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: All four groups (\( -CN \), \( -CHO \), \( -COOH \), \( -COOCH_3 \)) are attached to the stereocenter through a carbon atom, so the first point of comparison, the attached atom itself, is tied. Priority must be decided by comparing what is attached to that first carbon.

Step 2: Using the CIP duplicate atom convention for multiple bonds, list the atoms bonded to each group's first carbon: \( -CN \) gives (N, N, N) since the triple bond to nitrogen is duplicated twice; \( -CHO \) gives (O, O, H) since the double bond oxygen is duplicated once; \( -COOH \) gives (O, O, O); \( -COOCH_3 \) gives (O, O, O).

Step 3: Comparing the highest atom in each set first: (O, O, O) beats (O, O, H) beats (N, N, N), because the first point of difference already favors oxygen (atomic number 8) over nitrogen (atomic number 7). So \( -COOH \) and \( -COOCH_3 \) both outrank \( -CHO \), which in turn outranks \( -CN \).

Step 4: To break the tie between \( -COOH \) and \( -COOCH_3 \), compare their third oxygen branch (the single bonded oxygen, since the carbonyl oxygen branch is identical in both). In \( -COOH \) this oxygen is bonded to H; in \( -COOCH_3 \) it is bonded to a carbon (of the methyl group). Since carbon outranks hydrogen, \( -COOCH_3 \) has the higher priority branch.

Step 5: The overall priority order is \( -COOCH_3 > -COOH > -CHO > -CN \), so \( -COOCH_3 \) has the highest priority.

\[ \boxed{-COOCH_3} \]