Alc Ether Phenol

a

the carbocation is much more stable on the double Ph. Thus the OH from there gets removed and then we get ring expansion.

a

a

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

a

Iodoform doesnt even occur

Ch3- released takes away the H+ from the acid, thats it

A

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

b

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

b

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

b

a

c

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

c

tscl attacks the 2 degree alc as its more reactive and then OTs leaves and Oh from 1 degree alc forms ether

b

c

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

c

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

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)

c

b

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

ans is b

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

b

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

ans is b

och3, =Ch2, ch2-oh get attacked

B as acetals are unstable in acidic medium