Organic Chemistry Reaction Mechanisms
Terms in this set (104)
Alkyl halide (definition)
Organic molecule containing a halogen (X) bonded to an sp³ hybridized carbon atom.
Most important factor in determining reactivity of alkly halides ...
Is the alkyl halide primary, secondary, or tertiary?
Halogen bonded to a carbon-carbon double bond. Remember: this is an sp² hybridized carbon!
Halogen bonded to a benzene ring. Remember: this is an sp² hybridized carbon!
Halogen bonded to a carbon adjacent to a C-C double bond. Still an sp³ hybridized carbon!
Halogen bonded to a carbon adjacent to a benzene ring. Still an sp³ hybridized carbon!
Alkyl halides: physical property
(m.p. and b.p.) trends
1. compared to alkanes
2. as size of R increases
3. as size of X increases
1. Compared to alkanes, alkyl halides have a higher m.p. and b.p.
2. As the size of R increases, b.p. and m.p. increase due to larger surface area
3. As size of X (halogen) increases, b.p. and m.p. increase due to increased polarizability of halogen
Alkyl halides: intermolecular forces
C-X bond is weakly polar, so alkyl halides exhibit dipole-dipole interactions (as well as van der Waals). But they are incapable of intermolecular H bonding (rest of molecule is only C-C and C-H bonds).
What type/s of reactions do alkyl halides undergo?
1. Alkyl halides undergo substitution rxns with nucleophiles
2. Alkyl halides undergo elimination rxns with Bronsted-Lowry bases.
Nucleophilic substitution: 3 general features
1. R: alkyl group R containing sp³ hybridized carbon bonded to X.
2. X: good "leaving group" (atom or group of atoms).
3. Nu⁻: nucleophile (contains lone pair or pi bond; is electron rich)
What makes a good leaving group?
The better leaving group is the WEAKER base. I.e., all good leaving groups are weak bases with strong conjugate acids that have low pKa values. Examples: Cl, Br, I, H₂O.
Basicity trends in periodic table
Basicity increases up the periodic table, and left on the periodic table.
Nucleophilicity trends in periodic table
Opposite of basicity: Increases to the right and down on the periodic table.
When does equilibrium favor the products of nucleophilic subsitution?
When the leaving group is a weaker base than the nucleophile.
Difference between nucleophiles and bases
1. Nucleophiles attack electron-deficient atoms (usually carbons)
2. Bases attack protons (H)
Parallels between nucleophilicty and basicity (3)
1. For two nucleophiles with the same nucleophilic atom, the stronger base is the stronger nucleophile.
2. A negatively charged nucleophile is always stronger than its conjugate acid.
3. Nucleophilicity increases right-to-left on the periodic table (as basicity increases).
When does nucleophilicity NOT parallel basicity?
When steric hindrance comes into play. I.e., presence of bulky groups. Steric hindrance decreases nucleophilicity but NOT basicity. Sterically hindered bases that are poor nucleophiles (such as potassium tert-butoxide) are called nonnucleophilic bases.
Polar protic solvents
1. solvating effects
2. nucleophilicity trend
Polar protic solvents are capable of intermolecular H bonds (contain O-H or N-H bonds); i.e., water or ROH.
1. Cations solvated by ion-dipole interactions; anions solvated by H bonding.
2. Nucleophilicity: increases DOWN a column of the periodic table (opposite to basicity), as size of atom increases.
Polar aprotic solvents
1. solvating effects
2. nucleophilicity trend
Polar aprotic solvents have dipole-dipole interactions, but no H bonding. Examples: acetone, THF, DMSO, DMF, HMPA, acetonitrile.
1. Solvate only cations; anions not well solvated because solvent cannot H-bond to them.
2. Nucleophilicity parallels basicity (increases UP a column).
SN2 reaction: kinetics
Second order (bimolecular):
rate = k[RX][Nu⁻]
One step (concerted). Bond-breaking and bond-making occur at the SAME time: C-X bond is broken as the Nu⁻ attacks the C and makes a new bond (replaces X).
