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Chemistry 221 Final Exam > Exam 3 > Flashcards

Exam 3 Flashcards

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1
Q

Sequential reaction

A
  1. first reaction (to completion)
  2. Second reaction (no interaction with the first reaction)
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2
Q

Substitution Rxn

A

Replacing one group on a carbon with another

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3
Q

Addition Rxn

A

combining reactants to form new products (typically one pi bond is split to form two new sigma bonds)

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4
Q

Elimination Rxn

A

A Hydrogen and a Leaving Group are removed and replaced by a pi bond

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5
Q

Rearrangement Rxn

A

a reaction that changes orientation

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6
Q

Concerted

A

All bonds are broken and formed simultaneously - no intermediates with formal charges

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7
Q

Step-wise

A

bonds are broken and formed in specific steps creating intermediates with formal charges that can move

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8
Q

Homolydic cleavage

A

each atom gets one electron, typically seen in a free radical halogenation reaction

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9
Q

Heterolydic Cleavage

A

electrons travel together

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10
Q

Free energy equation

A

∆G= ∆H - (Tx∆S)
∆H = BFE - BDE

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11
Q

∆H is positive

A

Energy is absorbed - Endothermic

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12
Q

∆H is negative

A

Energy is released - exothermic

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13
Q

stronger bonds…

A

requires more energy to break

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14
Q

Energy Diagram

A

one peak represents a concerted reaction, multiple peaks represent each step in the reaction

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15
Q

intermediate stability

A

Carbon with the charge:
primary < secondary < tertiary < benzylic < Allylic (adjacent to a double bond) < Tertiary allylic

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16
Q

Nucleophilic addition

A

nucleophile uses a lone pair of e- to add itself to the molecule and push off another atom

17
Q

Leaving Group

A

atom or group comfortable with a negative charge

18
Q

Proton Transfer

A

addition of a hydrogen

19
Q

Pi bond formation

A

Transfer of a non bonded lone pair to move the positive charge

20
Q

Free radical halogenation

A

*Stereochemistry for chirol centers
1. homolytic cleavage of a halogen, one removes a hydrogen
2. a second homolytic cleavage of a halogen to bond with the radical carbon intermediate

21
Q

SN2 Reaction

A
  • Concerted substitution
  • Backside attack

Factors to consider:
1. Electrophile
methyl > primary > secondary, no tertiary addition
2. Leaving group
I > Br > Cl > F
3. Nucleophile
Must have an available lone pair
4. Solvent
Polar aprotic - no O-H Bonds
DMSO, HMPA, Acetone

22
Q

SN1 Reaction

A
  • Stepwise
  • Leaving groups leaves on its own first
  • Racemic
  • Cation rearrangement for best stability (Hydride/Methyl shift)

Factors to consider:
1. Electrophile
tertiary > secondary > primary > Methyl
2. Leaving group
I > Br > Cl > F
3. Nucleophile
Weak - Compounds with lone pairs but are neutral
4. Solvent
Polar protic - an O-H Bond helps with cation stability

23
Q

Elimination Reaction - E1

A
  • Stepwise
  • Hydride shift
  • double bond with more substituents is preferable
  • Leaving group leaves on its own
  • Trans double bond is more stable

Factors to consider:
1. Electrophile
tertiary > secondary > primary > Methyl
2. Leaving group
I > Br > Cl > F
3. Solvent
Polar protic - stabilizes the carbocation intermediate and one of its lone pairs are able to remove the hydrogen to form the double bond
4. Environment
When heat is applied E1 is favored over SN1

24
Q

Alcohol (OH) Leaving group

A
  • poor leaving group
  • can be activated with either an acid (SN1/E1) or with TsCl and pyr (SN1, SN2, E1, E2)
25
Q

Activation with Acid (SN1, E1)

A
  1. a hydrogen is taken from the acid to make the OH an H2O
  2. H2O is a good leaving group, so it does
  3. Hydrogens can shift if needed
  4. the H2O in solution can remove the second H which leaves its electrons to the electrophile
26
Q

Activation with TsCl and Pyr (SN1, SN2, E1, E2)

A
  1. TsCl turns the OH into OTs which is a good leaving group, akin to Br
  2. from there it can do any of the reactions
27
Q

Elimination reaction - E2

A
  • Concerted
  • Strong Base - no positive charges allowed
  • small bases -> less hydrogens
    -Bulky Bases -> more hydrogens
  • stereoselective: leaving group and hydrogen must be in the same plane, preferably anti planer

Factors to consider:
1. Electrophile
tertiary > secondary > primary > Methyl
2. Leaving group
I > Br > Cl > F
3. Nucleophile
strong base - any salt with an organic compound (NaOCH3, ect.)
4. Solvent
not important but typically Conj. Acid of the base

28
Q

E2 Elimination in rings

A

Normal ring:
- Trans diaxial formation (one up one down)
Bridged system:
- Syn elimination only, both down

Chemistry 221 Final Exam (5 decks)

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