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Chapter 3: Problem 109

Mononitration of biphenyl by treating with nitrating mixture gives mainly

Short Answer

Expert verified
Answer: 4-nitrobiphenyl

Step by step solution

01

Write the starting compound and reagents

First, we will write down the starting compound, biphenyl, and the reagents used for mononitration: a nitrating mixture (consisting of concentrated nitric acid (HNO₃) and concentrated sulfuric acid (H₂SO₄)). Biphenyl: `\text{C}_{12}\text{H}_{10}` Nitrating mixture: `\text{HNO}_{3}` and `\text{H}_{2}\text{SO}_{4}`
02

Formation of nitronium ion

The nitronium ion (NO2+), which is the electrophile in the nitration process, is formed through a reaction between concentrated nitric acid and concentrated sulfuric acid. The reaction is as follows: `\text{HNO}_{3} + \text{2H}_{2}\text{SO}_{4} \rightarrow \text{NO}_{2}^{+} + \text{2HSO}_{4}^{-} + \text{H}_{3}\text{O}^{+}`
03

Electrophilic aromatic substitution

The next step is for the electrophilic aromatic substitution to occur, where the nitronium ion (NO2+) generated in the previous step attacks the biphenyl (C12H10) molecule. Since biphenyl has two aromatic rings, the nitronium ion can attack either of them, causing the formation of an intermediate. For mononitration of biphenyl, the nitronium ion will attack the para position (relative to the other ring in biphenyl), as it is the least sterically hindered position. The intermediate will have a positive charge on the carbon where the nitro group is attached.
04

Removal of a proton and product formation

Finally, one of the hydrogens attached to the positively charged carbon in the intermediate is removed by the bisulfate ion (HSO4-) to restore aromaticity, leading to the formation of the mononitrated biphenyl product. The overall reaction is: `\text{C}_{12}\text{H}_{10} + \text{NO}_{2}^{+} \rightarrow \text{C}_{12}\text{H}_{9}\text{NO}_{2} + \text{H}^{+}` The major product after mononitration of biphenyl is 4-nitrobiphenyl.

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Most popular questions from this chapter

An aromatic compound having the molecular formula \(\mathrm{C}_{8} \mathrm{H}_{10}\) gave only one mono nitro derivative on nitration. The compound is (a) ethyl benzene (b) o-xylene (c) \(\mathrm{m}\) -xylene (d) p-xylene

Ethyl benzene \(\stackrel{\mathrm{NBS}}{\longrightarrow} \mathrm{A} \stackrel{\text { alc.KOH }}{\longrightarrow} \mathrm{B}\). The product \(\mathrm{B}\) in the above reaction sequence is (a) styrene (b) 1 -phenyl ethanol (c) 2 -phenyl ethanol (d) p-ethylphenol

When phenol reacts with dilute neutral \(\mathrm{FeCl}_{3}\) solution, a violet colouration is produced. The structure of the violet coloured species is (a) \(\left[\mathrm{Fe}\left(\mathrm{OC}_{6} \mathrm{H}_{5}\right)_{6}\right]^{4}\) (b) \(\mathrm{Fe}\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3}\) (c) \(\left[\mathrm{Fe}\left(\mathrm{OC}_{6} \mathrm{H}_{5}\right)_{6}\right]^{3-}\) (d) \(\left(\mathrm{FeCl}_{3}\right)\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\)

Identify the wrong statement. (a) Phenols are stronger acids than alcohols, but weaker than carboxylic acids (b) Phenols, being stronger acids, are soluble in aqueous \(\mathrm{NaOH}\) solution and in saturated \(\mathrm{NaHCO}_{3}\) solution. (c) Phenols are acidic because of resonance stabilization of the phenoxide ion formed by ionization. (d) The \(\mathrm{pK}_{\mathrm{a}}\) of phenol is about \(10 .\)

The conversion of cumene hydroperoxide to phenol and acetone involves (a) an oxidative process (b) a 1,2 -phenyl migration (c) Friedel-Crafts alkylation (d) keto-enol tautomerism
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