a) Interpretation: Starting from benzene and assuming that the ortho and para isomers can be separated how the compound can be synthesized is to be shown. Concept introduction: The compound required is 2-bromo-4-nitrotoluene. Benzene can be converted into alkyl benzene by Friedal-Crafts alkylation. The alkylbenzene can be nitrated in the ortho and para positions as the alkyl groups are ortho and para directors. Nitro group is deactivating while alkyl group is activating in nature. Hence bromonation of the nitrotoluene will introduce the bromine in the ortho position to alkyl group. To show: Starting from benzene and assuming that the ortho and para isomers can be separated how 2-bromo-4-nitrotoluene can be synthesized.

Question

Reagan is doing an atomic absorption experiment that requires a set of zinc standards in the 0.4- 1.6 ppm range. A 1000 ppm Zn solution was prepared by dissolving the necessary amount of solid Zn(NO3)2 in water. The standards can be prepared by diluting the 1000 ppm Zn solution. Table 1 shows one possible set of serial dilutions (stepwise dilution of a solution) that Reagan could perform to make the necessary standards. Solution A was prepared by diluting 5.00 ml of the 1000 ppm Zn standard to 50.00 ml. Solutions C-E are called "calibration standards" because they will be used to calibrate the atomic absorption spectrometer. Table 1: Dilutions of Zinc Solutions Solution Zinc Solution Volume Diluted Solution Concentration used volume (ppm Zn) (mL) (mL) concentration (ppm Zn) Solution concentration A 1000 5.00 50.00 1.00×10² (ppm Zn(NO3)2) 2.90×10² Solution concentration (M Zn(NO3)2 1.53×10-3 B Solution A 5.00 100.00 5.00 C Solution B 5.00 50.00 0.50 7.65×10-6 D Solution B 10.00 50.00…

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Question

Chapter 16.SE, Problem 71AP

Interpretation Introduction

a)

Interpretation:

Starting from benzene and assuming that the ortho and para isomers can be separated how the compound can be synthesized is to be shown.

Concept introduction:

The compound required is 2-bromo-4-nitrotoluene. Benzene can be converted into alkyl benzene by Friedal-Crafts alkylation. The alkylbenzene can be nitrated in the ortho and para positions as the alkyl groups are ortho and para directors. Nitro group is deactivating while alkyl group is activating in nature. Hence bromonation of the nitrotoluene will introduce the bromine in the ortho position to alkyl group.

To show:

Starting from benzene and assuming that the ortho and para isomers can be separated how 2-bromo-4-nitrotoluene can be synthesized.

Interpretation Introduction

b)

Interpretation:

Starting from benzene and assuming that the ortho and para isomers can be separated how the compound can be synthesized is to be shown.

Concept introduction:

The Friedal-Crafts acylation of benzene will yield an acyl benzene as the product. Chlorination of the acyl benzene will produce m-chloroacyl benzene as the acyl group is meta directing. Reduction of the acyl group to the corresponding alkyl group and sulfonation of the product will result in the product required.

To show:

Starting from benzene and assuming that the ortho and para isomers can be separated how the compound can be synthesized.

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Reagan is doing an atomic absorption experiment that requires a set of zinc standards in the 0.4- 1.6 ppm range. A 1000 ppm Zn solution was prepared by dissolving the necessary amount of solid Zn(NO3)2 in water. The standards can be prepared by diluting the 1000 ppm Zn solution. Table 1 shows one possible set of serial dilutions (stepwise dilution of a solution) that Reagan could perform to make the necessary standards. Solution A was prepared by diluting 5.00 ml of the 1000 ppm Zn standard to 50.00 ml. Solutions C-E are called "calibration standards" because they will be used to calibrate the atomic absorption spectrometer. Table 1: Dilutions of Zinc Solutions Solution Zinc Solution Volume Diluted Solution Concentration used volume (ppm Zn) (mL) (mL) concentration (ppm Zn) Solution concentration A 1000 5.00 50.00 1.00×10² (ppm Zn(NO3)2) 2.90×10² Solution concentration (M Zn(NO3)2 1.53×10-3 B Solution A 5.00 100.00 5.00 C Solution B 5.00 50.00 0.50 7.65×10-6 D Solution B 10.00 50.00…

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