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Chapter 9: Problem 40

Many important compounds in the chemical industry are derivatives of ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right) .\) Two of them are acrylonitrile and methyl methacrylate. Complete the Lewis structures, showing all lone pairs. Give approximate values for bond angles \(a\) through \(f\) . Give the hybridization of all carbon atoms. In acrylonitrile, how many of the atoms in the molecule must lie in the same plane? How many \(\sigma\) bonds and how many \(\pi\) bonds are there in methyl methacrylate and acrylonitrile?

Short Answer

Expert verified
In acrylonitrile (C3H3N), there are 6 σ bonds and 3 π bonds. Its C1 and C3 carbon atoms are sp2 hybridized, while the C2 carbon atom is sp hybridized. In methyl methacrylate (C5H8O2), there are 14 σ bonds and 2 π bonds. All carbons, except C1, are sp2 hybridized, and C1 is sp3 hybridized. Bond angles a, b, and c are approximately \(120^{\circ}\), while d, e, and f are around \(109.5^{\circ}\). In acrylonitrile, six atoms must lie in the same plane, including the three carbons, one nitrogen, and two hydrogens on C1 and C3.

Step by step solution

01

1. Build the Molecular Structures

Acrylonitrile's chemical formula is C3H3N. Methyl methacrylate's chemical formula is C5H8O2. The Lewis structures of the molecules can be drawn to show all lone pairs and bond angles a through f.
02

2. Determine Bond Angles

Based on the Lewis structures above, we identify bond angles a through f: a: H-C-H in acrylonitrile b: H-C-N in acrylonitrile c: H-C-C in acrylonitrile d: H-C-C in methyl methacrylate e: H-C-O in methyl methacrylate f: C-C-O in methyl methacrylate Now we can approximate these angles based on their molecular geometry determined by the electron domain of central atoms. If all the electron domains around an atom are bonded, then the angles are approximately: a) sp3 hybridized: \(109.5^{\circ}\) b) sp2 hybridized: \(120^{\circ}\) c) sp hybridized: \(180^{\circ}\)
03

3. Finding Hybridization

Analyze each carbon atom in both compounds and determine hybridization: Acrylonitrile's carbons: C1: sp2 hybridized C2: sp hybridized C3: sp2 hybridized (double bond to the nitrogen) Methyl methacrylate's carbons: C1: sp3 hybridized C2: sp2 hybridized (double bond to the oxygen) C3: sp2 hybridized C4: sp2 hybridized C5: sp2 hybridized (double bond to the oxygen) Now, we can approximate bond angles a through f based on the hybridization state mentioned before.
04

4. Find the Number of Atoms Lying in the Same Plane

In acrylonitrile, C2 carbon atom is sp hybridized, which means all the atoms directly bonded to this atom, as well as itself, must lie in the same plane. Therefore, six atoms are in the same plane: three carbon atoms, one nitrogen atom, and two hydrogen atoms on C1 and C3.
05

5. Counting σ bonds and π bonds

We need to analyze carbon-carbon, carbon-nitrogen, carbon-oxygen, and carbon-hydrogen bonds in both acrylonitrile and methyl methacrylate to count total σ and π bonds. Acrylonitrile (C3H3N): σ bonds: 2 C-C bonds + 1 C-N bond + 3 C-H bonds = 6 π bonds: 2 C=C bonds + 1 C≡N bond = 3 Methyl methacrylate (C5H8O2): σ bonds: 4 C-C bonds + 2 C-O bonds + 8 C-H bonds = 14 π bonds: 2 C=C bonds = 2 The molecules have the following number of σ and π bonds: - Acrylonitrile: 6 σ bonds and 3 π bonds - Methyl methacrylate: 14 σ bonds and 2 π bonds

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Most popular questions from this chapter

As the head engineer of your starship in charge of the warp drive, you notice that the supply of dilithium is critically low. While searching for a replacement fuel, you discover some diboron, B. a. What is the bond order in \(\mathrm{Li}_{2}\) and \(\mathrm{B}_{2} ?\) b. How many electrons must be removed from \(\mathrm{B}_{2}\) to make it isoelectronic with \(\mathrm{Li}_{2}\) so that it might be used in the warp drive? c. The reaction to make \(\mathrm{B}_{2}\) isoelectronic with \(\mathrm{Li}_{2}\) is generalized (where \(n=\) number of electrons determined in part \(\mathrm{b}\) ) as follows: $$ \mathrm{B}_{2} \longrightarrow \mathrm{B}_{2}^{n+}+n \mathrm{e}^{-} \quad \Delta H=6455 \mathrm{kJ} / \mathrm{mol} $$ How much energy is needed to ionize 1.5 \(\mathrm{kg} \mathrm{B}_{2}\) to the desired isoelectronic species?

Place the species \(\mathrm{B}_{2}^{+}, \mathrm{B}_{2},\) and \(\mathrm{B}_{2}^{-}\) in order of increasing bond length and increasing bond energy.

Why does the molecular orbital model do a better job in explaining the bonding in \(\mathrm{NO}^{-}\) and \(\mathrm{NO}\) the hybrid orbital model?

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.)

The molecules \(\mathrm{N}_{2}\) and \(\mathrm{CO}\) are isoelectronic but their properties are quite different. Although as a first approximation we often use the same MO diagram for both, suggest how the \(\mathrm{MOs}\) in \(\mathrm{N}_{2}\) and \(\mathrm{CO}\) might be different.
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