldehyde is the principal functional group and the alcohol becomes a substituent named with the prefix “hydroxy”.
The parent chain contains four carbons including the aldehyde carbon, so the parent name is butane and the suffix “al” gives butanal. The aldehyde carbon is C-1 by rule, so its locant is not written.
Numbering from that carbon places the \(\mathrm{-OH}\) on C-3. The IUPAC name is therefore 3-hydroxybutanal.
(question only — no solution) Write the IUPAC name of \(\mathrm{CH_3\!-\!CH(NH_2)\!-\!CH_2\!-\!CH_2\!-\!CHO}\), and state which of its two functional groups appears as the suffix and why.
14.4.8 IUPAC Nomenclature of Substituted Benzene
I. Monosubstituted benzene
The IUPAC name of a monosubstituted benzene is obtained by placing the name of the substituent as a prefix to the parent skeleton, which is benzene. Since all six positions of the ring are equivalent, no locant number is needed. Examples are drawn in Fig. 14.29: chlorobenzene, nitrobenzene, bromobenzene, iodobenzene and tert-butylbenzene.
Some monosubstituted benzenes have trivial names which may show no resemblance with the name of the attached substituent group. For example, methylbenzene is known as toluene, aminobenzene as aniline, hydroxybenzene as phenol, and so on. The common names written in the bracket are also used universally and are accepted by IUPAC. Six such compounds are shown in Fig. 14.30.
If the alkyl substituent is larger than the benzene ring (7 or more carbon atoms), the compound is named as a phenyl-substituted alkane. For example, the compound with a seven-carbon chain bearing a ring at C-2 is 2-phenylheptane, while a six-carbon chain on a ring remains hexyl benzene, since six carbons do not exceed the ring. Both are shown in Fig. 14.31.
The benzene ring can as well be considered as a substituent when it is attached to an alkane bearing a functional group. In such a case the functional group decides the parent name and the ring becomes a phenyl substituent. Figure 14.31 also shows 1-chloro-3-phenylpropane, phenylmethanol (benzyl alcohol) and 1-phenylethanone (acetophenone).
II. Disubstituted benzene derivatives
Common names of the three possible isomers of disubstituted benzene derivatives are given using one of the prefixes ortho (o-), meta (m-) or para (p-). These describe the relative positions of the two groups: adjacent, one carbon apart, and directly opposite.
The IUPAC system, however, uses numbering instead of the prefixes o-, m- or p-. Thus the three dichlorobenzenes shown in Fig. 14.32 are named 1,2-dichlorobenzene (o-dichlorobenzene), 1,3-dichlorobenzene (m-dichlorobenzene) and 1,4-dichlorobenzene (p-dichlorobenzene).
If the two substituents are different, then they enter the name in alphabetical order. For example, 1-bromo-4-iodobenzene (p-bromoiodobenzene) and 1-chloro-3-nitrobenzene (m-chloronitrobenzene), also drawn in Fig. 14.32.
If one of the two groups gives a special name to the molecule, then the compound is named as a derivative of that special compound. The carbon carrying that special group is then numbered 1. This gives 2-chloroaniline (o-chloroaniline), 3-nitrotoluene (m-nitrotoluene) and 4-bromophenol (p-bromophenol).
A benzene ring carries an \(\mathrm{-NH_2}\) group and a \(\mathrm{-Br}\) atom on opposite carbons of the ring. Write its IUPAC name and its common name.
Solution: First look for a group that gives the molecule a special name. Aminobenzene is universally known as aniline, so the compound is named as a derivative of aniline and the carbon bearing \(\mathrm{-NH_2}\) is numbered 1.
The two groups are on opposite carbons of the ring, which means they are separated by two carbons on either side. Counting from the amino carbon, the bromine therefore falls on C-4.
The IUPAC name is 4-bromoaniline. Since “opposite” corresponds to the para relationship, the common name is p-bromoaniline.
III. Trisubstituted benzene derivatives
If more than two substituents are attached to the benzene ring, numbers are used to indicate their relative positions, following the alphabetical order and the lowest locant rule. In some cases, the common name of the benzene derivative is taken as the parent compound.
Examples are drawn in Fig. 14.33: 1,2,4-tribromobenzene, 1-chloro-2,4-dinitrobenzene, 4-bromo-1,2-dimethylbenzene, 2-chloro-1-methyl-4-nitrobenzene, 2-chloro-4-nitrophenol, 2,4,6-tribromoaniline and 2-chloro-4-methylanisole.
Compare the last three names in Fig. 14.33 with the first four. Where no special-name group is present, the ring is numbered purely by the lowest locant and alphabetical rules. Where phenol, aniline or anisole is present, that group’s carbon is fixed as C-1 and the remaining substituents are numbered from it — which is why the nitro group in 2-chloro-4-nitrophenol takes the number 4 rather than a lower one.
(question only — no solution) A benzene ring carries a \(\mathrm{-CH_3}\) group at one position and \(\mathrm{-NO_2}\) groups at the two positions adjacent to it. Write the IUPAC name of the compound, and state which rule fixes the position numbered 1.
14.5 Isomerism
Two structural formulae are drawn in Fig. 14.34. a. Observe the structural formulae (i) and (ii). b. Find out their molecular formulae. c. What is the difference between them? d. What is the relation between the two compounds represented by these structural formulae?
The phenomenon of existence of two or more compounds possessing the same molecular formula is known as isomerism. Such compounds are known as isomers of each other.
Isomerism is the existence of two or more compounds that have the same molecular formula but are not identical; such compounds are called isomers.
Two types of isomerism are observed in organic compounds, namely structural isomerism and stereoisomerism.
When two or more compounds have the same molecular formula but different structural formulae, it is called structural isomerism. On the other hand, when different compounds have the same structural formula but a different relative arrangement of groups or atoms in space — that is, a different spatial arrangement of groups or atoms — it is called stereoisomerism.
Structural isomerism is isomerism in which the isomers have the same molecular formula but different structural formulae, that is, a different sequence of bonding.
Stereoisomerism is isomerism in which the isomers have the same structural formula but differ in the arrangement of their atoms or groups in space.
The distinction is worth fixing firmly, because the whole classification rests on it. In structural isomerism the atoms are joined up differently. In stereoisomerism the atoms are joined up in exactly the same order, and only their positions in three-dimensional space differ — which is why the wedge, Fischer, Newman and sawhorse formulae of Section 14.2.3 were needed at all. In this chapter we are going to look at some details of only structural isomerism.
14.5.1 Structural Isomerism
We have seen at the beginning of this chapter that the structural formula describes the sequence in which the constituent atoms of a molecule are bonded, and the nature of the bonds between them.
When two or more compounds have the same molecular formula and different structural formulae, they are said to be structural isomers of each other. The phenomenon is known as structural isomerism.
Structural isomerism is further classified as chain isomerism, position isomerism, functional group isomerism, metamerism and tautomerism. Figure 14.9 shows the various types of isomerism, structural and stereo, in one scheme.
a. Chain isomerism
When two or more compounds have the same molecular formula but a different parent chain, or a different carbon skeleton, it is referred to as chain isomerism, and such isomers are known as chain isomers.
For example, butane (\(\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!CH_3}\)) and 2-methylpropane [\(\mathrm{CH_3\!-\!CH(CH_3)\!-\!CH_3}\)] are chain isomers of each other, as they have different carbon skeletons and the same molecular formula \(\mathrm{C_4H_{10}}\).
Count the atoms in both formulae and the point becomes clear: four carbons and ten hydrogens in each, yet one is a straight chain of four while the other is a chain of three carrying a branch. This is why chain isomerism is also spoken of as skeletal isomerism.
b. Position isomerism
The phenomenon in which different compounds have the same functional groups at different positions on the parent chain is known as position isomerism.
