Funny & Stup O saka rutgers. O 307-Fall-2015 0 307-Fall 2014 O 307 Fall-2014307-
ID: 1051142 • Letter: F
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Funny & Stup O saka rutgers. O 307-Fall-2015 0 307-Fall 2014 O 307 Fall-2014307-Fall-l × 0 307-Fall-2014 nmr practice l if a new chiral sim city 2000 grand theft al + contentsakai .rutgers.edu/access/content/group/cf4a533 6-f1ec-44d7-806f-be87c351 cel 8/Practice%20Exams/2014/307-Fall-2014-Exam-2.pdf MC 4. Given only coupling through 2 or 3 bonds what is the maximum number of peaks you expect for the signals representing the denoted protons in the compound shown? 12 B.3 E. 4 12 3 MC 5. What is the color of your exam? A White B Green C Pink D Yellow OAsk me anything 5:03 PM 11/5/2016 q,Explanation / Answer
In NMR spectroscopy, number of couplinga is given by the formula (n+1) where n is the number of non-equivalent protons. If there is 2 non equivalent protons , then number of coupling is (n+1)(n'+1)
H(a) = (1+1)(1+1) =4
H(b) = (1+1)(1+1)
Ha and Hb are no equivalent as they are connected to a C=C , they can assume a cis or trans conformation and depending on that their chemical shift will change. So, they are magnetically non equivalant.
H(c) = (n+1)(n'+1)(n''+1) = (1+1)(1+1)(2+1) = 12
Hd = (2+1)(3+1) = 12
He = (2+1) = 3
answer is E
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