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Chapter 14: Q.21558-14-59P (page 565)

Question: Identify the structures of isomers E and F (molecular formula C4H802 ). Relative areas are given above each signal.

a.

b.

Short Answer

Expert verified

Answer

a.

b.

Step by step solution

01

Isomers  

The isomers (with the same molecular formula) have different chemical shifts due to the different orientations of the groups. The different orientations give varied chemical environments.

02

Explanation for a

IR absorption at 1743cm-1 : CO

1 degree of unsaturation

NMR data: Ha : quartet at 4.1 ppm

Hb: singlet at 2.0 ppm

Hc: triplet at 1.4 ppm

Totalintegrationofunits=23+29+308

=10

structure

03

Explanation for b

IR absorption at 1730cm-1 : CO

1 degree of unsaturation

Ha: singlet at 4.1 ppm

Hb: singlet at 3.4 ppm

Hc: singlet at 2.1 ppm

Totalintegrationofunits=18+30+312

=10

Structure

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Most popular questions from this chapter

Question: Identify the structures of isomers A and B (molecular formula C9H10O ).

Compound A: IR peak at 1742 cm-1 ; 1 H NMR data (ppm) at 2.15 (singlet, 3 H), 3.70 (singlet, 2 H), and 7.20 cm-1(broad singlet, 5 H). Compound B: IR peak at 1688 ; 1 H NMR data (ppm) at 1.22 (triplet, 3 H), 2.98 (quartet, 2 H), and 7.28鈥�7.95 (multiplet, 5 H).

Question: How many different types of protons are present in each compound?

When 2-bromo-3,3-dimethylbutane is treated with \({{\bf{K}}^{\bf{ + }}}{\bf{ - }}{}^{\bf{ - }}{\bf{OC}}{\left( {{\bf{C}}{{\bf{H}}_{\bf{3}}}} \right)_{\bf{3}}}\), a single product T having molecular formula \({{\bf{C}}_{\bf{6}}}{{\bf{H}}_{{\bf{12}}}}\) is formed. When 3,3-dimethylbutan-2-ol is treated with \({{\bf{H}}_{\bf{2}}}{\bf{S}}{{\bf{O}}_{\bf{4}}}\), the major product U has the same molecular formula. Given the following \({}^{\bf{1}}{\bf{H}}\)-NMR data, what are the structures of T and U? Explain in detail the splitting patterns observed for the three split signals in T. 1 H NMR of T: 1.01 (singlet, 9 H), 4.82 (doublet of doublets, 1 H, J = 10, 1.7 Hz), 4.93 (doublet of doublets, 1 H, J = 18, 1.7 Hz), and 5.83 (doublet of doublets, 1 H, J = 18, 10 Hz) ppm 1 H NMR of U: 1.60 (singlet) ppm.

Question: Identify the carbon atoms that give rise to the signals in the 13C NMR spectrum of each compound.

a. CH3CH2CH2CH2OH ; 13CNMR: 14, 19, 35, and 62 ppm

b. (CH3)2CHCHO ; 13C NMR: 16, 41, and 205 ppm

c. CH2=CHCHOHCH3 ; 13C NMR: 23, 69, 113, and 143 ppm

Question: Label the signals due to Ha ,Hb and Hc in the 1 H NMR spectrum of acrylonitrile (CH2=CHCN ). Draw a splitting diagram for the absorption due to the proton Hc.
See all solutions

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