For example, but-2-ene [\(\mathrm{CH_3\!-\!CH\!=\!CH\!-\!CH_3}\)] and but-1-ene [\(\mathrm{CH_2\!=\!CH\!-\!CH_2\!-\!CH_3}\)] are position isomers of each other, as they have the same molecular formula \(\mathrm{C_4H_8}\) and the same carbon skeleton, but the double bonds are located at different positions.
Here the double bond counts as the functional group, which is why moving it along an unchanged four-carbon chain produces position isomers rather than chain isomers. Notice also how the IUPAC names of Section 14.4 do the distinguishing work automatically: the locant in the name is precisely what differs.
c. Functional group isomerism
The phenomenon in which different compounds with the same molecular formula have different functional groups is known as functional group isomerism.
For example, dimethyl ether (\(\mathrm{CH_3\!-\!O\!-\!CH_3}\)) and ethyl alcohol (\(\mathrm{CH_3\!-\!CH_2\!-\!OH}\)) have the same molecular formula \(\mathrm{C_2H_6O}\), but the former has the ether (\(\mathrm{-O-}\)) functional group and the latter has the alcohol (\(\mathrm{-OH}\)) functional group.
Because the functional group decides chemical behaviour, functional group isomers belong to entirely different families and behave as chemically unrelated substances — even though a combustion analysis of the two would give identical results.
Q1. Two compounds can have exactly the same molecular formula and still behave completely differently. What is it about a molecule, other than the atoms it contains, that decides how it behaves?
A. Its structure — that is, which atom is bonded to which, and how those atoms are arranged in space. The molecular formula is only an inventory of parts. Ethanol and dimethyl ether both count as \(\mathrm{C_2H_6O}\), yet in one the oxygen carries a hydrogen and in the other it sits between two carbons, so one is an alcohol and the other an ether. Chemical behaviour follows the functional group and the skeleton, not the tally of atoms, which is why an analysis of composition alone can never identify a compound.
These two functional group isomers of \(\mathrm{C_2H_6O}\) are put to entirely different industrial uses. Ethanol is a liquid at room temperature and serves as a solvent, a disinfectant and a blended motor fuel. Dimethyl ether is a gas that liquefies easily under modest pressure, so it is used as an aerosol propellant — one of the substances that replaced the ozone-depleting chlorofluorocarbons — and is being developed as a low-soot diesel substitute. Identical composition; two different industries.
e. Tautomerism
The same compound exists as a mixture of two or more structurally distinct molecules which are in rapid equilibrium with each other. This phenomenon is referred to as tautomerism, and such interconverting isomers are called tautomers.
In nearly all the cases, it is the proton which shifts from one atom to another atom in the molecule to form its tautomer. The general change is shown in Fig. 14.35, where the keto form on the left carries a hydrogen on the carbon next to the carbonyl group and the enol form on the right carries it on oxygen.
Tautomerism is the existence of a compound as a mixture of two or more structurally distinct molecules in rapid equilibrium with each other, usually through the shift of a proton from one atom to another within the molecule.
For example, keto-enol tautomerism is a very common form of tautomerism. It is found in carbonyl compounds. Here a hydrogen atom shifts reversibly from the \(\alpha\)-carbon of the keto form to the oxygen atom of the enol.
One term used there needs stating plainly, since it recurs throughout organic chemistry. The \(\alpha\)-carbon is simply the carbon atom directly attached to the functional group — in a carbonyl compound, the carbon next to the \(\mathrm{C\!=\!O}\). A carbonyl compound with no hydrogen on its \(\alpha\)-carbon cannot form an enol at all, because there is no proton available to shift.
The word “enol” is a piece of chemical shorthand built out of two suffixes you already know from Section 14.4: “ene” for a carbon–carbon double bond and “ol” for an alcohol. An enol is exactly that — a double bond with an \(\mathrm{-OH}\) attached to one of its carbons. “Keto” simply marks the ketone-like carbonyl form.
chain vs position vs functional group vs metamerism vs tautomerism | Feature | Chain isomerism | Position isomerism | Functional group isomerism | Metamerism | Tautomerism | |—|—|—|—|—|—| | What differs | The carbon skeleton | Position of the same functional group | The functional group itself | Distribution of carbons about the functional group | Position of a proton and of a double bond | | Carbon skeleton | Different | Same | May be same or different | Rearranged about the group | Same set of atoms, rebonded | | Functional group | Same (or none) | Same | Different | Same | Interconverting forms | | Can the isomers be separated? | Yes | Yes | Yes | Yes | Usually not — they interconvert rapidly | | Example | Butane and 2-methylpropane | But-1-ene and but-2-ene | Ethyl alcohol and dimethyl ether | Ethoxyethane and methoxypropane | Keto and enol forms of a carbonyl compound |
To decide which type a given pair shows, work through the tests of FC 14.5 in order.
State the type of structural isomerism shown by each of the following pairs, with a reason: (i) pentane and 2-methylbutane; (ii) propan-1-ol and propan-2-ol; (iii) propanone and propanal; (iv) ethoxyethane and methoxypropane.
Solution: In every case first confirm that the molecular formulae agree, then apply the tests of FC 14.5.
Both are \(\mathrm{C_5H_{12}}\). Pentane is a straight chain of five carbons; 2-methylbutane is a four-carbon chain carrying a methyl branch. The skeletons differ, so this is chain isomerism.
Both are \(\mathrm{C_3H_8O}\) and both are alcohols on an unbranched three-carbon chain. Only the locant of the \(\mathrm{-OH}\) has changed, from C-1 to C-2, so this is position isomerism.
Both are \(\mathrm{C_3H_6O}\). Propanone contains the ketone group and propanal contains the aldehyde group, so the functional groups themselves are different. This is functional group isomerism — and these are exactly the two compounds drawn in Fig. 14.34.
Both are \(\mathrm{C_4H_{10}O}\) and both are ethers, so neither the formula nor the functional group distinguishes them. The oxygen carries two ethyl groups in one and a methyl plus a propyl group in the other, an unequal distribution of carbons about the same functional group. This is metamerism.
(question only — no solution) Draw all the chain isomers of \(\mathrm{C_5H_{12}}\), give the IUPAC name of each, and state how many of them contain a quaternary carbon atom.
(question only — no solution) Identify the type of structural isomerism in each pair and justify your answer in one line: (i) but-1-yne and buta-1,3-diene; (ii) methoxypropane and 2-methoxypropane; (iii) propan-2-one and prop-1-en-2-ol; (iv) N-methylethanamine and propan-1-amine.
14.6 Theoretical Basis of Organic Reactions
All the organic molecules primarily contain various types of covalent bonds between their constituent atoms. During an organic reaction, molecules of the reactant undergo a change in their structure.
A covalent bond at a carbon atom in the reactant is broken and a new covalent bond is formed at it, giving rise to the product. Often this change is brought about by means of another reactant.
Substrate, reagent, product and byproduct
The reactant that provides carbon to the new bond is called the substrate. The other reactant, which brings about this change, is called the reagent. In case both the reactants provide carbon atoms to a new bond, the reactant considered is called the substrate.
Apart from the product of interest, some other products are formed in a reaction. These are called byproducts. For example:
\[\mathrm{CH_4} + \mathrm{Cl_2} \xrightarrow{\ h\nu\ } \mathrm{CH_3Cl} + \mathrm{HCl}\]
In this reaction \(\mathrm{CH_4}\) is the substrate, \(\mathrm{Cl_2}\) is the reagent, \(\mathrm{CH_3Cl}\) is the product and \(\mathrm{HCl}\) is the byproduct.
The substrate is the reactant that supplies the carbon atom to the newly formed bond; the reagent is the other reactant that brings about the change.
