Question

The hybridization of carbon atoms in C-C single bond of \(\mathrm{HC} \equiv \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2}\) is (a) sp \(^{3}-s p^{3}\) (b) \(s p^{2}-s p^{3}\) (c) \(\mathrm{sp}^{2}-\mathrm{sp}\) (d) \(\mathrm{sp}^{3}-\mathrm{sp}\)

Short answer

The hybridization of the carbon atoms in the C-C single bond is (c) \(\text{sp}^2-\text{sp}\).

Step-by-step solution

Structure Analysis

To determine the hybridization of carbon atoms in a C-C single bond, understand the molecular structure. The given structure is \( ext{HC} \ ext{C}- ext{CH}= ext{CH}_2\). Identify each carbon: the first carbon in a triple bond, the second connecting triple and double bonds, the third in a double bond, and the fourth in an individual single bond.

Triple Bond Hybridization

The first carbon, which is part of the alkynyl group \( ext{HC} \ ext{C}-\), forms one triple bond. The hybridization around a carbon in a triple bond is \( ext{sp}\). Thus, the first carbon is \( ext{sp}\) hybridized.

Hybridization of C-C Single Bond Linked Carbon

The second carbon connects to both triple and double bonds \(- ext{CH}=\). For a carbon forming a triple bond with one partner and single or other bond types with the other, it is often \( ext{sp}\) if it's mainly involved in \(\sigma\) and \(\pi\) bond formations. Hence, in this part of the molecule, the hybridization for carbon forming the C-C single bond is \( ext{sp}\).

Connected Carbon Hybridization

The third carbon is part of a double bond \(= ext{CH}-\). The hybridization around carbon forming a double bond is \( ext{sp}^2\). Therefore, for the C=C segment, the hybridization of the carbon connecting via a double bond to \( ext{CH}_2\) is \( ext{sp}^2\).

Answer Extraction

From the analysis, the C-C single bond in question involves carbons that are \( ext{sp}\) and \( ext{sp}^2\) hybridized. Thus, the given hybridization option is \(\mathrm{sp}^2-\mathrm{sp}\). Therefore, the correct answer is (c).

Key concepts

Molecular Structure Analysis

Understanding molecular structure is crucial in determining hybridization states of atoms within a molecule. When analyzing a compound like \( \mathrm{HC} \equiv \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2} \), it is important to identify the types of bonds and the molecular geometry. Each carbon atom must be considered individually:

• The first carbon is involved in a triple bond.
• The second connects both the triple and double bonds.
• The third participates in a double bond.
• The fourth is bonded individually.

Through this step-by-step breakdown, each carbon's hybridization state can be understood by looking at how many other atoms each carbon is bonded to, and the types of bonds it forms. Breaking it down like this simplifies analyzing complex molecules.

Carbon-Carbon Single Bond

In organic chemistry, a carbon-carbon single bond refers to the connection between two carbon atoms with a single shared pair of electrons. This type of bond is a sigma (\( \sigma \)) bond, which is strong and allows for free rotation around the bond axis. The unique aspect of analyzing such a bond in the provided molecule is understanding which carbon atoms are involved and their hybridization states. For instance, in \( \mathrm{HC} \equiv \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2} \), the carbon-carbon single bond connects a carbon from a triple bond region (\( \mathrm{sp} \) hybridized) to a carbon in a double bond region (\( \mathrm{sp}^2 \) hybridized). This blend of geometry and bonding type is critical in molecular structure analysis and hybridization determination.

Triple Bond Hybridization

Triple bonds in organic molecules are indicative of \( \mathrm{sp} \) hybridization. Such bonds involve one \( \sigma \) bond and two \( \pi \) bonds, allowing the bonded atoms to form a linear shape. In the molecule \( \mathrm{HC} \equiv \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2} \), the presence of a triple bond at the beginning signifies that those carbon atoms utilize \( \mathrm{sp} \) hybridization. The linear geometry is due to the requirement of the three \( sp \) hybrid orbitals to overlap linearly, facilitating the strong triple bond formation. This arrangement is crucial for understanding not only the bond's strength but also its influence on the molecule's overall geometry and reactivity.

Double Bond Hybridization

In a double bond, such as the one found in the segment \( =\mathrm{CH}- \), the carbon atoms are \( \mathrm{sp^2} \) hybridized. This hybridization state involves the mixing of one \( s \) orbital and two \( p \) orbitals. The result is a planar triangle shape, with one \( \sigma \) bond and one \( \pi \) bond, restricting the bond's rotation. In the compound \( \mathrm{HC} \equiv \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2} \), the \( \mathrm{sp^2} \) hybridization at the double bond site contributes significantly to the molecule's planar sections, influencing its reactivity pattern and physical properties. Recognizing how this hybridization influences a molecule provides insights into molecular interaction, stability, and reactivity dynamics, which are keys to mastering structural chemistry.