Organic Chemistry Notes

Shorthand Notation in Organic Chemistry

• Shorthand notation is crucial for representing organic molecules.
• It simplifies the drawing of Lewis structures.

Hydrocarbons

• Basic hydrocarbons consist of hydrogen and carbon.
• They are divided into:
  • Alkanes: single bonds only.
  • Alkenes: at least one double bond.
  • Alkynes: at least one triple bond.
• Hydrocarbons are nonpolar, affecting the number of hydrogens they can have.

Saturation Formula

• The saturation formula dictates the maximum number of hydrogens in an alkane:
  • $C<em>nH</em>{2n+2}$
• Alkenes and ring structures have fewer hydrogens than the saturated number.

Hydrogen Loss

• Double bond: loses two hydrogens.
• Triple bond: loses four hydrogens.
• Ring structure: loses two hydrogens.

Representing Molecules

• Points/corners represent carbon atoms.
• Lines represent bonds.
  • Single bond: one line.
  • Double bond: two lines.
  • Triple bond: three lines.

Carbon Bonding

• Carbon must have four lines (bonds) coming out of it.
• These can be single, double, or triple bonds.
• The number of hydrogens can be determined by the number of remaining bonds needed to reach four.
• Example: A carbon with two lines has two hydrogens attached.

Nitrogen and Oxygen Bonding

• Nitrogen likes to have three lines (bonds).
• Oxygen likes to have two lines (bonds).
• This helps determine the placement of lone pairs of electrons.
• Hydrogens bonded to oxygen are usually written out in shorthand notation, while lone pairs on oxygen are often omitted but can be deduced.

Isomers

• Isomers are elements with the same number of protons but different numbers of neutrons. *Structural Isomers
  • Structural isomers have the same formula but different arrangements.
    *Millions of structural isomers are possible for many formulas.

Review Questions

Carbons Present

• Each point or bend represents a carbon atom.

Hydrogen Atoms

• Add the correct number of hydrogen atoms to complete each carbon's four bonds.

Bond Angles

• The central atom determines the bond angle.
• Assess the geometry of each carbon atom.
• Tetrahedral: four electron domains, 109.5 degrees.
• Linear: two electron domains, 180 degrees.

Functional Groups

• Memorize functional groups in both Lewis structure and shorthand notation.
• Carboxylic acid takes precedence over other groups (alcohol, aldehyde). *Ketone
  • Requires three carbons in a row with the middle one double bonded to oxygen.
    *Amine
  • Requires nitrogen single bonded to hydrogen.
    *Alcohol
  • Requires a carbon single bond to OH, and it must not be part of any other functional group like the carboxylic acid.
    *Aldehyde
  • Requires carbon double bonded to oxygen while single bonded to hydrogen.

Ranking Bond Angles

• Consider lone pairs and electronegativity when ranking bond angles.
• Lone pairs decrease the bond angle.
• More electronegative atoms make the angle smaller.
• Central atom's electron domain geometry determines angle.

Example 1 (Oxygen Atom)

• Two bonds, two lone pairs = four electron domains = tetrahedral = 109.5 degrees (ideal).

Example 2 (Central Carbon)

• Four bonds, no lone pairs = three electron domains = trigonal planar = 120 degrees (ideal).

Example 3 (Central Carbon)

• Requires a hidden hydrogen to complete four bonds = four electron domains = 109.5 degrees.

Comparing Angles

• Lone pairs push atoms closer, making the angle smaller than 109.5 degrees.
• Similar electronegativity around the central carbon results in angles close to the ideal.
• Double bonds act like super electronegative groups, increasing the angle.

Structural Isomers Example ($C<em>5H</em>{12}$)

• Must have single bonds (saturated).
• No rings allowed.
• Limited number of isomers.
• Isomers: Linear, branched (one methyl group on second carbon), and more branched (two methyl groups on central carbon).
• There are three structural isomers for $C<em>5H</em>{12}$.

Valence Bond Theory

• Covalent bonds form through the mixing (hybridization) of atomic orbitals.
• Hybrid orbitals: sp, sp2, sp3, sp3d, sp3d2.
• Predict hybridization using electron domain geometry.
• Two domains: sp.
• Three domains: sp2.
• Four domains: sp3.
• One electron domain: s orbital (for hydrogen).

Sigma and Pi Bonds

• Single bond = sigma bond.
• Double bond = one sigma bond + one pi bond.
• Triple bond = one sigma bond + two pi bonds.

Hybridization Example (Caffeine)

• Determine electron domains for each carbon atom.
• Three electron domains = sp2.
• Four electron domains = sp3.

Sigma and Pi Bond Counting

• Draw the molecule with all bonds (including hidden hydrogens).
• Count single, double, and triple bonds.
• Translate: Single bonds = sigma bonds; Double bond = one sigma + one pi; Triple bond = one sigma + two pi.

Intermolecular Forces (IMFs)

• Determine if molecules are ionic, polar, or nonpolar.
• Ion-ion: between two oppositely charged ions (ionic compound).
• Ion-dipole: between an ion and a polar molecule.
• Dipole-dipole: between two polar molecules.
• Hydrogen bonding: specific type of dipole-dipole with X-H and X (lone pair), where X is N, O, or F.
• Ion-induced dipole: weaker IMF.
• Dispersion forces (London forces, Van der Waals forces): always present, but only IMF in nonpolar molecules.

IMF Influence

• IMFs influence melting points, boiling points, vapor pressure.
• Stronger IMF = higher melting/boiling point.
• Memorize IMF order (strongest to weakest).

Qualitative Approach to Solubility

• Solute with similar IMF strength to solvent will be more soluble.

Intermolecular Forces Example (Acetonitrile in Water)

• Determine the intermolecular force in pure CH3CN.

• CH3CN is polar (nitrogen impurity in hydrocarbon).

• Lewis structure:

  $H_3C - C \equiv N$ (with lone pair on N).

• Dipole-dipole forces are present.

• No hydrogen bonding (H bonded to C, not N, O, or F).

• London dispersion forces are always present.

Side Note (Acetonitrile Dissolved in Water)

• Both dipole-dipole (both molecules are polar).
• Hydrogen bonding (O-H in water, N with lone pair in acetonitrile).
• London dispersion forces.

Boiling Point Comparison

• Compare intermolecular forces to determine relative boiling points.
• Lowest boiling point = weakest intermolecular force.
• Hydrocarbons (London forces) have low boiling points.

Dominant Intermolecular Force

• If tied (e.g., all London forces), compare molecular weight.
• Increased molecular weight = increased intermolecular force = higher boiling/melting point.

Surface Area

• If molecular weights are also tied, compare surface area.
• Increased surface area (more spread out/linear molecules) = increased intermolecular force.