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Chapter 9: Problem 39

Biacetyl and acetoin are added to margarine to make it taste more like butter Complete the Lewis structures, predict values for all \(\mathrm{C}-\mathrm{C}-\mathrm{O}\) bond angles, and give the hybridization of the carbon atoms in these two compounds. Must the four carbon atoms and two oxygen atoms in biacetyl lie the same plane? How many \(\sigma\) bonds and how many \(\pi\) bonds are there in biacetyl and acetoin?

Short Answer

Expert verified
The Lewis structures for biacetyl and acetoin are H₃C-C(=O)-C(=O)-H and H₃C-C(OH)-C(=O)-CH₃, respectively. In biacetyl, all carbon atoms have sp² hybridization and have a C-C-O bond angle of 120°. All atoms in biacetyl lie in the same plane. For acetoin, the C-C-O bond angles are approximately 109.5° and 120° due to the difference in hybridization. Biacetyl has 8 σ bonds and 2 π bonds, while acetoin has 11 σ bonds and 1 π bond.

Step by step solution

01

Lewis Structures

Draw Lewis structures for biacetyl and acetoin: - Biacetyl: H₃C-C(=O)-C(=O)-H - Acetoin: H₃C-C(OH)-C(=O)-CH₃
02

Bond Angles

Predict bond angles for all C-C-O bonds in biacetyl and acetoin: - Biacetyl: The carbons in biacetyl have sp² hybridization, so they have a trigonal planar shape. Hence, the C-C-O angle is 120°. - Acetoin: The first carbon in Acetoin (CH₃) has sp³ hybridization, so it has a tetrahedral shape. The central carbon with OH group is also sp³ hybridized with a tetrahedral shape. The C(=O) carbon atom has sp² hybridization with a trigonal planar shape. Hence, the C-C-O bond angles are approximately 109.5° and 120° respectively, due to the difference in hybridization.
03

Hybridization of Carbon Atoms

Give the hybridization of the carbon atoms in these two compounds: - Biacetyl: All carbon atoms in biacetyl have sp² hybridization. - Acetoin: The first and last carbon atoms have sp³ hybridization, the central carbon with OH group has sp³ hybridization, and the carbon with the carbonyl group has sp² hybridization.
04

Planar Structure of Biacetyl

Decide whether the four carbon atoms and two oxygen atoms in biacetyl lie in the same plane: Since all the carbon atoms in biacetyl have sp² hybridization and the molecule is in continuous conjugation, all the atoms lie in the same plane.
05

σ and π Bonds

Calculate the number of σ and π bonds in biacetyl and acetoin: - Biacetyl: * σ bonds: 8 (3 single bonds with hydrogen, 2 single bonds with carbon, and 2 single bond with oxygen) * π bonds: 2 (2 double bonds with oxygen) - Acetoin: * σ bonds: 11 (6 single bonds with hydrogen, 3 single bonds with carbon, and 2 single bonds with oxygen) * π bonds: 1 (1 double bond with oxygen)

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Most popular questions from this chapter

Using the molecular orbital model, write electron configurations for the following diatomic species and calculate the bond orders. Which ones are paramagnetic? Place the species in order of increasing bond length and bond energy. $$ \text {a} \mathrm{CO} \quad \text { b. } \mathrm{CO}^{+} \quad \text { c. } \mathrm{CO}^{2+} $$

Acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) can be produced from the reaction of calcium carbide \(\left(\mathrm{CaC}_{2}\right)\) with water. Use both the localized electron and molecular orbital models to describe the bonding in the acetylide anion \(\left(\mathrm{C}_{2}^{2-}\right)\)

Describe the bonding in the first excited state of \(\mathrm{N}_{2}\) (the one closest in energy to the ground state) using the molecular orbital model. What differences do you expect in the properties of the molecule in the ground state as compared to the first excited state? (An excited state of a molecule corresponds to an electron arrangement other than that giving the lowest possible energy.)

Values of measured bond energies may vary greatly depending on the molecule studied. Consider the following reactions: $$ \begin{array}{ll}{\mathrm{NCl}_{3}(g) \longrightarrow \mathrm{NCl}_{2}(g)+\mathrm{Cl}(g)} & {\Delta H=375 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{ONCI}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{Cl}(g)} & {\Delta H=158 \mathrm{kJ} / \mathrm{mol}}\end{array} $$ Rationalize the difference in the values of \(\Delta H\) for these reactions, even though each reaction appears to involve only the breaking of one \(\mathrm{N}-\mathrm{Cl}\) bond. (Hint: Consider the bond order of the NO bond in ONCl and in NO.)

Draw the Lewis structure for HCN. Indicate the hybrid orbitals, and draw a picture showing all the bonds between the atoms, labeling each bond as \(\sigma\) or \(\pi .\)
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