Organic Chemistry Exam I, Chapters 6 and (SN2 and E2)
SN2 - backside attack The nucleophile attacks from the back-side Why?
- Electron density repels the attacking nucleophile from the front-side - The nucleophile must approach the back-side to allow electrons to flow from the HOMO of the nucleophile to the LUMO of the electrophile. - Proper orbital overlap cannot occur with front-side attack because there is a node on the front-side of the LUMO
ΔG is the Gibbs Free Energy
A negative value of DG means the reaction is spontaneous, and a positive value is a nonspontaneous reaction.
DG > 0
Keq < 1
Lewis acid
a compound or ionic species that can accept an electron pair from a donor compound.
Lewis base
a compound or ionic species that can donate an electron pair to an acceptor compound.
Negative charges
attracted to positive charges
what do "nucleophiles" and "bases" have in common?
they donate a lone pair of electrons.
Electron-rich species attracted
to electron-deficient species
The Hammond Postulate: Endothermic rxn -
transition state resembles product(s)
The Hammond Postulate: Exothermic rxn -
transition states resembles reactant(s)
Bond dissociation energy (BDE)
ΔH for bond breaking corresponds to homolytic bond cleavage
Two main reasons why alkyl halides undergo substitution and elimination reactions:
1. The halogen is electron-withdrawing, creating a partial positive charge on the alpha carbon, making it susceptible to nucleophilic attack. 2. The halogen acts as a leaving group, and for a substrate to undergo a substitution/elimination reaction, it must possess a good leaving group.
Notice that increasing (or decreasing) the value of DG by about 6 kJ/mol corresponds to a
10x change in the equilibrium constant
SN2 - Rationalizing kinetic data
3° substrates react too slowly to measure.
Endothermic reaction
Energy needed for bonds broken exceeds the stability gained by the bonds formed Products less stable than reactants • Products higher in energy • Energy is consumed (KE converted to PE) • DH˚ is positive • Temp of surroundings consumed
_____ must both be considered when predicting whether a reaction will occur
Enthalpy (DH) and entropy (DS)
Bad leaving group examples:
F, HO, EtO, tBuO, NH2
Experimental data indicates that a bulky, sterically hindered base will favor the formation of the ______, but an unhindered base (like ethoxide) will favor the _____
Hofmann product, Zaitsev product
The greater the magnitude of a (-)ΔG, the greater
the equilibrium concentration of products
Rate: The lower the activation energy,
the faster the rate
Increasing the substitution increases the stability of a carbocation, due to
the increasing number of adjacent sigma bonds aligned with the empty p-orbita
If ΔStot is positive
the process is spontaneous.
Catalyst
- speeds up the rate of a reaction without being consumed Enzymes are catalysts Catalyst provides an alternate, and faster pathway of reaction Catalysts lower the activation energy
E2 - effect of the substrate Consider a reagent such as NaOH, which is a strong nucleophile (SN2) and a strong base (E2)...
... when the substrate is sterically hindered, E2 elimination will occur.
Rearrangements - Two types of carbocation rearrangement are common
1,2-hydride shift 1,2-methide shift Shifts can only occur from an adjacent carbon. Shifts only occur if a more stable carbocation results
DG =
DH - TDS
ANY MOLECULE WITH A GOOD LG WILL UNDERGO SN2 IN THE PRESENCE OF A
GOOD Nuc
Rate: Steric Considerations
Steric hindrance, and the geometry of a compound, affects the rate of reaction When molecules collide, they have to have the correct orientation for bonds to be made/broken. If the reactive conformation of a compound is high energy, it will spend less time in that conformation, and so the probability of collision resulting in a reaction is low
Because of steric strain, cis isomers are generally
less stable than trans
A substitution reaction requires the loss of a leaving group, and nucleophilic attack. There are two possible mechanisms: (1) concerted, and (2) stepwise.
1. The concerted mechanism involves breaking of the bond to the leaving group and making of the bond to the nucleophile at the same time: SN2 2. The stepwise mechanism the leaving groups leaves first, to give a carbocation intermediate, followed by nucleophilic attack
Why doesn't an exergonic process react 100% to give products? Why will some reactants still remain at equilibrium?
