Write the structure from the IUPAC name

The parent chain is hex-4-enoic acid:

• 6-carbon chain
• carboxylic acid at carbon 1
• double bond between carbon 4 and carbon 5
• formyl substituent at carbon 2

So the structure is:

$\mathrm{HOOC-CH(CHO)-CH_2-CH=CH-CH_3}$

Count all $\sigma$ and $\pi$ bonds

We count each single bond as one $\sigma$, each double bond as one $\sigma$ and one $\pi$.

A. Bonds between heavy atoms

In $\mathrm{HOOC-CH(CHO)-CH_2-CH=CH-CH_3}$

Heavy-atom connections are:

1. $\mathrm{C_1-O}$ in $\mathrm{-OH}$ : $1\,\sigma$
2. $\mathrm{C_1=O}$ : $1\,\sigma + 1\,\pi$
3. $\mathrm{C_1-C_2}$ : $1\,\sigma$
4. $\mathrm{C_2-C_3}$ : $1\,\sigma$
5. $\mathrm{C_3-C_4}$ : $1\,\sigma$
6. $\mathrm{C_4=C_5}$ : $1\,\sigma + 1\,\pi$
7. $\mathrm{C_5-C_6}$ : $1\,\sigma$
8. $\mathrm{C_2-C_{formyl}}$ : $1\,\sigma$
9. $\mathrm{C_{formyl}=O}$ : $1\,\sigma + 1\,\pi$

Therefore from heavy atoms:

• $\sigma$ bonds = $9$
• $\pi$ bonds = $3$

B. Hydrogen-containing bonds

Now count all $\mathrm{C-H}$ and $\mathrm{O-H}$ bonds.

Structure again:

$\mathrm{HOOC-CH(CHO)-CH_2-CH=CH-CH_3}$

• $\mathrm{-OH}$ gives $1$ bond: $1\,\sigma$
• $\mathrm{C_2 = CH}$ gives $1$ $\mathrm{C-H}$ bond
• formyl carbon $(\mathrm{-CHO})$ gives $1$ $\mathrm{C-H}$ bond
• $\mathrm{CH_2}$ gives $2$ $\mathrm{C-H}$ bonds
• $\mathrm{CH}$ at carbon 4 gives $1$ $\mathrm{C-H}$ bond
• $\mathrm{CH}$ at carbon 5 gives $1$ $\mathrm{C-H}$ bond
• terminal $\mathrm{CH_3}$ gives $3$ $\mathrm{C-H}$ bonds

Total hydrogen bonds:

$1+1+1+2+1+1+3 = 10\,\sigma$

Total bonds

Total $\sigma$ bonds: $9 + 10 = 19$

Total $\pi$ bonds: $3$

Hence total number of $\sigma$ and $\pi$ bonds is:

$19 + 3 = 22$

Final Answer

$\boxed{22}$

Comparison with stored correct answer

Stored correct answer = $22$

Our derived answer also = $22$, so they agree.