Nu⁻ does backside attack of the carbon (opposite side of the leaving group); leads to inversion of configuration if C is a stereogenic center. I.e., R→S or S→R.
SN2: Identity of R group and rate of rxn
As number of R groups on the carbon with the leaving group increases, the rate of SN2 decreases. Methyl and primary alkyl halides are best for SN2; secondary are slower. Tertiary alkyl halides NEVER undergo SN2.
First order (unimolecular): rate = k[RX]
CONCENTRATION OF NU⁻ HAS NO EFFECT ON RATE!
1. Heterolysis of C-X hond forms an intermediate carbocation (C⁺).
2. Nu⁻ attack of C⁺ forms the new C-Nu bond (substitution).
Nu⁻ attack occurs on BOTH sides of the C⁺ with equal frequency, leading to racemization (equal mixture of R/S enantiomers).
SN1: identity of R group and rate
As number of R groups on the C with the leaving group increases, the rate of SN1 increases (opposite of SN2). Tertiary alkyl halides are fastest, secondary go more slowly. Methyl and primary NEVER undergo SN1 because an unstable C⁺ intermediate would be formed.
What factors (2) affect carbocation stability?
1. Inductive effects: Alkyl groups are electron-donating groups that stabilize the C⁺ positive charge.
2. Hyperconjugation: Spreading out charge over a larger volume delocalizes positive charge on C⁺.
SN1 vs. SN2: alkyl halide
Methyl and primary: SN2 only
Secondary: both SN1 and SN2; must use other factors to determine whether it's SN1 or SN2
Tertiary: SN1 only
SN1 vs. SN2: identity of nucleophile
Strong nucleophile (usually negative): favors SN2
Weak nucleophile (usually neutral): favors SN1
SN1 vs. SN2: leaving group identity
Better leaving group increases rate of both SN1 and SN2
SN1 vs. SN2: solvent
Polar aprotic: favors SN2
Polar protic: favors SN1
Elimination reaction: definition
Alkyl halide loses -H and -X in a Bronsted-Lowry base reaction; an alkene (double C bond) is formed. Called dehydrohalogenation.
Steps for drawing the product/s of dehydrohalogenation (3 steps)
1. Find the alpha C (sp³ hybdridized C bonded to the leaving group)
2. Identify all beta carbonds with H atoms
3. Remove HX and form a pi bond
Alkene with 1 R group
Alkene with 2 R groups
Alkene with 3 R groups
Alkene with 4 R groups
Alkene stability (trends)
Increases as number of R groups bonded to the double bond carbons increases (i.e., tetrasubsituted is most stable; monosubstituted is least stable) because of electron-donating inductive effects from R groups.
Alkenes: assigning cis and trans
Stereoisomers (diastereomers) are possible when the two groups on EACH end of the carbon-carbon double bond are different from each other.
Cis: both groups are either up/down (see left image)
Trans: 1 group up, 1 group down (see right image)
cis vs. trans: stability
trans is more stable, because groups bonded to double bond carbons are farther apart (less steric hindrance)
2nd order (bimolecular): rate = k[HX][B⁻]
One step (concerted):
1. Base removes proton from beta C
2. e⁻ pair in C-H bond forms the new pi bond
3. leaving group comes off with the e⁻ pair from the C-X bond
Sterically hindered base used in E2 rxn
Sterically hindered base used in E2 rxn
E2 rxn: identity of alkyl halide
Rate increases as number of R groups on C with leaving group increases (more stable alkene is formed). Primary, secondary, and tertiary ALL undergo E2; tertiary is fastest.
E2: identity of base
Favored by strong bases
E2: identity of solvent
Favored by polar aprotic solvents
Major product in elimination rxn has the more subsituted double bond (regioselectivity). But minor products are formed, too (in smaller amounts).
Major product is the more stable trans product (stereoselectivity). But both products (cis and trans) are formed!
1st order (unimolecular): rate = k[HX]
1. Heterolysis of C-X bond (leaving group leaves), forming intermediate C⁺
2. Base removes proton from C adjacent to C⁺; e⁻ pair from C-H bond makes new pi bond
E1: identity of alkyl halide
Rate of E1 increases as number of R groups on the C with the leaving group increases (only secondary and tertiary alkyl halides will undergo E1, though, because a primary alkyl halide would form an unstable primary C⁺).