As [A] and [B] decrease collisions between A and B will occur less often As [C] and [D] increase, collisions between C and D will occur more often Eventually the forward and reverse reaction rates will be equal
DeltaH =
BDE (bonds broken) - BDE(bonds formed)
How to assign E or Z to alkene stereoisomers...
First, prioritize the groups attached to the C=C double bond based on atomic number If the important groups are on opposite sides of the double bond = E If the important groups are on same side of the double bond = Z
Stereospecificity of E2: Cyclohexyl Halides Consider the dehydrohalogenation of a cyclohexane derivative, where the leaving group is attached to the ring
Given the anti-periplanar requirement, E2 elimination can only occur when the leaving group is in the axial position.
It is very important to understand the difference between the terms stereospecific and stereoselective.
In a stereospecific rxn, the substrate is stereoisomeric and results in one stereoisomer as the product In a stereoselective rxn, the substrate can produce two stereoisomers as products, where one is the major product.
Nucleophiles:
Inductive effects (ie Lithium group ) Lone pairs (ie water) pi bonds (empty p orbitals)
Electrophiles:
Inductive effects (ie chlorine group) empty p orbital (ie carbocation, positive charge)
E2 regioselectivity
It is common for a substrate to have more than one b-carbon that can be deprotonated by a strong base, and so E2 elimination results in more than one alkene product.
DG = 0
Keq = 1
An equilibrium constant (Keq) is used to show the degree to which a reaction is product or reactant favored
Keq = [products]/[reactants] DG = -RTlnKeq
DG < 0
Keq > 1
Alkyl groups stabilize the C=C pi bond via hyperconjugation. So....
More alkyl groups = more stable alkene
There are four main ways that electrons move in polar reactions
Nucleophilic Attack Loss of a Leaving Group Proton Transfers (Acid/Base) Rearrangements
Polar aprotic solvents:
Protic solvents engage in H-bonding, and stabilize anionic species (such as good nucleophiles) turning them is poor nucleophiles. Polar Aprotic solvents stabilize prevalently cations
Rate: Temperature (T)
Raising temperature will result in a faster rxn. At higher T, molecules have more kinetic energy. At higher T, more molecules will have enough energy to produce a reaction
Both SN2 and E2 are second order rxns with a rate equation:
Rate = k [substrate] [reagent] the reagent can be a good nuc or base
SN2- Kinetics: Less sterically hindered electrophiles react more readily under
SN2 conditions. Tertiary halides are too hindered to react via SN2 mechanism.
Evidence suggests that a strict 180° angle is not necessary for E2 mechanisms.....
Similar angles (175-179°) are sufficient The term, anti-periplanar is generally used instead of anti-coplanar to account for slight deviations from coplanarity Although the E isomer is usually more stable because it is less sterically hindered, the requirement for an anti-periplanar transition state can often lead to the less stable Z isomer as the major product
Recall that a (-) sign for ΔG tells us a process is product favored (spontaneous)
That does NOT tell us anything about the RATE or kinetics for the process. In other words, DG says nothing about how fast a reaction will occur.
The reaction rate is a function of the number of molecular collisions that will occur in a given period of time, which is affected by the following factors:
The concentrations of the reactants The Activation Energy The Temperature Geometry and Sterics The presence of a catalyst
Activation Energy (Ea)
The energy barrier between reactants and products Ea is the minimum amt. of energy required for a molecular collision to result in a reaction As Ea increases, the number of molecules possessing enough energy to react decreases...
Exothermic reaction
The energy gained by bonds formed exceeds the energy needed for bonds broken Products more stable than reactants Products lower in energy • Energy is released as heat (PE converted to KE) • DH˚ is negative • Temp of surroundings increases
Neomenthyl chloride is 200 times more reactive toward an E2 process. Why?
The more stable chair conformation has the larger isopropyl group occupying an equatorial position. In this chair conformation, the chlorine occupies an axial position, which is set up nicely for an E2 elimination. In other words, neomenthyl chloride spends most of its time in the conformation neces- sary for an E2 process to occur. In contrast, menthyl chloride spends most of its time in the wrong conformation: In this case, an E2 process can only occur from the higher energy chair conformation, whose concentration is small at equilibrium. As a result, menthyl chloride will undergo an E2 reaction more slowly.
Carbocation rearrangements are generally irreversible
Thermodynamically, there is no driving force for a more stable carbocation to rearrange to a less stable one.
