Lecture Notes on Alkenes, Alkynes, and Structure Determination

Alkenes and Alkynes

Electrophilic Addition (Review)

  • Nucleophilic alkene reacts by breaking a pi bond.
  • Forms a carbon-electrophile bond.
  • A nucleophile then reacts with the electrophilic carbocation intermediate.
  • Forms a carbon-nucleophile bond.
  • Net result: breaking a pi bond and forming two new sigma bonds.
  • Electrophile examples: proton or brominium.
  • Nucleophile examples: bromide, water, chloride, etc.

Markovnikov's Rule

  • Applies to unsymmetrical alkenes.
  • The electrophile bonds to the carbon with more hydrogens, resulting in the major product.
  • The minor product results from the electrophile bonding to the carbon with fewer hydrogens.
  • Reason: More stable carbocation intermediate is formed when the hydrogen bonds to the carbon with more hydrogens.
  • Carbocation stability trend: methyl < primary < secondary < tertiary.

Hydrogenation of Alkenes

  • Reaction: Alkene + Hydrogen gas → Alkane (two new C-H bonds formed).
  • Catalyst: Palladium on carbon (Pd/C).
  • Heterogeneous catalysis: reaction occurs on the surface of the solid metal catalyst.
  • Industrial importance: Hydrogenation of liquid fats to form solid fats.
  • Mechanism (cartoon description):
    • Hydrogen gas adsorbs onto the palladium metal surface and becomes activated.
    • Alkene approaches the metal surface.
    • Sequential addition of hydrogen atoms to the carbons of the alkene.

Alkynes: Structure

  • Unsaturated hydrocarbons with a carbon-carbon triple bond.
  • General formula: C<em>nH</em>2n−2C<em>nH</em>{2n-2}
  • Simplest alkyne: ethyne (acetylene).
  • Linear shape: all four atoms lie in a line; bond angles are 180∘180^\circ
  • sp hybridized carbons.
  • Bond Strengths:
    • Adding a pi bond doesn't double the strength.
    • The second pi bond added to form an alkyne is even more reactive than the first.

Alkynes: Bonding Orbitals

  • Two sp hybridized carbons form a triple bond.
  • One sp orbital from each carbon overlaps to form a sigma bond.
  • The other sp orbital overlaps with the hydrogen 1s orbital to form a C-H bond.
  • Two unhybridized p orbitals on each carbon.
  • One set of p orbitals form one pi bond.
  • The second set of p orbitals (perpendicular to the first set) forms a second pi bond.
  • Results in a cylinder of pi electron density surrounding the carbon-carbon bond.

Alkynes: Reactivity

  • Six electrons forced between two carbon atoms cause repulsion.
  • Pi electrons are reactive and available for reaction.

Alkynes: Nomenclature

  • Ending: "-yne" instead of "-ene".
  • Stem must contain the triple bond.
  • Numbering prioritizes the triple bond to give it the lowest possible number.
  • Constitutional isomers: molecules with the same number of atoms but different connectivity.

Alkynes: Reactions

  • React similarly to alkenes due to the presence of pi electrons.
  • Electrophilic addition can occur twice, often proceeding to the alkane stage.

Alkynes: Reaction Overview

Electrophilic Addition
  • Halogenation: requires two equivalents of halogen to add four halogens across the alkyne.
  • Hydrohalogenation: addition of hydrogen and halogen.
Hydrogenation
  • Complete hydrogenation: Alkyne → Alkane (two equivalents of hydrogen gas).
  • Partial hydrogenation: Alkyne → Alkene (stopping at the alkene stage).

Alkynes: Hydrogenation Reactions

  • Palladium on carbon (Pd/C) catalyst.
  • Excess hydrogen leads to complete hydrogenation to the alkane.
  • Syn addition: Hydrogens add on the same side of the alkyne, leading to cis (Z) alkene formation.
  • Stereoselectivity: Reaction invokes stereochemistry.
  • Partial Hydrogenation: Use of a poisoned catalyst to stop at the alkene stage.
Lindlar's Catalyst:
  • Palladium on carbon treated with lead.
  • Lead blocks some catalytic sites, reducing reactivity.
  • Can only reduce alkynes to alkenes, not alkenes to alkanes.

Alkynes: Electrophilic Addition Mechanism

  • Alkyne acts as a nucleophile, pi electrons form a carbon-electrophile bond (e.g., with hydrogen).
  • Forms a carbocation intermediate with a double bond.
  • Markovnikov's rule applies: hydrogen adds to the carbon with the most hydrogens.
  • Carbon-nucleophile bond forms (e.g., with bromide ion).
  • Reaction often continues to the alkane product.

Alkynes: Interesting Examples

  • Acetylene: used to form polymers for semiconductors.
  • Conjugation: alternating single and double bonds leading to delocalized pi system (semiconductors).
  • Alkynes in medicinal chemistry: drugs like Effeverinase (HIV treatment) and steroids in oral contraceptives.
  • Benzyne: unstable molecule with a triple bond in place of a double bond in benzene; formed in situ.
  • Cyclooctyne: smallest isolable cycloalkyne, used in bio labeling, bio conjugation.

Electrophilic Addition Practice Question

  • Given alkene + H2O, H2SO4.
  • Part A: Draw major and minor products.
    • Sulfuric acid dissociates into H+ (electrophile) and HSO4-.
    • Pi bond breaks, forming C-H bond.
    • Markovnikov's rule: H bonds to carbon with most hydrogens.
  • Part B: Identify key intermediates.
  • Part C: Explain how intermediates influence product distribution.
    • Tertiary carbocation (major product pathway) is more stable than secondary carbocation (minor product pathway).

Structure Determination: Introduction

  • Tools to identify compounds, determine structure, assess purity.
  • Examples: Mass spectrometry, infrared spectrometry, ultraviolet and visible spectroscopy, NMR (most powerful).
  • Analytical Methods are needed to:
    • Identify unknown compounds.
    • Figure out purity.
    • Ensure food/drug safety.
    • Forensics.

Structure Determination: Classical Tests vs. Modern Tools

  • Classical tests (e.g., alkene test with bromine, Tollen's test for aldehydes) are specific to certain functional groups.
  • Modern tools are broadly applicable to a range of molecules.

Elemental Analysis

  • Determines empirical formula: simplest positive integer ratio of atoms.
  • Example:
    • Benzene: empirical formula CH, molecular formula C6H6.
    • Ethyne: empirical formula CH (same as benzene).
  • Quantitative technique: combusting a sample in excess oxygen and collecting products (CO2, H2O, NOx).
Calculating Empirical Formula:
  • Compound contains only C, H, O.
  • Elemental analysis results: C = 54.55%, H = 9.09%.
  • Determine oxygen percentage: 100% - (54.55% + 9.09%) = 36.36%.
  • Assume 100g sample: 54.55g C, 9.09g H, 36.36g O.
  • Convert to moles: divide by molar mass.
  • Find the smallest molar value and divide each mole value by this value. You will find the ratio between the elements and their indexes.
  • Important: When you get this values after dividing you must use use the lowest full integral integer values.