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JEE Chemistry 2024 Question with Solution

The total number of sigma (σ\sigma) and pi (π\pi) bonds in 22-formylhex-44-enoic acid is:

  • A

    2222

  • B

    1818

  • C

    2020

  • D

    2525

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The compound is 22-formylhex-44-enoic acid.

Find: The total number of sigma (σ\sigma) and pi (π\pi) bonds.

Structure of 2-formylhex-4-enoic acid showing carboxylic acid group, formyl substituent at carbon 2, double bond between carbons 4 and 5, and carbon numbering 1 to 6.

The molecular structure is represented as:

H–O–C(=O)–CH(CHO)–CH2–CH=CH–CH3\text{H–O–C(=O)–CH(CHO)–CH}_2\text{–CH=CH–CH}_3

Each single bond is a σ\sigma bond, and each double bond contains one σ\sigma bond and one π\pi bond.

From the structure:

  • The molecule contains a carboxylic acid group –COOH\text{–COOH}.
  • It also contains a formyl group –CHO\text{–CHO}.
  • There is one carbon-carbon double bond C=C\text{C=C} at position 4−54{-}5.

Thus, the pi bonds are:

π-bonds=1 from carboxyl C=O+1 from formyl C=O+1 from C=C=3\begin{aligned} \pi\text{-bonds} &= 1 \text{ from carboxyl } \text{C=O} \\ &\quad + 1 \text{ from formyl } \text{C=O} \\ &\quad + 1 \text{ from } \text{C=C} \\ &= 3 \end{aligned}

Counting Sigma and Pi Bonds

The solution states directly:

σ-bonds=19,π-bonds=3\sigma\text{-bonds} = 19, \quad \pi\text{-bonds} = 3

Therefore, total bonds:

19+3=2219 + 3 = 22

So, the total number of sigma and pi bonds is 2222.

The correct option is A.

Common mistakes

  • Counting each double bond as only a π\pi bond is incorrect because every double bond contains one σ\sigma bond and one π\pi bond. Always include both parts while counting total bonds.

  • Ignoring the formyl group –CHO\text{–CHO} is incorrect because its carbonyl bond contributes one additional π\pi bond and associated σ\sigma bonds. Include all functional groups present in the structure.

  • Adding only sigma bonds or only pi bonds is incorrect because the question asks for the total number of sigma and pi bonds. First count them separately, then add the two totals.

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