E1: identity of base
Favored by weak bases.
E1: identity of solvent
Favored by polar protic (to solvate the ionic intermediate formed).
E1 and SN1: competition
E1 and SN1 have same 1st step (formation of C⁺), so a mixture of products usually forms, making E1 less useful than E2.
E2 occurs in anti periplanar geometry (H and X on opposite sides of the molecule). For cyclic compounds, this means the C-H and C-X bonds must BOTH be axial.
trans diaxial geometry
Refers to the need (in cyclic compounds) for both the C-H and C-X bonds to be axial for an E2 reaction to occur.
E1 vs E2?
Look at the base: strong base favors E2; weak base favors E1.
Produced by two consecutive elimination (E2) reactions.
Degrees of unsaturation: steps to calculate
- and what to do w/heteroatoms (O, X, N)
1. Calculate the max number of H's possible (for n carbons = 2n+2)
2. Subtract the actual number of H's; divide by two.
3. What to do with heteroatoms:
- ignore the presence of O in a compound
- if a compound has an X (halogen), add 1 H to the actual H's
- if a compound has an N, subtract 1 H from the actual H's
Naming a compound with TWO double C-C bonds
Change -ane ending of alkane to -adiene (these are called "dienes"). Three double bonds = trienes, etc.
Used when there are 3 or 4 different groups on an alkene (unable to assign cis/trans).
- Assign priorities to the substituents on each end using rules for R/S
- Assign E or Z based on location of higher priority groups (E = opposite sides; Z = same sides)
Alkenes: physical properties
- low m.p. and b.p.
- m.p. and b.p. increases as number of C increases (more surface area)
- soluble in organic solvents; insoluble in water
- cis stereoisomer has higher b.p. because sp³ hybridized alkyl carbon donates electron density to sp² hybridized alkenyl carbon - i.e., has a small net dipole and is slightly polar (in trans stereoisomer, dipoles cancel out)
Addition rxn: definition
- The pi bond of an alkene is broken; two new sigma bonds are formed.
- Alkenes are electron rich (have a double bond), so do NOT react w/nucleophiles or bases; alkenes react with electrophiles!
1. syn addition
2. anti addition
1. syn: When both X and Y are added to the same side of an alkene.
2. anti: When X and Y are added from opposite sides.
(electrophilic addition of HX)
Addition of HX (X = Cl, Br, I) to alkenes to form an alkyl halide. Exothermic!
1. pi bond of alkene attacks the H atom of H-X (X leaves); forms an intermediate C⁺. Rate-determining.
2. Nucleophilic attack of X⁻ on the C⁺ forms the new C-X bond.
In addition to an unsymmetrical alkene, the H bonds to the less substituted C (i.e., the C that has more H to begin with). This forms a more substituted C⁺.
1,2-methyl and 1,2-H shifts
The rearrangement of a secondary C⁺ → tertiary C⁺ via shifting of either a methyl or a H to the C⁺ from an adjacent C atom. Forms more substituted (stable) product.
- Addition of H⁺ occurs from both above and below, which can generate enantiomers if a stereogenic center is formed.
- Nu attack of X⁻ occurs from both sides of the C⁺, which can again produce 2 stereoisomers for each enantiomer (for a total of 4 stereoisomers; 2ⁿ where n = # stereoisomers).
In other words: both syn and anti addition occur.
Hydration - definition
(electrophilic addition of water)
Addition of water to an alkene to form an alcohol. Requires presence of H₂SO₄ to protonate the H₂O.
Hydration - mechanism
THREE steps: (protonation of H₂O occurs prior to actual mechanism)
1. pi bond attacks H⁺ of H₃O⁺, forming a new C-H bond and breaking H-O bond; an intermediate C⁺ is formed.
