Study Notes

Organic Chemistry Exam - Cheatsheet

Martin Priess
0 imports

Free ยท 2 imports included

Study Notes Preview

4 sections locked
Section 1

Organic Chemistry Exam - Cheatsheet

STUDY GUIDE

๐ŸŽ“ Organic Chemistry Exam - Study Guide

๐Ÿ“‹ Course Structure

code
๐Ÿ“š Organic Chemistry โ”œโ”€โ”€ ๐Ÿ“– Chapter 1: Introduction to Radicals and Homolysis โ”‚ โ”œโ”€โ”€ ๐Ÿ”น Definition and Characteristics of Radicals โ”‚ โ”œโ”€โ”€ ๐Ÿ”น Homolytic Bond Dissociation Energies (BDE) โ”‚ โ””โ”€โ”€ ๐Ÿ”น Initiation of Homolysis โ”œโ”€โ”€ ๐Ÿ“– Chapter 2: Radical Reactions and Mechanisms โ”‚ โ”œโ”€โ”€ ๐Ÿ”น Radical Addition Reactions โ”‚ โ”œโ”€โ”€ ๐Ÿ”น Radical Chain Reactions โ”‚ โ””โ”€โ”€ ๐Ÿ”น Radical Substitution Reactions โ”œโ”€โ”€ ๐Ÿ“– Chapter 3: Factors Affecting Radical Stability and Selectivity โ”‚ โ”œโ”€โ”€ ๐Ÿ”น Radical Stability โ”‚ โ”œโ”€โ”€ ๐Ÿ”น Persistent Radicals โ”‚ โ””โ”€โ”€ ๐Ÿ”น Regioselectivity in Radical Reactions โ”œโ”€โ”€ ๐Ÿ“– Chapter 4: Radical Reactions in Synthesis โ”‚ โ”œโ”€โ”€ ๐Ÿ”น Barton-McCombie Deoxygenation โ”‚ โ”œโ”€โ”€ ๐Ÿ”น Radical Dehalogenation โ”‚ โ””โ”€โ”€ ๐Ÿ”น Radical Cyclization โ””โ”€โ”€ ๐Ÿ“– Chapter 5: Polymerization Reactions โ”œโ”€โ”€ ๐Ÿ”น Mechanism of Radical Polymerization โ””โ”€โ”€ ๐Ÿ”น Radical Initiators
Section 2

๐Ÿ“– Chapter 1: Introduction to Radicals and Homolysis

What this chapter covers: This chapter introduces the concept of radicals, their formation through homolytic bond cleavage, and the factors influencing this process. It covers the definition of radicals, their electronic structure, and the significance of bond dissociation energies (BDEs) in determining the ease of radical formation. Understanding these fundamentals is crucial for predicting and analyzing radical reactions.

๐Ÿ”‘ Essential Concepts & Formulas

Concept/FormulaDefinition/EquationWhen to UseQuick Check
RadicalSpecies with an unpaired electronIdentifying reactive intermediatesCheck for odd number of valence electrons
HomolysisAโˆ’Bโ†’Aโ‹…+Bโ‹…A-B \to A\cdot + B\cdotDescribing bond cleavage forming radicalsVerify symmetrical bond breaking
BDEEnergy required for homolytic bond cleavagePredicting ease of radical formationLower BDE indicates easier cleavage
Thermal InitiationHeating a molecule to induce homolysisReactions with weak bondsCheck for high temperatures
Photochemical InitiationIrradiating a molecule with light (hvhv) to induce homolysisReactions with light-sensitive bondsCheck for presence of light

๐Ÿ› ๏ธ Problem Types

Type A: Predicting Radical Formation

Setup: "Given a molecule and reaction conditions (heat or light), determine if radical formation is likely."

Method: "Analyze the bond strengths (BDEs) and the presence of initiating conditions (heat or light). Lower BDEs and presence of heat or light favor radical formation."

Example: "Predict whether homolysis of Cl-Cl is more likely under irradiation with light or at room temperature. Cl-Cl BDE is relatively high, so light (hv) is needed."

Type B: Comparing Bond Dissociation Energies

Setup: "Given a set of bonds, rank them in order of ease of homolytic cleavage."

Method: "Compare the BDEs of the bonds. The bond with the lowest BDE will cleave most easily."

Example: "Rank C-H, C-Cl, and C-Br bonds in order of ease of homolytic cleavage, given their respective BDEs. Lower BDE indicates easier cleavage."

๐Ÿงฎ Solved Example

Problem: Calculate the enthalpy change for the homolytic cleavage of methane (CH4) into a methyl radical and a hydrogen atom, given the BDE of the C-H bond in methane is 439 kJ/mol.

Given: BDE(C-H) = 439 kJ/mol Reaction: CH4 -> CH3. + H.

Steps:

  1. Identify the bond being broken: C-H bond in methane.
  2. Apply the definition of BDE: Enthalpy change = BDE.
  3. Substitute the given BDE value: Enthalpy change = 439 kJ/mol.
  4. State the result with units.
"
โœ…
Answer: The enthalpy change for the homolytic cleavage of methane is 439 kJ/mol.

โš ๏ธ Common Mistakes

โŒ Mistake 1: Confusing homolytic and heterolytic cleavage.

โœ… How to avoid: Remember homolytic cleavage forms radicals, while heterolytic cleavage forms ions.

โŒ Mistake 2: Ignoring the role of light or heat in initiating homolysis.

โœ… How to avoid: Recognize that weak bonds may still require energy input (light or heat) to undergo homolysis.

๐Ÿฆ Erik's Tip

Memorize common BDE values for frequently encountered bonds (e.g., C-H, halogen-halogen) to quickly assess the feasibility of radical formation.

