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Chapter 13: Q17P (page 675)

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 “equivalent by NMR”.

  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?

Short Answer

Expert verified

a.

b.

The products formed are enantiomers and hence, HCprotons are enantiotopic.

c.The Hd protons are enantiotopic.

Step by step solution

01

Step-1. Explanation of part (a):

a.

Allylic protons of allyl bromide are enantiotopic in nature. By replacing the protons HA and HBwith some atom “Z”, we get two products and these products are enantiomers of each other. Enantiomers are non-superimposable mirror images of each other and enantiotopic protons are not distinguishable by NMR.

02

Step-2. Explanation of part (b)

b.

Cyclobutanol has protons HC and Hdand both are enantiotopic protons. On replacing the protons Hc with some atom “Z”, we get the two products which are enantiomers of each other as they are non-superimposable mirror images of each other.

03

Step-3: Explanation of part  (c):

c.

The Hd protons are also enantiotopic in cyclobutanol.

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Most popular questions from this chapter

Determine the number of different kinds of protons in each compounds

  1. 1-bromopropane (b) 2-bromopropane
  2. 2,2-dimethylpropane (d) 2,2-dimethylpentane
  3. (e) 1-chloro-4-methylbenzene (f) 1-chloro-2-methylbenzene

Each of these four structures has molecular formula . Match the structure with its characteristic proton NMR signals. (Not all of the signals are listed in each case.)

(a) Sharpsinglet atδ8.0 andtriplet atδ4.0

(b) Sharpsinglet atδ2.0 andquartet atδ4.1

(c) Sharpsinglet atδ3.7 andquartet atδ2.3

(d) Broadsinglet atδ11.5 andtriplet atδ2.3

Phenyl Grignard reagent adds to 2-methylpropanal to give the secondary alcohol shown. The proton NMR of 2-methylpropanal shows the two methyl groups as equivalent (one doublet at δ 1.1), yet the product alcohol, a racemic mixture, shows two different 3H doublets, one at δ 0.75 and one around δ 1.0

(a)Draw a Newmann projection of the product along the C1-C2 axis.
(b)Explain why the two methyl groups have different NMR chemical shifts. What is the term applied to protons such as these?

Question: An unknown compound has the molecular formula C9H11Br. Its proton NMR spectrum shows the following absorptions: singlet, d7.1, integral 44 mm singlet, d2.3, integral 130 mm singlet, d2.2, integral 67 mm Propose a structure for this compound.

A laboratory student was converting cyclohexanol to cyclohexyl bromide by using one equivalent of sodium bromide in a large excess of concentrated sulfuric acid. The major product she recovered was not cyclohexyl bromide, but a compound of formula C6H10that gave the following 13CNMR spectrum:

  1. Propose a structure for this product.
  2. Assign the peaks in the 13CNMR spectrum to the carbon atoms in the structure.
  3. Suggest modifications in the reaction to obtain a better yield of cyclohexyl bromide.
See all solutions

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