Organic chemistry: functional groups

Organic chemists use several formula types, each showing a different level of detail:

• Empirical formula - simplest whole-number ratio of atoms, e.g. ethane is $\text{CH}_3$.
• Molecular formula - actual number of each atom, e.g. ethane is $\text{C}_2\text{H}_6$.
• Full (displayed) structural formula - shows *every* atom and *every* bond as a line.
• Condensed structural formula - groups atoms together, e.g. butane is $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3$.
• Skeletal formula - the most economical: carbons are line ends/vertices, C-H bonds are not drawn, and only heteroatoms and functional groups are labelled.

In a skeletal diagram a zig-zag line of 4 vertices/ends is butane; a vertex where two lines meet is a $\text{CH}_2$, a line end is a $\text{CH}_3$.

When drawing, remember each carbon makes four bonds and each hydrogen one. Always check the valences add up. Skeletal formulae are quickest for large molecules and make functional groups stand out, which is why they dominate at higher levels.

A functional group is the reactive part of a molecule - it determines the compound's chemistry, while the alkyl ($\text{R-}$) skeleton is comparatively inert.

Key groups to recognise:

• Alkene $\text{C=C}$ (double bond)
• Halogenoalkane $\text{C-X}$ ($X$ = F, Cl, Br, I)
• Alcohol $-\text{OH}$ (hydroxyl)
• Aldehyde $-\text{CHO}$ (carbonyl at chain end)
• Ketone $\text{C=O}$ (carbonyl in the chain), written $-\text{CO}-$
• Carboxylic acid $-\text{COOH}$
• Ester $-\text{COO}-$
• Amine $-\text{NH}_2$
• Amide $-\text{CONH}_2$
• Nitrile $-\text{CN}$
• Ether $-\text{O}-$
• Arene (benzene ring)

Because members of a series share the same group, they undergo the same type of reactions - for example all alcohols can be oxidised and all carboxylic acids are weak acids. Being able to spot a group from a structure (and vice versa) is a core exam skill.

A homologous series is a family of organic compounds that:

• share the same functional group,
• fit the same general formula (e.g. alkanes $\text{C}_n\text{H}_{2n+2}$),
• have successive members differing by $\text{CH}_2$ ($\Delta M = 14$),
• show a gradual trend in physical properties as chain length grows,
• show similar chemical reactions.

For example, the alcohols methanol, ethanol, propan-1-ol... each add a $\text{CH}_2$ and all react with sodium and can be oxidised.

The smooth trend lets you predict unknown members: if ethanol boils at $78\,^{\circ}\text{C}$ and propan-1-ol at $97\,^{\circ}\text{C}$, butan-1-ol will boil higher still. The shared functional group means once you know how one alcohol reacts, you know how the whole series behaves - this is why classifying compounds into series is so useful.

Within a homologous series, physical properties change gradually with chain length because the dominant intermolecular forces change.

• Boiling point increases down a series: longer chains have more electrons and a larger surface area, so stronger London (dispersion) forces need more energy to overcome.
• Volatility decreases as boiling point rises.
• Branching lowers boiling point between isomers: branched molecules pack less closely, reducing contact area and London forces.
• Solubility in water decreases with chain length: the non-polar hydrocarbon part grows while the polar functional group stays the same, so the molecule becomes more hydrophobic.

The functional group itself controls which intermolecular forces are present. Alcohols and carboxylic acids form hydrogen bonds, so they boil much higher and dissolve better than alkanes of similar mass. For example, ethanol ($M = 46$) boils at $78\,^{\circ}\text{C}$ while propane ($M = 44$) boils at $-42\,^{\circ}\text{C}$ - almost entirely due to hydrogen bonding.

IUPAC names are built systematically from three parts:

• Stem - the longest continuous carbon chain: meth- (1C), eth- (2C), prop- (3C), but- (4C), pent- (5C), hex- (6C).
• Suffix / prefix - the functional group: -ane (alkane), -ene (alkene), -ol (alcohol), -al (aldehyde), -one (ketone), -oic acid (carboxylic acid). Halogens are prefixes (chloro-, bromo-).
• Locants - numbers giving positions, chosen to be as low as possible.

