Org Ch 6

Halogens

heterolysis

As bond length increases, bond strength decreases

a halide bonded to a sp2, double bonded carbon

a halide bonded to an sp2 hybridized carbon aromatic ring

a halide bonded to sp hybridized, triple bonded carbon

Vinyl, Aryl, and Acetylenic halides

Nucleophilic Substitution

• The Nu- donates an e- pair to the substrate
• The bond between C and LG breaks, giving borh e- from the bond to LG
• The Nu- uses its e- pair to form a new covalent bond with the substrate C
• The LG gains the pair of e- originally bonded in the substrate

• S_N1
• S_N2

• A reagent that seeks a positive center
• Has an unshared pair of e-

The substituent must be able to leave as a relatively stable, weakly basic molecule or ion

The rate of the substitutio reaction is linearly dependent on the concentration of OH- and CH3Br

• The consumption of the reactants
• The appearance of the products

The rate of reaction is directly proportional to the concentration of either reactant

• Negative OH- brings an e- pair to partial positive C; partially negative Br begins to move away with an e- pair
• O-C bond partially formed; C-Br bond partially broken. Configuration of C begins to invert
• O-C bond formed; Br- departed. Configuration of C inverted

A reaction that proceeds with a negative free-energy change (releases energy to its surroundings)

A reaction that proceeds with a positive free-energy change (absorbs energy from its surroundings)

Most reactions should overcome an energy barrier known as activation energy (∆G) and form the transition state (T.S.)

• ∆G° is the standard free energy change in a reaction
• this is the overall energy change of a reaction
• it is directly related to the equilibrium constant of a reaction
• if ∆G° is negative, products are facored at equilibrium (Keq>1)
• If ∆G° is positive, reactants are favored at equilibirum (Keq<1)
• If ∆G° is 0, products and reactants are equally favored (Keq=0)

Products are favored at equilibrium (Keq>1)

Reactants are favored at equilibrium; Keq<1

Products and reactants are equally favored; Keq=0

Indicated the progress of the reaction, in terms of the conversion fo reactants to products

The top of the energy curve

The difference in energy between the reactants and the transition state

The difference in energy between the reactants and the products

A 10° temperature will cause the reaction rate to double for many reactions taking place near room temperature

the relationship is exponential

Absolute rate constant, which equals the rate at which all transition states proceed to products (at 25°C k₀=6.2*10¹²s^-1)

Inversion of configuration

• The rate of S_N1 reactions depends only on concentration of the alkyl halide and is independent on concentration of the Nu-
• Rate=k[RX]

In a multistep reaction, the rate of the overall reaction is the same as the rate of the slowest step, known as the rate-determining step (r.d.s.)

An S_N1 reaction where then ucleophile is a polar protic solvent

• Trigonal planar
• The central carbon atom in a carbocation is electron deficient; it is only six e- in its valence shell
• The p orbital of a carbocation contains no electrons, but it can accept an electron pair when the carbocation undergoes further reaction

3°>2°>1°>methl

The more stable the carbocation formed, the faster the S_N1 reaction

There is a 50:50 chance that the attack will come from the right and the left

• Structure of substrate
• Concentration and reactivity of the nucleophile (for S_N2 reactions only)
• The effect of the solvent
• The nature of the leaving group

Methyl>1°>2°>3°> vinyl or aryl

3°, vinyl and aryl

sp² and sp hybridized carbons

3°>2°>>1°>methyl

Yes, at reasonable rates

rate=k[RX][RX]

Measured in terms of the relative rate of its S_N2 reaction with a given substrate

• A negatively charged Nu- is always a more reactive Nu- than its conjugate acid
• In a group of Nu-s in which the nucleophilic atom is the same, nucleophilicities parallel basicities
• When the nucleophilic atoms are different, then nucleophilicities may not parallel basicities (type of solvent should be considered before comparing)

• Polar solvents
• Non-polar solvents

• Polar protic solvents
• Polar aprotic solvents

Polar aprotic solvents, which have strong dipoles but do not have OH or NH groups

• Aprotic solvents tend to solvate metal cations rather than nucleophilic anions, and this results in 'naked' anions of the Nu- and makes the e- pair of the Nu more available
• Tremendous acceleration in S_N2 reactions with polar aprotic solvents

In polar protic solvents, the Nu- anion is solvated by the surrounding protic solvent which makes the e- pair of the Nu- less available and thus less reactive in S_N2 reactions

I- is stronger than F- because its e- pair os more available to attack the substrate in the S_N2 reaction

They do not solvate anions, but solvate cations

Ace as the Nu-

• Polar aprotic solvents do not solvate anions, but solvate cations
• The 'naked' anions act as the Nu-
• Since F- is smaller in size and the charge per surface area is larger than I-, the nucleophilicty of F- in this environment is greater than I-

• Solvent effects in S_N1 reactions are due largely to stabilization or destabilization of the transition state
• Polar protic solvents stabilize the development of the polar transition state and thus accelerate this rate-determining step

The better a species can stabilize a negative charge, the better the LG in S_N2 and S_N1 reactions

• R-F
• R-OH
• R-NH₂
• R-OR'

Vinylic and Phenyl halides

Conjugate base of alcohols is often used as the base in dehydrohalogenations

Rate determining step involves both the alkyl halide and the alkoxide anion

• EtO- removes a beta proton; C-H breaks; new pi bond forms and Br begins to depart
• Partial bonds in the transition state: C-H and C-Br bonds break, new pi C-C bond forms
• C=C is fully formed and the other products are EtOH and Br-

Unimolecular elimination

• Aided by the polar solvent, a chlorine departs with the e- pair that bonded it to the carbon
• Produces relatively stable 3° carbocation and a Cl-. The ions are solvated (and stabilized) by surrounding H₂O molecules

• H₂O molecule removes one of the beta hydrogens which are acidic due to the adjacent positive charge. An e- pair moves in to form a double bond between the beta and alpha carbon atoms
• Produces alkene and hydronium ion

With a strong base, favors S_N2

With a strong base, favors E2

With a strong base, favors E2 highly

small bases are unhindered, wheras bulky bases are hindered

• A weak base will follow a S_N2 reaction to 100% completion
• A strong base will follow a S_N2 reaction to 20% and an E2 reaction to 80%