Reactions of pi bonds

Question 1

FOR THIS COURSE, double bonds will act as…

Answer 1

Nucleophiles – reacts by electrophilic addition to them
ie. electron-rich double bonds only

note that double bonds CAN be electrophilic / electron poor due to adjacent C=O withdrawing e- density

Question 2

main types of electrophilic addition reactions considered in this course

Answer 2

addition of H-X
bromination / iodination
Epoxidation
Hydroboration

Question 3

reagents for hydrogenation

Answer 3

H2
Pd/C (palladium on inert charcoal support)

note: because H’s are added from the same face (on the metal catalyst) the cis-product often dominates

Question 4

how to determine major product for addition of H-X to asymmetrical alkene

Answer 4

addition product has most stable carbocation intermediate (ie. rich get richer, and aromatic ring will stabilise significantly)

note: mech should be draw with arrow going from double bond THROUGH the C where the bond will be forming to the electrophile H

Question 5

product of bromination of alkene

Answer 5

a 1,2, dibrominated alkane product

Question 6

describe the two-step mechanism for bromination/iodination and the HOMO-LUMO interactions for each

Answer 6

1. creating bromonium or iodinium ion
   HOMO: C=C 𝜋-bond, LUMO: empty X-X σ*
   pi-bond attacks one X, lone pair on that X attacks at C –> bromonium bridge
2. remaining X- ion attacks at C in Sn2 mech, forming 1,2-dihalogenated product
   HOMO: e- in Nucleophilic halide anion
   LUMO: C-X σ*

Question 7

stereochemical outcome of bromination on a ring system

Answer 7

gives rise to trans product due to back face attack at bromonium bridge

Question 8

reaction conditions for halohydrin formation

Answer 8

bromination IN WATER

(vs bromination in an inert solvent will result in dibrominated product)

Question 9

mechanism for halohydrin formation

Answer 9

same first step to form bromonium bridge
water acts as the Nu and attacks the bromonium ion instead in the second step

H+ removed by another H2O molecule

Question 10

Two ways to form epoxides

Answer 10

1. by reacting a trans-halohydrin with a base
2. epoxidation: from alkene and the peroxy-acid m-CPBA

Question 11

steps for formation of epoxide from trans-halohydrin

Answer 11

1. alcohol is deprotonated by addition of base (eg. NaH)
2. Sn2 of attached O- onto C, expelling X- as a leaving group (works very well in trans configuration)

Question 12

functional group of peroxy-acids

Answer 12

acidic -OOH group
as in like a carboxylic acid group w extra O attached between C and -OH

Question 13

epoxidation mechanism
and HUMO-LUMO interaction

Answer 13

HOMO: e-rich C=C 𝜋-bond
LUMO: O-O σ* bond (parallels the X-X σ* for bromination)

LOTS of arrows for the mech so practice drawing it!!
but the first arrow goes from the 𝜋-bond through one C to the O closest to the H

Question 14

stereochemical outcome of epoxidation

Answer 14

depends on the configuration of the starting alkene

both new C-O bonds are formed on the same face of the alkene’s pi-bond, so cis alkene –> cis epoxide and trans–>trans

Question 15

main purpose of hydroboration

Answer 15

a process used to achieve opposite regioselectivity for addition to an alkene substrate

(ie. opposite to typical ‘Markovnikov selectivity’)

Question 16

reagents for hydroboration reaction

Answer 16

1. BH3, borane
2. H2O2, NaOH, H2O

Question 17

three main steps in hydroboration reaction mechanism

Answer 17

1. addition of BH3 across double bond, with the larger B group lining up with the LEAST hindered C=C end
   occurs three times –> trialkylated boron
2. hydroperoxide anion HO-O - adds into empty p-orbital on boron to form unstable intermediate
   weak O-O breaks, OH- leaves, and an alkyl group forms C-O bond
3. -OH attacks into B’s empty p-orbital, assisting the PRODUCT to leave

Question 18

why is BH3 added with opposite regioselectivity to the alkene in the first step of hydroboration?

Answer 18

1. electronics: pi-bond e- are being donated in to B’s empty p-orbital, which generates a partial positive charge. this should go on the C that is best able to STABILISE this positive charge
2. BH2 is larger, ends up where there is less steric hindrance

Question 19

stereochemical outcome of hydroboration

Answer 19

the SYN addition of H and OH to the double bond, as it must be delivered from the same face

Question 20

how do groups on a ring with a double bond affect stereochemical outcome of electrophilic attack?

Answer 20

if there is a group on the C adjacent to double bond, it may hinder the approach of the incoming electrophile

hence the less hindered face is favoured, and results in the major product