Rearrangements

Question 1

Describe the Beckmann rearrangement

Answer 1

N-O and C-C bonds antiperiplanar,

OH protonation for LG, C-C antiperiplanar to N attacks it,

H2O attacks carbocation in ring to add OH, OH converted to ketone, N converted to NH,

Ketone in place of =N (swap)

Question 2

Explain stereoselectivity in Beckmann rearrangment

Answer 2

Carbocation can be on either side of the N so the ketone can add on either side to form isomers

Question 3

To convert OH into a good leaving group?

Answer 3

Protonation or using TsCl to convert to TsO to leave

Question 4

Reagents for Beckmann?

Answer 4

H+ (acid) and water,

or NH2OH.HCl with MeOH and reflux

Question 5

Describe the Baeyer-Villiger rearrangement

Answer 5

Lone pairs on OH attack its own bond

Bond joining the RL to the centre where the O is attached, attacks the next O

Next bond in chain of attached O, attacks the next O R group

Question 6

What is RL?

Answer 6

Larger R group (more electron rich)

Question 7

Best to worst migrating groups in terms of carbocation?

Answer 7

tertiary = best
secondary, phenyl, benzyl, primary

Me = worst

Question 8

Best to worst migrating groups in terms of R group?

Answer 8

Best = OMe,
Me, H, Cl,

NO2 = worst

Question 9

Reagents for Curtius rearrangement?

Answer 9

DPPA and Et3N,

SOCl2 or (COCl)2 and NaN3

ClCO2Me and Et3N then NaN3

Question 10

Describe step 1 of Curtius and Schmidt rearrangement

Answer 10

from carboxylic acid, formation of acyl azide then isocyanate:

N-N antiperiplanar to R group, nearest N attacks bond inbetween it and ketone, e.d. shifts to O

Resonance: R group bond attacks nearest N, e.d. shifts to N triple bond, charge on O shifts back into bond

Loss of N2 forms R- N=C=O

Question 11

Describe Curtius rearrangement from isocyanate

Answer 11

Nu attack: H from Nu added to the N to form NH,
NH swaps with the C=O,
then rest of R group (minus 1H) is added to the end of the ketone

Question 12

DPPA?

Answer 12

P with 2PhO R groups, =O on P, N-NtripleN on P

Question 13

formation of isocyanate for Hoffmann rearrangement

Answer 13

starting with amide, base (NaOH) deprotonates NH2 to neg NH

Br adds onto NH using Br2

lone pair on N attacks bond inbetween it and C=O, R group bond attacks neg N, Br leaves

Question 14

Conditions for Hoffmann rearrangement

Answer 14

Base (NaOH) with Br2

NBS with base (DBU)

NaOCl with base (NaOH)

PhI(OAc)2 with base (pyridine)

Question 15

DMAP?

Answer 15

pyridine ring with NMe2 coming out of the ring (C-N-Me2) opposite position to the pyridine N

Question 16

THF?

Answer 16

5 ring with O in top position

Question 17

DMF?

Answer 17

aldehyde with NMe2 attached

Question 18

toluene?

Answer 18

benzene ring with Me coming out