Sir i am in short of words to praise u ....for d frst tm i undrstd it just only bcz of u ....plz sir help us when u get time ,a big thanks to u sir stay blessed n plz help us more
You taught the things in 14 mins to which my well educated professor was unable to do. If got a chance to get a guidance from you. I pay 10times more than my college. Thanks, you're a life saviour.❤
The problem came in Dec 2017, first one is the correct answer by simply counting the number of signals. Your teaching skill is very good and easy to understand. Thank you sir.
At 22 mins, you can't ignore the option b base on the chemical shift value and the no.of Hs giving singlet. In option B, the Hs in the methyl group will also give a singlet. Isn't it?
Yes you're right, I missed that one, But singlet at 3.59 is due to OMe group, a methyl attached to CO group will give peak around 2.8 due to electronegativity of Oxygen atom.
In last example the protons ortho to methoxy group appear below seven due to electron donating group hence more shielding, another aspect to reject option 2 is IR value as when electron donating group is present the value carbonyl shift below 1700 due decrease in double bond character of carbonyl. overall very nice explanation thank you very much
In last question in between two most probable options we can distinguish them by IR value given in question because for ester it will be near about 1750.
sir Very much useful video. super trick n tips we learn from you. but in December 2016 there will be a OCH3 group in option c. kindly check out n explain in detail. bcz we learn more from you
The answer would be option 1... because data is given for 3 different protons only option one have 3 different protons . .....other options have. 4 and 5 protons
You're welcome 😊 If you want more detailed explanation on problem solving based on NMR Spectroscopy. Then plz watch my videos on Unacademy too. unacademy.com/lesson/course-overview/TNULQWVZ And follow me there for more helpful videos 😊
thank you and all the best for Exam :) and you can remove that "if", and be confident enough that you are going to get CSIR JRF this time and GATE too :)
Sir, your explanation is good. But in the 4-acetyl anisole, the assignment is not correct. When methoxy group is present, the ortho to OMe protons always shielded than the meta to OMe protons.... please revise this.. Thank you
Sir if we conclude that 1st one is correct from the IR data because the IR value is 1690 nearer to the ketone(1710)..whereas the IR Vale is far apart from ester (1740).is it correct way to think?? Plz rply..
sir, i watched your ''stereo of bi phenyl video'' in that video you told that is org. chemistry of csir exam is not so easy to attempt although many expert also fails to predict right answer...sir, you are very right so is it impossible to do org chemistry acurately.....i am also survrving....plz guide
Basic Chemistry, it is not impossible, many people who are good in Organic Chemistry are doing great. But I feel it's difficult to attempt. And if you are also struggling with organic then don't waste much time on it... Try to give that time to the topic which you can solve in exam. It could be anything like physical chemistry or Inorganic Chemistry
sir, honestly speaking i attempt both paper of 2017 and try to self examine that how much i can understand the concept of given que. i think its 50- 50% chances in present ....so do you think is there any hope to do well?
Sir Ur lectures are really helping a lot..plz share Ur videos on coordination chemistry too including stereoisomers prediction and inner as well as outer sphere reactions concept along with expected questions
Very good video sir please make video on magnetic and chemical equivalnce and theory of nmr spectroscopy means how to remember chemical shift values and how to use
Sir csir net Dec 2016 question is not correct. The option you have written is wrong. The c option written by you is wrong.. plz correct that and then plz upload it again.
they both will show peaks at different positions. the proton of OH group in ohenol will show peak at 4.5 where as that in benzoic acid at 12.8 due to resonance and all other factors. I'll suggest you to read Organic Spectroscopy by Pavia, there you will find detailed explanantion.
