a
the carbocation is much more stable on the double Ph. Thus the OH from there gets removed and then we get ring expansion.
the double bond gets reduced as then the carbocation formed can be resonated with oxygen.
then ring expansion
b
first the OH gets oxidised to ketone
the heating leads to decarboxylation
in acidic medium, ester forms a more stable carbocation while opening. thus the ring opens such that +ve charge is in reso with db from ketone.
The lp from O stabilises the positive charge
then regular ring expansion
ans A
d
Lialh4 doesnt attack Och3
A
Resorcinol forms a resonation structure with OH such rhat we break beneze and form diketo
Iodoform attacks and converts a keto to carboxylic
Iodoform doesnt even occur
Ch3- released takes away the H+ from the acid, thats it
H+ from HBr attacks and removed OH from alc. then br- attacks
thus reactivity is proportional to electron density on alcholic O OR stabilityu of carbocation
Ch2OH more reactive due to I effect of CH3
Its just NGP
Same reasons why SoCl2 is preffered. products are gasses so you dont have to deal with them.
Plus with HBr, elimination may also occur
B
undergoes ANTI Elimination
hence a
RR
SS
RS=SR
Hence 3
The OH- attacks then the S-C bond breaks
similar happens in the other ester too
c
Grid converts to alc. Then H+ convcerts alc to alkene. Then KMNO4 forms diol and CroO3 oxidises the 2 degree alc only.
h2 liberation means acid/alcohol
Br2 no action means no reactive db or triple bond
cro3/H+ is chromic acid
Ans d
Plus it can't be C as it doesn't have any benzylic H on the upar waali side to convert it to benzylic acid with KMNO4
the other two do
NaBh4 doesnt fw acid
trans more stable
hence A
II and IV
II as it forms geminal dicarboxylic which means too much ewgs and thus easy to decarboxylate
IV as it forms Beta Keto acid
tscl attacks the 2 degree alc as its more reactive and then OTs leaves and Oh from 1 degree alc forms ether
end is just hoffman bromamide
the reagent selectivelt reduces db
A. b
B. b
Li+ attacks and converts acid to cb
grig converts acid to ester
ester + H+ forms acid again
Oh turns to OAc
Write ketone in its rs with +ve charge on carbon.
this gets attacked by alkynyl ion
Ans C
PCC converts alc to ald and ketone
glycol attacks aldehyde as ald more reactive than keto. thus now ald is protected
grig attacks keto to make 3 degree alc and H+ removes BOTH mgbr and glycol
NaBH4 then reduces ald to alcohol again
Hence ans b
in conc H2so4, the acid undergoes decarboxylation and forms Ph3C+ ion which trhen gets attacked by methanol
hence ans is c
first Ch3Li is just acid base
second one releases Ch3- ions which attack the acid
h3O removes the Li and the glycol
ans is c
mild oxidising agent. Converts 1 degree alc to ald
ans is b
NABH4 does no reduce nitriles
ans is A
the nucleophilicity of OR- is highest for primary alc as they suffer the least steric hinderance
Nucleophilicity is inversely proprotional to steric hinderance
3 degree akc and thus cant be oxidised
as conditions are cool, elimination cant occur either
as a result, no rxn occurs (d)
HIO4 cant break esters
Hence only the 3 OH get separated
b.
This is Phosgene
It decomposes readily to from CO and Cl2.
na2cr2o7 is a vv strong oxidising agent.
hence it converts primary alc to acid
thus we need a 2 degree alc to get ketone
thus ans is b
the positive charge resonates with o to form a db with O and thus o gets a positive charge
this breaks the left side C-O bond which forms anbother carbocation which now gets attacked by H20 to form alc
thus ans is c
Alcl3 attacks on O and then LAH attacks on C and gets rid of the O-Alcl3-
thus we are left with c
D
Chiral centre pe to kuch hua hi nahi D:
beta keto ester
hence it gets removed on heating
the acid is S
alcohol may be R or S
thus result may be SR or SS
Thus b
as both are beta keto esters, both esters get removed
ans is c which is the most stable enolic form of quinoline
just a test for primary amines
hence ans is a as it reduces amide to primary amine
in ii and iii, the carbonyl carbon gets electron from the N and O
N is a better electron donor than O and hence ii will be less reactive than iii
ans is d
och3, =Ch2, ch2-oh get attacked
which will cleave first? A or B
B as acetals are unstable in acidic medium
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