Ch. 7 Flashcards
What makes a good leaving group and why?
( Does this vary for SN1 vs SN2)
Weak Bases are the Best Leaving Groups
In order for a leaving group to leave, it must be able to accept electrons. A strong bases wants to donate electrons; therefore, the leaving group must be a weak base.
As Electronegativity Increases, The Ability of the Leaving Group to Leave Increases.
As Size Increases, The Ability of the Leaving Group to Leave Increases
resonance increases the ability of the LG to leave
How does electronegativity affect a leaving group?
As Electronegativity Increases, The Ability of the Leaving Group to Leave Increases. This is because an increase in electronegativity results in a species that wants to hold onto its electrons rather than donate them.
what are alkyl halides?
compounds in which a halogen (such as Cl, Br, or I) is connected to an sp3 hybridized carbon atom
what are Aryl halides and vinyl halides
a halogen is connected to an sp2 hybridized carbon atom
what are the two general types of reactions for alkyl halids?
substitution and elimiantion
substitution reaction
in which the nucleophile replaces the halogen
Reactions in which one group is replaced by another group.
elimination reaction
A reaction involving the loss of a leaving group and formation of a π bond.
substrate aka
alkyl halide
isn the carbon connected to the halogen electrophilic nor nucleophilic and why?
The halogen withdraws electron density via induction, rendering the adjacent carbon atom electrophilic, and therefore subject to attack by a nucleophile.
what makes a good leaving group in regards to acids
Good leaving groups are the conjugate bases of strong acids. For example, iodide (I−) is the conjugate base of a very strong acid (HI),
ie idodide is a very weak base
As a rule of thumb, a good leaving group is the conjugate base of an acid with a pKa < 0. For this reason, chloride, bromide, and iodide are all good leaving groups, while fluoride is not (the pKa of HF is 3.2)
is hydroxide a good or bad leaving group?
hydroxide is a bad leaving group, because it is not a stabilized base. In fact, hydroxide is a relatively strong base, and therefore, it rarely functions as a leaving group
among halides which is the best leaving group
iodide is the best leaving group bc it is the weakest base
alpha position
the position immediately adjacent to the functional group
there is only one
beta position
the position immediately adjacent to the alpha position
max of 3
steps to name alkyl halides
- identify and name parent
- id and name the substituents
- number the parent chain and assign a locant to each substitute
- assemble subsituents alphabetically
names for halogens
fluoro -
chloro-
bromo-
iodo-
what does SN2 stand for?
bimolecular nucleophilic substitution
Rate law for SN2 and explain why
Rate =K (alkyl halid) (nucleophile)
rate law for SN1 and explain why
Rate = K( substrate)
rate is linearly dependent on the concentration of only one compound (the substrate)
What is the mechanism for Sn2?
- Nucleophilic Attack
- loss of leaving group
a nucleophile attacks the alkyl halides, causing the loss of a leaving group in a CONCERTED
what is the stereospecificity of SN2
inversion of configuration
During a reaction, when the configuration of a chiral center is changed.
what is the requirement for inversion of configuration?
The requirement for inversion of configuration means that the nucleophile can only attack from the back side (the side opposite the leaving group) and never from the front side
6 making it stereospecific : a reaction in which the configuration of the product is dependent on the configuration of the starting material
back-side attack
In SN2 reactions, the side opposite the leaving group, which is where the nucleophile attacks.
why does a backside attack occur?
- The lone pairs of the leaving group create regions of high electron density that effectively block the front side of the substrate, so the nucleophile can only approach from the back side.
- Molecular orbital theory
homo and lumo
If a nucleophile attacks methyl bromide from the front side, the nucleophile will encounter a node, and as a result, no net bonding will result from the overlap between the HOMO of the nucleophile and the LUMO of the electrophile