I can't make a video on that bt I'll try to help you out here only... See firstly from the IR data 3400 is for OH group and 1680 could be for aromatic ketone, as the resonance effect of phenyl ring will weaken the bond strength on C=O and its IR will decrease from usual 1715 to 1680. So it is confirmed that it contains OH and ketone group. That means option 3 and 4 could be probable answers. Now see the ¹H signals having J value same, that means those two protons are adjacent to each other one shows peak at 7.8 other at 7.0 which is greater and lower than usual phenylic proton at 7.1. Now if you'll see option 3, we have two adjacent protons to OH and OMe both of the groups enhance electron density at their ortho positions respectively. But in case of option 4 we have two protons adjacent to OH and CO group and we know OH enhance electron density at its ortho position while CO decreases electron density at its ortho, so option 4 becomes correct choice.
Thanks A lot sir n sir plse make video on organometallic unique rxn ( oxidative addition ; reduction etc Plse sir my humble request) before net exm beocz whole organometallic rxn based on these rxns
Praveen, if you'll watch the video carefully, I've mentioned that the splitting is due to protons present on adjacent carbon butnot the signal. lets take this, in chloro ethane we have two types of prtons one is CH3 and other is CH2, now CH2 will show chemical shift at much higher ppm value than the CH3 as it is directly attached to Cl (around 3.8). Now lets talk about splitting, so CH2 will give quartet due to CH3 which is adjacent to it, but in order to write the its chemical shift value we will write "3.8 (q, 2H). now here q denotes that it gives quartet due to adjacent CH3 and 2H denotes that the 3.8 is given by carbon that has 2 protons, I hope I made it clear to you. P.S. : Please watch the video again I've explained it there
आज जारी की गई csir फाइनल ans key में xeF2O2 ki geometry trigonal bipyramidal बताई गई हैं लेकिन xeF2O2 ki geometry distorted tetrahedral होती हैं प्लीज कन्फर्म
DINESH KUMAR SAINI, aap shayad shape aur geometry me confuse ho rhe h... Geometry Hybridisation se decide hoti h, us question me SNF3 ka Hybridisation sp³ h yani tetrahedral aur XeO2F2 ka Hybridisation sp3d h yani trigonal bipyramidal. Distorted tetrahedral uska shape hoga geometry nhi.
By using chemical shift only, how to detection of organic compounds in NMR..Why I ask the question means "In both H1 NMR and C13 NMR some chemical shift values are simultaneously same..(For Ex: C-O : 50-80 & C-N ; 40-60) Yes, this is not understand explain it.
Hi Kamesh Very nice question, See for Spectroscopy and analysis of data to elucidate structure of compound, you cannot rely on single type of NMR value. Say if you're getting two functional groups having chemical shift overlapping like you described, then look for the other one that is 1H NMR, moreover other data like IR, Mass etc are also given in the question paper.
U did a mistake @28:30 first of all u didn't write the option c correctly and after that u match the data forcefully to option 'd' which is not possible answer so please take care about it. Thank you
Sir i am in short of words to praise u ....for d frst tm i undrstd it just only bcz of u ....plz sir help us when u get time ,a big thanks to u sir stay blessed n plz help us more
thank you for your kind words :)
it means a lot.
You taught the things in 14 mins to which my well educated professor was unable to do.
If got a chance to get a guidance from you. I pay 10times more than my college.
Thanks, you're a life saviour.❤
The problem came in Dec 2017, first one is the correct answer by simply counting the number of signals. Your teaching skill is very good and easy to understand. Thank you sir.
Thank you 😇
Option 1 is correct ... It has three types of protons . 2 has 4 types, 3 has 5 types and 4 has 4 types of protons.
This method cleared only for u .. thank you so much sir... always i watch ur all videos...give clear information ...
Amazing sir.and i love you and love you your teaching. Basically i am slow learner. Now i understand very well
We all are so thanful to you Sir
It was impossible for me to go fr any coaching centre.
Thnks to Almighty Allah You are there.
May Allah bless you.
IR helps. IR band at 1690 cm-1 corresponds to C=O of Structure 1 when we have confusion between structures 1 and 2
still feels amazing to watch after 2 years after JRF ...thanks Noorul Sir :)
You're most welcome 🥰
How's your life after qualifying JRF?
