Problem 64 - Chemistry The Central Science

Question

Ethyl acetate, \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2},\) is a fragrant substance used both as a solvent and as an aroma enhancer. Its Lewis structure is (a) What is the hybridization at each of the carbon atoms of the molecule? (b) What is the total number of valence electrons in ethyl acetate? (c) How many of the valence electrons are used to make \(\sigma\) bonds in the molecule? (d) How many valence electrons are used to make \(\pi\) bonds? (e) How many valence electrons remain in nonbonding pairs in the molecule?

Step-by-Step Solution

Answer

(a) C1: sp^3, C2: sp^2, C3: sp^2 (b) 36 valence electrons (c) 20 electrons in σ bonds (d) 2 electrons in π bonds (e) 14 electrons in nonbonding pairs

Step 1: Drawing the Lewis Structure

First, let's draw the Lewis structure of ethyl acetate molecule. Here is the Lewis structure: O || H - C - O - C - H | || | H C H | H

Step 2: Identifying the Hybridization of Carbon Atoms

To determine the hybridization of each carbon atom, we will count the σ bonds and electron pairs in the structure. - The left carbon atom (C1) is bonded to three hydrogen atoms and one oxygen atom, forming four σ bonds. Therefore, its hybridization is sp^3. - The central carbon atom (C2) is double bonded to the oxygen atom and single bonded to the other two carbon atoms, forming three sigma bonds in total. Therefore, its hybridization is sp^2. - The right carbon atom (C3) is bonded to one oxygen atom and two hydrogen atoms, forming three σ bonds. Therefore, its hybridization is sp^2.

Step 3: Counting the Total Number of Valence Electrons

Now, let's calculate the total number of valence electrons in the ethyl acetate molecule. - Each of the 4 carbon atoms has 4 valence electrons, so 4 × 4 = 16 valence electrons from carbon atoms. - Each of the 8 hydrogen atoms has 1 valence electron, so 8 × 1 = 8 valence electrons from hydrogen atoms. - The 2 oxygen atoms each have 6 valence electrons, so 2 × 6 = 12 valence electrons from oxygen atoms. Thus, the total number of valence electrons in ethyl acetate is 16 + 8 + 12 = 36 valence electrons.

Step 4: Counting the Valence Electrons used in σ Bonds

In ethyl acetate, for each bond, there are two electrons involved. - C1 forms 4 σ bonds with 3 hydrogen atoms and 1 oxygen atom, thus using 8 electrons. - C2 forms 3 σ bonds with 1 oxygen atom and 2 other carbon atoms, using 6 electrons. - C3 forms 3 σ bonds with 1 oxygen atom and 2 hydrogen atoms, using 6 electrons. So, there are a total of 8 + 6 + 6 = 20 electrons making σ bonds in the molecule.

Step 5: Counting Valence Electrons used in π Bonds

Examining the Lewis structure, we see that there is a double bond between C2 and the oxygen atom (O1). A double bond consists of one σ bond and one π bond, which means there are 2 electrons involved in the π bond. So, there are 2 valence electrons used to make π bonds in the molecule.

Step 6: Counting Nonbonding Pairs of Valence Electrons

Finally, we need to count the number of valence electrons not involved in the bonding process. We can calculate this by subtracting the total number of electrons used in σ and π bonds from the total number of valence electrons. Total nonbonding pairs of valence electrons = Total valence electrons - Electrons used in σ bonds - Electrons used in π bonds = 36 - 20 - 2 = 14 electrons Thus, there are 14 valence electrons in nonbonding pairs in the ethyl acetate molecule. (a) The hybridization at each carbon atom in ethyl acetate is: - C1: sp^3 - C2: sp^2 - C3: sp^2 (b) The total number of valence electrons in ethyl acetate is 36. (c) The number of valence electrons used to make σ bonds in the molecule is 20. (d) The number of valence electrons used to make π bonds in the molecule is 2. (e) The number of valence electrons remaining in nonbonding pairs in the molecule is 14.