IOM 2017, Теория — Задача 2. Big adventure of a small alkyl

Solutions

1

A, B and F: 1 point for each structure.

Other alkyl bromides which allow for the synthesizing F as a single oxidation product: 2 compounds, 1 point for each.

Calculations proving that only A satisfies the problem conditions: See the detailed solution. (2 × 0.5 points) [6 т.]

2

C, D and E: 1 point for each structure. [3 т.]

3 Calculations. According to the problem conditions, the final mixture contains alkylbenzenes only as organic compounds. Therefore, $1.56:78=0.02$ mol of benzene reacted with 0.06 mol of ethyl bromide producing 0.02 mol of B and 0.06 mol of HBr. Moreover, 0.001 mol of AlBr₃ is present in the reaction mixture.

To achieve pH 7.00 it is required to add $0.06+3*0.001=0.063$ mol of NaHCO₃. This corresponds to 5.29 g.

$m(\mathrm{NaHCO_3})=5.29\ \mathrm g$ — 2 points

(1 point if AlBr₃ is not taken into account) [2 т.]

4

G–K: 1 point for each structure. [5 т.]

Detailed solution

1 It is clear that compound B belongs to the class of alkylbenzenes which has the general formula of $\mathrm{C_6H_n(C_mH_{2m+1})_{6-n}}$. Heating of alkylbenzenes with KMnO₄ in the presence of acid leads to the oxidation of side alkyl chain to the corresponding carboxylic acid. So, the product of the oxidation of F can be described as $\mathrm{C_6H_n(COOH)_{6-n}}$. The neutralization of 0.63 g F requires $6.84\times1.054=7.2$ g of 5% aq. NaOH solution. This corresponds to 9 mmol of NaOH. If F is monoacid ($n=5$), its molecular weight should be equals to $0.63:0.009=70$. This is impossible. Only tricarboxylic acid ($n=3;M_\mathrm F=210$) satisfies the problem conditions. As compound F has three types of carbon atoms, this can be only benzene-1,3,5-tricarboxylic acid. Therefore, B is 1,3,5-trialkylbenzene. The formation of F as a single organic product of the oxidation of B is possible for 1,3,5-trimethylbenzene,1,3,5-triethylbenzene,1,3,5-triisopropylbenzene. So, possible alkyl bromides are ethyl bromide (A), methyl bromide (A′) and isopropyl bromide (A″).

$1.56:78=0.02$ mol of benzene were introduced into the reaction. Weight of alkyl bromide (6.54 g) corresponds to 0.0688 mol of CH₃Br,0.06 mol of C₂H₅Br and 0.053 mol of isopropyl bromide. If methyl bromide was a started alkylation agent,0.06 mol of CH₃Br is consumed for the formation of 0.02 mol of trialkylbenzene; after that 0.0088 mol of CH₃Br should further react with trialkylbenzenes producing the mixture of 0.0112 mol of trimethylbenzenes and 0.0088 mol of tetramethylbenzenes (in the reaction mixture some quantities of di- and pentamethylbenzenes can also be present but this does not affect the general conclusion). Evidently, it is impossible to isolate 1,3,5-trimethylbenzene in 90% yield from this mixture. In the case of isopropyl bromide,0.04 mol of alkyl halide is consumed for the formation of dialkylbenzenes; after that 0.013 mol of isopropyl bromide should react with dialkylbenzenes producing trialkylbenzenes. In other words, after the reaction will finished it should contain approximately 65% ($0.013\times100\%/0.02$) of trialkylbenzene and 35% of trialkylbenzene. Again, this is impossible to isolate 1,3,5-triisopropylbenzene in 90% yield from this mixture. On the contrary, ethyl bromide is the ideal substrate from this point of view as in this case the theoretical yield of triethylbenzene is 100%. Evidently, in this case it is possible to isolate 1,3,5-triethylbenzene in 90% yield. Therefore, B is 1,3,5-triethylbenzene, A is ethyl bromide.

2 Compounds C, D and E are not isomers. Therefore, they are products of the sequential alkylation of benzene with ethyl bromide Therefore, C is ethylbenzene. Compound D is diethylbenzene. Ethyl group is electron-donating substituent and directs electrophilic attack to ortho- and para-positions. Steric demands prevent ortho-attack. Therefore, the major product in the second probe should be 1,4-diethylbenzene (compound D). Minor product is 1,2-diethylbenzene (D′). Compound E is the product of D alkylation. This is 1,2,4-triethylbenzene (the same compound is formed from D′). Under the prolonged keeping in the presence of Lewis acid and HBr evolved during the previous alkylation steps, E undergoes isomerization to the more stable isomer (1,3,5-triethylbenzene).

4 Compound B has three types of hydrogen atoms (aromatic CH, methylene protons and methyl groups protons) while in compound G all protons are identical. This shows that during bromination only protons of methyl groups left intact. On the contrary, other protons were substituted by bromine atoms (aromatic protons via aromatic electrophilic substitution catalyzed by FeBr₃; methylene protons via radical bromination which demonstrate high selectivity in regard to benzylic C-H bonds. Therefore, G is 1,3,5-tribromo-2,4,6-(1,1-dibromoethyl)benzene. Diluted alkali solution hydrolyzed selectively Ar–C–Br but not Ar-Br bonds. As a result, every CBr₂ moiety is transformed into C=O group. The reaction of H with phenylhydrazine is the formation of hydrazone. All three carbonyl groups are transformed in hydrazones, this is supported by nitrogen content in I. Under heating in polyphosphoric acid, phenylhydrazones are transformed into indoles (Fischer reaction). This is supported by both bromine content in J and (this is more important and conclusive) structure of the final product which has three indole fragments. The comparison of J with the final product shows that J has bromine atoms in the central benzene ring and hydrogen atoms at the C(3) position of indole moieties but in the target product these atoms are substituted by CH=CH fragments. Despite school students do not know reactions used for these transformations (Sonogashira coupling and Au-catalyzed alkyne cyclization), it is possible to understand that the source of two carbon atoms is alkyne used at the J-to-K step. So, during this step it is needed to substituted bromines with Me₃SiCC fragments (Si content supports this conclusion), the last two steps are Au-induced cyclization and hydrolysis for removal of silyl groups.