Sir, 1st option is right na it has 3 types of proton(2 doublet and one singlet) and other have 4 types.
Nidhi Singh, yes 👍
Well done
At 22 mins, you can't ignore the option b base on the chemical shift value and the no.of Hs giving singlet. In option B, the Hs in the methyl group will also give a singlet. Isn't it?
Yes you're right, I missed that one,
But singlet at 3.59 is due to OMe group, a methyl attached to CO group will give peak around 2.8 due to electronegativity of Oxygen atom.
sir nice explination ..can u please solve the problems related with mass spectroscopy ...
Yes I'll make one video on Mass spectra as well.. 👍
In last example the protons ortho to methoxy group appear below seven due to electron donating group hence more shielding, another aspect to reject option 2 is IR value as when electron donating group is present the value carbonyl shift below 1700 due decrease in double bond character of carbonyl.
overall very nice explanation thank you very much
From just the types of protons... We have option a as the correct answer
THANKS A lot
Great sir, Hats off u. Thanks for videos which will help me to get more & more confidence.
In last question in between two most probable options we can distinguish them by IR value given in question because for ester it will be near about 1750.
thank you sir, you're a life saver
sir Very much useful video. super trick n tips we learn from you. but in December 2016 there will be a OCH3 group in option c. kindly check out n explain in detail. bcz we learn more from you
yes you're correct, sorry i missed that.
And option 3is right
May be
Very nice explanation , thanks a lot but please make video to calculate nmr and r , s
Only Chemistry, thank you
Both videos are there please search in our videos list.
The answer would be option 1... because data is given for 3 different protons only option one have 3 different protons . .....other options have. 4 and 5 protons
Aayushi Mishra yes options A is correct
Chemical shifts we have to learn or will be given to us?
Sir u are really doing best.....touch wood
Option 1is correct..for the question u given for solving 😊
Have you made the detailed video of NMR? If not plz do it
the value will be given in question paper but you have to learn about different functional groups then only you can match and predict the structure
finally can answer in my upcoming GSET exam which is on 30/9/18... thank u
You're welcome 😊
If you want more detailed explanation on problem solving based on NMR Spectroscopy. Then plz watch my videos on Unacademy too.
unacademy.com/lesson/course-overview/TNULQWVZ
And follow me there for more helpful videos 😊
sir how you are considering plane ? where and in which molecule there is plane how we came to know?
this you can only understand by imaging the molecule, knowing symmetry and having basic knowledge of group theory.
watch my video on group theory.
OMe E.D.G group hence it increase electron density to Ortho position.therfore Ortho proton some shielded so it's valu will be 6.9 in 1st option
PANKAJ KUMAR,
Thank you for correcting brother 😊, I think I got confused there...
PANKAJ KUMAR can you explain it again what you are trying to say...is it for last Question?
@@simranvasir1443 yes
Thanks so much Sir. You are simply the BEST!!!
Sir you r telling very good
The answer will be the 1st one as there is two types of aromatic protons and one aliphatic CH3 protons.
I am just Speechless.......
Hats off SIR....
The credit will go only for you if I get NET or GATE
The answer for this question might be option 1
thank you
and
all the best for Exam :)
and you can remove that "if", and be confident enough that you are going to get CSIR JRF this time and GATE too :)
Sir, your explanation is good. But in the 4-acetyl anisole, the assignment is not correct. When methoxy group is present, the ortho to OMe protons always shielded than the meta to OMe protons.... please revise this.. Thank you
how beautiful u explain keep it up
Thank you 😇
excellent sir..................very much useful
Sir if we conclude that 1st one is correct from the IR data because the IR value is 1690 nearer to the ketone(1710)..whereas the IR Vale is far apart from ester (1740).is it correct way to think?? Plz rply..
And thanks a lot sir for making this video... It is very helpful to me..
Plz specify the question so that i can answer it.
You're welcome :)
The last question...
