Draw the NMR spectra you expect for the following complexes.

Short Answer

Expert verified

The values of NMR spectra of the complexes given below:

The nmr values of each compounds of this compounds with respect to Tms are given here.

Step by step solution

01

Nmr of (a)

Protocol of the H-1 NMR Prediction (Lib=SU Solvent=DMSO 300 MHz):

The nmr values of each compounds of this compounds are given here.

Node Shift Base + Inc. (ppm rel. to TMS)

CH 7.24 7.26 1-benzene

0.04 1 -C=C

-0.06 general corrections

CH 7.24 7.26 1-benzene

0.04 1 –C=C

-0.06 general corrections

CH 7.24 7.26 1-benzene

-0.05 1 -C=C

0.03 general corrections

CH 7.24 7.26 1-benzene

-0.05 1 -C=C

0.03 general corrections

CH 7.33 7.26 1-benzene

-0.12 1 -C=C

0.19 general corrections

CH3 0.82 0.86 methyl

0.02 1 beta -C=C

0.10 2 beta -C

-0.16 general corrections

CH3 0.82 0.86 methyl

0.02 1 beta -C=C

0.10 2 beta -C

-0.16 general corrections

CH3 0.82 0.86 methyl

0.02 1 beta -C=C

0.10 2 beta -C

-0.16 general corrections

H 6.44 5.25 1-ethylene

1.38 1 -1:C*C*C*C*C*C*1 gem

-0.22 1 -C cis

0.03 general corrections

H 6.06 5.25 1-ethylene

0.36 1 -1:C*C*C*C*C*C*1 cis

0.45 1 -C ge

02

Nmr of (b)

Protocol of the H-1 NMR Prediction (Lib=SU Solvent=DMSO 300 MHz):

The nmr values of each compounds of this compounds are given here.

Node Shift Base + Inc. Comment (ppm rel. to TMS)

CH3 3.51 0.86 methyl

2.64 1 alpha -O-C=C

0.01 general corrections

CH3 2.05 0.86 methyl

0.85 1 alpha -C=C

0.34 general corrections

H 4.67 5.25 1-ethylene

0.18 1 -Cl cis

-1.21 1 -O-C trans

0.45 1 -C gem

1H NMR Coupling Constant Prediction

shift atom index coupling partner, constant and vector

3.51 7

2.05 3

5 6.4 H-CH2-C(sp2)-H

03

Step  3: Nmr of (c)

Protocol of the H-1 NMR Prediction (Lib=SU Solvent=DMSO 300 MHz):

The nmr values of each compounds of this compounds are given here.

Node Shift Base + Inc. Comment (ppm rel. to TMS)

CH2 4.06 1.37 methylene

0.00 1 alpha -C

2.82 1 alpha -OC(=O)C=C

-0.13 general corrections

CH3 3.80 0.86 methyl

2.64 1 alpha -O-C=C

0.30 general corrections

CH3 1.21 0.86 methyl

0.44 1 beta -OC(=O)-C=C

-0.09 general corrections

H 6.93 5.25 1-ethylene

1.22 1 -O-C gem

0.46 1 -C(=O)O-R trans

H 4.82 5.25 1-ethylene

-1.21 1 -O-C trans

0.78 1 -C(=O)O-R gem

04

Nmr of (d)

Protocol of the H-1 NMR Prediction (Lib=SU Solvent=DMSO 300 MHz):

The nmr values of each compounds of this compounds are given here.

Node Shift Base + Inc. Comment (ppm rel. to TMS)

OH 12.71 11.00 carboxylic acid

1.00 1 -C*R

0.71 general corrections

CH3 2.41 0.86 methyl

1.40 1 alpha -1:C*C*C*C*C*C*1

0.15 general corrections

H 8.04 7.26 1-benzene

0.87 1 -C(=O)O

-0.05 1 -C

-0.04 general corrections

H 7.30 7.26 1-benzene

0.21 1 -C(=O)O

-0.12 1 -C

-0.05 general corrections

H 8.04 7.26 1-benzene

0.87 1 -C(=O)O

-0.05 1 -C

-0.04 general corrections

H 7.30 7.26 1-benzene

0.21 1 -C(=O)O

-0.12 1 -C

-0.05 general corrections

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Most popular questions from this chapter

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 using 13C NMR

(a)Describe how carbon NMR distinguishes these three isomers.
(b)Explain why they are difficult to distinguish using proton NMR.

Hexamethylbenzene undergoes radical bromination with N-bromosuccinimide to give one monobrominated product (C12H17Br) and four dibrominated products (C12H16Br2). These products are easily separated by GC-MS, but the dibrominated products are difficult to distinguish by their mass spectra. Draw the monobrominated product and the four dibrominated products and explain how 13C NMR would easily distinguish among these compounds .

Show how you would distinguish among the following threecompounds

(a)Using infrared spectroscopy and no other information.
(b)Using proton NMR spectroscopy and no other information.
(c)Using 13C-NMR, including DEPT, and no other information.

A compound was isolated as a minor constituent in an extract from garden cress. Its spectra are shown here.

  1. Look at each spectrum individually and list the structural characteristics you can determine from that spectrum.
  2. Look at the set of spectra as a group and propose a tentative structure.
  3. Verify that your proposed structure accounts for the major features of each spectrum.

Sets of spectra are given for two compounds. For each set,

(1) Look at each spectrum individually and list the structural characteristics you can determine from that spectrum.

(2) Look ar the set of spectraas a group and purpose a tentative structure.

(3) Verify that your proposed structure accounts for the major features of each spectrum. The solution for compound 1 is given after the problem but go as far as you can before looking at the solution.

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