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The three isomers of dichlorobenzene are commonly named ortho-chlorobenzene, meta-chlorobenzene, and para-chlorobenzene. These three isomers are difficult to distinguish using proton NMR, but they are instantly identifiable usingNMR.

  1. Describe how carbon NMR distinguishes these three isomers.
  2. Explain why they are difficult to distinguish using proton NMR.

Short Answer

Expert verified
  1. Considering the symmetry of structures, ortho-xylene would have 3 carbon signals from the ring, meta-xylene would have 4 carbon signals from the ring and para-xylene would have only 2 carbon signals from the ring.
  2. Unless the substituent on the benzene ring is electron withdrawing or electron donating, ring protons absorb at roughly the same position as methyl group has no electronic effect on the ring hydrogens, thus, proton NMR will not be able to distinguish between these isomers.

Step by step solution

01

Explanation of part (a):

Considering the symmetry of structures, ortho-xylene would have 3 carbon signals from the ring and total of 4 peaks whereas meta-xylene would have 4 carbon signals from the ring and total of 5 peaks and para-xylene would have only 2 carbon signals from the ring and total of 3 peaks. Thus, these compounds would be instantly identifiable by number of peaks in carbon-NMR.

02

Explanation of part (b):

The proton-NMR would not be able to distinguish these isomers as there is no powerful electron withdrawing or donating substituent attached to the ring and methyl group has essentially no electronic effect on the ring hydrogens, thus all ring protons will absorb at roughly same chemical shift. Para isomer would give clean singlet and ortho and meta isomers would have only slightly broadened singlets for their protons. Very high field NMR would only be able to distinguish these isomers.

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Most popular questions from this chapter

A new chemist moved into an industrial lab where work was being done on oxygenated gasoline additives. Among the additives that had been tested, she found an old bottle containing a clear, pleasant-smelling liquid that was missing its label. She took the quick NMR spectrum shown and was able to determine the identity of the compound without any additional information. Propose a structure and assign the peaks. (Hint: This is a very pure sample.)

If the imaginary replacement of either of two protons forms enantiomers, then those protons are said to be enantiotopic.The NMR is not a chiral probe, and it cannot distinguish between enantiotopic protons. They are seen to be 鈥渆quivalent by NMR鈥.

  1. Use the imaginary replacement technique to show that the two allylic protons (those on) of allyl bromide are enantiotopic.
  2. Similarly, show that the two HCprotons in cyclobutanol are enantiotopic.
  3. What other protons in cyclobutanol are enantiotopic?

The following off-resonance-decoupled carbon NMR was obtained from a compound of formula. Propose a structure for this compound and show which carbon atoms give rise to which peaks in the spectrum.


When 2-chloro-2-methylbutane is treated with a variety of strong bases, the products always seem to contain two isomers (A and B) of formula. When sodium hydroxide is used as base, isomer A predominates. When potassium tert-butoxide is used as the base, isomer B predominates. TheandNMR spectra of A and B are given below.

  1. Determine the structures of isomers A and B.
  2. Explain why A is the major product when using sodium hydroxide as the base and why B is the major product when using potassium tert-butoxide as the base.

Question:In a 300-MHz spectrometer, the protons in bromomethane absorb at a position 660 Hz downfield from TMS.

(a) What is the chemical shift of these protons?

(b) What is the chemical shift of the bromomethane protons in a 60-MHz spectrometer?

(c) How many hertz downfield from TMS would they absorb at 60 MHz?

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