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Chapter 23: Problem 90

Show reagents and conditions to convert toluene to 3-hromo-4.methylphenol.

Short Answer

Expert verified
Question: Describe the step-by-step process to convert toluene to 3-bromo-4-methylphenol. Answer: To convert toluene to 3-bromo-4-methylphenol, follow these steps: 1. Perform electrophilic aromatic substitution with bromine (Br2) using N-bromosuccinimide (NBS) to form an isomeric mixture of ortho-bromo-toluene and para-bromo-toluene. 2. Introduce a nitro (-NO2) group at the 4th position through nitration using concentrated nitric (HNO3) and sulfuric acid (H2SO4) to obtain 3-bromo-4-nitrotoluene. 3. Reduce the nitro group (NO2) in 3-bromo-4-nitrotoluene to an amino group (NH2) with iron (Fe) and hydrochloric acid (HCl), resulting in 3-bromo-4-aminotoluene. 4. Replace the amino group (NH2) in 3-bromo-4-aminotoluene with a hydroxyl group (OH) by reacting it with sodium nitrite (NaNO2) and hydrochloric acid (HCl) to form a diazonium salt, which then reacts with water (H2O) to produce the desired 3-bromo-4-methylphenol.

Step by step solution

01

Halogenation

To introduce a bromine atom at the 3rd position in toluene, perform electrophilic aromatic substitution with bromine (Br2). Use N-bromosuccinimide (NBS) as the brominating agent and catalytic amount of light or heat for the reaction. The presence of the methyl group on the benzene ring in toluene directs the bromination at the ortho or para positions. The resulting compound will be isomeric mixture of ortho-bromo-toluene and para-bromo-toluene. "\\[ \\ce{C6H5CH3 + Br2 ->[NBS, hv] ortho-BrC6H4CH3 + para-BrC6H4CH3} \\]"
02

Nitration

Next, introduce a nitro (-NO2) group at the 4th position through nitration. Mix the isomeric mixture from step 1 with a mixture of concentrated nitric (HNO3) and sulfuric acid (H2SO4) at low temperature (0 °C). The ortho-bromo-toluene would result in the desired 3-bromo-4-nitrotoluene product. The para- and the meta-nitro derivatives are also formed in trace quantities but can be separated. "\\[ \\ce{ortho-BrC6H4CH3 + HNO3 ->[H2SO4, 0^\\circ C] 3-Br-4-NO2C6H4CH3} \\]"
03

Reduction of Nitro Group

The nitro group (NO2) in 3-bromo-4-nitrotoluene needs to be reduced to an amino group (NH2). This can be achieved by treating the 3-bromo-4-nitrotoluene with iron (Fe) and hydrochloric acid (HCl). The product obtained will be 3-bromo-4-aminotoluene. "\\[ \\ce{3-Br-4-NO2C6H4CH3 + Fe + 3 HCl -> 3-Br-4-NH2C6H4CH3 + FeCl3 + 2 H2O} \\]"
04

Substitution of Amino Group with Hydroxyl Group

Replace the amino group (NH2) in 3-bromo-4-aminotoluene with a hydroxyl group (OH) to form 3-bromo-4-methylphenol. Treat the 3-bromo-4-aminotoluene with sodium nitrite (NaNO2) in presence of hydrochloric acid (HCl) at low temperature (0 °C) to form diazonium salt. Then react the diazonium salt with water (H2O) to obtain the desired 3-bromo-4-methylphenol. "\\[ \\ce{3-Br-4-NH2C6H4CH3 ->[NaNO2, HCl, 0^\\circ C]}); \\[ \\ce{3-Br-4-N2^+C6H4CH3 + H2O -> 3-Br-4-OHC6H4CH3 + N2} \\]"

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Most popular questions from this chapter

Heterocyclic aromatic amines are weaker bases than aliphatic heterocyclic amines. Campare, for example, the \(\mathrm{p} K_{\mathrm{a}}\) values for the conjugate acids of piperidine, pyri-

Write structural formulas for these amines. (a) Isobutylamine (b) Triphenylamine (c) Diixspropylamine

Proton transfer from water or other acid to pyridine does not involve the electrons of the aromatic sextet. Why, then, is pyridine a considerably weaker base than aliphatic amines? The answer is that the unshared pair of electrons on the pyridine nitrogen lies in a relatively electronegative \(x f^{2}\) hybrid orbital, whereas in aliphatic amines, the unshared pair lies in an spr hybrid orbital. This effect decreases markedly the basicity of the electron pair on an sp²-hybridized nitrogen compared with that on an \(s p^{3}\) hytvridized nitrogen. There are two nitrogen atoms in imidazole, each with an unshared pair of electrons. One unshared pair lies in a \(2 p\) orbital and is an integral part of the \((4 n+2)\) \(\pi\) electrons of the aromatic system. The other unshared pair lies in an spr hybrid orbital and is mot a part of the aromatic sextet; this pair of electrons functions as the proton acceptor.

We discussed the structure and bonding in pyridine and imidazole in Section 21.2D. In accounting for the relative basicities of these and other heterocyclic aromatic amines, it is important to determine first if the unshared pair of electrons on nitrogen is or is not a part of the \((4 w+2) \pi\) electrons giving rise to aromaticity. In the case of pyridine, the unshared pair of electrons is not a part of the aromatic sextet. \(23.5\) Basicity Rather, it lies in an \({x p^{2}}^{2}\) hybrid orbital in the plane of the ring and perpendicular to the six \(2 p\) orbitals containing the aromatic sextet.

The geometry of a nitrogen atom bonded to three other atoms or groups of atoms is trigonal pyramidal (Section 1.4). The sp, bybridized nitrogen atom is at the apex of the Pyramid, and the three groups bouded to it extend densaward to form the triangular base of the prramid. If we consider the unshared pair of electrons on nitrogen as a foeurth grosp, then the amrangement of "groups" around nitrogen is approximately tetrahedral. Berause of this geoenetry, an ansine with three different groups bonded to nitrogen is chiral and can exist as a pair of enantiomers, as illustrated by the nonsuperporable mirror images of ethyimethylamine. In assigning configuration to these enantiomers, the groxp of lowest prioxity on nitrogen is the unshared pair of electrons.
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