You have three bottles containing the three isomeric dibromobenzenes; they have the melting points \(+87^{\circ},+6^{\circ}\), and \(-7^{\circ} .\) By a great deal of work, you prepare six dibromonitrobenzenes \(\left(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{Br}_{2} \mathrm{NO}_{2}\right)\) and find that, of the six, one is related to (derived from or convertible into) the dibromobenzene of m.p. \(+87^{\circ}\), two to the isomer of m.p. \(+6^{\circ}\), and three to the isomer of m.p. \(-7^{\circ}\). Label each bottle with the correct name of ortho, meta, or' para. (This work was actually carried out by Wilhelm Körner, of the University of Milan, and was the first example of the Korner method of absolute orientation.)

Short Answer

Expert verified
The ortho-dibromobenzene has a melting point of +87°C, the meta-dibromobenzene has a melting point of +6°C, and the para-dibromobenzene has a melting point of -7°C.

Step by step solution

01

Ortho, Meta, and Para positions overview

These terms are used to describe the relative position of substituents on a benzene ring. The ortho position refers to two adjacent carbon atoms on the ring, the meta position describes a gap between two substituents, and the para position has the substituents directly opposite to one another on the benzene ring. Now we can analyze the effect of these positions on the number of possible dibromonitrobenzenes:
02

Ortho-dibromobenzene

In ortho-dibromobenzene, there are two bromine atoms located at adjacent carbon atoms. When we add a nitro group, there is only one possible position for it to be attached without being in direct contact with the adjacent bromine atoms. Therefore, there can only be one ortho-dibromonitrobenzene derivative.
03

Meta-dibromobenzene

In meta-dibromobenzene, there are two bromine atoms located on the benzene ring with one carbon atom separating them. When we add a nitro group to this isomer, there are actually two possible positions where the nitro group can be added, creating two distinct meta-dibromonitrobenzenes.
04

Para-dibromobenzene

In para-dibromobenzene, there are two bromine atoms located directly opposite each other on the benzene ring. When we add a nitro group to the para-dibromobenzene, there are three possible positions where the nitro group can be added, which leads to the formation of three distinct para-dibromonitrobenzenes. With this information, we can now confidently label the bottles:
05

Label the bottles

Based on the analysis, we can conclude that the bottle with the melting point of +87°C is the ortho-dibromobenzene, since one dibromonitrobenzene is related to it. The bottle with the melting point of +6°C is the meta-dibromobenzene, as two dibromonitrobenzenes are related to it. Finally, the bottle with the melting point of -7°C is the para-dibromobenzene, as three dibromonitrobenzenes are related to it.

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Most popular questions from this chapter

The anion of benzene \(\mathrm{C}_{6} \mathrm{H}_{5}-\), is aromatic although it contains a total of eight "free" electrons. How do you account for this?

Draw structures for each of the following: (a) m-Bromonitrobenzene (e) 2,4 -Dibromomesitylene (b) o-Ethyltoluene (f) Pheny 1 cyclohexane (c) \(p\) -Xylene (g) Biphenyl (d) 3-Nitrobenzene (h) Benzyl cyanide

Calculate from appropriate bond and stabilization energies the heats of reaction of chlorine with benzene to give (a) chlorobenzene and (b) 1,2 -dichloro-3,5-cyclohexadiene. Your answer should indicate that substitution is energetically more favorable than addition. Assume the bond dissociation energy for a \(\mathrm{C}=\mathrm{C} \pi\) bond to be \(65 \mathrm{kcal}\); the resonance stabilization energy of benzene to be \(36 \mathrm{kcal}\), and that of dichloro-3, 5 -cyclohexadiene to be \(3 \mathrm{kcal}\).

For a time the prism formula VI, proposed in 1869 by Albert Ladenburg of Germany, was considered as a possible structure for benzene, on the grounds that it would yield one monosubstitution product and three isomeric disubstitution products. (a) Draw Ladenburg structures of three possible isomeric dibromobenzenes. (b) On the basis of the Korner method of absolute orientation, label each Ladenburg structure in (a) as ortho, meta, or para. (c) Can the Ladenburg formula actually pass the test of isomer number? (Derivatives of Ladenburg "benzene," called prismanes, have actually been made.)

Biphenyl, \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{C}_{6} \mathrm{H}_{5}\), has a conjugation energy of 71 \(\mathrm{kcal} / \mathrm{mole}\) (a) Draw an atomic orbital picture of biphenyl, (b) What are its most important resonance contributing structures? (c) Estimate the heat of hydrogenation of biphenyl.

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