Steps to name a molecule:

1. Find and number the longest chain to give the functional group the lowest locant.

2. Identify the functional group and choose its suffix.

3. Name and number side-groups (substituents) alphabetically.

4. Combine, using hyphens between numbers and letters and commas between numbers.

For example $\text{CH}_3\text{CH(CH}_3)\text{CH}_2\text{OH}$ is 2-methylpropan-1-ol: 3-carbon stem (propan), -ol for the alcohol on carbon 1, a methyl branch on carbon 2.

Structural (constitutional) isomers have the *same molecular formula* but a *different connectivity* of atoms. There are three types to know:

• Chain isomers - different carbon-skeleton arrangement, e.g. butane vs 2-methylpropane (both $\text{C}_4\text{H}_{10}$).
• Position isomers - the functional group is on a different carbon, e.g. propan-1-ol vs propan-2-ol (both $\text{C}_3\text{H}_8\text{O}$).
• Functional-group isomers - same formula but a *different* functional group, e.g. ethanol (alcohol) vs methoxymethane (ether), both $\text{C}_2\text{H}_6\text{O}$; or propanal (aldehyde) vs propanone (ketone), both $\text{C}_3\text{H}_6\text{O}$.

Isomers can have markedly different physical and chemical properties: branched chain isomers boil lower than straight chains, and functional-group isomers react completely differently (an aldehyde is oxidised easily, a ketone is not).

When counting isomers, work systematically: vary the chain first, then the position of the group, then the group type. Don't double-count the same structure drawn a different way.

Stereoisomers have the same structural formula (same connectivity) but a different arrangement of atoms in space. Two types:

Cis-trans isomerism arises when rotation is restricted - typically around a $\text{C=C}$ double bond - *and* each doubly bonded carbon carries two different groups. It is defined by where the like (matching) groups sit:

• cis: the two like groups are on the *same* side of the double bond.
• trans: the two like groups are on *opposite* sides.

Z/E nomenclature is a separate, more general system based on CIP priority (atomic number): on each carbon you find the higher-priority group, then:

• Z (*zusammen*): the two higher-priority groups are on the *same* side.
• E (*entgegen*): they are on *opposite* sides.

The two systems give the same label only in simple cases where the like groups are also the higher-priority groups (e.g. but-2-ene, where cis = Z and trans = E). In general they are *not* synonymous: when the matching groups are not the highest-priority ones, cis can correspond to E and trans to Z, so do not treat cis = Z and trans = E as a universal rule.

Cis-trans (and Z/E) isomers have different physical properties (e.g. different melting/boiling points and polarity), e.g. cis- and trans-but-2-ene.

Optical isomerism occurs at a chiral carbon - one bonded to four different groups. The molecule and its mirror image are non-superimposable; these mirror images are called enantiomers.

• Enantiomers have *identical* physical and chemical properties in an ordinary environment.
• They differ in rotating plane-polarised light in opposite directions (one $+$, one $-$).
• A 50:50 mixture is a racemate and shows no net rotation.

To spot chirality, look for a carbon with four distinct substituents. Lactic acid and most amino acids (except glycine) are chiral.

In mass spectrometry (MS), molecules are ionised and the resulting ions are separated by their mass/charge ($m/z$) ratio. For structure determination two features matter:

• The molecular ion peak ($\text{M}^+$) at the highest $m/z$ gives the relative molecular mass $M_r$ of the compound.
• Fragmentation peaks at lower $m/z$ arise when the molecular ion breaks apart. The *difference* between $\text{M}^+$ and a fragment peak tells you what neutral piece was lost.

The data booklet's *mass spectral fragments lost* table lists common losses, for example:

• loss of $15 \Rightarrow$ a methyl group $\text{CH}_3$
• loss of $17 \Rightarrow$ a hydroxyl $\text{OH}$
• loss of $29 \Rightarrow \text{CHO}$ (aldehyde) or $\text{C}_2\text{H}_5$
• loss of $45 \Rightarrow \text{COOH}$ (carboxylic acid)

So if $\text{M}^+ = 60$ and a strong peak appears at 15 below it (45), the molecule has likely lost $\text{CH}_3$, consistent with ethanoic acid. Always read the *difference* between peaks, and use the data-booklet table rather than memorising every value.

