
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
- Draw the Haworth projection of the disaccharide made by joining D-glucose and D-mannose with a ẞ(1-4) glycosidic bond. If the disaccharide has more than one anomer, you can draw any of them. Click and drag to start drawing a structure. Xarrow_forwardEpoxides can be opened in aqueous acid or aqueous base to produce diols (molecules with two OH groups). In this question, you'll explore the mechanism of epoxide opening in aqueous acid. 2nd attempt Be sure to show all four bonds at stereocenters using hash and wedge lines. 0 0 Draw curved arrows to show how the epoxide reacts with hydronium ion. 100 +1: 1st attempt Feedback Be sure to show all four bonds at stereocenters using hash and wedge lines. See Periodic Table See Hint H A 5 F F Hr See Periodic Table See Hintarrow_forward03 Question (1 point) For the reaction below, draw both of the major organic products. Be sure to consider stereochemistry. > 1. CH₂CH₂MgBr 2. H₂O 3rd attempt Draw all four bonds at chiral centers. Draw all stereoisomers formed. Draw the structures here. e 130 AN H See Periodic Table See Hint P C Brarrow_forward
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- Alcohols can be synthesized using an acid-catalyzed hydration of an alkene. An alkene is combined with aqueous acid (e.. sulfuric acid in water). The reaction mechanism typically involves a carbocation intermediate. > 3rd attempt 3343 10 8 Draw arrows to show the reaction between the alkene and hydronium ion. that 2nd attempt Feedback 1st attempt تعمال Ju See Periodic Table See Hint F D Ju See Periodic Table See Hintarrow_forwardDraw the simplified curved arrow mechanism for the reaction of acetone and CHgLi to give the major product. 4th attempt Π Draw the simplified curved arrow mechanism T 3rd attempt Feedback Ju See Periodic Table See Hint H -H H -I H F See Periodic Table See Hintarrow_forwardSelect the correct reagent to accomplish the first step of this reaction. Then draw a mechanism on the Grignard reagent using curved arrow notation to show how it is converted to the final product. 4th attempt Part 1 (0.5 point) Select the correct reagent to accomplish the first step of this reaction. Choose one: OA Mg in ethanol (EtOH) OB. 2 Li in THF O C. Li in THF D. Mg in THF O E Mg in H2O Part 2 (0.5 point) Br Part 1 Bri Mg CH B CH, 1 Draw intermediate here, but no arrows. © TE See Periodic Table See Hint See Hint ין Harrow_forward
- Select the product for the following reaction. HO HO PCC OH ○ OH O HO ○ HO HO HOarrow_forward5:45 Х Select the final product for the following reaction sequence. O O 1. Mg. ether 2.D.Oarrow_forwardBased on the chart Two similarities between the molecule with alpha glycosidic linkages. Two similarities between the molecules with beta glycosidtic linkages. Two differences between the alpha and beta glycosidic linkages.arrow_forward
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