Exam 2

Question 1

complexation reaction

Answer 1

making a new sigma bond

Question 2

decomplexation reaction

Answer 2

breaking a sigma bond

Question 3

substitution reaction

Answer 3

sigma bond broken and replaced

Question 4

addition reaction

Answer 4

breaking a pi bond and forming two new sigma bonds

Question 5

elimination reactions

Answer 5

forming a pi bond and breaking two sigma bonds

dehydrobromination
dehydration
dehydrogenation

Question 6

ΔG‡

Answer 6

activation energy
- fast or slow reaction?
- distance from ground state energy to transition state
- higher ΔG‡, few molecules can react/overcome barrier/requirements –> slower reaction

Question 7

ΔG°

Answer 7

Standard free energy
- favorable or unfavorable?
- ΔG°: products are more stable than reactants (Keq > 1)
ΔG° = 0: products and reactants energies are equal; Keq = 1; reaction at equilibrium
+ΔG°: products less stable than reactants; Keq < 1; reverse reaction is favored

Question 8

transition state

Answer 8

highest energy point between steps on an energy diagram

Question 9

rate determining step

Answer 9

highest “hill” on rxn coordinate; height determines the reaction speed –> higher hill, slower reaction

Question 10

pKa and reaction stability

Answer 10

Higher pKa –> weaker acid –> more stable bond –> lower energy –> harder to break

Question 11

thermodynamic control

Answer 11

• if all steps are reversible; relative stability dictates the outcome
• thermodynamic product: more stable/lower energy product favored and formed when system reaches equilibrium

Question 12

kinetic control

Answer 12

• irreversible reactions: when the reaction doesn’t “know” which path leads to the most stable product
• kinetic product: formed faster because it requires less energy, even if less stable (higher energy)

Question 13

what do you consider for stability of a molecule

Answer 13

• aromaticity
• resonance contributors
• delocalization

Question 14

aromaticity

Answer 14

• aromatic compounds more stable and lower energy
• criteria for aromaticity: must be closed ring structure (cyclic), every atom in the ring must have a p-orbital (no sp3 atoms), molecule must be planar/flat, number of delocalizable pi electrons in 4n + 2 equation is even and NOT multiple of 4

Question 15

antiaromaticity

Answer 15

• high energy: antiaromatic compounds are high energy and very unstable
• criteria: satisfies all the criteria for aromaticity, except the number of deocalizable pi electrons in 4n + 2 equation is a multiple of 4
• slow and difficult reaction due to high energy and instability

Question 16

nonaromaticity

Answer 16

• energy level: between aromatic and antiaromatic compounds; typically more energetically favorable and stable
• nonplanar
• faster and easier reactions because more energetically favorable

Question 17

disruption of aromaticity

Answer 17

makes the reaction thermodynamically unfavorable

Question 18

inductive effect

Answer 18

EN atoms without resonance involvement destabilize carbocations due to the inductive effect

Question 19

weak acids

Answer 19

pKa 5-18 need a catalyst for EAR

Question 20

EAR steps

Answer 20

1. any acid base chemistry
2. protonate alkene with strongest acid around
3. form carbocation (rate of protonation faster is carbocation intermediate is more stable
4. carbocation capture: CC is quickly captured by the original weak acid (acting as a Lewis base) ,forming a cationic intermediate
5. deprotonation: cationic intermediate is deprotonated, yielding final product

Question 21

major/minor products for EAR

Answer 21

• more substituted carbocation (more carbons attached) is more stable
• resonance stabilization (resonance trumps substitution in determining stability)

Question 22

carbocation stability

Answer 22

• resonance stability and delocalization trumps degree of substitution
• degree of substitution: 1 degree (bonded to one other carbon), and so on
• hyperconjugation: good consideration when only degree of sub of a CC is the difference being compared; more sp3 alkyl groups = more electrons available for hyperconjugation = more stable carbocation; bonding orbital donates electron density to empty p orbital of the carbocation, stabilizing it

Question 23

regioisomers

Answer 23

structural isomers that differ in position of their functional groups
- major regioisomer: protonation occurs on less substituted end of the double bond

Question 24

markovnikov’s rule

Answer 24

• predicts regioselectivity
• the H atom from the reagent adds to the end of the C=C bond with MORE H ATOMS
• regioisomer formation: two possible regioisomers and choose major/minor product

Question 25

HNMR integration

Answer 25

size/height of the signal
- each signal height represents the number of H atoms in each equivalent group
- taller signals –> more equivalent Hs

Question 26

molecular symmetry

Answer 26

super symmetrical molecules = fewer NMR signals

Question 27

chemical shift

Answer 27

position of signals on horizontal axis
- Left (downfield, high δ) = High chemical shift; Closer to an electronegative (EN) atom; Higher deshielding (less electron density around the atom)
- Right (upfield, low δ) = Low chemical shift; Further from an EN atom, more shielded