Question

Which among the following four substituent groups has the highest CIP (Cahn-Ingold-Prelog) priority?

• A. \( -CN \)
• B. \( -CHO \)
• C. \( -COOH \)
• D. \( -COOCH_3 \)

Hint:

Compare atoms attached to the first carbon of each group, then break ties by going one bond further.

Answer 1

The Correct Option is D

Step 1: Compare the groups two at a time instead of all at once. First compare $-CHO$ with $-CN$. The first carbon of $-CHO$ is attached to $(O,O,H)$ while the first carbon of $-CN$ is attached to $(N,N,N)$. Since the highest atom O (8) beats N (7) right at the first comparison, $-CHO$ outranks $-CN$, eliminating $-CN$ from contention for the top spot.

Step 2: Next compare $-COOH$ with $-CHO$. Both start their comparison at a carbon, but $-COOH$'s carbon is attached to $(O,O,O)$ while $-CHO$'s carbon is attached to $(O,O,H)$. The third position, O versus H, favors $-COOH$, so $-COOH$ outranks $-CHO$, eliminating $-CHO$ as well.

Step 3: Only $-COOH$ and $-COOCH_3$ remain, both giving $(O,O,O)$ at the first carbon. Explore one branch deeper: the carbonyl oxygen branch is identical for both (a real oxygen duplicated by an equivalent phantom carbon), so it cannot break the tie.

Step 4: The deciding branch is the remaining single bonded oxygen. In $-COOH$ this oxygen carries an H (atomic number 1); in $-COOCH_3$ it carries a real methyl carbon whose own substituents are three hydrogens, clearly outranking a bare phantom or a lone H. This makes the $-COOCH_3$ branch senior.

Step 5: Having survived every pairwise comparison, $-COOCH_3$ is the group of highest priority.

\[ \boxed{-COOCH_3 \text{ wins every pairwise comparison}} \]