Chapter 9 Flashcards
Substation reaction
Electronegative group is replaced by another group
Done with poor base usually
Elimination group
Electronegative group is eliminated along with a hydrogen
Also creates double bond/ triple bond
Drone with strong base
Sn2
Substitution nucleophile bimolecular
Rate=k[alkyl halide][nuvleophile]
Tells us what molecules are involved in the transition state
Inverts configuration if asymmetrical center because of steric hynderance
S usually becomes R, unless the group added is not the same priority as the leaving group
Stereo specific
No carbocation, 1 step
Stronger a base, worse as a leaving group and better as nucleophile
Relative rate Sn2
Methyl halide most most quick (Xch3)
For quickest reaction, primary alkyl halide used
For slowest, tertiary alkyl halide (most likely to do elimination instead of substitution) (because of steric hynderance, more alkyl group in the way of nucleophile addition) (also more free energy in tertiary, so more activation energy)
Pka and Sn2
Lower pka of Hx, quicker reaction when X nucleophile
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Polarizability
Larger, more polarizable
Bigger makes it better nucleophile
If they are in a protein polar solvent, it makes up for their decreased basicity
Protic polar solvent
H2O
HOR
RNH2
If does hydrogen bonds, hydrogen bonds to oxygen or hydrogen
Protocol solvent makes strong bases weaker as it attaches to it easily, trapped in a solvent cage, especially if smaller
E2 Sn2 are usually better in aprotkc Solvent
Protic polar solvents Best at stabilizing cations, (made in e1 and sn1)
Steric hynderance and nucleophile Ty
More steric hyderance on nucleophile slows down substation
Will usually shift to elimination
Sn1
Only 1 reactants in rate law, only alkyl halide in slowest step (carbocation formation)
Best done with tertiary alkyl halide and poor nucleophile, especially one that can double as the solvent
Rate = k [alkyne halide]
Breaker the halogen. Weaker the bond, faster the reaction
Rate decreases with stability of carbocation
Sn1 stereospecificity
Even if start with pure product, you get 2 product, one with retained configuration and one with inverted
More of inverted, because leaving group in the way of nucleophile in retained configuration, stereoselective
Partial racemization
E2
Rate=k[alkyl halide][base]
Need a base, better if strong
Major product of secondary alkyl halide with strong base is E2
Substitution can increase if there was less crowing at the carbon that has the leaving group
Removes beta hydrogen, usually beta carbon with fewest hydrogens unless another double bond or benzene ring present or bulky base (steric Hyderance) or alkyl fluoride (antizandskeif)
Mixture of E and Z
Tertiary alkyl halide reacts faster than primary because products ore stable
STRONG BASE
RO-, HO-
Leads to anti elimination: beta hydrogen and leaving group are anti to each other when reaction occurs
This leads the double bond to lock the molecule in a trans or cis conformation after the reaction depending on initial molecule
If chair conformation, then beta hydrogen and leaving group must both be axial, only one that allows for anti elimination
If equatorial, then chair flip and reaction slows down
Fast if cyclohexane molecule is most stable
E1
Rate = k[alkyl halide ]
Weak base
Tertiary alkyl halide
Also does a sn1 reaction
High temp favors E1, low favoris Sn1
Strong base
OH
OR
NH2
Check for potential double bonds
Ignore Na or K attached
Tertiary reactions
So bulky that elimination is the major reaction
Except solvolysis which does substitution
Intermolecular reactions
Complimentary sides of the reaction reacts with the same molecule, producing a polymer
