Class 16: Carbonyl Reactivity

Question 1

Predict the relative order of Lewis acidity or electrophilicity based on the carbonyl compound’s structure.

Answer 1

• More electrophilic = more Lewis acidic
• Depends on:
  
  • Electronegativity of atoms bonded to carbonyl carbon
  • Resonance stabilization of the resulting oxyanion
• General trend:
  Acid chlorides > Aldehydes > Ketones > Esters > Amides
• Effects of substituents:
  
  • Electron-withdrawing groups increase electrophilicity
    
    • Example: CF3CO > CH3CO
  • Resonance donation decreases electrophilicity
    
    • Example: CH3CO > CH3COCH3
• Steric factors play a minor role
  
  • More hindered = slightly less reactive
• Exceptions:
  
  • α,β-unsaturated > saturated due to conjugation
  • Aryl ketones less reactive than alkyl due to resonance

Bottom line:
Draw resonance structures
Analyze electron-withdrawing/donating effects
Find most stable/resonance-stabilized oxyanion

Question 2

Draw an energy diagram for a multi-step mechanism showing how free energy changes as a function of reaction progress.

Answer 2

Here are the concise bullet points for drawing an energy diagram for a multi-step mechanism showing free energy changes:

• Y-axis: Free Energy or Gibbs Energy (G)
• X-axis: Reaction Coordinate or Reaction Progress
• Plot the free energies of:
  
  • Reactants
  • Transition states (Highest energy points, peaks)
  • Intermediates
  • Products
• Each step has an activation barrier (ΔG‡)
  
  • Height of peak from reactants/intermediates
• Overall ΔG for reaction = Gproducts - Greactants
  
  • Shown by final G difference
• For stepwise mechanisms:
  
  • Each step raises G to transition state
  • Then lowers to next intermediate
  • Final product is lowest G state
• Identify rate-determining step
  
  • Highest activation barrier (ΔG‡)
• More favorable ΔG of reaction = more product formed at equilibrium

Question 3

Apply what you know about equilibrium positions to an energy diagram.

Answer 3

• Energy diagrams show relative free energy levels
• Equilibrium favors the state with lowest free energy
• For an exergonic reaction (ΔG < 0):
  
  • Products are lower in free energy than reactants
  • At equilibrium, products are favored over reactants
• For an endergonic reaction (ΔG > 0):
  
  • Reactants are lower in free energy than products
  • At equilibrium, reactants are favored over products
• For ΔG = 0:
  
  • Reactants and products are equal in free energy
  • No free energy driving force, 50:50 at equilibrium
• Relative heights of activation barriers impact rate
  
  • Higher barriers mean slower rates
• Can estimate equilibrium constants from free energy differences
  
  • ΔG = -RT ln(K)
• Highly exergonic reactions are essentially irreversible
  
  • All reactants go to products

So in summary, the lowest free energy state is the most favored at equilibrium based on the energy diagram.

Question 4

Use your newly gained vocabulary, structures, and curved arrows in the context of reactions: mechanisms, reaction pathway, and transition state.

Answer 4

• Mechanisms show the step-by-step pathway of bond-breaking and bond-forming
• Each elementary step is represented with curved arrow(s)
  
  • Nucleophilic attack, electrophilic addition, etc.
• Molecules are shown as Lewis structures
  
  • Charges, partial charges, radical electrons indicated
• Transition states are high-energy species
  
  • Represent maximum energy along the reaction coordinate
  • Drawn with partial bonds, distorted geometries
• Intermediates are formed after each step
  
  • Often reactive species like carbocations, radicals
• Energy diagrams map out enthalpy/free energy
  
  • Y-axis: ΔG or ΔH
  • X-axis: Reaction coordinate
  • Transition states are energy maxima
• Rate-determining step has highest energy barrier
• Curved arrows track electron movement
  
  • Show how bonds break/form in each step
• Resonance structures can contribute in some steps
  
  • Distributes charge, increases stability

In summary, curved arrows, molecular structures, transition state drawings, and energy diagrams are all unified in showing a detailed picture of a reaction mechanism and pathway.

Question 5

Use pKa’s as a method to predict the position of equilibrium in a nucleophilic acyl substitution reaction.

Answer 5

• Nucleophilic acyl substitutions are reversible reactions
• Equilibrium position depends on relative strengths of nucleophiles/bases
• Compare pKa values of conjugate acids
  
  • Lower pKa = stronger conjugate base = better nucleophile
• For reaction: Nu1- + R2C=OR’ ⇌ R2C(=O)Nu1 + R’O-
  
  • If pKa(Nu1H) < pKa(R’OH), equilibrium favors products
  • If pKa(Nu1H) > pKa(R’OH), equilibrium favors reactants
• Larger the pKa difference, more equilibrium is driven that way
• Can use pKa values to:
  
  • Predict if reaction is favorable
  • Estimate equilibrium constant
  • Guide synthesis by favoring formation of desired product
• This method applies to other nucleophilic substitutions too

So in summary, comparing nucleophile and leaving group pKa’s allows predicting which way the substitution equilibrium will lie.

