Question

Match the column Column I (General formula)

• A. Naphthalene
• B. Anthracene-9-carboxylic acid
• C. Bicyclo[2.2.1]hept-2-ene (Norbornene)
• D. Biphenyl

Answer 1

A-PRS, B-PR, C-PS, D-QRS

Answer 2

To solve this problem, we need to analyze each given organic compound for its Index of Hydrogen Deficiency (IHD), the number of secondary (2°) carbon atoms, and the number of tertiary (3°) carbon atoms.

IHD Calculation Formula:
For a compound with formula $C_c H_h N_n O_o X_x$ (where X is a halogen), the IHD is given by:
$IHD = c - \frac{h}{2} + \frac{n}{2} - \frac{x}{2} + 1$
For hydrocarbons ($C_c H_h$), it simplifies to:
$IHD = c - \frac{h}{2} + 1$ or $IHD = \frac{2c + 2 - h}{2}$

Carbon Classification:

• Primary (1°): Carbon atom bonded to one other carbon atom.
• Secondary (2°): Carbon atom bonded to two other carbon atoms.
• Tertiary (3°): Carbon atom bonded to three other carbon atoms.
• Quaternary (4°): Carbon atom bonded to four other carbon atoms.

Let's analyze each compound:

(A) Naphthalene

• Structure: Two fused benzene rings.
• Molecular Formula: C₁₀H₈
• IHD: $IHD = \frac{2(10) + 2 - 8}{2} = \frac{22 - 8}{2} = \frac{14}{2} = 7$.
  Since IHD is 7 (odd), it matches (P).
• Carbon Classification:
  Naphthalene has 10 carbon atoms. The two carbons at the fusion points are bonded to 4 other carbons (4°). The remaining 8 carbons are bonded to 2 other carbons within the ring and 1 hydrogen atom each. Therefore, these 8 carbons are 2°.
  • Number of 2° carbons = 8 (Even) $\implies$ Matches (R).
  • Number of 3° carbons = 0 (Even) $\implies$ Matches (S).
• Conclusion for (A): (P), (R), (S)

(B) Anthracene-9-carboxylic acid

• Structure: Anthracene (three fused benzene rings) with a -COOH group attached at position 9.
• Molecular Formula: C₁₅H₁₀O₂ (Anthracene is C₁₄H₁₀. Replacing one H with -COOH gives C₁₄H₉COOH, so C₁₅H₁₀O₂)
• IHD: For $C_{15}H_{10}O_2$, oxygen atoms do not affect the IHD calculation for hydrocarbons.
  $IHD = \frac{2(15) + 2 - 10}{2} = \frac{30 + 2 - 10}{2} = \frac{22}{2} = 11$.
  Since IHD is 11 (odd), it matches (P).
• Carbon Classification:
  In anthracene ($C_{14}H_{10}$), there are:
  • 4 quaternary (4°) carbons (at the fusion points).
  • 2 tertiary (3°) carbons (at positions 9 and 10, each bonded to 3 other carbons and 1 H).
  • 8 secondary (2°) carbons (in the outer rings, each bonded to 2 other carbons and 1 H).
    When the -COOH group is attached at position 9, the carbon at position 9 (which was 3°) becomes 4° (bonded to 3 ring carbons and the carbon of -COOH). The other carbons remain unchanged.
  • Number of 2° carbons = 8 (Even) $\implies$ Matches (R).
  • Number of 3° carbons = 1 (C10) (Odd) $\implies$ Does NOT match (S).
• Conclusion for (B): (P), (R)

(C) Bicyclo[2.2.1]hept-2-ene (Norbornene)

• Structure:
  (Note: C2=C3 is a double bond)
• Molecular Formula: C₇H₁₀ (7 carbons, 2 rings, 1 double bond)
• IHD: $IHD = \text{number of rings} + \text{number of double bonds} = 2 + 1 = 3$.
  Since IHD is 3 (odd), it matches (P).
• Carbon Classification:
  • C1 (bridgehead): Bonded to C2, C6, C7. It has 1 H. $\implies$ 3° carbon.
  • C4 (bridgehead): Bonded to C3, C5, C7. It has 1 H. $\implies$ 3° carbon.
  • C2 (alkene): Bonded to C1, C3. It has 1 H. $\implies$ 2° carbon.
  • C3 (alkene): Bonded to C2, C4. It has 1 H. $\implies$ 2° carbon.
  • C5 (CH₂): Bonded to C4, C6. It has 2 H. $\implies$ 2° carbon.
  • C6 (CH₂): Bonded to C1, C5. It has 2 H. $\implies$ 2° carbon.
  • C7 (CH₂, bridge): Bonded to C1, C4. It has 2 H. $\implies$ 2° carbon.
  • Number of 2° carbons = 5 (Odd) $\implies$ Does NOT match (R).
  • Number of 3° carbons = 2 (Even) $\implies$ Matches (S).
• Conclusion for (C): (P), (S)

(D) Biphenyl

• Structure: Two benzene rings directly bonded by a single bond.
• Molecular Formula: C₁₂H₁₀
• IHD: $IHD = \frac{2(12) + 2 - 10}{2} = \frac{24 + 2 - 10}{2} = \frac{16}{2} = 8$.
  Since IHD is 8 (even), it matches (Q).
• Carbon Classification:
  Each benzene ring has 6 carbons. When two rings are joined, one carbon from each ring is bonded to the other ring.
  • The two carbons directly involved in the inter-ring bond (one from each ring) are each bonded to 2 carbons within their own ring and 1 carbon of the other ring. Thus, these two carbons are 3°.
  • The remaining 5 carbons in each ring (total 10 carbons) are each bonded to 2 carbons within their own ring and 1 hydrogen. Thus, these 10 carbons are 2°.
  • Number of 2° carbons = 10 (Even) $\implies$ Matches (R).
  • Number of 3° carbons = 2 (Even) $\implies$ Matches (S).
• Conclusion for (D): (Q), (R), (S)

Final Match:

• (A) $\rightarrow$ (P), (R), (S)
• (B) $\rightarrow$ (P), (R)
• (C) $\rightarrow$ (P), (S)
• (D) $\rightarrow$ (Q), (R), (S)

The question asks to match the column, and the format implies listing all correct matches for each item in Column I.