Yes, you're thinking correctly based on IR data...
thank you very much fir your immediate reply sir
Muthu Lakshmi, you're welcome 😇
Sir will you plz upload IR spectroscopy .
sir, i watched your ''stereo of bi phenyl video'' in that video you told that is org. chemistry of csir exam is not so easy to attempt although many expert also fails to predict right answer...sir, you are very right so is it impossible to do org chemistry acurately.....i am also survrving....plz guide
Basic Chemistry, it is not impossible, many people who are good in Organic Chemistry are doing great. But I feel it's difficult to attempt. And if you are also struggling with organic then don't waste much time on it... Try to give that time to the topic which you can solve in exam. It could be anything like physical chemistry or Inorganic Chemistry
yes sir....
sir, honestly speaking i attempt both paper of 2017 and try to self examine that how much i can understand the concept of given que.
i think its 50- 50% chances in present ....so do you think is there any hope to do well?
16:50 the symmetry plane should pass horizontally, then only it will show mirror image..
Really awesome trick to solve questions
29:30 option a is correct ✅... Total no. Of signal 3
Thanku sir ji the way u teach sir its amazing very helpful sir
In your question at 28:48 about CSIR Dec 2017 Answer option (1) might be correct... Thank you Sir...
superb explanation trick sir
thank you :)
Sir Ur lectures are really helping a lot..plz share Ur videos on coordination chemistry too including stereoisomers prediction and inner as well as outer sphere reactions concept along with expected questions
Very good video sir please make video on magnetic and chemical equivalnce and theory of nmr spectroscopy means how to remember chemical shift values and how to use
Very very helpful me and other thanks for that sir
you're welcome :)
Sir , the question which came in csir dec 2016 , there is some mistake in option no 3 , there is OCH3 gp at para position .
p
I think he can't differentiate then, so he deliberately missed that.....
Sir ans is 1st for dat questionnaire is it correct
Sir at 26:46 you took some other molecule of dec 2016 in c option ,methoxy group is present in both C AND D dn how we solve it
Very nice video sir , plz upload one video on esr and massbaeur questions of csir net gate
Thank you :)
I'll try to make one
Sir options 1 for CSIR D 2017 is right ans.plz confirm
Please a video on basics of NMR please
Sir csir net Dec 2016 question is not correct. The option you have written is wrong. The c option written by you is wrong.. plz correct that and then plz upload it again.
if i have phenol and benzoic acid.How to explain/distinguish with the help of NMR-spectroscopy.Question is asked for 10marks.plz tell me.
they both will show peaks at different positions. the proton of OH group in ohenol will show peak at 4.5 where as that in benzoic acid at 12.8 due to resonance and all other factors. I'll suggest you to read Organic Spectroscopy by Pavia, there you will find detailed explanantion.
Nice explanation sir..
thank you :)
very nice explanation
Thanku sir.very helpful for us
You're welcome 😊
is the answer to the question at 29:20 option a??
Thank you very much
answer will be (a)for dec 2017 question. Is it correct sir?
Sir very nice explanation
sir Kindly solve the dec 2012 nmr question its very tough to find out
I can't make a video on that bt I'll try to help you out here only...
See firstly from the IR data 3400 is for OH group and 1680 could be for aromatic ketone, as the resonance effect of phenyl ring will weaken the bond strength on C=O and its IR will decrease from usual 1715 to 1680. So it is confirmed that it contains OH and ketone group.
That means option 3 and 4 could be probable answers.
Now see the ¹H signals having J value same, that means those two protons are adjacent to each other one shows peak at 7.8 other at 7.0 which is greater and lower than usual phenylic proton at 7.1.
Now if you'll see option 3, we have two adjacent protons to OH and OMe both of the groups enhance electron density at their ortho positions respectively.
