2/3b new Flashcards
a more stable cc+ can be due to what
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!!
does going from a tertiary cc+ to a secondary cc+ always mean the cc+ is less stable
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.
if we migrate a group,, where will the cc+ be
the migrated groups original place!!!
whats a pinacol
a molecule with 2 OH on adjacent carbonds
whats a pinacolone
a ketoneeee
with an added R group to it,, one that was migrated? usually a ketone with a CH3
whats the purpose of a pinacol rearrangement
formation of a pinacolone,, aka formation of the carbonyl.
steps in a pinacol reaction + how to identify one
- 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
in the pinacol reaction,, what orbital is the lumo // what orbital is th eempty one
the cc+ p
the one we are putting e- into
in the pinacol reaction, what orbital is the homo,, // filled one
the c-c sigma bond
epoxide with MgBr2
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 O is bonded to will attack the cc+. (aka the c bonded to the O 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.
whats MgBr2
a lewis acid
accepts e-
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
other lewis acids
cyclohexane epoxide + H+ + RLi
cyclohexane with R wedged on one C and OH dashed on the other C
epoxide + H+ + RMgBr
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
aldehyde formed on the ring.
o - Mgbr
c bonded to o will attack cc+ formed,, then the lone pair on o will stabilise the new cc+
when we have an asymmetric pinacol rearangement aka 2 OH’s on adjacent C’s,, what do we look for
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.
pinacol reaction usually occurs in which conditions + how do we know this
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+
a semipinacol reaction occurs in what reaction conditions + why does this differ from a normal pinacol reaction
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
describe a semipinacol reaction
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!!
what thing is used to make conditions basic in a semipianacol reaction
CaCO3
TsCl
in a semipinacol reaction,, does the lg have to be OTs formed from OH and TsCl originally
nope!! u just need a good LG on the C adjacent to the one with the OH.
OH on C,, adjacent C has an OH,, acidic conditions
pinacol reaction.
protonate most stable OH and form CC+
migrate a group
form a carbonyl
OH on C,, adjacent C has a OH.
TsCl + pyridine//base + CaCO3
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
OH on C, C adjacent has a good LG : I iodine etc
semipinacol reaction.
loss of good LG
formation of CC+
migration
carbonyl formation
in the pinacol and semipinacol reaction whats the homo and what the lumo
homo = C C sigma bond (the one that migrates
lumo = C - LG antibonding orbital
amine NH2 + NaNO2 + HCl
forms diazonium salts!!
N(+) triple bond N
this is a very good LG
how to make a diazonium salt LG
NH2
+ NaNO2 + HCl
forms N(+) triple N
if NH2 forms a N(+)tripleN what can happen
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.
Tiffeneau–Demjanov rearrangements is when u have what fg
diazonium salt FG with an OH on the adjacent C.
lose NtripleN
CC+
migration
carbonyl formation
NaNO2 + HCl // H+ reaction to form what we need (NO+) to then form a diazonium salt
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+
NO+ and an amine
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!!!
whats good about the diazonium salt fg
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.
pinacol rearangement brief
loss of most OH that leads to most stable cc+
migration of a group bonded to the C the other OH id bonded to
carbonyl forms
semipinacol reaction breif
less stable OH needs to be removed
react it with TsCl and a base
forms an Ots on most stable OH
this is removed and the r group from the other oh migrates,
thrn the other oh forms a carbonyl
