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Chapter 9: Problem 102

Butadiene, \(\mathrm{C}_{4} \mathrm{H}_{6},\) is a planar molecule that has the following carbon-carbon bond lengths: $$ \mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{134 \mathrm{pm}} \mathrm{CH}=\mathrm{CH}_{2} $$ (a) Predict the bond angles around each of the carbon atoms and sketch the molecule. (b) From left to right, what is the hybridization of each carbon atom in butadiene? (c) The middle \(\mathrm{C}-\mathrm{C}\) bond length in butadiene \((148 \mathrm{pm})\) is a little shorter than the average \(\mathrm{C}-\mathrm{C}\) single bond length (154 pm). Does this imply that the middle \(\mathrm{C}-\mathrm{C}\) bond in butadiene is weaker or stronger than the average \(\mathrm{C}-\mathrm{C}\) single bond? (d) Based on your answer for part (c), discuss what additional aspects of bonding in butadiene might support the shorter middle \(\mathrm{C}-\mathrm{C}\) bond.

Short Answer

Expert verified
In butadiene, all carbon atoms are \(\mathrm{sp}^2\) hybridized, resulting in bond angles close to \(120^{\circ}\). The middle C-C bond length (148 pm) is shorter than the average C-C single bond length (154 pm) due to the presence of a pi bond between the carbon atoms and electron delocalization in the conjugated system. This results in a stronger bond as well as a shorter bond length.

Step by step solution

01

Sketch the structure of butadiene

First, we will draw the structure of butadiene based on the given bond lengths: $$ \text{H}_2\text{C}=\text{CH}_{\text{(134 pm)}}-\text{CH}=\text{CH}_2 $$ Here is a sketch of the planar butadiene molecule: H H | | H2C = C - C = CH2 H H
02

Predict the bond angles around each carbon atom

In order to predict the bond angles around each carbon atom, we need to determine the hybridization of each carbon atom. For the first and last carbon atoms (with two hydrogens attached), they are \(\mathrm{sp}^2\) hybridized, forming three \(\mathrm{sp}^2\) orbitals - two with hydrogens and one with a carbon atom. Hence, the bond angles around these carbon atoms are approximately \(120^{\circ}\). The middle carbon atoms in the molecule are also \(\mathrm{sp}^2\) hybridized as there are two sigma bonds and one pi bond attached to it. Therefore, the bond angles around these carbon atoms should also be close to \(120^{\circ}\).
03

Determine the hybridization of each carbon atom in butadiene

As mentioned in Step 2, all carbon atoms in butadiene are \(\mathrm{sp}^2\) hybridized. So the hybridization of each carbon atom from left to right is: $$ \mathrm{sp}^2 - \mathrm{sp}^2 - \mathrm{sp}^2 - \mathrm{sp}^2 $$
04

Analyse the middle C-C bond length, and discuss its strength compared to an average C-C single bond

The middle C-C bond length in butadiene is 148 pm, which is shorter than the average C-C single bond length of 154 pm. A shorter bond length usually implies a stronger bond.
05

Discuss additional aspects of bonding in butadiene that might support the shorter middle C-C bond

One factor that contributes to the shorter C-C bond is the presence of the pi bond between the middle carbon atoms. This leads to a double bond character in butadiene, which not only increases the bond strength but also leads to a shorter bond length. This double bond character results from electron delocalization in the conjugated system of alternating single and double bonds, enhancing the stability of the molecule.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hybridization in Butadiene
Butadiene is an organic compound consisting of four carbon atoms and six hydrogen atoms, written chemically as \(\mathrm{C}_4\mathrm{H}_6\). In butadiene, each carbon atom is \(\mathrm{sp}^2\) hybridized. This hybridization involves the mixing of one \(s\) orbital and two \(p\) orbitals to form three equivalent \(\mathrm{sp}^2\) hybrid orbitals. These orbitals are oriented at 120 degrees to each other, which is characteristic of \(\mathrm{sp}^2\) hybridization.

In butadiene, the first and last carbon atoms each form two sigma bonds with hydrogen atoms and one sigma bond with another carbon atom, confirming their \(\mathrm{sp}^2\) hybridization. Similarly, the two middle carbon atoms participate in two sigma bonds with adjacent carbon atoms and maintain one pi bond in the conjugated chain, which also aligns them in the \(\mathrm{sp}^2\) hybridized state.
Bond Angles in Molecules
In a molecule, bond angles are determined by the hybridization of the atoms. In butadiene, all carbon atoms are \(\mathrm{sp}^2\) hybridized, which naturally aligns them in a planar structure with bond angles of approximately 120掳. These angles arise from the trigonal planar arrangement of the \(\mathrm{sp}^2\) hybrid orbitals, which aim to minimize repulsion by spreading out as equally as possible.

