037. In 13C NMR, peak splitting is caused by which spin-spin interaction? a carb
ID: 547054 • Letter: 0
Question
037. In 13C NMR, peak splitting is caused by which spin-spin interaction? a carbon nucleus with the surrounding electrons b) carbon nucleolus with surrounding proton nuclei c) carbon nucleus and surrounding carbon nuclei d) carbon nucleus and electrons on adjacent nuclei Q38. Which of the following will NOT give a strong peak around 1700 wavenumbers? a) b) c) d) Q39. An aromatic ring usually has a 'HNMR spectrum containing a) peaks around 7 ppm b) a very broad ill-defined peak c) peaks around 1-2 ppm d) a single peak from all ronatic proton signals 040. In a proton NMR spectrum, a "doublet of triplets" indicates a) adjacent non-identical methyl groups b) two protons on a terminal alcohol c) the compound contains a labile proton d) a bi-substituted aromatic ringExplanation / Answer
Q37
Answer: b) Carbon nucleolus with surrounding proton nuclei
The heteronuclear coupling (carbon-hydrogen) is common in carbon NMR as the probability of finding the two 13C atoms adjecent to each other in a molecule is very less.
Q38
Answer: a)
all the other options have carbonyl group (aldhyde & ketone) whose streching freqency will be around 1700 in the IR spectroscopy
alkene C=C or the aromatic C=C will appear in the range of 1450-1650
Q39
Answer: a) Peaks around 7 ppm
aromatic protons are deshielded due to Magnetic Anisotropy and appear around 7 in the 1H NMR spectrum
Q40
Answer: a) a bi-substituted aromatic ring
The OH group does not give the splitting.
The labile protons are fast exchanged and also does not give splitting.
methyl groups will give qurtet
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.