Discussion
Number of signals observed in the 1H NMR and proton decoupled 13C NMR spectrum of the following compound, respectively, are :
![A bicyclo[2.2.1]hept-2-ene (norbornene) skeleton with a bromine on each of the two saturated ring carbons C5 and C6, both Br drawn pointing down and both H pointing right](/uploads/question-images/sppu-mhset-2018-01-28-chemical-sciences-paper-2/q21.png)
Answer: B
The compound gives 5 1H signals and 4 13C signals.
The drawing is a 5,6-dibromonorborn-2-ene with both bromines on the same face (cis). Such a molecule has a mirror plane through C7 and the midpoints of the C2=C3 and C5-C6 bonds, which makes each carbon on one side equivalent to its partner on the other.
Carbons: C2/C3 (alkene), C1/C4 (bridgeheads), C5/C6 (CHBr) and C7, so 4 signals. Protons: the two vinyl H, the two bridgehead H and the two CHBr H each form one signal, while the two hydrogens on C7 are inequivalent (one faces the double bond, the other the bromines), so they give two signals: 5 in all.
8 and 7 would be the counts with no symmetry at all, and 4, 4 misses the inequivalence of the two C7 hydrogens.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development