Which reactant is called the substrate is therefore decided by what the chemist is trying to make, not by quantity. In the reaction above both molecules are consumed, but only methane contributes the carbon to the new C–Cl bond.
14.6.1 Types of Cleavage of Covalent Bond
A covalent bond can undergo cleavage or fission in two ways: (i) homolytic cleavage and (ii) heterolytic cleavage. A covalent bond comprises two electrons — the bond pair of electrons — shared between the two bonded atoms, and the two modes of cleavage differ simply in how that pair is divided.
Homolytic cleavage
In homolytic cleavage of a covalent bond, one of the two electrons goes to one of the bonded atoms and the other is bound to the other atom. The bond pair is thus split equally.
The movement of a single electron is represented by a half-headed curved arrow, or fish hook. The tail of the curved arrow indicates the place from where the electron is removed, while the head of the curved arrow indicates the place where the electron reaches as the result of the movement.
Such homolytic cleavage of a covalent bond gives rise to two neutral species carrying one unpaired electron each. A species with a single unpaired electron is called a ‘free radical’, or sometimes just a ‘radical’.
A free radical is unstable and thereby a reactive species, having a tendency to seek an electron for pairing. A homolytic cleavage is thus represented as shown in Fig. 14.36.
\[\mathrm{A\!-\!B} \longrightarrow \mathrm{A^{\bullet}} + \mathrm{B^{\bullet}}\]
For example,
\[\mathrm{H_3C\!-\!Cl} \xrightarrow{\ h\nu\ } \mathrm{H_3C^{\bullet}} + \mathrm{Cl^{\bullet}}\]
where \(\mathrm{H_3C^{\bullet}}\) is the methyl free radical and \(\mathrm{Cl^{\bullet}}\) is the chlorine radical, or chlorine atom.
A free radical is a species carrying a single unpaired electron; it is electrically neutral, unstable and highly reactive.
Homolytic fission is favoured in the presence of ultraviolet radiation or suitable peroxides, or at high temperatures. Each of these supplies energy or a ready-made radical, so the bond pair separates evenly instead of being pulled towards one atom.
The organic reactions which proceed by homolytic fission are known as free radical or non-polar reactions. Free radicals have only a transitory existence. These are formed as reaction intermediates which subsequently react with another radical or molecule to restore a stable bonding pair.
A carbon free radical is \(sp^2\) hybridized and reveals a planar trigonal geometry, as drawn in Fig. 14.37. The single unpaired electron occupies the \(p_z\) orbital, which stands perpendicular to the plane of the three sigma bonds.
Alkyl free radicals are classified as primary (\(1^0\)), secondary (\(2^0\)) and tertiary (\(3^0\)), in the same way as the carbons of Section 14.3.2 were classified. The observed stability order of alkyl free radicals is shown in Fig. 14.38:
\[\mathrm{(CH_3)_3C^{\bullet}} > \mathrm{CH_3\!-\!\overset{\bullet}{C}H\!-\!CH_3} > \mathrm{CH_3\!-\!\overset{\bullet}{C}H_2} > \mathrm{\overset{\bullet}{C}H_3}\]
that is, tert-butyl free radical (\(3^0\)) > isopropyl free radical (\(2^0\)) > ethyl free radical (\(1^0\)) > methyl free radical.
Free radicals are not merely laboratory curiosities. The polyethene of carry bags, bottles and cable insulation is made by a free radical chain reaction: a peroxide initiator breaks homolytically, the resulting radical attacks an ethene molecule, and the new radical attacks the next one, thousands of times over. The same peroxides that start this useful chain are the reason old, air-exposed ether bottles are handled with great care in a laboratory — an ether that has slowly built up peroxides can start a reaction of its own on being disturbed.
Heterolytic cleavage
In heterolytic cleavage of a covalent bond, both the shared electrons go to one of the two bonded atoms. This turns out to be the more electronegative atom of the two.
The movement of an electron pair is represented by a full-headed curved arrow. Heterolytic cleavage of a covalent bond gives rise to two charged species, one with a negative charge and the other with a positive charge.
The negatively charged (anionic) species has the more electronegative atom, which has taken away the shared pair of electrons with it. A heterolytic cleavage is represented as in Fig. 14.39:
\[\mathrm{A\!-\!B} \longrightarrow \mathrm{A^{\oplus}}\ (\text{cation}) + \mathrm{B^{\ominus}}\ (\text{anion})\]
where B is more electronegative than A. For example,
\[\mathrm{H_3C\!-\!Br} \longrightarrow \mathrm{H_3C^{\oplus}} + \mathrm{Br^{\ominus}}\]
Here \(\mathrm{H_3C^{\oplus}}\) is the methyl carbocation and \(\mathrm{Br^{\ominus}}\) is the bromide ion, since Br is more electronegative than C.
Heterolysis is favoured in polar solvents, because the separated ions are stabilised by the solvent that surrounds them. In methyl bromide, bromine is more electronegative; thus heterolysis of methyl bromide results in the formation of a methyl carbocation and a bromide anion.
Carbocations
A carbon atom having a sextet of electrons and a positive charge is known as a carbocation (earlier called a carbonium ion). Carbocations are highly unstable and reactive species, formed as intermediates in many organic reactions.
A carbocation is a species in which a carbon atom carries a positive charge and only six valence electrons — a sextet — around it.
The central carbon atom of a carbocation is in the \(sp^2\) hybrid state and has trigonal planar geometry, as shown in Fig. 14.40. The \(p_z\) atomic orbital is vacant and is perpendicular to the plane containing the three sigma bonds at the carbon.
Compare Fig. 14.37 with Fig. 14.40. Both species are \(sp^2\) hybridized and trigonal planar; the difference lies only in the \(p_z\) orbital, which holds one electron in the free radical and is completely empty in the carbocation.
Carbocations are classified as primary (\(1^0\)), secondary (\(2^0\)) and tertiary (\(3^0\)). The observed stability of alkyl carbocations follows the order shown in Fig. 14.41:
\[\mathrm{(CH_3)_3\overset{\oplus}{C}} > \mathrm{CH_3\!-\!\overset{\oplus}{C}H\!-\!CH_3} > \mathrm{CH_3\!-\!\overset{\oplus}{C}H_2} > \mathrm{\overset{\oplus}{C}H_3}\]
that is, t-butyl carbocation > iso-propyl carbocation > ethyl carbocation > methyl carbocation.
Notice that the same order — tertiary above secondary above primary above methyl — holds for both free radicals and carbocations. The reason for it is taken up in Section 14.6.8.
Carbanions
Heterolytic bond fission also generates a species called a carbanion, in which carbon gets the shared pair of electrons. This happens when a carbon atom is bonded to an electropositive atom, as drawn in Fig. 14.42.
\[\mathrm{H_3C\!-\!Z} \xrightarrow{\ \text{Heterolysis}\ } \mathrm{H_3C^{\ominus}} + \mathrm{Z^{\oplus}}\]
where Z is more electropositive than C.
The central carbon atom of a carbanion has a completed octet of electrons and a negative charge. The carbanion is unstable and reactive, and is formed as a reaction intermediate.
A carbanion is a species in which a carbon atom carries a negative charge and a complete octet of valence electrons, having retained both electrons of a broken bond.
