Exam 1 Material Flashcards
SN2 reactions
- one step
- leaving group leaves as nucleophile attacks (opposite side of LG).
- If LG is on a chiral carbon, you will see inversion of stereochemistry (R vs. S)
- Rate= [substrate][nucleophile]
- VERY sensitive to nature of starting materials (LG).
- methyl>primary>secondary>tertiary
R vs. S
R= clockwise S= counterclockwise
The elimination reaction
The beta proton is removed by the base, the electrons from that bond form a double bond, and the leaving group gets kicked off.
*** concerned with strong bases not strong nucleophiles!
E2 Reaction
- second order (one step)
- Rate = [substrate][base]
- can proceed quickly with tertiary substrates because the steric hindrance is not as important since a proton is being targeted.
Cis vs. trans
Trans is always more stable because there is les steric hindrance.
E vs. Z
- E means that the priority groups are on different sides of the double bond.
- Z means that they are on the same side.
Stability of alkenes
- Trans is more stable than cis
- E is more stable than Z
- substitution= how many groups are attached to the double bond.
- substitution stability= tetra> tri> di> mono
Regioselectivity of E2 reactions
- when small bases (NaOEt, NaOMe) are used, the more substituted product ZAITSEV will be favored.
- when bulky bases (t-Bu- Ok, diisopropyl amine, triethyl amine) are used, the less substituted HOFMANN product will be favored.
- this is because hydrogens on terminal carbons are easier for a sterically hindered base to attack.
Stereoselectivity of E2 Reactions
Need to consider that rotation can occur at double bonds, therefore the trans product will be the major product over the cis.
Stereospecificity of E2 reaction
- In Newman projections and chair conformations (carbon rings), the beta proton must be anti- periplanar to the leaving group ( in the same plane)
- ** if the leaving group is on a wedge then the beta proton must be on a dash and vice versa.
SN1 Reaction
- stepwise process
- Rate= k[substrate]
- the leaving group must first leave, resulting in a carbocation. The nucleophile then attacks.
- if a chirality center is involved, can result in a race ic mix because the nucleophile can attack either side of the molecule.
- tertiary action> secondary> primary
Carbocation Rearrangement
- common for SN1
- goal is to make a more stable carbocation ( tertiary over secondary)
1) hydride shift- movement of a hydrogen and the electrons from its bond from one carbon to another.
2) methyl shift- migration of a methyl group and its bonded electrons from one carbon to another.
3) alkyl shift- migration of an alkyl group ( something other than a methyl) often happens to alleviate ring strain.
Nucleophile for SN1
Rate is not dependent on the nucleophile, therefore it DOES NOT need to be strong.
Ex) H2O, MeOH, EtOH.
*** if water attacks, always assume that another water molecule is in solution and takes the extra H to form an alcohol.
E1 Reaction
- Stepwise Process
- Leaving group leaves and a carbocation is formed. THEN the base attacks the beta proton and a double bond is formed.
- Rate=k[substrate]
- substrate trend is the same as with SN1… tertiary>secondary> primary
- ** because most nucleophiles can also be good bases, a mixture of SN1 and E1 can be obtained.
Regioselectivity for E1 reactions
- Due to carbocation formation, the more substituted alkene (zaitsev) is generally favored.
- **If conc. H2SO4 and heat are reactants, you automatically know that the elimination product is formed!
Solvent effects for ALL reactions
SN2 + E2: polar aprotic ( no H connected to electronegative atom)
** ex) DMSO, DMF, THF
- this is because these mechanisms use strong nucleophiles and strong bases, so you don’t want them to become surrounded by a solvent shell if they need to attack an electrophilic center.
SN1 + E1: polar protic ( has an H attached to an electronegative atom)
*** ex) H2O, MeOH, EtOH, acetic acid, NH3
- this is because when the leaving group leaves, you want it to be surrounded by a solvent shell so that it doesnt re-attach to substrate.
Alkene
- incorporates a Carbon- carbon double bond
- 3 separate bonds ( 2 sigma and one pi)
- each carbon is sp2 hybridized
Alkyne
- Has a carbon- carbon triple bond
- 3 separate bonds: one sigma and two pi
- each carbon is sp hybridized
Geminal dihalides
When there are two halide atoms on the same carbon
Vicinal halides
When There are two halide atoms in a molecule but they are on different carbons.
Addition reactions
The addition of two groups/ atoms across a pi bond.
Hydrohalogenation
- the addition of a hydrogen and a halogen across ( Br, CL, etc.) a pi bond
- the halogen will attach to the more substituted side of the double bond (markovnikov product)
- ** if a chirality center is produced, then we will get a racemic mixture of products.
- reactants: an alkene or alkyne and H-Cl, H-Br, etc.
- with an alkyne, if the reactant is present in excess, then it will react twice to form a single bond with two halogen atoms attached to the markovnikov position.
Acid- catalyzed hydration
- the addition of H and OH across a pi bond.
- reactants: an alkene or alkyne and H3O+
- **must have an additional water molecule take the extra H off of water to form OH
Acid- catalyzed hydration of alkyne
- Reactants: H2SO4, H2O… HgSO4 increases the rate of the reaction (catalyst)
- the transition state of this reaction is an enol and then tautomerization occurs to form a ketone.
Acid- catalyzed tautomerization
Start with an enol and form a ketone. The double bond in the enol attacks a hydrogen on H3O+ , this hydrogen adds to the end and there is no longer a carbon- carbon double bond. Next, a lone pair from the O-H forms a double bond between the C and the O and the electrons shared between the O-H move to the O, removing the H and forming a ketone.
Base- catalyzed tautomerization
Enol to ketone using OH-… The O in OH- attacks the H on the OH group in the enol, those electrons are left on the O as the H is taken. Next, these electrons form a double bond between the O and the C, this causes the c-c double bond to break forming an anion on the terminal carbon. An H attached here and forms a ketone.
Hydroboration Oxidation
- anti-markovnikov addition of H2O… places OH group at less substituted carbon
- reactants: 1)BH3 and THF AND 2) H2O2 and NaOH
- OH and H add syn (same side).
Hydroboration- oxidation of alkyne
- enol- aldehyde
- use 9-BBN instead of BH3, THF
- this is because you want to prevent 2 additions on BH3 across an alkyne, so you use a sterically hindered borane reagent.
Catalytic hydrogenation
- addition of molecular hydrogen (H2) across a double bond
- use metal catalyst (Pd, Ni, Pt)
- **reduce alkene to alkane
- both hydrogens add to the same face of pi bond (syn)
- ** if you have an alkyne and you see Pd, Ni, Pt… you reduce alkyne to alkane
Reduction of alkynes
1) catalytic hydrogenation: reduces all the way to alkane (no stereochemistry since you are adding two sets of H’s.
2) poisoned catalyst: H2 and Lindlar catalyst…forms a cis alkene (like a dog playing dead)
3) dissolving metal reduction: Na and NH3(l) forms a trans alkene.
Halogenation
- Addition of X2 across pi bond
- happens with Cl2 and Br2 with CCl4 as solvent
- results in anti- addition (opposite faces of double bond… one on wedge and one on dash)
- Br2 and Cl2 are not polar but they are polarizable when placed by a nucleophile(double bond) making it an electrophile for the double bond to attack.
- transition state is bromonium cation (VERY UNSTABLE )
Halogenation of alkynes
- Can add one or two rounds of X2
- if one round, result is an alkene ( no stereochemistry because only three things attached)
- if two rounds (xs),result is an alkane
Halohydrin formation
-Adding X and OH across a double bond
Reactants: X2 and H2O
- OH and X2 add anti ( opposite face of double bond)
- transition state is bromonium cation
- Br anion that is left gets surrounded by solvent shell so that it cannot continue to react
- after water adds to the molecule, another water molecule deprotonates it so that it forms OH
- result is a racemic mix.
