Question

Углеводород А хлорирует соя на свету, образуя моногалогенпроизводное В. Последнее легко взаимодействует с водным раствором гидроксида натрия, не образуя алкена в качестве побочного продукта. При добавлении избытка безводного хлорида алюминия с последующим нагреванием вещество В образует два монохлорпроизводных С и Д и углеводород Е, который при нагревании в присутствии платинового катализатора образует антрацен. Назовите указанные вещества, приведите уравнения реакций

Answer 1

Углеводород А - это толуол C6H5-CH3, кот-й при хлор-ии на свету обр-т (ввиду радикального механизма р-ии) бензилхлорид C6H5-CH2Cl (B): C6H5-CH3 + Cl2 (+hv) =  C6H5-CH2Cl + HCl;

 

далее, C6H5-CH2Cl реаг-т с водным р-ом NaOH с обр-ем бензилового спирта: C6H5-CH2Cl + NaOH = C6H5-CH2OH (бензиловый спирт) + NaCl;

 

 

в присутствии избытка AlCl3 протекают р-ии с обр-ем орто-хлортолуола C6H4(Cl)CH3 (монохлорпроизводное C), пара-хлортолуола Cl-C6H4-CH3 (монохлорпроизводное D) и углеводорода C6H4(CH2)2C6H4 (E), кот-й при дегидрировании в присутствии Pt обр-т антрацен C14H10:

 

 

1) 2C6H5-CH2Cl (+T, AlCl3 (изб.)) = C6H4(Cl)CH3 (C) + Cl-C6H4-CH3 (D) и 

 

 

2) 2C6H5-CH2Cl (+T, AlCl3 (изб.)) = C6H4(CH2)2C6H4 (E) + 2HCl;

 

 

3) C6H4(CH2)2C6H4 (+T, Pt) = C14H10 + H2.

Answer 2

Chemical Reaction Analysis

The given chemical reaction involves the chlorination of a hydrocarbon A to form a monohalogen derivative B, which further reacts with a sodium hydroxide solution and an excess of anhydrous aluminum chloride to form two monochloro derivatives C and D, and a hydrocarbon E. Finally, the hydrocarbon E, when heated in the presence of a platinum catalyst, forms anthracene.

Identified Substances and Reactions

The substances and reactions involved in the given chemical process are as follows:

Substances: - Hydrocarbon A - Monohalogen derivative B - Monochloro derivatives C and D - Hydrocarbon E - Anthracene

Reactions: 1. Chlorination of Hydrocarbon A to form Monohalogen Derivative B: - This reaction involves the chlorination of hydrocarbon A in the presence of light to form monohalogen derivative B.

2. Reaction of Monohalogen Derivative B with Sodium Hydroxide Solution: - Monohalogen derivative B reacts with a sodium hydroxide solution without forming an alkene as a byproduct.

3. Reaction of Monohalogen Derivative B with Excess Anhydrous Aluminum Chloride: - Monohalogen derivative B reacts with an excess of anhydrous aluminum chloride, followed by heating, to form monochloro derivatives C and D, and hydrocarbon E.

4. Heating of Hydrocarbon E in the Presence of a Platinum Catalyst: - Hydrocarbon E, when heated in the presence of a platinum catalyst, forms anthracene.

Equations for the Reactions

1. Chlorination of Hydrocarbon A to form Monohalogen Derivative B: ``` A + Cl2 (in the presence of light) -> B ```

2. Reaction of Monohalogen Derivative B with Sodium Hydroxide Solution: ``` B + NaOH -> (product without alkene formation) ```

3. Reaction of Monohalogen Derivative B with Excess Anhydrous Aluminum Chloride: ``` B + AlCl3 (excess, followed by heating) -> C + D + E ```

4. Heating of Hydrocarbon E in the Presence of a Platinum Catalyst to form Anthracene: ``` E (in the presence of Pt catalyst, heating) -> Anthracene ```

These reactions illustrate the series of transformations involved in the given chemical process. If you have any further questions or need additional details, feel free to ask!