2/3b new

Question 1

a more stable cc+ can be due to what

Answer 1

more substituted: tertiary cc+ rather than a primary or a methyl cc+.

when there is less ring strain!! from a 4 membered ring to a 5 membered ring!!

Question 2

does going from a tertiary cc+ to a secondary cc+ always mean the cc+ is less stable

Answer 2

nope!!

if its incuded // near a ring,, and the ring is going from being 4 membered with lots of strain to 5 membered,, its more stable to expand the ring and reduce the cc+ substitution.

Question 3

if we migrate a group,, where will the cc+ be

Answer 3

the migrated groups original place!!!

Question 4

whats a pinacol

Answer 4

a molecule with 2 OH on adjacent carbonds

Question 5

whats a pinacolone

Answer 5

a ketoneeee

with an added R group to it,, one that was migrated? usually a ketone with a CH3

Question 6

whats the purpose of a pinacol rearrangement

Answer 6

formation of a pinacolone,, aka formation of the carbonyl.

Question 7

steps in a pinacol reaction + how to identify one

Answer 7

• if theres 2 OH’s on adjacent carbons in the reactant
• and a ketone in the product.
• protonate the OH and remove it to form a CC+
• migrate a group to the cc+ (normally alpha to the C the other OH is bonded to)
• use the lone pair on OH to form a double bond for the carbonyl!!

this is dont bc O is very good at stabilising + charge (only when theyre close together tho!!) the cc+ pulls the migration group towards it,, and the OH group pushes the migrating group away from itself

Question 8

in the pinacol reaction,, what orbital is the lumo // what orbital is th eempty one

Answer 8

the cc+ p

the one we are putting e- into

Question 9

in the pinacol reaction, what orbital is the homo,, // filled one

Answer 9

the c-c sigma bond

Question 10

epoxide with MgBr2

Answer 10

the o attacks the MgBr with its lone pair and the MgBr bonds to the epoxide making the o+.

one of the epoxide bonds break and go to the o+ making the o neutral and the c a cc+.

a c,, adjacent to the c the OH is bonded to will attack the cc+. (aka the c bonded to the OH is bonded to other stuff,, its one of these bonds that wil be the homo and attack the cc+). the o lone pairs will then attack the newly formed cc+ on the new carbon forming a carbonyl.

Question 11

whats MgBr2

Answer 11

a lewis acid

accepts e-

Question 12

if epoxides can be opened with MgBr2 to give a reaction similar to a pinacol reaction (o+ formed,,o+ neutralised,, cc+ formed, c migration, cc+ formation,, carbonyl formation),, what else can be used

Answer 12

other lewis acids

Question 13

cyclohexane epoxide + H+ + RLi

Answer 13

cyclohexane with R wedged on one C and OH dashed on the other C

Question 14

epoxide + H+ + RMgBr

Answer 14

cyclopentane with an OH and R on a branched C

cyclopentane with Me on the side,, but instead of 3H’s its an R and an OH group instead

Question 15

when we have an asymmetric pinacol rearangement aka 2 OH’s on adjacent C’s,, what do we look for

Answer 15

if we remove each OH,, which one will give the most stable C++

if theres phenyl groups near,, the + can be delocalised into it so its favoured lowkey.

also look at the products.

• stability of CC+
• product stability - ring strain etc.

Question 16

pinacol reaction usually occurs in which conditions + how do we know this

Answer 16

acidic conditions.
bc the OH that leaves to give the most stable CC+ is first protonated by attacking a H+.

to form H2O+

which is then removed to give a CC+

Question 17

a semipinacol reaction occurs in what reaction conditions + why does this differ from a normal pinacol reaction

Answer 17

usually in neutral contitions or BASIC conditions.

this is why the most stable CC+ cannot be formed,, bc that OH can not be protonated to give H2O.

a different OH will therefore be removed,, giving a different product

Question 18

describe a semipinacol reaction

Answer 18

pinacol - 2 OH’s on adjacent C’s.

one will need to be removed , a group will migrate and a carbonyl will be made

in a semipinacol we use tosylate - TsCl - which bonds to the most avalable OH,, aka the least sterically hindered OH. to form a superrrr good LG. bc its OTS - super good.

this is then removed to give a CC+ (not as stable but its okay) then a group will migrate,, then the other OH will form a carbonyl!!

Question 19

what thing is used to make conditions basic in a semipianacol reaction

Answer 19

CaCO3

TsCl

Question 20

in a semipinacol reaction,, does the lg have to be OTs formed from OH and TsCl originally

Answer 20

nope!! u just need a good LG on the C adjacent to the one with the OH.

Question 21

OH on C,, adjacent C has an OH,, acidic conditions

Answer 21

pinacol reaction.

protonate most stable OH and form CC+

migrate a group

form a carbonyl

Question 22

OH on C,, adjacent C has a OH.

TsCl + pyridine//base + CaCO3

Answer 22

semipinacol reaction.

least hindered // most available OH is turned into OTs,, this is a very good lg.

this leaves and a CC+ is formed

migration occurs

carbonyl is formed

Question 23

OH on C, C adjacent has a good LG : I iodine etc

Answer 23

semipinacol reaction.

loss of good LG
formation of CC+
migration
carbonyl formation

Question 24

in the pinacol and semipinacol reaction whats the homo and what the lumo

Answer 24

homo = C C sigma bond (the one that migrates

lumo = C - LG antibonding orbital

Question 25

amine NH2 + NaNO2 + HCl

Answer 25

forms diazonium salts!!

N(+) triple bond N

this is a very good LG

Question 26

how to make a diazonium salt LG

Answer 26

NH2

+ NaNO2 + HCl

forms N(+) triple N

Question 27

if NH2 forms a N(+)tripleN what can happen

Answer 27

NtripleN is a very good LG.
if theres an OH on the adjacent C.

a pinacol reaction can occur

loss of LG,, migration, carbonyl formation.

Question 28

Tiffeneau–Demjanov rearrangements is when u have what fg

Answer 28

diazonium salt FG with an OH on the adjacent C.

lose NtripleN
CC+
migration
carbonyl formation

Question 29

NaNO2 + HCl // H+ reaction to form what we need (NO+) to then form a diazonium salt

Answer 29

NO2: -O-N=O attacks the H+ to give HO-N=O,, this attacks another H+ to give H2O-N=O,, which the lone pair on O then uses to kick off the H2O,, giving N triple O+

Question 30

NO+ and an amine

Answer 30

amine attacks the NtripleO+
attacks the N,, triple bond moves to O to neutralise it.

amine is now N+ so H is given to form adouble bond between the N’s and the double bond between NO attacks a H to give OH.

OH is protonated again to form H2O+

N is positive still so H is gived to form a triple bodn betweenthe N’s and H2O+ is kicked off.

u now have N triple N!!!

Question 31

whats good about the diazonium salt fg

Answer 31

its a good lgggg

so obvs ur gona lose it,, form a CC+,, migration and then formation of a carbonyl to give u ur product xox.