Heterolytic fission is favoured in polar solvents. The organic reactions that proceed via heterolytic fission are known as heteropolar, polar or ionic reactions. Most organic reactions taking place in solutions follow an ionic mechanism.
homolytic vs heterolytic cleavage | Feature | Homolytic cleavage | Heterolytic cleavage | |—|—|—| | Division of the bond pair | One electron to each atom | Both electrons to one atom | | Arrow used | Half-headed curved arrow (fish hook) | Full-headed curved arrow | | Species produced | Two neutral free radicals | One cation and one anion | | Which atom keeps the electrons | Neither; the pair is split | The more electronegative atom | | Favoured by | Ultraviolet radiation, peroxides, high temperature | Polar solvents | | Reactions so produced | Free radical or non-polar reactions | Heteropolar, polar or ionic reactions |
free radical vs carbocation vs carbanion | Feature | Free radical | Carbocation | Carbanion | |—|—|—|—| | Charge on the carbon | None | Positive | Negative | | Valence electrons at that carbon | Seven, one of them unpaired | Six, a sextet | Eight, a complete octet | | Hybridization and shape | \(sp^2\), planar trigonal | \(sp^2\), trigonal planar | Carries a lone pair on carbon | | The \(p_z\) orbital | Holds one electron | Vacant | — | | Formed by | Homolytic cleavage | Heterolytic cleavage, carbon bonded to a more electronegative atom | Heterolytic cleavage, carbon bonded to a more electropositive atom |
The two modes of cleavage and the three intermediates they generate are gathered together in FC 14.6.
Some bond fissions are described in the following table. For each of them, show the movement of electron/s using curved arrow notation. Classify them as homolysis or heterolysis, and identify the intermediate species produced as carbocation, carbanion or free radical.
| \(\mathrm{CH_3\!-\!Cu} \longrightarrow \mathrm{CH_3^{\ominus}} + \mathrm{Cu^{\oplus}}\) |
heterolysis |
carbanion |
| \(\mathrm{C_6H_5\!-\!O\!-\!O\!-\!C_6H_5} \longrightarrow 2\,\mathrm{C_6H_5O^{\bullet}}\) |
……………… |
……………… |
| \(\mathrm{CH_3\!-\!CO\!-\!CH_2\!-\!H} + \mathrm{{}^{\ominus}OH} \longrightarrow \mathrm{CH_3\!-\!CO\!-\!CH_2^{\ominus}} + \mathrm{H_2O}\) |
……………… |
……………… |
| \(\mathrm{CH_3\!-\!C(CH_3)_2\!-\!Br} \longrightarrow \mathrm{(CH_3)_3C^{\oplus}} + \mathrm{Br^{\ominus}}\) |
……………… |
……………… |
| \(\mathrm{CH_3\!-\!CN} \longrightarrow \mathrm{CH_3^{\oplus}} + \mathrm{{}^{\ominus}CN}\) |
……………… |
……………… |
Arrange the following carbocations in decreasing order of stability and give the reason for your order: \(\mathrm{CH_3^{\oplus}}\), \(\mathrm{(CH_3)_2CH^{\oplus}}\), \(\mathrm{CH_3CH_2^{\oplus}}\), \(\mathrm{(CH_3)_3C^{\oplus}}\).
Solution: First classify each cation by the number of carbon atoms attached to the positively charged carbon. \(\mathrm{CH_3^{\oplus}}\) has none, \(\mathrm{CH_3CH_2^{\oplus}}\) has one and is primary, \(\mathrm{(CH_3)_2CH^{\oplus}}\) has two and is secondary, and \(\mathrm{(CH_3)_3C^{\oplus}}\) has three and is tertiary.
A carbocation is electron deficient, since the positive carbon has only six valence electrons. Anything that pushes electron density towards that carbon will therefore stabilise it. Alkyl groups are electron releasing groups, exerting a \(+\mathrm{I}\) effect on the carbon chain, so the more alkyl groups attached, the more the positive charge is relieved.
The order of decreasing stability is therefore
\[\mathrm{(CH_3)_3C^{\oplus}} > \mathrm{(CH_3)_2CH^{\oplus}} > \mathrm{CH_3CH_2^{\oplus}} > \mathrm{CH_3^{\oplus}}\]
that is, tertiary > secondary > primary > methyl, which is exactly the observed order given in Fig. 14.41.
(question only — no solution) For the fission \(\mathrm{(CH_3)_3C\!-\!Cl} \longrightarrow \mathrm{(CH_3)_3C^{\oplus}} + \mathrm{Cl^{\ominus}}\), state (i) the type of cleavage, (ii) the type of arrow used to show the electron movement, (iii) the hybridization and shape of the cation formed, and (iv) whether a polar or a non-polar solvent would favour this fission.
14.6.2 Types of Reagent
- What is meant by ‘reagent’?
- Identify the ‘reagent’, ‘substrate’, ‘product’ and ‘byproduct’ in the following reaction.
\[\mathrm{CH_3COCl} + \mathrm{NH_3} \longrightarrow \mathrm{CH_3CONH_2} + \mathrm{HCl}\]
The polar organic reactions are brought about by two types of reagents. Depending upon their ability to accept or donate electrons from or to the substrate, reagents are classified as (i) electrophiles (\(\mathrm{E^{\oplus}}\)) and (ii) nucleophiles (\(\mathrm{Nu:}\)).
Electrophiles, meaning electron loving species, accept electrons from the substrate. Thus electrophiles are electron seeking species. This is because they themselves are electron deficient. For example, a positively charged ion such as \(\mathrm{Br^{\oplus}}\) or \(\mathrm{{}^{\oplus}CH_3}\), or a neutral species having vacant orbitals such as \(\mathrm{AlCl_3}\).
Nucleophiles, meaning nucleus seeking species, give away electrons to the substrate. This is because nucleophiles are electron rich species. For example, negatively charged species such as \(\mathrm{OH^{\ominus}}\), or neutral species such as \(\mathrm{H_2O}\), having a lone pair of electrons.
An electrophile is an electron-deficient reagent that accepts a pair of electrons from the substrate.
A nucleophile is an electron-rich reagent that donates a pair of electrons to the substrate.
The two names describe what each species is looking for. An electrophile is short of electrons, so it seeks out the electron-rich part of a molecule. A nucleophile has electrons to spare, so it seeks a positively polarised centre — a nucleus, in the older description.
Identify the nucleophile and electrophile from \(\mathrm{NH_3}\) and \(\mathrm{{}^{\oplus}CH_3}\). Also indicate the nucleophilic and electrophilic centres in them. Justify.
Solution: The structural formulae of the two reagents showing all the valence electrons are drawn in Fig. 14.43.
Thus \(\mathrm{NH_3}\) contains N with a lone pair of electrons, which can be given away to another species. Therefore \(\mathrm{NH_3}\) is a nucleophile and ‘N’ in it is the nucleophilic centre.
The \(\mathrm{{}^{\oplus}CH_3}\) is a positively charged, electron deficient species having a vacant orbital on the carbon. It is an electrophile. The ‘C’ in it is the electrophilic centre.
Electrophilic and nucleophilic centres
A polyatomic electrophile has an electron deficient atom in it called the electrophilic centre, while a polyatomic nucleophile has an electron rich atom in it, called the nucleophilic centre.
For example, the electrophilic centre of the electrophile \(\mathrm{AlCl_3}\) is Al, which has only 6 valence electrons. The nucleophilic centre of the nucleophile \(\mathrm{H_2O}\) is ‘O’, which has two lone pairs of electrons.
