Organic Chemistry Reactions Flashcards
What is the rate law for an SN2 reaction?
An SN2 reaction is bimolecular. The rate depends on both the substrate and the nucleophile. The formation of the transition complex (which involves both the substrate and the nucleophile) is the rate-limiting step.
What is the rate law for an SN1 reaction?
An SN1 reaction is unimolecular. The rate depends on just the substrate. Specifically, the formation of the carbocation is the rate-limiting step.
What type of solvents are best suited for an SN1 reaction? Why?
Polar protic solvents are best for SN1 reactions. This is becuase they help stabilize carbocations, and aid in removal of the leaving group.
Below are the most common polar protic solvents
Examples of polar protic solvents: alcohols, carboxylic acids, ammonia, Bronsted acids (typically aqueous), and of course water.
What type of solvents are best suited for an SN2 reaction?
Polar aprotic solvents are best for SN2 reactions, which allow for the nucleophile to react with the substrate without interacting with the solvent.
Below are the most common polar aprotic solvents
Examples of polar aprotic solvents: acetone, dimethyl sulfoxide (DMSO), ethers, alkane (i.e. hexane), and chloroform.
For an E2 reaction, what orientation must the substrate be in?
The leaving group and cleaved proton must be anti-coplanar in respect to one another.
List the thermodynamic stability trend for alkenes
Unsubstituted < monosubstituted < cis-disbustituted < trans-disubstituted < trisubstituted < tetrasubstituted
For an E2 reaction, why do we consider the thermodynamic stability of the product?
The most thermodynamically stable product (the Zaitsev product) will be favored when using a nonbulky base (minimizing steric hindrance) and the less thermodynamically stable alkene (the Hoffman product) will be formed when using a bulky base.
List the stability trend for carbocations
The alkyl groups around a carbocation will help stabilize the positive charge due to their positive inductive effect (increasing electron density at the carbocation center). Also, remember that mesomeric effects (resonance stabilization) is stronger than inductive effects.
What is the rate-limiting step for an SN1 reaction?
The formation of the carbocation
What is a Lewis base?
A substance that can donate a pair of nonbonding electrons (such as the lone pair of an oxygen or nitrogen atom). Many Lewis bases are utilized for this chemical trait due to their nucleophilicity.
What is a Lewis acid?
A substance that can accept an electron pair (an electrophile) due to a vacant orbital. Many cationic metals are Lewis acids (Ag+</sub>,Cu+, Al3+), also a common Lewis acid is BF3.
What is a nucleophile?
A molecular or atomic species that can attack a nucleus (nucleophile - nucleus loving) typically a positively charged or partially positive center.
What is the rate-limiting step for an E2 reaction?
The formation of the transition-state complex.
What is an electrophile?
Electrophiles are chemical species that are attract/attracted to electron dense species (electrophile - electron loving).
How does a positive inductive effect impact the acidity of a carboxylic acid?
Will the pKa increase or decrease?
The pKa will increase, the positive inductive effect will increase the electron density of the O-H bond and consequently strengthen the bond (decreasing the acidity).
How does a positive inductive effect impact the acidity of a carboxylic acid?
Will the pKa increase or decrease?
The pKa will decrease, the negative inductive effect will decrease the elctron density of the O-H bond, weaknening the bond so that the proton is more likely to dissociate (becoming more acidic).
What does the reagent mCPBA form when reacting with alkenes?
Epoxides. If the alkene is trans the epoxide’s substituents will also be trans relative to the ring structure. If the alkene substrate is in the cis conformation, the epoxide will likewise be cis.
What is Markovnikov addition?
A Markovnikov addition, such as the case with HBr added to an alkene, states the inductive effect of the alkyl groups of an asymmetric alkene will increase the electron density at one carbon and consequently decrease the electron density at the adjacent carbon. The proton, H+, will be added to the carbon with increased electron density and bromide to the carbon with decreased electron density.
Some use the rule that the proton will be added to the carbon with the most number of hydrogens and the bromide will be added to the carbon with the least number of bonded hydrogens (this rule does not always work so we must consider the inductive effects).
Is halohydrin addition Markovnikov or anti-Markovnikov addition?
Reagents: Br2 and H2O
Markovnikov addition. Where the brominium cation will form and end up on the most electron dense carbon, and water will act as a nucleophile and attack the least electron dense carbon.
Note that the regioselectivity is Markovnikov addition and the stereoselectivity is an anti-addition (just as if adding Br2 accross a double bond). It is also important to note that the brominium cation formation prevents the need for a carbocation to form, without a carbocation there are ** no rearrangements**.