This uniform bond angle of 120掳 not only contributes to the planarity but also to the stability of the butadiene molecule. Understanding bond angles helps us predict the three-dimensional shape of molecules, crucial for anticipating their reactions and interactions in chemical processes.
Bond Length and Strength
In chemistry, the length of a bond can suggest its strength. In butadiene, the middle \(\mathrm{C}-\mathrm{C}\) bond is 148 pm, which is shorter than the typical \(\mathrm{C}-\mathrm{C}\) single bond length of 154 pm. Shorter bond lengths are generally associated with stronger bonds because the bonded atoms are closer together, enhancing the force of attraction between them.

The shorter length of the middle bond in butadiene suggests a higher bond energy and strength. This is due to additional interactions beyond a simple single bond, leading to a partial double bond character, making it stronger and shorter. Recognizing the correlation between bond length and strength helps in understanding the stability and reactivity of molecules.
Pi Bonds and Electron Delocalization
Delocalization of electrons in pi bonds plays a significant role in the structure of butadiene. In a conjugated system like butadiene, alternating single and double bonds allow pi electrons to spread over multiple atoms. This electron delocalization contributes to a "resonance" effect, where the true structure of a molecule is an average of possible configurations.

In butadiene, the pi bonds are not confined to a single pair of carbon atoms but extend across the entire conjugated system. This delocalization stabilizes the molecule, reducing the energy and giving rise to a partial double bond character in the middle \(\mathrm{C}-\mathrm{C}\) bond. It is this electron sharing over the entire molecule that enhances both the structural rigidity and the chemical uniqueness of butadiene.

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Most popular questions from this chapter

Antibonding molecular orbitals can be used to make bonds to other atoms in a molecule. For example, metal atoms can use appropriate \(d\) orbitals to overlap with the \(\pi_{2}^{*}\), orbitals of the carbon monoxide molecule. This is called \(d-\pi\) backbonding. (a) Draw a coordinate axis system in which the \(y\) -axis is vertical in the plane of the paper and the \(x\) -axis horizontal. Write \({ }^{4} \mathrm{M}^{\prime \prime}\) at the origin to denote a metal atom. (b) Now, on the \(x\) -axis to the right of \(\mathrm{M}\), draw the Lewis structure of a CO molecule, with the carbon nearest the \(\mathrm{M}\). The CO bond axis should be on the \(x\) -axis. (c) Draw the \(\mathrm{CO} \pi_{2 p}^{*}\) orbital, with phases (see the "Closer Look" box on phases) in the plane of the paper. Two lobes should be pointing toward M. (d) Now draw the \(d_{x y}\) orbital of \(\mathrm{M}\), with phases. Can you see how they will overlap with the \(\pi_{2 p}^{*}\) orbital of \(\mathrm{CO} ?\) (e) What kind of bond is being made with the orbitals between \(\mathrm{M}\) and \(\mathrm{C}, \sigma\) or \(\pi ?\) (f) Predict what will happen to the strength of the CO bond in a metal-CO complex compared to CO alone.

Ethyl propanoate, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOCH}_{2} \mathrm{CH}_{3},\) gives a fruity pineapple-like smell. (a) Draw the Lewis structure for the molecule, assuming that carbon always forms four bonds in its stable compounds. (b) How many \(\sigma\) and how many \(\pi\) bonds are in the molecule? (c) Which CO bond is shortest in the molecule? (d) What is the hybridization of atomic orbitals around the carbon atom associated with that short bond? (e) What are the approximate bond angles around each carbon atom in the molecule?

The Lewis structure for allene is Make a sketch of the structure of this molecule that is analogous to Figure \(9.25 .\) In addition, answer the following three questions: (a) Is the molecule planar? (b) Does it have a nonzero dipole moment? (c) Would the bonding in allene be described as delocalized? Explain.

The oxygen atoms in \(\mathrm{O}_{2}\) participate in multiple bonding, whereas those in hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2},\) do not. (a) Draw Lewis structures for both molecules. (b) What is the hybridization of the oxygen atoms in each molecule? (c) Which molecule has the stronger \(\mathrm{O}-\mathrm{O}\) bond?

The energy-level diagram in Figure 9.36 shows that the sideways overlap of a pair of \(p\) orbitals produces two molecular orbitals, one bonding and one antibonding. In ethylene there is a pair of electrons in the bonding \(\pi\) orbital between the two carbons. Absorption of a photon of the appropriate wavelength can result in promotion of one of the bonding electrons from the \(\pi_{2 p}\) to the \(\pi_{2 p}^{*}\) molecular orbital. (a) Assuming this electronic transition corresponds to the HOMO-LUMO transition, what is the HOMO in ethylene? (b) Assuming this electronic transition corresponds to the HOMO-LUMO transition, what is the LUMO in ethylene? (c) Is the \(\mathrm{C}-\mathrm{C}\) bond in ethylene stronger or weaker in the excited state than in the ground state? Why? (d) Is the \(\mathrm{C}-\mathrm{C}\) bond in ethylene easier to twist in the ground state or in the excited state?
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