A nucleophile attacks the electrophilic centre in the substrate and brings about a nucleophilic reaction. On the other hand, an electrophile attacks a nucleophilic centre in the substrate and brings about an electrophilic reaction. This principle is illustrated in the following simple reaction, though it is not an organic reaction.
\[\mathrm{NH_3} + \mathrm{BF_3} \longrightarrow \mathrm{H_3N^{\oplus}\!-\!BF_3^{\ominus}}\]
Here the nucleophilic centre N in the nucleophile \(\mathrm{NH_3}\) attacks the electrophilic centre ‘B’ in the electrophile \(\mathrm{BF_3}\) and gives the product.
electrophile vs nucleophile | Feature | Electrophile (\(\mathrm{E^{\oplus}}\)) | Nucleophile (\(\mathrm{Nu:}\)) | |—|—|—| | Meaning of the name | Electron loving | Nucleus seeking | | Electron status | Electron deficient | Electron rich | | Action on the substrate | Accepts a pair of electrons | Donates a pair of electrons | | Charge | Positive, or neutral with a vacant orbital | Negative, or neutral with a lone pair | | The reactive centre | Electrophilic centre — the electron deficient atom | Nucleophilic centre — the atom bearing the lone pair | | Examples | \(\mathrm{Br^{\oplus}}\), \(\mathrm{{}^{\oplus}CH_3}\), \(\mathrm{AlCl_3}\), \(\mathrm{BF_3}\) | \(\mathrm{OH^{\ominus}}\), \(\mathrm{H_2O}\), \(\mathrm{NH_3}\) |
anhydrous aluminium chloride, \(\mathrm{AlCl_3}\) (What it does → the aluminium atom has only six valence electrons and a vacant orbital, so it accepts an electron pair and behaves as a Lewis acid, generating a strong electrophile from an otherwise sluggish reagent. The conditions it needs → strictly anhydrous, since traces of moisture destroy its activity. What it is typically used for → catalysing electrophilic substitution on aromatic rings, where the electron-rich ring supplies the electrons and \(\mathrm{AlCl_3}\) creates the attacking electrophile.)
The words electrophile and nucleophile were introduced in the 1930s by the English chemist Christopher Ingold, who built much of the modern language of reaction mechanisms. Both are literal descriptions: “electro-phile” is electron-loving, “nucleo-phile” is nucleus-seeking. Once reagents could be sorted into just these two classes, a very large number of separate reactions could be seen as variations of two patterns.
(question only — no solution) Classify each of the following as an electrophile or a nucleophile and name the reactive centre in each: (i) \(\mathrm{CN^{\ominus}}\), (ii) \(\mathrm{BF_3}\), (iii) \(\mathrm{NH_3}\), (iv) \(\mathrm{{}^{\oplus}NO_2}\), (v) \(\mathrm{H_2O}\).
14.6.3 Electronic Effects in Organic Reaction
We noted earlier that in polar organic reactions the nucleophilic centre in a nucleophile attacks the electrophilic centre in the substrate, whereas the electrophilic centre in an electrophile attacks the nucleophilic centre in the substrate.
This raises an obvious question: how is an electrophilic or nucleophilic centre generated in a neutral substrate? A neutral molecule carries no charge anywhere, so it is not obvious why one of its atoms should be attacked rather than another.
This happens because of the displacement of valence electrons within some organic molecules, which results in their polarization. The displacement of valence electrons resulting in polarization of an organic molecule is called an electronic effect.
An electronic effect is the displacement of valence electrons within an organic molecule that results in the polarization of that molecule.
The electronic effect that occurs in the ground state is a permanent effect. This takes place due to the influence of a certain atom or substituent, or a certain structural feature present in the molecule. Inductive effect and resonance effect are two examples of permanent electronic effect.
On the other hand, the electron displacement effect occurring in a substrate due to the approach of the attacking reagent is a temporary effect. This type of electronic effect is called the electromeric effect or polarisability effect.
The distinction between permanent and temporary is worth holding on to. A permanent effect is present in the bottle, before any reagent arrives. A temporary effect exists only while the reagent is approaching, and vanishes when it is withdrawn. We consider the various types of electronic effects in the following sections.
14.6.4 Inductive Effect
- How is a covalent bond formed between two atoms?
- Consider two covalently bonded atoms Q and R, where R is more electronegative than Q. Will these atoms share the electron pair equally between them?
- Represent the above polar covalent bond between Q and R using fractional charges \(\delta^{\oplus}\) and \(\delta^{\ominus}\).
We learnt earlier that a covalent bond between two atoms differing in their electronegativity is a polar covalent bond. For example, the covalent bond between carbon and the more electronegative chlorine is a polar covalent bond.
When an organic molecule has a polar covalent bond in its structure, polarity is induced in the adjacent carbon–carbon single bonds too. This is called the inductive effect. Consider, for example, the chloroethane molecule:
\[\overset{2}{\mathrm{C}}\mathrm{H_3}\!-\!\overset{1}{\mathrm{C}}\mathrm{H_2}\!-\!\mathrm{Cl}\]
In this molecule a positive polarity is developed on \(\mathrm{C^2}\) by the inductive effect of the Cl atom bonded to \(\mathrm{C^1}\). Normally the \(\mathrm{C^2}\!-\!\mathrm{C^1}\) bond would not be expected to be polar, as it is a covalent bond between two atoms of the same element, carbon.
Yet this bond acquires some polarity, because of the presence of the chlorine atom in the molecule. Chlorine being more electronegative than carbon, the \(\mathrm{C^1}\!-\!\mathrm{Cl}\) bond is a polar covalent bond with a partial negative charge on Cl and an equal amount of partial positive charge on \(\mathrm{C^1}\).
As the \(\mathrm{C^1}\) is further bonded to \(\mathrm{C^2}\), the positive polarity of \(\mathrm{C^1}\) pulls the shared pair of electrons of the \(\mathrm{C^2}\!-\!\mathrm{C^1}\) bond more towards itself. As a result, a part of the partial positive charge produced on \(\mathrm{C^1}\) is transferred to \(\mathrm{C^2}\). This results in developing a still smaller positive charge on \(\mathrm{C^2}\).
In other words, the electron density gets displaced towards the chlorine atom not only along the \(\mathrm{C^1}\!-\!\mathrm{Cl}\) bond, but also along the \(\mathrm{C^2}\!-\!\mathrm{C^1}\) bond, due to the inductive effect of Cl. This is represented with arrowheads placed in the centre of the bonds, as drawn in Fig. 14.44.
The inductive effect is the permanent polarity induced in the adjacent carbon–carbon single bonds of a molecule by the presence of a polar covalent bond in it.
The arrow head put in the centre of the bond is used to represent the inductive effect. The direction of the arrow head indicates the direction of the permanent electron displacement taking place along the sigma bond in the ground state.
The inductive effect of an influencing group is transferred to subsequent carbon atoms along the chain of C–C bonds. It decreases rapidly as the number of intervening C–C bonds increases, and becomes negligibly small beyond three bonds. This can be represented as
\[\overset{\delta_3\oplus}{\mathrm{C}}\mathrm{H_3} \longrightarrow \overset{\delta_2\oplus}{\mathrm{C}}\mathrm{H_2} \longrightarrow \overset{\delta_1\oplus}{\mathrm{C}}\mathrm{H_2} \longrightarrow \overset{\delta\ominus}{\mathrm{Cl}}\]
where \(\delta_3\oplus < \delta_2\oplus < \delta_1\oplus\).
Electron withdrawing and electron donating groups
The direction of the inductive effect of a bonded group depends upon whether the electron density of the bond is withdrawn from the bonded carbon or donated by the bonded carbon. According to this ability, the group substituents are classified as either electron withdrawing (accepting) or electron donating (releasing) groups, with respect to hydrogen.
In the above example, electron displacement takes place towards chlorine. This means that Cl withdraws electron density from the carbon chain and is electron withdrawing. Chlorine is said to exert an electron withdrawing inductive effect, or negative inductive effect (\(-\mathrm{I}\) effect), on the carbon chain.
Some other groups having \(-\mathrm{I}\) effect are \(\mathrm{-NO_2}\) (nitro), \(\mathrm{-CN}\) (cyano), \(\mathrm{-COOH}\) (carboxy), \(\mathrm{-COOR}\) (ester) and \(\mathrm{-OAr}\) (aryloxy).
On the other hand, alkyl groups (\(\mathrm{-CH_3}\), \(\mathrm{-CH_2\!-\!CH_3}\)) are considered as groups exerting an electron releasing inductive effect (\(+\mathrm{I}\) effect) on the carbon chain.
