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Chapter 9: Problem 92

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?

Short Answer

Expert verified
The Lewis structure for ethyl propanoate is: \(H - C - C - C = O\) \(| | |\) \(H H H - O - C - H\) \( | |\) \( H H H\) There are 11 sigma (\(\sigma\)) bonds and 1 pi (\(\pi\)) bond in the molecule. The C=O bond is the shortest CO bond in the molecule. The carbon atom associated with the shortest CO bond has sp hybridization. The approximate bond angles around each carbon in the molecule are: 1. CH3: 109.5° (sp3) 2. CH2: 109.5° (sp3) 3. C=O: 180° (sp) 4. CH3: 109.5° (sp3)

Step by step solution

01

a) Drawing the Lewis structure

To draw the Lewis structure for ethyl propanoate, follow these steps: 1. Write the skeletal structure: From the formula, we have \(\mathrm{CH}_{3} \mathrm{CH}_{2} \text{COOCH}_{2}\mathrm{CH}_{3}\). 2. Fill in the electron pairs around each atom to complete the octets for carbon and oxygen atoms and the duet for hydrogen atoms. The Lewis structure for ethyl propanoate is: \(H - C - C - C = O\) \(| | |\) \(H H H - O - C - H\) \( | |\) \( H H H\)
02

b) Counting \(\sigma\) and \(\pi\) bonds

In the ethyl propanoate molecule, we have the following bonds: - 10 single (σ) bonds: 8 C-H bonds, 1 C-C bond, and 1 C-O bond. - 1 double (σ and π) bond: 1 C=O bond, which has one σ bond and one π bond. So, there are 11 sigma (\(\sigma\)) bonds and 1 pi (\(\pi\)) bond in the molecule.
03

c) Identifying the shortest CO bond

The molecule has two CO bonds: one single bond (C-O) and one double bond (C=O). Double bonds are generally shorter than single bonds because they contain both a sigma and a pi bond, which pull the atoms closer together. Therefore, the C=O bond is shortest in the molecule.
04

d) Determining the hybridization of the carbon atom

For the carbon atom associated with the shortest CO bond (the one with a double bond to oxygen), we can determine the hybridization by counting the electron domains around the atom: 1. One double bond (C=O): count as one electron domain. 2. One single bond (C-C): count as one electron domain. Thus, there are 2 electron domains around that carbon atom, so it undergoes sp hybridization.
05

e) Approximate bond angles around each carbon atom

To determine the bond angles, we need to consider the hybridization of each carbon atom in the molecule: 1. First carbon (CH3): sp3 hybridization (four single bonds) - the bond angles around this carbon are approximately 109.5°. 2. Second carbon (CH2): sp3 hybridization (three single bonds and one free electron pair) - the bond angles around this carbon are also approximately 109.5°. 3. Third carbon (C=O): sp hybridization (one double bond and one single bond) - the bond angles around this carbon are approximately 180°. 4. Last carbon (CH3): sp3 hybridization (four single bonds) - the bond angles around this carbon are approximately 109.5°.

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Most popular questions from this chapter

The structure of borazine, \(\mathrm{B}_{3} \mathrm{~N}_{3} \mathrm{H}_{6},\) is a six-membered ring of alternating \(\mathrm{B}\) and \(\mathrm{N}\) atoms. There is one \(\mathrm{H}\) atom bonded to each \(B\) and to each \(\mathrm{N}\) atom. The molecule is planar. (a) Write a Lewis structure for borazine in which the formal charge on every atom is zero. (b) Write a Lewis structure for borazine in which the octet rule is satisfied for every atom. (c) What are the formal charges on the atoms in the Lewis structure from part (b)? Given the electronegativities of \(B\) and \(N,\) do the formal charges seem favorable or unfavorable? (d) Do either of the Lewis structures in parts (a) and (b) have multiple resonance structures? (e) What are the hybridizations at the \(\mathrm{B}\) and \(\mathrm{N}\) atoms in the Lewis structures from parts (a) and (b)? Would you expect the molecule to be planar for both Lewis structures? (f) The six \(\mathrm{B}-\mathrm{N}\) bonds in the borazine molecule are all identical in length at \(144 \mathrm{pm} .\) Typical values for the bond lengths of \(\mathrm{B}-\mathrm{N}\) single and double bonds are \(151 \mathrm{pm}\) and \(131 \mathrm{pm},\) respectively. Does the value of the \(\mathrm{B}-\mathrm{N}\) bond length seem to favor one Lewis structure over the other? (g) How many electrons are in the \(\pi\) system of botazine?

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?

Consider a molecule with formula \(\mathrm{AX}_{2}\). Supposing the \(\mathrm{A}-\mathrm{X}\) bond is polar, how would you expect the dipole moment of the \(\mathrm{AX}_{2}\) molecule to change as the \(\mathrm{X}-\mathrm{A}-\mathrm{X}\) bond angle decreases from \(180^{\circ}\) to \(100^{\circ} ?\)

Azo dyes are organic dyes that are used for many applications, such as the coloring of fabrics. Many azo dyes are derivatives of the organic substance azobenzene, \(\mathrm{C}_{12} \mathrm{H}_{10} \mathrm{~N}_{2}\) A closely related substance is hydrazobenzene, $\mathrm{C}_{12} \mathrm{H}_{12} \mathrm{~N}_{2}$ (Recall the shorthand notation used for benzene.) (a) What is the hybridization at the \(\mathrm{N}\) atom in each of the substances? (b) How many unhybridized atomic orbitals are there on the \(\mathrm{N}\) and the \(\mathrm{C}\) atoms in each of the substances? (c) Predict the \(\mathrm{N}-\mathrm{N}-\mathrm{C}\) angles in each of the substances. (d) Azobenzene is said to have greater delocalization of its \(\pi\) electrons than hydrazobenzene. Discuss this statement in light of your answers to (a) and (b). (e) All the atoms of azobenzene lie in one plane, whereas those of hydrazobenzene do not. Is this observation consistent with the statement in part (d)? (f) Azobenzene is an intense red-orange color, whereas hydrazobenzene is nearly colorless. Which molecule would be a better one to use in a solar energy conversion device? (See the "Chemistry Put to Work" box for more information about solar cells.)

How would we describe a substance that contains only paired electrons and is weakly repelled by a magnetic field? Which of the following ions would you expect to possess similar characteristics: $\mathrm{H}_{2}^{-}, \mathrm{Ne}_{2}^{+}, \mathrm{F}_{2}, \mathrm{O}_{2}^{2+} ?$
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