IOM 2018, Теория — Задача 2. Raspberry chemistry

Official answers and grading scheme

1

1 pt for each correct structure. [8 т.]

2

2 pts for each correct structure. [18 т.]

3

1 pt for each correct structure, 1 pt for the name. [3 т.]

4

1 pt. [1 т.]

Comments from the author

1 Some of these compounds are well known. It is not difficult to determine the structure of others from the given molecular formulae. Thus, according to the formula $\mathrm{C_4H_6O_5}$, it is easy to conclude that malic acid is a diacid, two carboxyl groups being connected by two carbon atoms, one of which is linked to a hydroxy group. There are two variants of the structure under this limitation: $\mathrm{HO_2CCH_2CH(OH)CO_2H}$ and $\mathrm{HO_2CC(CH_3)(OH)CO_2H}$. The first seems more likely. This is a true structure, which is estimated by a full score (1 mark). The second option is an incorrect structure, but it satisfies the condition, therefore it is estimated as 0.5 marks. Similarly, we can consider the remaining structures. In all cases, the correct structure is estimated at 1 mark, wrong, but chemically possible and satisfying the condition structure as 0.5 or 0.25 marks, depending on the degree of difference from the correct structure. Total – 8 marks.

2 According to the formula of compound H, it is possible to establish that this is the product of oxidation of the methyl group to the aldehyde. A comparison of the formulae of the compounds H and B shows that the chlorine atom is replaced by an OH group at this stage. So, B is 4-hydroxybenzaldehyde. The formula of the raspberry ketone (10 carbon atoms) shows that it is formed via the reaction of 1 eq. of B with 1 eq. of acetone. Further analysis allows to conclude that C is the product of aldol condensation, and the reduction of the C=C bond in C gives a saturated ketone A (by name it is clear that the keto group did not reduced). The nitration of 4-chlorotoluene yields two isomeric products: 3-nitro-4-chlorotoluene and 2-nitro-4-chlorotoluene. However, only in the first case the aromatic nucleophilic substitution of chlorine for methoxy group can occur under the treatment with a solution of sodium methoxide in methanol under heating, since this reaction is facilitated by an ortho-nitro group, but not by a meta-nitro group. Therefore, D is 3-nitro-4-chlorotoluene, and E is 4-methyl-2-nitroanisole. Further, the reduction of the nitro group to the amine takes place (the reduction of the benzene ring does not proceed at room temperature) and the sulfonation of the resulting aniline. Here, in principle, the reaction can take place both in the para- and ortho-position with respect to amino group (which is a stronger donor than methoxy group). However, in the latter case, there are considerably larger steric requirements for the approach of the bulk electrophile to an atom surrounded on both sides by substituents, so 4-amino-2-methyl-5-methoxybenzenesulfonic acid G is formed (it exists mainly as an internal salt). This compound, under the action of sodium nitrite in the presence of an acid, is converted into a diazonium salt, which attacks the $\beta$-naphthol derivative on the C1 atom to form an azo dye E129.

3 It is known that heating of non-enolizable aldehydes with strong alkali induces their disproportionation to the corresponding alcohols and acids (their salts). This process is referred to as a Cannizzaro reaction.

4 Chlorine has negative inductive effect but positive mesomeric effect. The latter should be more important than the combination of inductive effects of chlorine atom and methyl group for the formation of 3-nitro-4-chlorotoluene as the major product.