Note that both labels are comparisons against hydrogen, not absolute statements. A group is called \(-\mathrm{I}\) if it pulls electron density more strongly than an H atom in the same position would, and \(+\mathrm{I}\) if it releases electron density better than that H atom.
Consider the following molecules and answer the questions: \(\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!Cl}\), \(\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!Br}\), \(\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!I}\).
(i) What type of inductive effect is expected to operate in these molecules?
(ii) Identify the molecules from these three having the strongest and the weakest inductive effect.
Solution:
The groups responsible for the inductive effect in these molecules are \(\mathrm{-Cl}\), \(\mathrm{-Br}\) and \(\mathrm{-I}\), respectively. All these are halogen atoms which are more electronegative than carbon, and therefore all of them exert a \(-\mathrm{I}\) effect, that is, an electron withdrawing inductive effect.
The \(-\mathrm{I}\) effect of halogens is due to their electronegativity. A decreasing order of electronegativity in these halogens follows \(\mathrm{Cl} > \mathrm{Br} > \mathrm{I}\). Therefore the strongest \(-\mathrm{I}\) effect is expected in \(\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!Cl}\), while the weakest \(-\mathrm{I}\) effect is expected for \(\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!I}\).
Which of \(\mathrm{CH_3\!-\!CHCl_2}\) and \(\mathrm{CH_3CH_2Cl}\) is expected to have a stronger \(-\mathrm{I}\) effect?
Solution: The group exerting the \(-\mathrm{I}\) effect in \(\mathrm{CH_3CH_2Cl}\) is one \(\mathrm{-Cl}\), while in \(\mathrm{CH_3\!-\!CHCl_2}\) there are two \(\mathrm{-Cl}\) atoms. Therefore \(\mathrm{CH_3\!-\!CHCl_2}\) is expected to have a strong \(-\mathrm{I}\) effect.
Two electron withdrawing atoms pull on the same chain, so their effects add. This is the general principle behind such comparisons: for identical groups, more groups mean a stronger effect.
(question only — no solution) Arrange \(\mathrm{CH_3CH_2CH_2Cl}\), \(\mathrm{CH_3CHClCH_3}\) and \(\mathrm{CCl_3CH_2CH_3}\) in increasing order of the \(-\mathrm{I}\) effect experienced by the terminal methyl carbon, and give a reason for the position of the last compound in your order.
14.6.5 Resonance Structures
- Draw a bond line structure of benzene (\(\mathrm{C_6H_6}\)).
- How many C - C and C = C bonds are there in this structure?
- Write down the expected values of the bond lengths of the carbon carbon bonds in benzene (refer to Table 5.7).
The cyclic structure of benzene incorporating three alternating C-C single bonds and C=C double bonds implies two distinct bond lengths of 154 pm and 133 pm respectively, as drawn on the left of Fig. 14.10.
Experimental measurements show that benzene has a regular hexagonal shape, and all the six carbon bonds have the same bond length of 138 pm, which is intermediate between a C-C single bond and a C=C double bond. This means that all the six carbon carbon bonds in benzene are equivalent, unlike what appears from the single Lewis structure.
the expected and observed carbon carbon bond lengths of benzene in Fig. 14.10 (i) By how much does the observed value of 138 pm differ from each of the two expected values? (ii) Is 138 pm closer to the single bond value or to the double bond value, and what does that suggest about the amount of double bond character in each C–C bond of benzene? (iii) If the alternating single-and-double-bond picture were correct, what shape would you expect the ring to have, and how does the measured regular hexagon contradict it?
- Write down two Lewis structures for ozone. (Refer to chapter 5)
- How are these two Lewis structures related to each other?
- What are these two Lewis structures called?
When the Lewis structure of a compound has two or more multiple bonds alternating with single bonds, it is called a conjugated system of \(\pi\) bonds. In such a system, or in species having an atom carrying a p orbital attached to a multiple bond, resonance theory is applicable.
A conjugated system of \(\pi\) bonds is an arrangement in which two or more multiple bonds alternate with single bonds.
Points understood from resonance theory
We understand the following points from the resonance theory:
The \(\pi\) electrons in a conjugated system of \(\pi\) bonds are not localized to a particular \(\pi\) bond.
For a compound having a conjugated system of \(\pi\) bonds (or similar other systems), two or more Lewis structures are written by showing the movement of \(\pi\) electrons — that is, the delocalization of \(\pi\) electrons — using curved arrows.
The Lewis structures so generated are linked by a double headed arrow and are called resonance structures, or contributing structures, or canonical structures of the species. Thus two resonance structures can be drawn for benzene by delocalizing or shifting the \(\pi\) electrons, as shown in Fig. 14.45.
The positions of the carbon atoms in the conjugated system of \(\pi\) bonds remain unchanged, but the positions of the \(\pi\) electrons are different in different resonance structures. For example, in the resonance structure I of benzene there is a single bond between \(\mathrm{C^1}\) and \(\mathrm{C^2}\), while in the resonance structure II there is a double bond between \(\mathrm{C^1}\) and \(\mathrm{C^2}\).
Any resonance structure is hypothetical; it does not by itself represent any real molecule, and by itself it cannot explain all the properties of the compound. The real molecule has, however, the character of all the resonance structures that can be written. The real or actual molecule is said to be the resonance hybrid of all the resonance structures.
For example, an actual benzene molecule is the resonance hybrid of structures I and II and exhibits the character of both these structures. Its approximate representation can be shown as a dotted circle inscribed in a regular hexagon, also drawn in Fig. 14.45. Thus each carbon carbon bond in benzene has single as well as double bond character, and the ring has a regular hexagonal shape.
- The hypothetical energy of an individual resonance structure can be calculated using bond energy values. The energy of the actual molecule is, however, lower than that of any one of the resonance structures.
In other words, the resonance hybrid is more stable than any of the resonance structures. The difference in the actual energy and the lowest calculated energy of a resonance structure is called the resonance stabilization energy, or just the resonance energy. Thus resonance leads to stabilization of the actual molecule.
Resonance: where a conjugated system allows more than one Lewis structure to be written, the actual molecule is a single resonance hybrid of all of them, and is more stable than any individual resonance structure.
resonance energy (SI unit \(\text{kJ mol}^{-1}\); it is quoted as a positive quantity, and the larger it is, the more the molecule has been stabilised by resonance — the actual molecule always lies lower in energy than the most stable resonance structure, never higher.)
Two cautions follow from point (iv), and both are examined regularly. First, resonance structures are not isomers: benzene is not a mixture of two compounds interconverting, and this is the essential difference between resonance and the tautomerism of Section 14.5.1. Second, only electrons move between resonance structures; the nuclei stay exactly where they are.
Q4. A single Lewis structure of benzene shows three single and three double carbon–carbon bonds, so we would expect two different bond lengths in the ring. Measurements show all six bonds are identical. What must be wrong with the single-structure picture?
A. The single structure is not so much wrong as incomplete. Benzene’s \(\pi\) electrons are not localised in three particular double bonds; they are delocalised over the whole conjugated ring. Two resonance structures can be written that differ only in where those \(\pi\) electrons are placed, and the real molecule is the resonance hybrid of both. Each carbon carbon bond therefore has partial single and partial double bond character, giving one intermediate bond length of 138 pm all round and a regular hexagonal ring — along with the extra resonance stabilization that makes benzene so unusually stable.
The ring itself was proposed in the nineteenth century by the German chemist August Kekulé, who suggested that the six carbons of benzene close into a ring with alternating single and double bonds. His structure accounted for benzene’s formula and for the number of its isomers, but not for the equality of the six bonds; that had to wait for the resonance description, developed later and shaped in large part by the work of Linus Pauling. Kekulé’s ring was not discarded — it became one of the two contributing structures of Fig. 14.45.
