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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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Most popular questions from this chapter

The highest occupied molecular orbital of a molecule is abbreviated as the HOMO. The lowest unoccupied molecular orbital in a molecule is called the LUMO. Experimentally, one can measure the difference in energy between the HOMO and LUMO by taking the electronic absorption (UV-visible) spectrum of the molecule. Peaks in the electronic absorption spectrum can be labeled as \(\pi_{2 \mathrm{p}}-\pi_{2 \mathrm{p}}{ }^{*}\), $\sigma_{2 s}-\sigma_{2 s}{ }^{*},$ and so on, corresponding to electrons being promoted from one orbital to another. The HOMO-LUMO transition corresponds to molecules going from their ground state to their first excited state. (a) Write out the molecular orbital valence electron configurations for the ground state and first excited state for \(\mathrm{N}_{2}\). (b) Is \(\mathrm{N}_{2}\) paramagnetic or diamagnetic in its first excited state? (c) The electronic absorption spectrum of the \(\mathrm{N}_{2}\) molecule has the lowest energy peak at \(170 \mathrm{nm}\). To what orbital transition does this correspond? (d) Calculate the energy of the HOMO-LUMO transition in part (a) in terms of \(\mathrm{kJ} / \mathrm{mol}\). (e) Is the \(\mathrm{N}-\mathrm{N}\) bond in the first excited state stronger or weaker compared to that in the ground state?

(a) What is the difference between hybrid orbitals and molecular orbitals? (b) How many electrons can be placed into each MO of a molecule? (c) Can antibonding molecular orbitals have electrons in them?

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?

Give the electron-domain and molecular geometries of a molecule that has the following electron domains on its central atom: (a) four bonding domains and no nonbonding domains, (b) three bonding domains and two nonbonding domains, (c) five bonding domains and one nonbonding domain, (d) four bonding domains and two nonbonding domains.

Shown here are three pairs of hybrid orbitals, with each set at a characteristic angle. For each pair, determine the type of hybridization, if any, that could lead to hybrid orbitals at the specified angle.
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