Concept explainers
a) Dinitrobenzene
Interpretation:
The possible isomers for dinitrobenzene are to be drawn and their names are to be given.
Concept introduction:
Disubstituted benzenes can exist as three isomers such as ortho, meta and para. They are named using the prefixes ortho (o), meta (m) and para (p). An ortho-disubstituted benzene has its two substituent groups in a 1,2-relationship on the ring. A meta-disubstituted benzene has its two substituent groups in a 1,3-relationship on the ring. A para-disubstituted benzene has its two substituent groups in a 1,4-relationship on the ring. While writing the name the substituent groups are arranged alphabetically.
To draw:
The possible isomers for dinitrobenzene and to name them.
b) Bromodimethylbenzene
Interpretation:
The possible isomers for bromodimethylbenzene are to be drawn and their names are to be given.
Concept introduction:
Trisubstitutedbenzenes can exist as six isomers with the three substituents in 1,2,3; 1,2,4; 1,2,5; 1,2,6; 1,3,4 and 1,3,5 relationship.
Benzene with more than two substituent groups are named choosing a point of attachment as carbon 1 and numbering the substituent groups on the ring so that the second substituent has as low number as possible. If ambiguity still exists, numbering is done such that the third and fourth substituent groups have a number as low as possible, until a point of difference is obtained. While writing the name the substituent groups are arranged alphabetically.
To draw:
The possible isomers for bromodimethylbenzene and to name them.
c) Trinitrophenol
Interpretation:
The possible isomers for trinitrophenol are to be drawn and their names are to be given.
Concept introduction:
Trinitrophenol can exist as six isomers with the four substituents in 1,2,3,4; 1,2,3,5; 1,2,3,6; 1,2,4,5; 1,2,4,6 and 1,3,4,5 relationship.
Benzene with more than two substituent groups are named choosing a point of attachment as carbon 1 and numbering the substituent groups on the ring so that the second substituent has as low number as possible. If ambiguity still exists, numbering is done such that the third and fourth substituent groups have a number as low as possible, until a point of difference is obtained. While writing the name the substituent groups are arranged alphabetically.
To draw:
The possible isomers for trinitrophenol and to name them.

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Chapter 15 Solutions
ORGANIC CHEMISTRY-EBOOK>I<
- Assign all the carbonsarrow_forward9 7 8 C 9 8 200 190 B 5 A -197.72 9 8 7 15 4 3 0: ང་ 200 190 180 147.52 134.98 170 160 150 140 130 120 110 100 90 90 OH 10 4 3 1 2 -143.04 140. 180 170 160 150 140 130 120 110 100 90 CI 3 5 1 2 141.89 140.07 200 190 180 170 160 150 140 130 120 110 100 ៖- 90 129. 126.25 80 70 60 -60 50 40 10 125.19 -129.21 80 70 3.0 20 20 -8 60 50 10 ppm -20 40 128.31 80 80 70 60 50 40 40 -70.27 3.0 20 10 ppm 00˚0-- 77.17 30 20 20 -45.36 10 ppm -0.00 26.48 22.32 ―30.10 ―-0.00arrow_forwardAssign all the carbonsarrow_forward
- C 5 4 3 CI 2 the Righ B A 5 4 3 The Lich. OH 10 4 5 3 1 LOOP- -147.52 T 77.17 -45.36 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 ppm B -126.25 77.03 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 ppm 200 190 180 170 160 150 140 130 120 110 100 90 80 TO LL <-50.00 70 60 50 40 30 20 10 ppm 45.06 30.18 -26.45 22.36 --0.00 45.07 7.5 1.93 2.05 -30.24 -22.36 C A 7 8 5 ° 4 3 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 ppm 9 8 5 4 3 ཡི་ OH 10 2 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 5 4 3 2 that th 7 I 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 115 2.21 4.00 1.0 ppm 6.96 2.76 5.01 1.0 ppm 6.30 1.00arrow_forwardCurved arrows were used to generate the significant resonance structure and labeled the most significant contribute. What are the errors in these resonance mechanisms. Draw out the correct resonance mechanisms with an brief explanation.arrow_forwardWhat are the: нсе * Moles of Hice while given: a) 10.0 ml 2.7M ? 6) 10.ome 12M ?arrow_forward
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- 7.5 1.93 2.05 C B A 4 3 5 The Joh. 9 7 8 1 2 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 ppm 9 7 8 0.86 OH 10 4 3 5 1 2 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 ppm 9 7 8 CI 4 3 5 1 2 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 2.21 4.00 1.5 2.00 2.07 1.0 ppm 2.76arrow_forwardAssign the functional group bands on the IR spectra.arrow_forwardFind the pH of a 0.120 M solution of HNO2. Find the pH ignoring activity effects (i.e., the normal way). Find the pH in a solution of 0.050 M NaCl, including activityarrow_forward
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