cap. 4

Question 1

what happens to the rate of E2 reaction if the concentration of the base is doubled?

Answer 1

the rate is doubled

Question 2

what kind of base will best generate a hoffman product?

Answer 2

strong base that is sterically large. it will deprononate the less hindered carbon atom and form a less substituted double bond

Question 3

hoffman product

Answer 3

the less substituted alkene forms the major product in an elimination reaction

Question 4

zaitzev product

Answer 4

the more substituted alkene forms the major product in an elimination reaction

Question 5

what happens to the rate of an sn2 reaction if the concentration of the nucleophile is doubled?

Answer 5

the rate is doubled

Question 6

rate law for E1 reaction

Answer 6

rate= k[substrate]

Question 7

what happens to the rate of an sn1 reaction if the concentration of the nucleophile is cut in half?

Answer 7

the rate is unchanged

Question 8

characteristics of bonding in alkenes

Answer 8

• the double bond acts as a nucleophile and attacks electrophilic species
• the reactions are electrophilic additions

Question 9

steps in electrophilic addition

Answer 9

1. Pi electrons attack electrophile
2. nucleophile attacks the carbocation

Question 10

markovnikov’s rule

Answer 10

1. the acidic proton will add to the carbon that already holds the greatest number of hydrogens
   ** in electrophilic addition, the electrophile will add itself in a way that generates the most stable intermediate (i.e to the most substituted carbon)

Question 11

anti markovnikov product

Answer 11

the H proton will add to the more substituted carbon, and the electrophile will add to the least substituted carbon (opposite of regular rule)

Question 12

free radical addition of HBr

Answer 12

• in the presence of peroxides, HBr will add to form the anti markovnikov product
• peroxides produce free radicals
• not seen with HCl or HI (too endothermic)

Question 13

anti-markovnikov stereochemistry

Answer 13

the intermediate tertiary radical forms faster because it is more stable

Question 14

hydration of alkenes

Answer 14

• addition of water to the double bond will form an alcohol
  -markovnikov product
• uses dilute solutions of H2SO4 or H3PO4 to drive equilibrium toward hydration

Question 15

alkyl halide

Answer 15

halogen is directly bonded to sp3 carbon

Question 16

vinyl halide

Answer 16

halogen is bonded to sp2 carbon of an alkene

Question 17

aryl halide

Answer 17

halogen is bonded to sp2 carbon on benzene ring

Question 18

stereochemistry of anti-markovnikov

Answer 18

the intermediate tertiary radical forms faster because it is more stable

Question 19

physical properties of alkynes

Answer 19

• nonpolar
• soluble in most organic solvents
• bp similar to alkenes of same size
• less dense than water
• up to 4 carbons; gas at room temperature

Question 20

triple bonds are (shorter/longer) than double & single bonds

Answer 20

shorter. they have 2 pi overlapping orbitals

Question 21

terminal alkynes are (more/less) acidic than other hydrocarbons

Answer 21

more acidic. they have higher s character

Question 22

how can terminal alkynes be deprotonated?

Answer 22

strong bases can deprotonate (-NH2).
- hydroxide (HO- and alkoxide (RO-) are not strong enough to deprotonate the alkyne

Question 23

acetylide is a (strong/weak) nucleophile

Answer 23

strong. it can do addition and substitution reactions

Question 24

lindlar’s catalyst

Answer 24

produces alkenes with cis stereochemistry

Question 25

characteristics of E1 reaction

Answer 25

- 2 groups are lost: a hydrogen and a halide - nucleophile acts as a base and creates a carbocation intermediate - a mix of E1 and SN1 products will be created

Question 26

rate law for E1

Answer 26

k[RX]

Question 27

what is the first step for both E1 and SN1 reactions?

Answer 27

formation of carbocations (it is also the rate-determining step

Question 28

what orientation do alcohols form in osmium tetroxide dihydroxylation?

Answer 28

cis

Question 29

what is the product of osmium tetroxide dihydroxylation?

Answer 29

formation of 2 alcohols that have cis orientation

Question 30

what product forms from ozonolysis (o3 and (CH3)2S)?

Answer 30

cleavage of an alkene that forms 2 aldehydes

Question 31

stereochemistry of E2

Answer 31

leaving group must be anti to the hydrogen removed

Question 32

syn addition

Answer 32

molecules will add to the same side (both wedged or both dashed)

Question 33

anti addition

Answer 33

molecules will add to opposite faces (one wedged one dashed)