Apply IUPAC rules to name and represent the structural formula of organic compounds including alkanes, alkenes, alkynes, alcohols, aldehydes, ketones, carboxylic acids, esters, and amines

What this dot point is asking

NESA wants you to take any organic structural formula in the HSC scope and produce the IUPAC name, and to take any IUPAC name and draw the structural formula. The compounds in scope are alkanes, alkenes, alkynes, alcohols, aldehydes, ketones, carboxylic acids, esters and amines, with branches up to about C6. This is the foundation for every other Module 7 dot point.

The answer

The five-step IUPAC algorithm

1. Identify the principal functional group to set the suffix. Priority (high to low): carboxylic acid > ester > amide > nitrile > aldehyde > ketone > alcohol > amine > alkene/alkyne > alkane. Only the highest-priority group becomes the suffix; others become prefixes.
2. Find the longest continuous carbon chain that contains the principal functional group. The carbon count sets the parent name root (meth, eth, prop, but, pent, hex, hept, oct, non, dec).
3. Number the chain so that the principal functional group gets the lowest possible locant. If the principal group is at a chain end (acid, ester, aldehyde), it is automatically C1. For unsaturation, give the first $C=C$ or $C \equiv C$ the lowest locant. For substituents only, use the lowest set of locants.
4. Identify substituents (branches and lower-priority groups), assign their locants, and list them alphabetically.
5. Assemble the name: locants of substituents, substituent names (alphabetised), parent root, locant of principal group, suffix.

Suffix and general formula for each series

Locant rules in detail

Alcohols, aldehydes, ketones, amines, alkenes, alkynes all take a locant before the suffix: butan-2-ol, pent-2-ene, hex-3-yne, butan-2-one, butan-2-amine. Aldehydes and carboxylic acids do not need a locant because they are always C1.

Lowest locant rule. When two numbering directions are possible, choose the one that gives the lowest locant to the principal group. If the principal group has the same locant in both directions, choose the direction that gives the lowest locants to the substituents, taken as a set (compare term by term).

Substituent prefixes. Halogens are $fluoro$, $chloro$, $bromo$, $iodo$. Alkyl groups are $methyl$, $ethyl$, $propyl$, etc. An $-OH$ becomes a $hydroxy-$ prefix only when not the principal group (e.g. in 2-hydroxypropanoic acid, where the acid outranks the alcohol). An $-NH_2$ becomes $amino-$ when downgraded.

Worked example: name a complex structure

Structure: $CH_3CH_2CH(CH_3)CH_2CH_2OH$.

1. Principal group: $-OH$ (alcohol), suffix $-ol$.
2. Longest chain containing $-OH$: 5 carbons (pent).
3. Number to give $-OH$ lowest locant: from the OH end, OH at C1.
4. Substituent: methyl at C3.
5. Assemble: 3-methylpentan-1-ol.

Worked example: draw from name

Name: 4-amino-2-methylpentanoic acid.

• Parent: pentanoic acid means a 5 carbon chain with $-COOH$ at C1.
• Methyl at C2, amino at C4.

The acid outranks the amine, so $-NH_2$ is a prefix.

Naming esters specifically

Esters are named alkyl alkanoate in two words.

• Alkyl part: comes from the alcohol, named as a substituent off the ester oxygen. Count the carbons attached to the single-bonded $O$.
• Alkanoate part: comes from the carboxylic acid, includes the $C=O$. Count those carbons.

For $CH_3COOCH_2CH_3$: $CH_3CO-$ has 2 carbons (ethanoate), $-OCH_2CH_3$ has 2 carbons (ethyl). Name: ethyl ethanoate.

For $CH_3CH_2CH_2COOCH_3$: 4 carbons on the acid side (butanoate), 1 carbon on the alcohol side (methyl). Name: methyl butanoate.

Naming amines

A primary amine $R-NH_2$ takes the suffix $-amine$ with a locant for the nitrogen-bearing carbon. Secondary and tertiary amines are named by the largest parent amine, with the other substituents named with the $N-$ locant prefix. Example: $CH_3 - N(CH_3) - CH_2CH_3$ is $N,N$-dimethylethanamine (parent is ethanamine, two methyls on nitrogen).

Common traps

Forgetting the lowest locant rule. Always renumber from both ends and choose the direction that minimises the principal group locant first.

Naming the longest chain when it does not contain the functional group. The chain must contain the principal group. A longer chain that bypasses the $-OH$ is not the parent.

Ester order confusion. Alkyl (from alcohol) is named first, alkanoate (from acid) second. The alkanoate carbons include the carbonyl.

Including the carbonyl carbon twice. When counting carbons in an aldehyde or carboxylic acid, the $C$ in $-CHO$ or $-COOH$ is C1, not an additional carbon.

Alphabetising with prefixes. Use only the letter of the substituent name itself, ignoring multipliers (di, tri, tetra). So $ethyl$ comes before $dimethyl$ alphabetically (e by e).

In one sentence

Identify the principal functional group to set the suffix, find the longest chain that contains it, number that chain so the principal group gets the lowest locant, then list substituents alphabetically with their locants in front of the parent name.