Identify the species that contains a conjugated system of \(\pi\) bonds. Explain your answer.
I. \(\mathrm{CH_2\!=\!CH\!-\!CH_2\!-\!CH\!=\!CH_2}\)
II. \(\mathrm{CH_2\!=\!CH\!-\!CH\!=\!CH\!-\!CH_3}\)
Solution: I does not contain a conjugated system of \(\pi\) bonds, as the two C = C double bonds are separated by two C - C single bonds.
II contains a conjugated system of \(\pi\) bonds, as the two C = C double bonds are separated by only one C - C single bond.
The test is therefore mechanical: count the single bonds lying between the two multiple bonds. Exactly one means conjugation; two or more means the two \(\pi\) systems are insulated from each other.
Rules to be followed for writing resonance structures
Resonance structures can be written only when all the atoms involved in the \(\pi\) conjugated system lie in the same plane.
All the resonance structures must have the same number of unpaired electrons.
Resonance structures contribute to the resonance hybrid in accordance with their energy or stability. More stable, that is lower energy, resonance structures contribute largely and are thus important.
Which resonance structure is more important?
Thus, between the compared resonance structures, that resonance structure is more stable — and therefore important — which has (a) a greater number of covalent bonds, (b) a greater number of atoms with a complete octet or duplet, (c) less separation, if any, of opposite charges, (d) a negative charge, if any, on the more electronegative atom and a positive charge, if any, on the more electropositive atom, and (e) more dispersal of charge.
These five tests are applied in the order set out in FC 14.7 whenever two resonance structures have to be compared.
Write resonance structures of \(\mathrm{H\!-\!COO^{\ominus}}\) and comment on their relative stability.
Solution: First the detailed bond structure of \(\mathrm{H\!-\!COO^{\ominus}}\) showing all the valence electrons is drawn, and then other resonance structures are generated using a curved arrow to show the movement of \(\pi\)-electrons. Two resonance structures are written for \(\mathrm{H\!-\!COO^{\ominus}}\), shown as I and II in Fig. 14.46.
Both the resonance structures I and II are equivalent to each other, and therefore, are equally stable.
When two resonance structures are mirror images of each other in this way, neither can be preferred, so both contribute equally to the hybrid. The negative charge is then shared equally by the two oxygen atoms instead of sitting on one of them.
This equal sharing explains a great deal of everyday chemistry. Once a carboxylic acid loses its proton, the resulting carboxylate ion spreads the negative charge over two equivalent oxygen atoms, so the anion is far more stable than the alkoxide ion obtained from an alcohol — and the acid is correspondingly stronger. Ordinary soap is the sodium salt of a long chain carboxylic acid, and it is this stable, charge-dispersed head that dissolves in water while the hydrocarbon tail clings to grease.
Identify the species which has resonance stabilization. Justify your answer.
(i) \(\mathrm{CH_3\!-\!O\!-\!H}\) (ii) \(\mathrm{CH_3\!-\!NO_2}\) (iii) buta - 1, 3 - diene
Solution:
The bond structure shows that there is no \(\pi\) bond. Therefore there is no resonance and no resonance stabilization.
In nitromethane the N = O double bond is attached to ‘O’ which carries a lone pair of electrons in a p orbital. Therefore resonance structures can be written, as drawn in Fig. 14.47, and the species is resonance stabilized.
In buta-1,3-diene the Lewis structure shows two C=C double bonds alternating with a C-C single bond. Therefore resonance structures can be written as shown in Fig. 14.47, and the species is resonance stabilized.
When all the resonance structures of a species are equivalent to each other, the species is highly resonance stabilized. For example, \(\mathrm{R\!-\!COO^{\ominus}}\) and \(\mathrm{CO_3^{2-}}\) — which is why both of these ions are so commonly met as stable anions.
Write three resonance structures for \(\mathrm{CH_3\!-\!CH\!=\!CH\!-\!CHO}\). Indicate their relative stabilities and explain.
Solution: The three resonance structures I, II and III are drawn in Fig. 14.48, generated by moving the \(\pi\) electrons of the conjugated system with curved arrows.
Stability order: I > II > III
I: Contains a greater number of covalent bonds, each carbon atom and the oxygen atom has a complete octet, and it involves no separation of opposite charges. Therefore it is the most stable resonance structure.
II: Contains one covalent bond less than in I, one carbon (\(\mathrm{C^{\oplus}}\)) has only 6 valence electrons, and it involves separation of opposite charges. The resonance structure II has the \(-\mathrm{ve}\) charge on the more electronegative ‘O’ and the \(+\mathrm{ve}\) charge on the more electropositive ‘C’. It has intermediate stability.
III: Contains one covalent bond less than in I, oxygen has only 6 valence electrons, and it involves separation of opposite charge. It has the \(-\mathrm{ve}\) charge on the more electropositive ‘C’ and the \(+\mathrm{ve}\) charge on the more electronegative ‘O’. All these factors are unfavourable for stability. Therefore it is the least stable.
Compare II and III carefully, because they differ in one respect only — where the two charges have been placed. That single difference is enough to separate them, which shows how much weight test (d) carries in practice.
(question only — no solution) Write the resonance structures of the acetate ion, \(\mathrm{CH_3COO^{\ominus}}\), and state, with a reason, whether one of them contributes more than the other to the resonance hybrid.
14.6.6 Resonance Effect
The existence of resonance often results in developing polarity in a molecule. The polarity produced in the molecule by the interaction between conjugated \(\pi\) bonds — or that between a \(\pi\) bond and a p orbital on an attached atom — is called the resonance effect or mesomeric effect.
The resonance effect, also called the mesomeric effect, is the polarity produced in a molecule by the interaction between conjugated \(\pi\) bonds, or between a \(\pi\) bond and a p orbital on an attached atom.
The effect is transmitted through a chain of conjugated \(\pi\) bonds. There are two types of resonance effects, or mesomeric effects, designated as positive resonance effect (\(+\mathrm{R}\) or \(+\mathrm{M}\)) and negative resonance effect (\(-\mathrm{R}\) or \(-\mathrm{M}\)).
Positive resonance effect
Positive resonance effect, or electron donating / releasing resonance effect (\(+\mathrm{R}\) effect): if the substituent group has a lone pair of electrons to donate to the attached \(\pi\) bond or conjugated system of \(\pi\) bonds, the effect is called the \(+\mathrm{R}\) effect.
Groups such as \(\mathrm{-OH}\), \(\mathrm{-OR}\), \(\mathrm{-O^{\ominus}}\), \(\mathrm{-NHR}\) and halogen, etc., having a lone pair of electrons, show the \(+\mathrm{R}\) effect. The \(+\mathrm{R}\) effect increases electron density at certain positions in a molecule. Figure 14.11 shows how the \(+\mathrm{R}\) effect in aniline increases the electron density at the ortho and para positions.
Negative resonance effect
Negative resonance effect, or electron withdrawing resonance effect (\(-\mathrm{R}\) effect): if the substituent group has a tendency to withdraw electrons from the attached \(\pi\) bond or conjugated system of \(\pi\) bonds towards itself, the effect is called the \(-\mathrm{R}\) effect.
Groups such as \(\mathrm{-COOH}\), \(\mathrm{-CHO}\), \(\mathrm{-CO-}\), \(\mathrm{-CN}\), \(\mathrm{-NO_2}\) and \(\mathrm{-COOR}\), etc., show the \(-\mathrm{R}\) effect. The \(-\mathrm{R}\) effect results in developing a positive polarity at certain positions in a molecule. Figure 14.12 shows how the \(-\mathrm{R}\) effect in nitrobenzene develops positive polarity at the ortho and para positions.