But in case of option 4 we have two protons adjacent to OH and CO group and we know OH enhance electron density at its ortho position while CO decreases electron density at its ortho, so option 4 becomes correct choice.
very nice explation
It's good.but I have a problem in these questions pls help me .how to solve it.step by step for beginners
If we have value of hnmr than how to make structure
Nice explanation
Thank you 😊
need not to say sorry sir. thank u you very much for your reply. kindly explain difference between c and d in that question if u have free time
What to do if value comes in tau?
Very helpfull , i like it
Thank you, 😊
Thanks A lot sir n sir plse make video on organometallic unique rxn ( oxidative addition ; reduction etc Plse sir my humble request) before net exm beocz whole organometallic rxn based on these rxns
alright... I'll try
Thanku sir
Sir vedios not clearly visible sir.....
Sir very good teaching I need nuclear chemistry topic , where I see
Organic is very easy and interesting
There is a confusion sir.. If we get quartet due to methyl protons then why do we write its (q, 2H) instead of (q, 3H)...plz give me explanation...
About which question you're asking, plz specify so that it becomes easier for me yo explain
All 'Bout Chemistry sir I'm asking about nmr spectra of chloro ethane
Praveen, if you'll watch the video carefully, I've mentioned that the splitting is due to protons present on adjacent carbon butnot the signal. lets take this, in chloro ethane we have two types of prtons one is CH3 and other is CH2, now CH2 will show chemical shift at much higher ppm value than the CH3 as it is directly attached to Cl (around 3.8). Now lets talk about splitting, so CH2 will give quartet due to CH3 which is adjacent to it, but in order to write the its chemical shift value we will write "3.8 (q, 2H). now here q denotes that it gives quartet due to adjacent CH3 and 2H denotes that the 3.8 is given by carbon that has 2 protons,
I hope I made it clear to you.
P.S. : Please watch the video again I've explained it there
Sir nice explanation
Thank you, 😇
Sir 1st option is correct
hats off to you sir
Thank You Sir
That's a good job
Thank you :)
Outstanding ....
Sir option 1 is correct
Dec 2017 ans is A n sir things are dis much simple can't imagine
You're welcome :)
how to deduce and draw the structure on the basis of spectral data
nice explanation sir
Superb class ..
Thank you 😇
Option 1 correct sir
Plz make more on IIT jam 🙏🙏🙏
In first question you are numbering 2nd carbon which doesn't contain proton
In 2017 dec how can 1 give pic@16.4
Thncuu so much sir for this
You're welcome 😇
Very easy method
share with your friends and don't forget to subscribe the channel.
Thank u Sooooooo much sir
आज जारी की गई csir फाइनल ans key में xeF2O2 ki geometry trigonal bipyramidal बताई गई हैं लेकिन xeF2O2 ki geometry distorted tetrahedral होती हैं प्लीज कन्फर्म
DINESH KUMAR SAINI, aap shayad shape aur geometry me confuse ho rhe h... Geometry Hybridisation se decide hoti h, us question me SNF3 ka Hybridisation sp³ h yani tetrahedral aur XeO2F2 ka Hybridisation sp3d h yani trigonal bipyramidal.
Distorted tetrahedral uska shape hoga geometry nhi.
Thanks sir
You're welcome 😊
Thanks a lot sir
Sir plzz solve dec 18 nmr problem
Thanks sir ji
By using chemical shift only, how to detection of organic compounds in NMR..Why I ask the question means "In both H1 NMR and C13 NMR some chemical shift values are simultaneously same..(For Ex: C-O : 50-80 & C-N ; 40-60) Yes, this is not understand explain it.
Hi Kamesh
Very nice question,
See for Spectroscopy and analysis of data to elucidate structure of compound, you cannot rely on single type of NMR value. Say if you're getting two functional groups having chemical shift overlapping like you described, then look for the other one that is 1H NMR, moreover other data like IR, Mass etc are also given in the question paper.
Thanq Sir
you're welcome :)
Best video
U did a mistake @28:30 first of all u didn't write the option c correctly and after that u match the data forcefully to option 'd' which is not possible answer so please take care about it.
Thank you