Valence, formal charges, isomerism, and units of unsaturation (DBE)

Valence electrons and formal charges

  • There are two approaches to figure out the charge on atoms in molecules:
    • Approach 1 (memorization): Certain patterns. For example, nitrogen with four bonds tends to be positive; nitrogen with two bonds tends to be negative. The more you work with it, the more intuitive this becomes.
    • Approach 2 (valence-electron ownership): Use the valence-electron count and how many electrons the atom "owns" in the structure.
  • Charge calculation rule (as described): charge = (valence electrons) − (electrons owned by the atom).
  • How to count "owned" electrons:
    • Each bond contributes one electron owned by the atom (shared pair).
    • A nonbonding lone pair contributes both electrons to the atom (owned by that atom).
  • Examples from the transcript (illustrative, from the second-row elements discussion):
    • Carbon (valence = 4): If carbon has four bonds, it would own 4 electrons in bonds; charge = $4 - 4 = 0$ (neutral).
    • Carbon with five bonds (hypothetical in the discussion): owning 5 electrons would give charge = $4 - 5 = -1$ (negative).
    • Oxygen (valence = 6): If neutral with six electrons owned in bonds/locations, charge = $6 - 6 = 0$ (neutral).
    • Oxygen with seven electrons owned: charge = $6 - 7 = -1$.
  • A notable example discussed: a five-carbon ring with a carbonyl where one carbon (the carbonyl carbon) has no hydrogens, and the other four ring carbons each have hydrogens. There is mention of a positively charged carbon in that ring context, contrasted with the rest having access to eight electrons.
  • Isomerism mentioned: simple rearrangements (e.g., moving oxygen around a ring) can produce isomers; this is not an exhaustive list of possible structures.
  • Takeaway: the two approaches (pattern recognition vs. valence-electron accounting) are complementary; the ownership method underpins understanding of formal charge more generally.

Units of unsaturation and drawing structures

  • When a molecule has rings or multiple bonds, it has units of unsaturation (also called double bond equivalents, DBE).
  • In the example discussion, six structures each have two units of unsaturation (two DBEs):
    • They can be a six-membered ring plus a carbonyl or other substitution, or
    • Two double bonds, or
    • A combination yielding two DBEs in various placements.
  • Consequence for molecular formula: a structure with two DBEs is no longer the fully saturated formula C nH n+saturated. For example, a C nH
    h saturated alkane would normally have H determined by the saturated formula; introducing DBEs reduces the hydrogen count.
  • Specific rule highlighted: for every additional carbon, you would add two hydrogens if you were forming a fully saturated open-chain alkane; this intuition is used to compare to unsaturated cases.
  • Saturated hydrocarbon intuition: the saturated formula is not simply 2 × (number of carbons) for all positions; it is governed by the open-chain alkane formula including ends. The standard saturated hydrocarbon formula is:
    • Hextsat=2C+2.H_{ ext{sat}} = 2C + 2.
    • For methane ($C1$), $H{ ext{sat}} = 2(1) + 2 = 4$ → CH$_4$.
    • For ethane ($C2$), $H{ ext{sat}} = 2(2) + 2 = 6$ → C$2$H$6$.
  • Important counting idea: a saturated hydrocarbon’s hydrogen count is determined by the number of carbons; adding rings or multiple bonds reduces hydrogens accordingly.