Question 6

Define what is meant by a “leaving group”

Answer 6

• A leaving group is an atomic or molecular species that departs from a substrate in a chemical reaction
• It is the group that is displaced/substituted by the incoming nucleophile
• Good leaving groups are:
  
  • Weak bases (low pKa)
  • Weakly nucleophilic
  • Stabilized by resonance or induction
  • Easily accommodated as a stable anion
• Common leaving groups:
  
  • Halides (Cl-, Br-, I-)
  • Weak oxyanions (H2O, ROH, RCOO-)
  • Resonance-stabilized (tosylate, mesylate)
• The bond to the leaving group must be sufficiently weak to allow heterolytic cleavage
• Stereoelectronic factors also impact leaving group ability
• A good leaving group facilitates the forward reaction by its departure
• The ease of leaving group departure strongly influences the reaction rate

So in summary, a leaving group is the atomic/molecular fragment that is displaced by a nucleophile, with weak basicity and stabilization favoring good leaving ability.

Question 7

Use pKa’s as a method to predict the position of equilibrium in a nucleophilic acyl substitution reaction.

Answer 7

the weaker the base, the better it is as a leaving group
Therefore, you can use pkas of conj acids to determine what is the better leaving group, by which is the strongest acid
Ex: Cl is a good leaving group because HCl is a strong acid

Question 8

Define what is meant by a “leaving group”

Answer 8

The leaving group is the group that is no longer attached to the initial molecule and becomes its own separate molecule in the products

Question 9

When determining whether reactants or products are favored in a carbonyl reactivity reaction, there are several factors to consider.

Answer 9

Here’s an overall guide or “cheat sheet” to help you analyze the reactivity:

1. Stability of the carbonyl compound:
   
   • Aldehydes are generally more reactive than ketones due to the lack of steric hindrance.
   • Ketones with bulky substituents are less reactive due to increased steric hindrance.
2. Nucleophilicity and basicity of the attacking species:
   
   • Strong nucleophiles and bases favor the formation of products.
   • Weak nucleophiles and bases favor the reactants.
3. Stability of the leaving group:
   
   • Good leaving groups (e.g., halides, weak bases) favor the formation of products.
   • Poor leaving groups (e.g., alkoxides, strong bases) favor the reactants.
4. Stability of the formed carbanion or enolate:
   
   • Resonance-stabilized carbanions or enolates favor the formation of products.
   • Non-stabilized carbanions or enolates favor the reactants.
5. Steric effects:
   
   • Bulky substituents on the carbonyl compound can hinder the approach of the nucleophile, favoring the reactants.
6. Solvent effects:
   
   • Protic solvents can solvate anions, favoring the formation of products.
   • Aprotic solvents can stabilize carbonyl compounds, favoring the reactants.
7. Thermodynamic and kinetic factors:
   
   • Highly exergonic reactions favor the formation of products.
   • Reactions with high activation energy barriers favor the reactants.

In the example you provided, the reaction of an acyl chloride with hydroxide ion, the products are favored because:
- Chloride (Cl⁻) is a good leaving group.
- The formed carboxylate anion is resonance-stabilized, making it a stable product.

Remember, these are general guidelines, and specific reactions may have additional factors to consider. It’s always a good idea to analyze the reactants, products, and reaction conditions carefully.

Question 10

To determine if a leaving group is weak or strong, and to understand the general principles of leaving groups and their role in organic reactions

Answer 10

1. Leaving group strength:
   
   • Strength is determined by the stability of the conjugate base formed when the leaving group departs.
   • Stronger leaving groups form more stable conjugate bases.
   • The order of leaving group strength is: I⁻ > Br⁻ > Cl⁻ > F⁻ > H₂O > ROH > RO⁻ > R₃N
   • Halides (I⁻, Br⁻, Cl⁻) and tosylates (OTs⁻) are generally good leaving groups.
   • Alkoxides (RO⁻) and amines (R₃N) are poor leaving groups.
2. Leaving groups and basicity:
   
   • Leaving groups are not always bases, but they often act as bases in their conjugate form.
   • Halides (X⁻) are weak bases, but their conjugate acids (HX) are strong acids.
   • Alkoxides (RO⁻) and amines (R₃N) are strong bases.
   • Water (H₂O) is a neutral leaving group, but its conjugate base (OH⁻) is a strong base.
3. Factors affecting leaving group ability:
   
   • Stability of the conjugate base: More stable conjugate bases make better leaving groups.
   • Polarizability: More polarizable groups (e.g., I⁻) are better leaving groups.
   • Steric effects: Bulkier leaving groups (e.g., tert-butoxide) are more difficult to displace.
4. Drawing reactions with leaving groups:
   
   • Identify the electrophilic center (e.g., carbonyl carbon) and the nucleophile.
   • Determine the leaving group and its strength based on the principles above.
   • Draw the nucleophilic attack and the departure of the leaving group.
   • Consider the stability of the formed products or intermediates.
5. General considerations:
   
   • Solvents can stabilize or destabilize leaving groups and intermediates, affecting the reaction.
   • Steric effects from substituents can hinder the approach of the nucleophile or the departure of the leaving group.
   • Resonance stabilization of the formed products or intermediates can drive the reaction forward.
   • Kinetic and thermodynamic factors also play a role in determining the feasibility and favorability of the reaction.

By understanding these principles, you can better predict the reactivity of different leaving groups, draw reasonable reaction mechanisms, and analyze the factors that influence the overall process.