Valence electrons and Lewis structures for C2H6O (ethanol)

Valence electrons and Lewis structures

• The transcript discusses determining the correct Lewis structure for a molecule with two carbons, six hydrogens, and an oxygen—i.e., C2H6O. The specific example being reasoned about is ethanol (CH3-CH2-OH), where hydrogens attach to carbons and an OH group is present.

• Key idea: count valence electrons for the entire molecule, then arrange bonds and lone pairs to satisfy octets where possible and minimize formal charges.

• Important takeaway from the dialogue: the order in which you place atoms or hydrogens around a carbon does not affect the final Lewis structure.

Case study: Ethanol, C2H6O (likely represented as CH3-CH2-OH)

• Empirical formula discussed: two carbons, six hydrogens, and one oxygen.

• Skeletal arrangement inferred: C–C–O chain with hydrogens completing the valence of each atom and an OH group.

Step-by-step construction of the Lewis structure for CH3-CH2-OH

• Step 1: Determine total valence electrons

  • Carbons: 2 imes 4 = 8 valence electrons

  • Hydrogens: 6 imes 1 = 6 valence electrons

  • Oxygen: 1 imes 6 = 6 valence electrons

  • Total valence electrons: V_{ ext{tot}} = 8 + 6 + 6 = 20

• Step 2: Draw the skeletal structure

  • Arrange atoms in a chain: C–C–O with the appropriate number of hydrogens attached to each carbon and the OH group:

  • First carbon (C1) is CH3: three C–H bonds

  • Second carbon (C2) is CH2: two C–H bonds

  • Oxygen (O) forms an O–H bond and a C–O bond

  • Hydrogens attached: CH3 (3 H), CH2 (2 H), OH (1 H) → total H = 6

• Step 3: Distribute electrons to satisfy octets

  • Bonds present: C1–C2 (1 bond), C2–O (1 bond), O–H (1 bond) plus the C–H and O–H bonds as appropriate

  • Count of bonds for the electron distribution: 8 bonds total (3 C–H on C1, 2 C–H on C2, 1 C1–C2, 1 C2–O, 1 O–H) = 8 bonds

  • Electrons in bonds: 8 ext{ bonds} imes 2 ext{ e−/bond} = 16 electrons

  • Lone pairs on atoms to reach octets: Oxygen in alcohol has two lone pairs (4 electrons)

  • Nonbonding electrons: 4 on O; Bonding electrons: 16; Total: 4 + 16 = 20 (matches $V_{ ext{tot}}$)

• Step 4: Check octets and formal charges

  • Carbon atoms: each forms four bonds (no lone pairs) → octet satisfied; Formal charge on each C: FC = V - (N_ ext{nonbond} + frac{1}{2}N_ ext{bond}) = 4 - (0 + frac{1}{2} imes 8) = 0

  • Oxygen: two lone pairs and two bonds (to C and H) → octet satisfied; Formal charge: FC = 6 - (4 + frac{1}{2} imes 4) = 6 - (4 + 2) = 0

  • Hydrogens: no lone pairs, each forms one bond; Human inspection shows no formal charges on H

• Step 5: Recognize alternative isomers with same formula

  • The same molecular formula C2H6O can also be dimethyl ether (CH3–O–CH3); differs in connectivity (no OH group). The transcript’s context suggests the presence of an OH group, i.e., ethanol (CH3-CH2-OH), rather than dimethyl ether.

• Step 6: Interpretations and practical notes from the transcript

  • “Order doesn’t matter” refers to not worrying about the sequence in which you list hydrogens around a carbon when counting to satisfy valence; the final structure and counts are invariant.

  • The moment-by-moment dialogue shows the process of counting hydrogens attached to each carbon (3 on the first carbon, 2 on the second) and adding the OH hydrogen to reach the total of 6 hydrogens.

Key formulas and numbers to remember

• Total valence electrons: V_{ ext{tot}} = ext{(number of C)} imes 4 + ext{(number of H)} imes 1 + ext{(number of O)} imes 6

  • For C2H6O: V_{ ext{tot}} = 2 imes 4 + 6 imes 1 + 1 imes 6 = 8 + 6 + 6 = 20

• Bonding electrons in the Lewis structure: N_ ext{bonding} imes 2 = 8 ext{ bonds} imes 2 = 16

• Lone pairs on O in ethanol: two lone pairs = 4 electrons

• Formal charge check (for each atom):

  • Carbon: FC = 4 - (0 + frac{1}{2} imes 8) = 0

  • Oxygen: FC = 6 - (4 + frac{1}{2} imes 4) = 0

• Ethanol connectivity (illustrative): CH3–CH2–OH

Why this is significant

• Demonstrates practical application of valence electron counting to build correct Lewis structures.

• Shows how to verify octets and zero formal charges in a neutral organic molecule.

• Highlights the possibility of constitutional isomers with the same formula and how functional groups (like OH) determine the specific structure.

• Reinforces that the exact order of presenting atoms or hydrogens in the sketch does not affect the final electronic structure.

Common pitfalls (as reflected in the transcript)

• Miscounting valence electrons due to confusion about which atoms contribute what electrons (e.g., not including O’s valence electrons).

• Thinking that the order of counting hydrogens around a carbon changes the structure; in reality, only the total counts and connectivity matter.

• Confusion between isomers with the same empirical formula (ethanol vs dimethyl ether) when an OH group is or isn’t present.

Connections to broader chemistry concepts

• Valence shell electron pair repulsion (VSEPR) theory implicitly supports octet-based Lewis structures and geometry.

• Formal charge concept helps assess the most stable resonance form and the distribution of electrons in covalent molecules.

• Real-world relevance: ethanol (C2H6O) is a common example in introductory chemistry for practicing Lewis structures and understanding functional groups (alcohols).

Practice prompts for self-check

• Draw the Lewis structure for ethanol and verify the octets and formal charges.

• Determine the total valence electrons for C2H6O and show how they are partitioned into bonding and lone pairs.

• Explain why the two possible isomers (ethanol and dimethyl ether) have the same molecular formula but different connectivity and properties.