Infrared (IR) spectroscopy detects the bonds present in a molecule. IR radiation makes covalent bonds stretch and bend; a bond absorbs IR only if the vibration changes the molecule's dipole moment.

Each bond type absorbs at a characteristic wavenumber $\tilde{\nu}$ (in $\text{cm}^{-1}$, where $\tilde{\nu} = 1/\lambda$). The data booklet's *IR absorption* table gives the ranges, including:

• $\text{O-H}$ (alcohol): broad, $\approx 3200-3600\ \text{cm}^{-1}$
• $\text{O-H}$ (carboxylic acid): very broad, $\approx 2500-3000\ \text{cm}^{-1}$
• $\text{N-H}$ (amine): $\approx 3300-3500\ \text{cm}^{-1}$
• $\text{C=O}$ (carbonyl): strong, $\approx 1700-1750\ \text{cm}^{-1}$
• $\text{C-O}$: $\approx 1050-1410\ \text{cm}^{-1}$

The region below about $1500\ \text{cm}^{-1}$ is the fingerprint region - a complex pattern unique to each compound, useful for confirming identity by comparison rather than for picking out individual bonds.

To use IR: look for the *presence or absence* of a strong $\text{C=O}$ and broad $\text{O-H}$ to distinguish, say, an alcohol from a carboxylic acid. Always quote the wavenumber from the data booklet to justify your assignment.

Proton ($^1\text{H}$) NMR probes the hydrogen atoms in a molecule. Hydrogen nuclei in different chemical surroundings absorb at slightly different frequencies, measured as chemical shift $\delta$ in ppm relative to the standard TMS ($\delta = 0$).

Three pieces of information are read from a (low-resolution) spectrum:

• Number of peaks = number of different chemical (hydrogen) environments.
• Chemical shift ($\delta$) = the *type* of environment; the data booklet's *$^1\text{H}$ NMR* table lists typical ranges, e.g. $\text{R-CH}_3 \approx 0.9$, $-\text{O-CH} \approx 3.3-4.0$, $-\text{COOH} \approx 9-13\ \text{ppm}$.
• Relative area under each peak (integration trace) = the ratio of hydrogens in each environment.

For example, ethanol $\text{CH}_3\text{CH}_2\text{OH}$ gives three peaks in a $3:2:1$ ratio for the $\text{CH}_3$, $\text{CH}_2$ and $\text{OH}$ hydrogens.

To deduce structure: count environments from the number of peaks, assign each using the data-booklet shift table, and use the integration ratio to count hydrogens. Always justify shifts with the booklet values rather than guessing.

No single technique gives a full structure - chemists combine them. A typical workflow for an unknown organic compound:

1. Mass spectrometry - find $\text{M}^+$ for the molar mass, and use fragment losses (the data-booklet fragments-lost table) to spot key groups.

2. Index of hydrogen deficiency - from the molecular formula, work out rings + $\pi$ bonds using $\text{IHD} = \frac{(2C+2+N)-H-X}{2}$.

3. Infrared - identify functional groups from absorption bands (the data-booklet IR table), e.g. a strong $\text{C=O}$.

4. $^1\text{H}$ NMR - count hydrogen environments, assign shifts (the data-booklet NMR table) and read the integration ratio.

The pieces must be self-consistent: if IR shows $\text{C=O}$ and NMR shows a peak near $\delta \approx 9.7$, both point to an aldehyde. Cross-checking is what makes the deduction reliable.

Example logic: $M_r = 60$ from MS; IR shows broad $\text{O-H}$ at $2500-3000$ and strong $\text{C=O}$ at $1715\ \text{cm}^{-1}$ (carboxylic acid); NMR shows a $\text{CH}_3$ near $\delta 2$ and a $-\text{COOH}$ near $\delta 11$ in a $3:1$ ratio. Together these confirm ethanoic acid, $\text{CH}_3\text{COOH}$.