The IUPAC name of the following compound is
(a) 3-bromo-5-methylcyclopentane carboxylic acid
(b) 3-b

Q: Why is the parent name 'cyclopentanecarboxylic acid' and not 'cyclopentanoic acid'?

When a $-COOH$ group is directly attached to a cycloalkane ring, the carbon atom of the carboxyl group is not counted as part of the parent ring. To account for this extra carbon, the special suffix '-carboxylic acid' is used instead of '-oic acid'.

Q: Bromine is heavier and more electronegative than a methyl group. Shouldn't we number towards it first?

No, IUPAC numbering rules prioritize the 'Lowest Locant Rule' over mass, electronegativity, or alphabetical order. We must choose the numbering path that gives the lowest possible set of numbers to the substituents (2,4 is lower than 3,5).

Q: What if the locant sets were identical in both directions?

If numbering in both directions gave the exact same set of locants (for example, if the substituents were at positions 2 and 5 either way), then and only then would you use alphabetical order to break the tie, giving the lower number to the substituent that comes first alphabetically.

Q: Is the carbon of the $-COOH$ group counted as Carbon-1?

No. In cyclic compounds named with the '-carboxylic acid' suffix, Carbon-1 is the carbon atom *of the ring* that the $-COOH$ group is attached to. The carboxyl carbon itself is not numbered as part of the ring.

Q: Can I write the name as 2-methyl-4-bromocyclopentanecarboxylic acid since 2 comes before 4?

No, that would be incorrect. While the numbers are determined by the lowest locant rule, the final written order of the substituents is strictly dictated by alphabetical order. 'Bromo' must always precede 'methyl' in the written name.

Question

The IUPAC name of the following compound is

Options:

(A) 3-bromo-5-methylcyclopentane carboxylic acid

(B) 3-bromo-5-methylcyclopentanoic acid

(C) 5-bromo-3-methylcyclopentanoic acid

(D) 4-bromo-2-methylcyclopentane carboxylic acid

Sigma Percentile · Accepted Answer

### Decoding the IUPAC Name of a Substituted Cycloalkane

Naming organic compounds can sometimes feel like solving a puzzle, but once you know the rules, it becomes a highly logical and satisfying process. Let's break down the IUPAC nomenclature for the given cyclic compound step-by-step.

### Identifying the Principal Functional Group

The very first step in naming any organic molecule is to scan it for functional groups. In our molecule, we have three distinct features attached to a five-membered carbon ring:
1.  A carboxylic acid group ($-COOH$)
2.  A methyl group ($-CH_3$)
3.  A bromine atom ($-Br$)

According to the IUPAC priority rules, the **carboxylic acid group** reigns supreme. It has the highest priority among these three. Because this principal functional group is directly attached to a cycloalkane ring, a special naming convention applies. Instead of using the suffix '-oic acid' (which is used for open chains where the carboxyl carbon is part of the main chain), we use the suffix **'-carboxylic acid'**. The parent ring is a five-membered alkane, so it is named **cyclopentane**.

Combining these gives us our parent name: **cyclopentanecarboxylic acid**.

### Numbering the Ring: The Lowest Locant Rule

Now that we have our parent name, we need to specify the locations of our substituents (the methyl and bromo groups). The rule is simple: the carbon atom of the ring that is directly attached to the principal functional group (the $-COOH$ group) is automatically designated as **Carbon-1**.

From Carbon-1, we must decide whether to number the ring clockwise or anti-clockwise. The goal is to assign the lowest possible numbers (locants) to the remaining substituents. Let's test both paths:

*   **Clockwise Path:** The methyl group lands on Carbon-2, and the bromo group lands on Carbon-4. Our locant set is **2, 4**.
*   **Anti-clockwise Path:** The bromo group lands on Carbon-3, and the methyl group lands on Carbon-5. Our locant set is **3, 5**.

Comparing the two sets, the first point of difference is between 2 and 3. Since 2 is lower than 3, the clockwise path wins. Therefore, we have a **2-methyl** group and a **4-bromo** group.

### Alphabetical Ordering and Final Assembly

We have all the pieces; now we just need to assemble them. When writing the final IUPAC name, substituents must be listed in strict **alphabetical order**, regardless of their locant numbers.

Comparing our substituents, 'bromo' starts with 'b' and 'methyl' starts with 'm'. Since 'b' comes before 'm' in the alphabet, we write the bromo group first.

Putting it all together:
Prefixes: **4-bromo-2-methyl**
Parent Name: **cyclopentanecarboxylic acid**

Merging them seamlessly gives us the final, correct IUPAC name: **4-bromo-2-methylcyclopentanecarboxylic acid**.

Animated Solution for Chemistry - Organic Chemistry: The IUPAC name of the following compound is

Select Answer:

Visualized Solution

\text{Molecular Structure}

\text{Principal Functional Group}

\text{Parent Name \& C1}

\text{Lowest Locant Rule}

\text{Alphabetical Order}

\text{Final IUPAC Name}

The Sigma Insight: Nomenclature and Characterisation

Decoding the IUPAC Name of a Substituted Cycloalkane

Identifying the Principal Functional Group

Numbering the Ring

Alphabetical Ordering and Final Assembly

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