1. Anthracene - aromatic, Fluoranthene - antiaromatic, Aceanthrylene - antiaromatic. Total: (10 points)

## TASK № 5

The compound $\mathrm{C}_{9} \mathrm{H}_{10}$ decolorizes bromine water and a solution of $\mathrm{KMnO}_{4}$ at room temperature. Upon heating with an aqueous solution of $\mathrm{KMnO}_{4}$, para-phthalic acid is formed:



Determine the structure of the original hydrocarbon and name it.
# Solution:

para-Phthalic acid is an aromatic compound with side chains. The aromatic ring does not decolorize bromine water and does not react with potassium permanganate solution at room temperature. These properties are characteristic of an unsaturated compound. An unsaturated fragment built from two carbon atoms is the vinyl radical $\mathrm{CH}_{2}=\mathrm{CH}-.(\mathbf{6}$ points)

The molecule of para-phthalic acid contains two side chains in para-positions. The molecular formula $\mathrm{C}_{9} \mathrm{H}_{10}$ corresponds to a structure consisting of an aromatic ring with a vinyl radical and a methyl radical in para-position.

1-vinyl-4-methylbenzene, para-methylstyrene (6 points)

Strong oxidizers oxidize the side chains to carboxylic acids.

![](https://cdn.mathpix.com/cropped/2024_05_06_cb2098284a8b01f58754g-09.jpg?height=186&width=1516&top_left_y=2060&top_left_x=356)

(6 points)

Answer: The initial hydrocarbon is 1-vinyl-4-methylbenzene, para-methylstyrene. Total: (24 points)

TASK № 6

Red phosphorus was burned in excess chlorine. The resulting product was treated with water, and zinc was added to the resulting solution upon heating. The gas released was passed over heated iron(II) oxide.

Write down the four equations for the transformations indicated.

Balance the redox reactions using the electron balance method.

Solution: