Module 6: Addition Reactions Flashcards
Addition is the opposite of
Elimination
The C=C is converted to…?
Two new sigma bonds!
The pi bond is an _ _ _ _ _ donor and can act as a _____ or a ______
Electron pair donor.
Nucleophile or Base.
Addition and Elimination are equilibrating reactions. Which side is favored depends on….?
Temperature.
Entropy Favours Elimination so _____ temperatures favor elimination
higher!
Addition reactions are _______ favored so they occur at ____ temperatures.
Enthalpically
Low
HX Addition Reaction Mechanism
1) Double bond attacks the hydrogen
2) alkene is protonated forming a carbocation intermediate
3) bromide ion functions as a nucleophile and attacks the carbocation intermediate
Stereochemistry of HX addition: do we form an enantiomer?
Yes! Our nucleophile can attack from either side because the carbocation is trigonal planar.
Acid Catalyzed (H2SO4) Hydration (H2O) to Form an alcohol.
Addition of water with an acid catalyst to form an alcohol.
1) Water acts as a base to deprotonate the acid
2) Water is deprotonated by the pi bond
3) water attacks the carbocation to form an alcohol
4) water in solution functions as a base to deprotenate the alcohol.
Stereochemistry of Hydration: Do we form enantiomers?
Yes! if we form a carbocation (sp2 hybridized/planar) the addition of H2O/HX generates two stereoisomers as the nucleophile can attack from either side
Thermodynamics of Hydration
If we are synthesizing an alcohol from an alkene, we would use excess water
If we are synthesizing an alkene from an alcohol, we would only use acid, and not add water to the reaction
What are some problems with hydration reactions
1) Alcohols React Similarly
2) Intramolecular
What are some problems with hydration reactions
1) Alcohols React Similarly
2) Intramolecular
Hydroboration Oxidation [(1)BH3 *THF, (2)H2O2, NAOH)] places the OH on the….?
Least substituted carbon.
1) The less substituted carbon attacks the boron, and the more substituted carbon develops a delta+ which triggers a hydride shift
2) the delta negative hydrogen attacks the carbocation as its being formed.
3) The boron ends up on the least substituted carbon and the hydrogen ends up on the most subsititued carbon.
4) The H2O2, the sodium hydroxide replaces the BH2 and converts it into an OH, so the position of the OH on the least substituted carbon using these two reactions is a direct result of the first step of the reaction which we can predict
Can occur up to three different times
Hydroboration Oxidation- Stereoselective:
H and OH are added in a ____ fashion across the double bond
the electron rich alkene attacks the boron, then the delta negative hydrogen attacks the carbocation that forms, all this occurs at the same time, the boron and hydrogen are added and delivered into the same face.
(SYN)
Hydroboration Oxidation- Stereoselective:
If ___ or ___ chirality center is formed, a pair of enantiomers is formed by addition to either side of the alkene
one, two.
both H and OH can be added to one face or the other face.
Hydrogenation (H2, Pt, Pd, Ni)
the addition of H2 across a C=C double bond.
Requires a metal catalyst
Alkene is converted to the corresponding alkane
SYN addition
Hydrogenation: is stereospecific only ____ addition is observed.
Syn.
Two chirality centers are formed only the stereoisomers resulting from syn addition are obtained
Catalytic Hydrogenation
1) The hydrogen molecule absorbs onto the metal surface and when it does it breaks the H-H bond.
2) The alkene binds to the metal surface as well through the pi bond, this facilitates the transfer of the hydrogens to the alkene, and because the alkene is absorbed on the surface its basically coming face to face with the hydrogens that are also bound on the surface.
3) Those to hydrogens are transferred to the same face of the alkene.
The metal surface is required to weaken the H-H bond of H2 and because the reaction is happening on the metal surface the hydrogens are transferred to the same face of the alkene.
Halogenation: Is only praticle with…?
Br2 and Cl2
Halogentation Stereoselective: We only get products from the _____ addition.
Anti.
Halogenation (Br2)
1) Br2 (or Cl2) is nonpolar, but polarizable. Approach of a nucleophile will induce a dipole
Think of Br2 as a bromine atom bonded to a good leaving group
2) An electron rich alkene will attack the delta + bromine.
3) The attacked bromine forms bonds with both carbons preventing the formation of the carbocation forming a halonium ion.
4) the bromonium ion is electron withdrawing leaving both carbons delta +. The bromine ion will attack either on of the carbons and attacks from the opposite side of the first bromine.
Halogenation is stereospecific
the stereochemistry of the starting alkene determines the stereochemistry of the product(s)
Halohydrin Formation: Br2 and H2O
- Halohydrins – formed when halogenation is conducted in water
- Water acts as the nucleophile that attacks the bromonium ion
- There are many more H2O molecules compared to Br - ions, so H2O outcompetes Br - for the bromonium ion
Halohydrin Formation: Reaction is Regioselective
The preference is an attack at the carbon who is best able support the positive charge (most substituted carbon), attacks from the opposite side
Syn-Dihydroxylation (addition of OH and OH across the double bond).
The key reagent is _ _ _ _
Key reagent is OsO4 the other reagents are co-oxidants
Use of the co-oxidants generate only syn addition
Syn dihydroxylation can also be achieved with _ __ _ _ but only under mild conditions (cold temperatures)
KMnO4
-Delivers both oxygen atoms concertedly added to the same face.
Intermediate is not isolated and NaOH is added to lead the syn addition of both groups.
Very strong oxidizing agent, more messy there is a byproduct that needs to be isolated.
Used a visual indicators for the presence of alkenes.
Anti-Dihydrogenation is a two step process with 1) R_ _ _ _ and 2) _ _ _
RCO3H
H2O
Anti dihydroxylation of an alkene is a two-reaction process
1)In the first step the alkene is reacted with a peroxyacid, The electron rich alkene attacks the electron deficient oxygen, as that carbon oxygen bond forms the delta negative oxygen takes the proton and those two electrons can form the other bond.
we form an epoxide in the first step
2)the epoxide can be isolated and in the second step reacted with H3O and then water acts as a nucleophile to open up this epoixde to form a trans diol
Oxidative Cleavage of Alkenes (O3 with DMS or Zn/H2O)
If you take an alkene and react with ozone followed by a reducing agent DMS, the outcome of this reaction is to split that alkene into to two carbonyl compounds.
Fragment the pi bond and replace it with two carbonyl bonds.