๐Ÿ“– Chapter 2: Radical Reactions and Mechanisms

What this chapter covers: This chapter delves into the mechanisms of radical reactions, including addition, substitution, and chain reactions. It emphasizes the initiation, propagation, and termination steps, and the regioselectivity of radical additions, particularly the anti-Markovnikov addition of HBr to alkenes in the presence of peroxides.

๐Ÿ”‘ Essential Concepts & Formulas

Concept/FormulaDefinition/EquationWhen to UseQuick Check
Radical AdditionAddition of a radical to a ฯ€\pi bondPredicting products of alkene/alkyne reactionsCheck for radical initiator
Chain InitiationFormation of radicals from non-radical speciesStarting a radical reactionPresence of initiator (e.g., peroxide)
Chain PropagationRadical reacts to form a new radicalSustaining a radical reactionCheck for cyclic reaction steps
Chain TerminationTwo radicals combine to form a non-radical speciesStopping a radical reactionLoss of radicals
Anti-Markovnikov AdditionAddition to alkene with reversed regioselectivityHBr addition with peroxidesPresence of peroxides

๐Ÿ› ๏ธ Problem Types

Type A: Predicting Products of Radical Addition

Setup: "Given an alkene and HBr with peroxides, predict the product of the radical addition reaction."

Method: "Apply anti-Markovnikov's rule: the bromine atom adds to the less substituted carbon."

Example: "Predict the product of the reaction of propene with HBr in the presence of benzoyl peroxide."

Type B: Drawing Radical Chain Mechanisms

Setup: "Given a radical reaction, draw the complete chain mechanism, including initiation, propagation, and termination steps."

Method: "Identify the initiator, show the formation of radicals, then show the cyclic propagation steps, and finally, the termination steps."

Example: "Draw the mechanism for the chlorination of methane."

๐Ÿงฎ Solved Example

Problem: Draw the mechanism for the addition of HBr to ethene in the presence of peroxides.

Given: Reactants: Ethene (CH2=CH2), HBr, Peroxides (ROOR)

Steps:

  1. Initiation: ROOR -> 2RO. and RO. + HBr -> Br. + ROH
  2. Propagation: Br. + CH2=CH2 -> BrCH2-CH2. and BrCH2-CH2. + HBr -> BrCH2-CH3 + Br.
  3. Termination: 2Br. -> Br2
"
โœ…
Answer: The product is bromoethane (CH3CH2Br).

โš ๏ธ Common Mistakes

โŒ Mistake 1: Forgetting the anti-Markovnikov rule in radical addition.

โœ… How to avoid: Remember that in the presence of peroxides, HBr adds anti-Markovnikov to alkenes.

โŒ Mistake 2: Incorrectly drawing the propagation steps in a chain reaction.

โœ… How to avoid: Ensure that each propagation step generates a new radical to continue the chain.

๐Ÿฆ Erik's Tip

Always draw out the full mechanism for radical reactions, paying close attention to electron flow and radical intermediates.

๐Ÿ“– Chapter 3: Factors Affecting Radical Stability and Selectivity

What this chapter covers: This chapter focuses on the factors influencing radical stability, including resonance, hyperconjugation, and inductive effects. It also covers persistent radicals and the regioselectivity of radical reactions, emphasizing how radical stability dictates product distribution.

๐Ÿ”‘ Essential Concepts & Formulas

Concept/FormulaDefinition/EquationWhen to UseQuick Check
Resonance StabilizationDelocalization of unpaired electronPredicting stability of allylic/benzylic radicalsDraw resonance structures
HyperconjugationStabilization through sigma bond overlapPredicting stability of alkyl radicalsMore alkyl substituents = more stable
Persistent RadicalRadical with long lifetimeIdentifying stable radicalsSterically hindered or electronically stabilized
RegioselectivityPreference for reaction at one site over anotherPredicting major productMore stable radical intermediate

๐Ÿ› ๏ธ Problem Types

Type A: Ranking Radical Stability

Setup: "Given a series of radicals, rank them in order of stability."

Method: "Consider resonance, hyperconjugation, and inductive effects. More substituted radicals are generally more stable."

Example: "Rank the stability of methyl, primary, secondary, and tertiary radicals."

Type B: Predicting Regioselectivity

Setup: "Given a radical reaction, predict the major product based on the stability of the intermediate radical."

Method: "Determine which radical intermediate is more stable and predict the product derived from it."

Example: "Predict the major product of the bromination of toluene."

๐Ÿงฎ Solved Example

Problem: Rank the following radicals in order of stability: benzyl, tertiary, secondary, and primary.

Given: Radicals: benzyl, tertiary, secondary, primary

Steps:

  1. Benzyl radical is stabilized by resonance.
  2. Tertiary radical is stabilized by hyperconjugation (3 alkyl groups).
  3. Secondary radical is stabilized by hyperconjugation (2 alkyl groups).
  4. Primary radical is stabilized by hyperconjugation (1 alkyl group).
"
โœ…
Answer: Stability order: benzyl > tertiary > secondary > primary.

โš ๏ธ Common Mistakes

โŒ Mistake 1: Overlooking resonance stabilization in allylic and benzylic radicals.

โœ… How to avoid: Always consider resonance when assessing radical stability.

โŒ Mistake 2: Ignoring the effect of hyperconjugation on alkyl radical stability.

โœ… How to avoid: Remember that more alkyl substituents lead to greater hyperconjugation and increased stability.

๐Ÿฆ Erik's Tip

Draw resonance structures for allylic and benzylic radicals to visualize the delocalization of the unpaired electron and understand their enhanced stability.

4 more sections

Create a free account to import and read the full study notes โ€” all 6 sections.

No credit card ยท 2 free imports included

    Organic Chemistry Exam - Cheatsheet โ€” Cheatsheet | Evrika | Evrika Study