IChO 1979 · Theory problem 5

Structure of a benzene derivative C9H12 from its bromination products

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Benzene derivative X has the empirical formula \(\ce{C9H12}\). Its bromination in the light leads to the formation of two monobromo derivatives in approximately identical yield. Bromination in the dark in the presence of iron also gives two monobromo derivatives. If the reaction is carried out to a higher degree, the formation of four dibromo derivatives may occur.

Suggest the structure for compound X and for the bromination products. Write schemes for the reactions.

Official English text of the problem.

Answers

The problem has no numbered subparts.

  • Compound X: 4-ethyltoluene (1-ethyl-4-methylbenzene), Cc1ccc(CC)cc1.
  • Bromination in light: the two side-chain products \(L_1\) and \(L_2\) below, in approximately equal yields. \(L_2\) is racemic.
  • Bromination in the dark with iron: the two ring-monobromination products \(M_2\) and \(M_3\).
  • Further bromination in the dark with iron: the four ring-dibromination products \(D_{23}\), \(D_{25}\), \(D_{26}\), and \(D_{35}\).

For all ring-product labels, keep the original methyl group at ring position 1 and the ethyl group at position 4.

Label Product name SMILES
\(L_1\) 1-(Bromomethyl)-4-ethylbenzene CCc1ccc(CBr)cc1
\(L_2\) Racemic 1-(1-bromoethyl)-4-methylbenzene C[C@H](Br)c1ccc(C)cc1 and C[C@@H](Br)c1ccc(C)cc1, in equal amounts
\(M_2\) 2-Bromo-4-ethyltoluene Cc1c(Br)cc(CC)cc1
\(M_3\) 3-Bromo-4-ethyltoluene Cc1cc(Br)c(CC)cc1
\(D_{23}\) 2,3-Dibromo-4-ethyltoluene Cc1c(Br)c(Br)c(CC)cc1
\(D_{25}\) 2,5-Dibromo-4-ethyltoluene Cc1c(Br)cc(CC)c(Br)c1
\(D_{26}\) 2,6-Dibromo-4-ethyltoluene Cc1c(Br)cc(CC)cc1Br
\(D_{35}\) 3,5-Dibromo-4-ethyltoluene Cc1cc(Br)c(CC)c(Br)c1

Solution

Identification of X.

The degree of unsaturation is \[ \mathrm{DBE}=\frac{2(9)+2-12}{2}=4. \] The benzene ring accounts for all four degrees of unsaturation. Thus the three carbons outside the ring belong to saturated, acyclic alkyl substituents. They can form one propyl group, an ethyl group plus a methyl group, or three methyl groups.

The possible structures and their numbers of constitutionally distinct ring-monobromination products are:

Candidate SMILES Ring-monobromination products
Propylbenzene CCCc1ccccc1 3
Isopropylbenzene (cumene) CC(C)c1ccccc1 3
2-Ethyltoluene Cc1ccccc1CC 4
3-Ethyltoluene Cc1cc(CC)ccc1 4
4-Ethyltoluene Cc1ccc(CC)cc1 2
1,2,3-Trimethylbenzene Cc1c(C)c(C)ccc1 2
1,2,4-Trimethylbenzene Cc1c(C)cc(C)cc1 3
1,3,5-Trimethylbenzene Cc1cc(C)cc(C)c1 1

These counts follow from the equivalence of ring positions under molecular symmetry. In particular, the ethyl and methyl groups are different, so a symmetry operation cannot interchange them.

The observation of two ring-monobromination products leaves 4-ethyltoluene and 1,2,3-trimethylbenzene.

For 1,2,3-trimethylbenzene, the available ring positions are 4, 5, and 6, with 4 and 6 equivalent. Introducing two bromines therefore gives only two arrangements: \[ \{4,5\}\equiv\{5,6\}, \qquad \{4,6\}. \] It cannot account for four ring-dibromination products.

For 4-ethyltoluene, the four available positions give four distinct dibromo arrangements, as demonstrated below. Hence \[ \boxed{X=\text{4-ethyltoluene}.} \]

Bromination in light.

Under the selective benzylic bromination assumed in the problem, light promotes radical substitution at a carbon directly attached to the aromatic ring. X has two different benzylic sites:

  • the carbon of its methyl group;
  • the methylene carbon of its ethyl group.

The two balanced reactions are \[ \ce{CH3-C6H4-CH2CH3 + Br2 ->[h\nu] BrCH2-C6H4-CH2CH3 + HBr} \qquad (L_1) \] and \[ \ce{CH3-C6H4-CH2CH3 + Br2 ->[h\nu] CH3-C6H4-CHBrCH3 + HBr}. \qquad (L_2) \] The two side chains remain para to one another in both equations.

Here molecular bromine is BrBr, and hydrogen bromide is Br.

In \(L_2\), the carbon bearing bromine has four different substituents and is stereogenic. Bromination under achiral conditions gives equal amounts of its two enantiomers, whose SMILES appear in the answer table. They constitute one constitutional product, so the light reaction gives two constitutional monobromo derivatives overall.

The approximately equal yields of \(L_1\) and \(L_2\) are supplied experimental information; their ratio cannot be calculated from symmetry alone.

Monobromination in the dark with iron.

With iron, [Fe], bromination in the dark proceeds by electrophilic aromatic substitution.

Numbering the ring with methyl at C1 and ethyl at C4, reflection through these two positions gives \[ 2\longleftrightarrow6, \qquad 3\longleftrightarrow5. \] Consequently, bromination at C2 or C6 gives \(M_2\), whereas bromination at C3 or C5 gives \(M_3\).

The following balanced equation applies separately to each product, with the bromine at the position specified in the answer table: \[ \ce{C6H4(CH3)(C2H5) + Br2 ->[Fe] C6H3Br(CH3)(C2H5) + HBr} \qquad (M_2\ \text{or}\ M_3;\ \text{dark}). \]

Further bromination: the four dibromo derivatives.

Two ring hydrogens must now be replaced among positions \(2,3,5,6\). There are initially \[ \binom{4}{2}=6 \] pairs, but reflection makes two pairs of arrangements equivalent:

Distinct product Equivalent choices of brominated positions
\(D_{23}\) \(\{2,3\}\equiv\{5,6\}\)
\(D_{25}\) \(\{2,5\}\equiv\{3,6\}\)
\(D_{26}\) \(\{2,6\}\)
\(D_{35}\) \(\{3,5\}\)

Thus there are exactly four constitutional ring-dibromination products.

Each further substitution obeys \[ \ce{C6H3Br(CH3)(C2H5) + Br2 ->[Fe] C6H2Br2(CH3)(C2H5) + HBr} \qquad (\text{dark}). \] The possible pathways are:

  • From \(M_2\), substitution at C3, C5, or C6 gives \(D_{23}\), \(D_{25}\), or \(D_{26}\), respectively.
  • From \(M_3\), substitution at C2, C5, or C6 gives \(D_{23}\), \(D_{35}\), or \(D_{25}\), respectively.

The balanced overall reaction, applying separately to each of the four dibromo isomers, is \[ \ce{C6H4(CH3)(C2H5) + 2Br2 ->[Fe] C6H2Br2(CH3)(C2H5) + 2HBr} \qquad (\text{dark}). \]

The solution was checked against the official solution.