The Hammond Postulate
Two points on an energy diagram that are close in energy should be similar in structure Based on this assumption, we can generalize the structure of a transition state, depending on whether the reaction is exothermic or endothermic
E2 elimination is a regioselective:
When constitutional isomers are formed as the products of a reaction, with one of them as the major product, the reaction is regioselective The regioselectivity of an E2 reaction can be controlled by carefully choosing the strong base used.
ΔSsys is affected most significantly by two factors, and will be positive...
When there are more moles of product than reactant When a cyclic compound becomes acyclic
There are many factors to consider in order to correctly predict the product(s) of an E2 rxn and decide what the major product will be.
Will the substrate react stereospecifically? or will it be a stereoselective E2 rxn? Will the substrate produce several regioisomeric alkenes? If so, what will be the major product, given the steric hindrance of the base that is used?
Nucleophiles are less stable, thus more reactive in ____
aprotic solvent. The activation energy will be lower and the reaction faster Aprotic solvents are best for SN2 reactions
When an alkyl halide is treated with a strong base, it can undergo ___
beta elimination (1,2-elimination) to form an alkene: A strong base will react in a concerted mechanism, called an E2 elimination.
entropy
can be described as molecular disorder, randomness, or freedom is actually the number of vibrational, rotational, and translational states the energy of a compound is distributed
Recall that kinetics and thermodynamics are
completely different concepts
E2 elimination is
concerted, where the base removes a b-proton, causing the loss of the leaving group and the formation of the C=C bond. So, concerted elimination is bimolecular and follows second-order kinetics:
Good leaving groups are the
conjugate bases of strong acids. (are weak bases) ie. I, Br, Cl, sulfate stuff, water
Electrophile -
electron deficient species, can accept a pair of electrons Electrophiles are Lewis acids Carbocations and partially-positive atoms are electrophilic
Nucleophile -
electron rich species, can donate a pair of electrons Nucleophiles are Lewis bases More polarizable nucleophile = stronger nucleophile
heterolytic cleavage
electrons go to one atom and make ions (cation and anionO
Homolytic cleavage
electrons split and make two radicals (arrow has one "barb")
If a process has a positive ΔG, the process is nonspontaneous, and the process is
endergonic
If a process has a negative ΔG, the process is spontaneous, and the process is
exergonic
The head of a curved arrow shows either the
formation of a bond, or the formation of a lone pair
Rearrangements Carbocations can be stabilized by neighboring groups through slight orbital overlapping called
hyperconjugation
If the attacking nucleophile is a poor leaving group, it will essentially be an irreversible attack
ie: CH3 as a nucleophile
Polar reactions
involve ions as reactants, intermediates, and/or products
All proton transfers are reversible. But, if the pKa difference is
is 10 units or more, it can be considered irreversible
An intermediate
is an intermediate species formed during the course of a reaction. They are energy minima on the diagram. Intermediates are observable. They are an actual chemical species that exists for a period of time before reacting further
Enthalpy (ΔH or q)
is the heat energy exchange between the reaction and its surroundings Breaking a bond requires the system to absorb energy
A transition state
is the high energy state a reaction passes through Transition states are fleeting; they cannot be observed On an energy diagram, transition states are energy maxima, and represent the transition as bonds are made and/or broken
Rate =
k [reactants] The degree to which a change in [reactant] will affect the rate is known as the order.
Some reactions are drawn as equilibria, and others are drawn as irreversible. The question of reversibility is both a
kinetic and a thermodynamic question.
The curved arrow starts on a
pair of electrons (a shared pair in a bond, or a lone pair)
SN2 - solvent effects Need a ______ solvent for SN2 rxns
polar aprotic
Consider an exergonic process with a (-) ΔG, which means the
products are favored to form (spontaneous)
A spontaneous process means there will be more
products than reactants
An endergonic reaction favors the
reactants
The loss of a leaving group is virtually always
reversible
If the attacking nucleophile is also a good leaving group, it will be a
reversible attack ie, water as a nucleophile
Why does a sterically hindered base favor the Hofmann product?
since the bulky base has to reach into a tight spot to rip off the proton if it tries to take the Zaitsev proton, its easier to take the Hofmann proton with more space
SN2 - Nucleophilicity A ______ is needed for an SN2 rxn
strong nucleophile
Alkyl halides undergo
substitution and elimination rxns