+R effect vs −R effect | Feature | \(+\mathrm{R}\) (positive resonance) effect | \(-\mathrm{R}\) (negative resonance) effect | |—|—|—| | What the substituent does | Donates a lone pair into the conjugated system | Withdraws electrons from the conjugated system | | Effect on the ring | Increases electron density at the ortho and para positions | Develops positive polarity at the ortho and para positions | | Requirement in the group | A lone pair available for donation | A multiple bond able to accept electrons | | Examples | \(\mathrm{-OH}\), \(\mathrm{-OR}\), \(\mathrm{-O^{\ominus}}\), \(\mathrm{-NHR}\), halogen | \(\mathrm{-COOH}\), \(\mathrm{-CHO}\), \(\mathrm{-CO-}\), \(\mathrm{-CN}\), \(\mathrm{-NO_2}\), \(\mathrm{-COOR}\) | | Illustrated by | Aniline, Fig. 14.11 | Nitrobenzene, Fig. 14.12 |
The ortho and para positions appear in both rows for a structural reason worth noticing: those are precisely the ring positions that carry the charge in the contributing structures, so a group’s resonance effect is felt there and not at the meta positions.
(question only — no solution) For each of \(\mathrm{-OCH_3}\), \(\mathrm{-CN}\), \(\mathrm{-NH_2}\) and \(\mathrm{-NO_2}\) attached to a benzene ring, state whether it shows the \(+\mathrm{R}\) or the \(-\mathrm{R}\) effect, and name the ring positions at which the electron density is altered.
14.6.7 Electromeric Effect
This is a temporary electronic effect exhibited by multiple-bonded groups in the excited state in the presence of a reagent.
When a reagent approaches a multiple bond, the electron pair gets completely shifted to one of the multiply bonded atoms, giving a charge separated structure, as drawn in Fig. 14.49. For example, when \(\mathrm{H^{\oplus}}\) approaches the carbonyl group:
\[\mathrm{R_2C\!=\!O} + \mathrm{H^{\oplus}} \rightleftharpoons \mathrm{R_2\overset{\oplus}{C}\!-\!\overset{\ominus}{O}H}\]
The electromeric effect, or polarisability effect, is the complete transfer of a \(\pi\) electron pair to one of the multiply bonded atoms on the approach of an attacking reagent; it is a temporary effect.
This effect is temporary and disappears when the reagent is removed from the reacting system. That single property separates it from every other effect in this section: the inductive, resonance and hyperconjugative effects are all present in the ground state, whether or not a reagent is anywhere near.
14.6.8 Hyperconjugation
Hyperconjugation is a permanent electronic effect which explains the stability of a carbocation, a free radical or an alkene.
It is the delocalization of \(\sigma\) (sigma) electrons of a C - H bond of an alkyl group directly attached to a carbon atom which is part of an unsaturated system, or has an empty p-orbital, or has a p-orbital with an unpaired electron. The three situations are shown in Fig. 14.13.
Hyperconjugation is the delocalization of the \(\sigma\) electrons of a C–H bond of an alkyl group into an adjacent empty p orbital, \(\pi\) bond, or p orbital carrying an unpaired electron.
Hyperconjugation in the ethyl cation
Let us consider the ethyl cation (\(\mathrm{CH_3\overset{\oplus}{C}H_2}\)). Here the positively charged carbon atom with an empty p-orbital has an adjacent methyl group, that is an \(\alpha\)-(alpha) methyl group.
One of the C-H bonds of this \(\alpha\) methyl group is in alignment with the empty p-orbital. The \(\sigma\) electrons of this C-H bond are delocalized into the empty p-orbital and thereby stabilize the cation, as drawn in Fig. 14.14.
Hyperconjugation is, thus, a \(\sigma\)-\(\pi\) conjugation. The delocalization can be represented by the contributing structures I to IV of Fig. 14.50.
The contributing structures II, III and IV, involving delocalization of the \(\sigma\) electrons of the C-H bond, show no covalent bond between carbon and one of the \(\alpha\)-hydrogens. Hyperconjugation is, therefore, called ‘no bond resonance’.
The more the number of such \(\alpha\)-hydrogens — that is, hydrogen on the \(\alpha\)-carbon — the more are the no bond resonance structures, and the more is the stability. Counting \(\alpha\)-hydrogens is therefore the working method: the tert-butyl cation has nine, the isopropyl cation six, the ethyl cation three and the methyl cation none.
The relative stability of carbocations therefore decreases in the order:
\[\mathrm{CH_3\!-\!\overset{\oplus}{C}(CH_3)\!-\!CH_3} > \mathrm{CH_3\!-\!\overset{\oplus}{C}H\!-\!CH_3} > \mathrm{CH_3\!-\!\overset{\oplus}{C}H_2} > \mathrm{\overset{\oplus}{C}H_3}\]
Q5. A carbocation carries the same single positive charge whether it is a methyl cation or a tert-butyl cation. Why, then, should one of them be so much more stable than the other?
A. Because stability depends not on the size of the charge but on how far that charge can be spread out. Alkyl groups attached to the positive carbon release electron density towards it by the \(+\mathrm{I}\) effect, and, more importantly, the \(\sigma\) electrons of their \(\alpha\) C–H bonds delocalise into the empty p orbital by hyperconjugation. Each additional \(\alpha\)-hydrogen adds another no bond resonance structure. The tert-butyl cation has nine such hydrogens and the methyl cation has none, so the charge is heavily dispersed in the first and completely concentrated in the second.
Hyperconjugation in propene
Let us now look at propene, which has a methyl group attached to C = C. Delocalization of electrons by hyperconjugation in propene can be depicted as shown in Fig. 14.51.
Figure 14.15 shows the orbital diagram of hyperconjugation in propene, in which the C–H sigma orbital of the methyl group overlaps sideways with the \(\pi\) orbital of the double bond.
The effect of hyperconjugation is usually stronger than the inductive effect. This is why hyperconjugation, rather than the \(+\mathrm{I}\) effect alone, is quoted as the principal explanation of carbocation, free radical and alkene stability.
inductive vs resonance vs electromeric vs hyperconjugative effect | Feature | Inductive effect | Resonance effect | Electromeric effect | Hyperconjugation | |—|—|—|—|—| | Permanent or temporary | Permanent | Permanent | Temporary | Permanent | | Electrons involved | \(\sigma\) electrons of C–C bonds | \(\pi\) electrons of a conjugated system, or a lone pair | \(\pi\) electrons of a multiple bond | \(\sigma\) electrons of an \(\alpha\) C–H bond | | Operates through | Sigma bonds | Conjugated \(\pi\) bonds | The multiple bond itself | \(\sigma\)–\(\pi\) or \(\sigma\)–p overlap | | How far it reaches | Falls off rapidly, negligible beyond three bonds | Transmitted along the whole conjugated chain | Only while the reagent is present | The adjacent \(\alpha\) position | | Types | \(+\mathrm{I}\) and \(-\mathrm{I}\) | \(+\mathrm{R}\) and \(-\mathrm{R}\) | — | Also called no bond resonance | | Explains | Relative strength of the pull on a chain, stability of ions | Equal bond lengths in benzene, ortho and para polarity | Addition of a reagent to a multiple bond | Stability of carbocations, free radicals and alkenes |
(question only — no solution) Count the number of \(\alpha\)-hydrogens available for hyperconjugation in (i) \(\mathrm{CH_3\!-\!\overset{\oplus}{C}H\!-\!CH_3}\), (ii) \(\mathrm{(CH_3)_3C^{\oplus}}\) and (iii) \(\mathrm{CH_3\!-\!CH\!=\!CH_2}\), and use your answers to place the two cations in order of stability.
- Read more about the basic principles of organic chemistry at https://authors.library.caltech.edu/25034
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