JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Chemical Bonding and Molecular Structure

2-Formylhex-4-enoic acid is $\mathrm{HOOC}-\mathrm{CH}(\mathrm{CHO})-\mathrm{C}{\mathrm{H}}_2-\mathrm{CH}=\mathrm{CH}-\mathrm{C}{\mathrm{H}}_3$: a hex-4-enoic acid chain (C1 to C6) carrying a formyl ($-\mathrm{CHO}$) branch on C2, with the C4=C5 double bond.

Every bond in the molecule is either a $\sigma$ bond, or, where a double bond exists, a $\sigma$ bond plus one $\pi$ bond. Counting them:

• 1 O–H $\sigma$ bond and 1 C1–OH $\sigma$ bond
• 2 C=O $\sigma$ bonds (one in $-\mathrm{COOH}$, one in $-\mathrm{CHO}$) with 2 accompanying $\pi$ bonds
• 6 C–C $\sigma$ bonds: C1–C2, C2–C(formyl), C2–C3, C3–C4, C4–C5, C5–C6, one of which (C4–C5) also carries a $\pi$ bond
• 9 C–H $\sigma$ bonds: formyl C–H, C2–H, two on C3, C4–H, C5–H, three on C6

Summing: $\sigma$ bonds $=1+1+2+6+9=19$; $\pi$ bonds $=1(\mathrm{COOH})+1(\mathrm{CHO})+1(\mathrm{C}=\mathrm{C})=3$.

The total number of $\sigma$ and $\pi$ bonds is $19+3=22$.