2. Nucleophilic attack of H₂O on the C⁺ forms a new C-O bond
3. Deprotonation by H₂O (yielding the ROH)
Hydration - stereochemistry
Addition of H and OH occurs in BOTH syn and anti directions (potentially forming stereoisomers)
Halogenation - definition
Addition of X₂ (X = Cl or Br) to an alkene, forming a vicinal dihalide (X atoms are on the same side).
Halogenation - mechanism
1. Formation of a bridged halonium ion
2. Nucelophilic attack of X⁻
(see left product - vicinal dihalide - in image)
Halogenation - stereochemistry
Forms a racemic mixture because Nu⁻ attack must occur from the backside (anti addition occurs)
Halohydrin formation - definition
Treatment of an alkene with X₂ and H₂O to add -X and -OH
Halohydrin - mechanism
1. formation of bridged halonium ion
2. Nucleophilic attack of H₂O
(see RIGHT product in image)
Halohydrin - stereochemistry / regioselectivity
Stereochemistry: anti addition occurs
Regioselectivity: electrophile X⁺ binds to less substituted C (-OH binds to MORE substituted C)
NBS in DMSO
Forms a bromohydrin (halohydrin); NBS is a source of Br₂
Hydroboration-Oxidation - definition
A rxn that converts an alkene to an alcohol via the addition of BH₃ (borane) when treated with H₂O₂ and HO⁻
Hydroboration-oxidation - mechanism
1. Hydroboration: syn addition of H and BH₂ to the alkene with treatment of borane (concerted rxn; no C⁺ formed; intermediate alkylborane is formed)
2. Oxidation: Substitution of BH₂ with ⁻OH; retention of stereochemistry configuration
Hydroboration transition state
"four-centered": four atoms are involved; pi bond and H-BH₂ bond are broken and two new sigma bonds are formed (syn addition)
With unsymmetrical alkanes, the Boron atom bonds to the LESS substituted carbon atom (and therefore, the OH group also bonds to the less substituted carbon).
1. Hydroboration step: occurs with syn addition (but with cycloalkenes, occurs both above and below the double bond, forming enantiomers!)
2. Oxidation step: occurs with retention of configuration
What is the electrophile in hydroboration?
BH₂ (because H is more electronegative than B)
Undergoes hydroboration in the same manner as BH₃ (general structure: R₂BH)
Which mechanism is anti-Markovnikov?
Hydroboration-oxidation (the OH eventually bonds to the LESS substituted carbon atom)
In which mechanism/s does SYN addition occur?
In which mechanism/s does ANTI addition occur?
- Halohydrin formation
In which mechanisms does BOTH syn and anti addition occur?
In which mechanism/s are carbocation intermediates formed?
In which mechanism/s is a bridged halonium ion formed?
- Halohydrin formation
In which mechanism/s is carbon rearrangement (1,2 methyl or 1,2 H shifts) possible?
Which mechanisms have 1 step?
- Hydroboration (oxidation is a separate mechanism)
- Oxidation (hydroboration is a separate mechanism)
Which mechanisms have 2 steps?
- Hydroboration-oxiation (when considering BOTH mechanisms)
Which mechanisms have 3 steps?
- Halohydrin formation
Synthesis: How do you prepare an alkene?
Synthesis: How do you prepare an alcohol?
Synthesis: How do you prepare an ether?
- Addition of an alcohol, ROH (RO is the good leaving group). Mechanism: same as hydration.
Synthesis: How do you form a vicinal dihalide?
Synthesis: How do you form an alkyl halide?
- Hydrohalogenation (addition of H-X)
Synthesis: How do you form a halohydrin (alkane chain with an -OH and an -X)?
- Halohydrin formation
OTHER SETS BY THIS CREATOR
KLEIN Chapter 15 - IR Spectroscopy & Mass Spectroscopy
KLEIN Chapter 9- Addition Reactions of Alkenes
KLEIN Chapter 8 - Alkenes: Structure & Prep via Elimination Reactions
THIS SET IS OFTEN IN FOLDERS WITH...
MCAT - Organic Chemistry - Lesson 1
MCAT Examkrackers Math and Verbal
Chapter 7 - Substitution Reactions