Introduction to Organic Chemistry: Question 5

Worked solution

Step 1: Place the substituents on the ring

Number the six ring carbons $1$ to $6$ going around the ring. A “meta” relationship means the two substituted carbons are separated by exactly one unsubstituted carbon on the shorter path between them, for example, substituents on carbons $1$ and $3$.

Step 2: Compare the two possible numbering directions

Since a benzene ring is symmetric, the carbon bearing the chlorine atom can always be labelled $\text{C1}$. From there, the carbon bearing the methyl group (which is meta to it) can be reached in two possible ways round the ring:

• Going the short way: methyl is reached at locant $3$, locant set $\{1,3\}$.
• Going the long way: methyl is reached at locant $5$, locant set $\{1,5\}$.

Comparing these two locant sets at the first point of difference, $3 < 5$, so $\{1,3\}$ is the lower locant set. IUPAC rules require the numbering that gives the lowest possible set of locants, so the ring must be numbered to place methyl at $3$, not $5$.

Step 3: Check the alphabetical tie-break (not needed here, but worth confirming)

Here the two substituents are not tied on the same locant set, since $\{1,3\}$ is unambiguously lower than $\{1,5\}$, so the “lowest locants” rule alone decides the numbering; chlorine keeps locant $1$ and methyl is at locant $3$. (The alphabetical tie-break rule, giving the lower locant to whichever substituent comes first alphabetically, would only be needed if two different directions produced the same locant set, which is not the case here since chlorine is already fixed at the lowest possible locant, $1$.)

Step 4: Assemble the name

Substituents are cited alphabetically: “chloro” before “methyl”:

$\textbf{1-chloro-3-methylbenzene}$

Why the other options are wrong

• B, 1-chloro-5-methylbenzene, numbers the ring the “long way” round, giving methyl the higher locant $5$ instead of the lower locant $3$ that is available. This does not give the lowest possible locant set.
• C, 3-chloro-1-methylbenzene, uses the correct locant set $\{1,3\}$ but assigns locant $1$ to methyl instead of to chlorine. Since chlorine is not otherwise fixed by the ring’s connectivity, the numbering should place the substituent that is cited first alphabetically (chloro) at the lower locant, giving chlorine (not methyl) the locant $1$.
• D, 2-chloro-4-methylbenzene, fails to start the numbering at a substituted carbon at all. Locant $1$ is left for an unsubstituted ring carbon, which unnecessarily raises both locants above the achievable minimum of $\{1,3\}$.

Final answer

A, 1-chloro-3-methylbenzene.