Chapter 9 Flashcards

1
Q

Substation reaction

A

Electronegative group is replaced by another group

Done with poor base usually

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2
Q

Elimination group

A

Electronegative group is eliminated along with a hydrogen

Also creates double bond/ triple bond

Drone with strong base

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3
Q

Sn2

A

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

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4
Q

Relative rate Sn2

A

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)

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5
Q

Pka and Sn2

A

Lower pka of Hx, quicker reaction when X nucleophile

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6
Q

Polarizability

A

Larger, more polarizable

Bigger makes it better nucleophile

If they are in a protein polar solvent, it makes up for their decreased basicity

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7
Q

Protic polar solvent

A

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)

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8
Q

Steric hynderance and nucleophile Ty

A

More steric hyderance on nucleophile slows down substation

Will usually shift to elimination

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9
Q

Sn1

A

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

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10
Q

Sn1 stereospecificity

A

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

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11
Q

E2

A

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
If equatorial, then chair flip and reaction slows down

Fast if cyclohexane molecule is most stable

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12
Q

E1

A

Rate = k[alkyl halide ]

Weak base
Tertiary alkyl halide
Also does a sn1 reaction
High temp favors E1, low favoris Sn1

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13
Q

Strong base

A

OH
OR
NH2

Check for potential double bonds

Ignore Na or K attached

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14
Q

Tertiary reactions

A

So bulky that elimination is the major reaction

Except solvolysis which does substitution

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15
Q

Intermolecular reactions

A

Complimentary sides of the reaction reacts with the same molecule, producing a polymer

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16
Q

Intranolecular

A

Bites it’s own tail, molecule reacts with itself producing a cyclic ether

Prioritized by making it a longer chain, or by decreasing concentration

17
Q

Dbn or dbu

A

Bulky strong bases, favors elimination

18
Q

Allylic and benzylic

A

Double bond 1 bond before halide, better than tertiary does e2 and sn1 and e1
Benzylic does e2 and e1o

19
Q

Halide directly on double bond

A

Only do elimination, never substitution

Only e2

20
Q

E1 and E2 next to benzene

A

Not Benzylic if X is not attached directly

Pit double bond closest to benzyl in order to do resonnance

21
Q

NaNH2

A

So strong it often does elimination anyway, even if primary

22
Q

DBN/DNU

A

Strong bulky bases

23
Q

NaH

A

Really good at removing protons from alcohols