sir for question number one initially formed cation since it is tertiary that to it contsins cyclo propyl and propargyl group it should be more stable so that the product should be option number one
Sir in the axial conformer can't it be ome coz they asked where b will be more stable. oMe will not go to equatorial due to rabbit ear effect so it is more unstable in equatorial than F.I guess
1H nmr signals in the case of bridged-hexadiene molecule would be 4 since it is symmetrical and achiral (so CH2 protons are equivalent). ..Too much disappointed with d score, did many blunder mistakes in Organic section. When is d next call for CSIR-NET?
Why option 3 won't be the answers for the dithian question?? And last qu will lda abstract proton from the side adjacent to the isopropyl group ..I guess that side is more sterically hindered??
Sir on unlock chemistry channel they have found a ref which shows signals are 4 for h1 nmr if u have any ref for 5 signals than do show ...plz give reply
@@geetanjalinegi1482 jo bhai bol rha option 4 uske paas abhi tk koi explanation nahi hai It may be option 4 will be right ans But when I solved it option 2nd comes And as we know after borton their is always a possibility of backmann rearrangement csir asked this type of question in June 18 or dec 17
@@ankush2511 yes sir but in presence of ACCl cyanide type of product will form but in presence of Acid backmann type rearrangement amide product will form according to the concept sir please check it's correct or not
But sir starting formed carbocation which is stable then Br^- attack occer here
The question whose product Cyclooctyne is from nantz
No. Of 1Hnmr should be 4. They homotopic bcz both the hydrogens are interchangeable by C2 axis
Yes, I also think so.....
yess
Yes
sir for question number one initially formed cation since it is tertiary that to it contsins cyclo propyl and propargyl group it should be more stable so that the product should be option number one
Video was not clear
Sir pericyclic rxn trans product, i think.....Y Bcz 4n rule thermally allowed con rotate is stable.... So streochemistry retains in the product...
Bro it is from clayden solution manual go and check
It is a 4n+2 system
Bromine wala glt hai bhai....cyclohexane ring bna k kr option c aae ga.....
Sir, there is one paper or two papers we are going to write in exam
Sir in the axial conformer can't it be ome coz they asked where b will be more stable. oMe will not go to equatorial due to rabbit ear effect so it is more unstable in equatorial than F.I guess
1H nmr signals in the case of bridged-hexadiene molecule would be 4 since it is symmetrical and achiral (so CH2 protons are equivalent). ..Too much disappointed with d score, did many blunder mistakes in Organic section. When is d next call for CSIR-NET?
Hello sir
If it possible pl. make a video on csir net dec. 2019
Sir, for an electrocyclic reaction, isn`t it a 4n+2 system dis-rotation
Sir bromine wala wrong hai...option c will be correct ans
Sir if possible provide paper 1 questions also.
Sir in elctrocyclic reaction option 4 is correct in official answer key.....how sir....
Why option 3 won't be the answers for the dithian question??
And last qu will lda abstract proton from the side adjacent to the isopropyl group ..I guess that side is more sterically hindered??
Sir on unlock chemistry channel they have found a ref which shows signals are 4 for h1 nmr if u have any ref for 5 signals than do show ...plz give reply
Plz explain part B 41question
Paper A
Sir please make a video on part c (physical and inorganic answers)....
sir, I hv 80 marks in general.can I expect anything for is ?
No dear :(
Thank you sir for sharing
Sir mre 83 marks h in obc catagrey I am.expect anithing or not
Yeah for net
Safe ho tum yr
83 to accurate bn rahe h or five to six questions have confusion
May be jrf ..this time
Thakyou so much sir!
Que no. 100 from booklet C....sir
For last question option will be 2nd
After photo chemistry
Backmann rearrangement will occure
Beckmann fragmentation
Barton rearrangement ya fragmentation?
@@geetanjalinegi1482 jo bhai bol rha option 4 uske paas abhi tk koi explanation nahi hai
It may be option 4 will be right ans
But when I solved it option 2nd comes
And as we know after borton their is always a possibility of backmann rearrangement csir asked this type of question in June 18 or dec 17
@Rhl Ydv i also do that but R u sure 100% with reference if yes then just give the reference..
@@ankush2511 yes sir but in presence of ACCl cyanide type of product will form but in presence of Acid backmann type rearrangement amide product will form according to the concept sir please check it's correct or not