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Difficulty: Very hardTetrahedral Carbon, Bonding, and Hybridization

Consider the cumulated diene compound penta-1,2-diene, represented by the condensed structure CH2=C=CH−CH2−CH3\text{CH}_2=\text{C}=\text{CH}-\text{CH}_2-\text{CH}_3. What are the hybridization states of the central carbon atom (C2) and the methylene carbon atom (C4), respectively?

  1. spsp and sp3sp^3Answer
  2. B
    sp2sp^2 and sp3sp^3
  3. C
    spsp and sp2sp^2
  4. D
    sp2sp^2 and sp2sp^2

Answer

The central carbon atom (C2) is spsp hybridized and the methylene carbon atom (C4) is sp3sp^3 hybridized.
In penta-1,2-diene, the central allene carbon atom at position 2 (C2) forms two double bonds (2 σ2\ \sigma bonds and 2 π2\ \pi bonds), which necessitates spsp hybridization. The methylene carbon atom at position 4 (C4) forms four single σ\sigma bonds with two hydrogen atoms and two carbon atoms, which corresponds to sp3sp^3 tetrahedral hybridization.

Step-by-Step Solution

1
Determine the number of σ\sigma and π\pi bonds on carbon-2 (C2).
C2 forms two double bonds, which consists of 2 σ2\ \sigma bonds and 2 π2\ \pi bonds.
Carbon atoms involved in two double bonds (cumulated dienes/allenes) use two spsp hybrid orbitals to form σ\sigma bonds at an angle of 180∘180^\circ.
2
Determine the hybridization state of C2.
C2 is spsp hybridized with linear geometry.
Two σ\sigma bonding domains correlate to spsp hybridization.
3
Determine the bonding domains and hybridization of carbon-4 (C4).
C4 is bonded to two hydrogen atoms, C3, and C5 via single covalent bonds, giving 4 σ4\ \sigma bonds.
Four single σ\sigma bonding domains require sp3sp^3 hybridization with tetrahedral geometry.

Key Concept

Hybridization in Cumulated Dienes and Saturated Carbons
Estimated Time:1m 30s
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Topics

Tetrahedral Carbon, Bonding, and Hybridization

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