alkene starting Ph H3C C C CH3 H
Chapter 8 and 9 review -- the reagents Nolta 2009 Electrophilic addition Reactions - the General Scheme E = Electrophile (something lacking electrons/positive in nature) it is attacked by the pi electrons in the initial step in the reaction, the pi electrons form a new bond from the hinge carbon to the electrophile. N = Nucleophile (negative in nature) provides electrons for a new bond to one of the carbons of the pi bond; this occurs in a different step or at the same time the electrophile is attacked. The timing impacts the stereoselection (or not) of the reaction.
two carbons hold the pi bond one acts as "hinge" other receives nucleophilic electrons
Timing: E and N come in at two different steps with a C+ intermediate forming in between = no stereoselection Stereoselection: E and N get final positions at the same time (their positions are dependent on each other: syn (coming from the same side) or anti (from opposite sides), but it can only be observed if stereochemical label(s) result. For instance, if the two sp2 carbons of the double bond become chiral carbons, or if an E or Z stereochemistry is selected, rather than both resulting. Stereoselection shows the dependency of the positions of E and N around the Cs of the pi bond. Focus: Are the two carbons identical (one possible product connectivity), equally substituted (regioisomers are possible), unequally substituted (which carbon is more able to be positive in nature? Regioselection!) Is the more positive carbon receiving the nucleophile - if so, Markovnikov regioiselection leads to a favored isomer...the only option is going against this rule: "anti-Markovnikov".
E/N / H OH
Reagents H3O+ or H2O/H2SO4
Regioselectivity Markovnikov Markovnikov
Stereoselectivity H3C C+ -- not stereoselective C C+ intermediate H H3C H C H C H
Product(s) OH o r X C Ph CH3
+ enantiomer one chiral carbon formed
Strong acid additions (from chapter 4) H/X H-X Other additions involving OH H / OH 1) BH3 (9-BBN) 2) NaOH, H2O2, H2O KMnO4 or OsO4 O O R Ozonolysis O/O C OOH
"Anti-Markovnikov" syn Hydroboration Reaction None Identical E and N None Only one atom adding
OH / OH Peroxidation
H3C HO CH3 OH H syn C C + enantiomer Ph HO CH3 O Ph H3C C Ph C H3C C
+ enantiomer Ph two chiral carbons formed Note a staggered conformer is drawn How E and N add is different from the conformations that the freely rotating bond can now take!
None
1) O3/ 2) Zn or (CH3)2S
reductive workup
None
None
CH3 + enantiomer H H3C O H OH, for oxidative C O
1) O3/ 2) H2O2, NaOH, H2O
oxidative workup
Hydrogenation H/H H2/Pt (metal) None syn H H3C C
Halonium ion intermediates - reactions beginning with electrophilic halogens anti Markovnikov X / OH X /H O
2 2
X/X X / OR For alkynes like:
H3C C C H
X2 X2/ROH
H/H H/H H/H H/X H / OH
None Markovnikov
H2/metal H2/Pd/BaSO4/quinoline Na/NH3 H-X (2 equivalents) H3O+/H2O
anti anti
Note that the conformational anti arrangement of the E C Ph and N just happens Br CH3 to coincide with HOW they added... + enantiomer
OH
all the way to the alkane with syn addition is a stereogenic double syn addition to form the double bond bond formed? anti addition to form the double bond Stereoselection... C+ intermediate twice, Markovnikov, not stereoselective X Both X end up on more substituted C H3C C CH3 Ketone at most substituted C O H3 C C CH3
X