How many hydrogens per formula given unsaturation

  • For a given carbon count, the fully saturated hydrocarbon would have Hextsat=2C+2.H_{ ext{sat}} = 2C + 2. If the actual hydrogen count is H, then the number of DBEs is:
    • DBE = rac{H_{ ext{sat}} - H}{2} = rac{(2C + 2) - H}{2}.
  • When you have heteroatoms, you adjust the hydrogen count to reflect typical valence rules:
    • Halogens (X, e.g., Cl, Br) occupy one bond like hydrogen; in DBE calculations you treat X as if it were hydrogen for counting purposes. The standard formula becomes:
    • DBE = rac{2C + 2 + N - H - X}{2}.
    • Example: C$5$H${11}$Cl (X = 1, N = 0): DBE = rac{2(5) + 2 + 0 - 11 - 1}{2} = rac{12 - 12}{2} = 0.
  • Oxygen (O) does not enter the DBE formula (it does not affect the hydrogen deficit with respect to unsaturation): O can be present without changing the DBE count.
  • Practical implications:
    • If a formula is C ext{n}H ext{m}O ext{p}N ext{q}X ext{r}, you can compute DBE with the modified formula above and then deduce possible structural features (rings, double bonds).
    • For C nH n with no heteroatoms (H = saturated count): DBE = 0 (fully saturated, no rings or multiple bonds).

Examples tying everything together

  • Example 1: five carbons, saturated open-chain alkane:
    • C$5$H${12}$ → DBE = 0 (saturated).
  • Example 2: two degrees of unsaturation:
    • C$5$H${12}$ would become C$5$H${8}$ if there are two DBEs (since $H = H_{ ext{sat}} - 2 imes DBE = 12 - 4 = 8$).
    • This could correspond to a structure with a ring and a double bond, or two double bonds, within C$_5$.
  • Example 3: ring/double-bond arrangement:
    • If you have a cyclopentene or cyclopentenone-type structure, you can place a six-membered ring or six-membered ring with a carbonyl; one DBE may come from the ring, another from the carbon-carbon or carbon-oxygen multiple bond, etc.
  • Example 4: carbonyl-containing ring discussed in the transcript:
    • In the described C$_5$ ring with a carbonyl, the carbonyl carbon has no hydrogens, while the other four ring carbons bear hydrogens.
  • Example 5: halogen substitution example from the transcript:
    • Replacing a hydrogen with chlorine does not by itself change the saturation; to compute DBE, treat X as H in H-count when using the DBE formula: e.g., C$2$H$5$Cl → DBE = 0 (saturated).
  • Example 6: oxygen’s role:
    • The presence of oxygen in the molecule does not alter the DBE count directly; it changes the formula but not the DBE formula’s structure as presented.

Practical tips and mental checks

  • When given a molecular formula and asked to draw possible structures:
    • First determine the number of DBEs using the formula
    • DBE = rac{2C + 2 + N - H - X}{2}.
    • Use DBE to decide how many rings or pi bonds are present.
    • If DBE = 0, the structure is fully saturated (no rings or multiple bonds).
    • If DBE > 0, distribute the unsaturation into rings and/or double bonds (e.g., zero, one, two, etc., depending on the value).
  • If the formula has halogens, count them as hydrogens for the purpose of DBE calculation (i.e., include X in the subtraction term).
  • Oxygen does not impact the DBE count in the standard expression above; focus on C, H, N, X.
  • The question of how many hydrogens “should” be present for a given skeleton is governed by the saturated hydrocarbon rule; remember H<em>extsat=2C+2.H<em>{ ext{sat}} = 2C + 2. For any real molecule, actual H = H{ ext{sat}} − 2 × DBE.
  • The transcript emphasizes the conceptual idea that you can think about unsaturation as losses in hydrogen count relative to a fully saturated hydrocarbon, and you can represent these losses through rings and multiple bonds.

Quick recap: key formulas to memorize

  • Saturated hydrocarbon hydrogen count for C carbons:
    • Hextsat=2C+2.H_{ ext{sat}} = 2C + 2.
  • General DBE formula (accounts for N and X as well):
    • DBE = rac{2C + 2 + N - H - X}{2}.
  • Hydrogen count in a molecule with DBE:
    • H=Hextsat−2imesDBE=(2C+2)−2imesDBE.H = H_{ ext{sat}} - 2 imes DBE = (2C + 2) - 2 imes DBE.
  • Halogens substitution example (as a check): for C nH nX ext{…}, replace X with H for the purpose of DBE counting; e.g., C$2$H$5$Cl has DBE 0 as shown above.