Question

В пробирку добавляют несколько капель олеиновой кислоты и добавляют небольшое количество раствора брома в тетрахлорметане. Раствор брома обесцвечивается. Написать реакцию бромирования олеиновой кислоты.

Answer 1

С17Н33СООН+Br2=C17H33Br2COOH

Answer 2

Пишу с помощью полуструктурной формулы:
CH₃-CH₂-CH₂-CH₂-CH₂-CH₂-CH₂-CH₂-CH=CH-CH₂-CH₂-CH₂-CH₂-CH₂-CH₂-CH₂-COOH + Br₂ --> CH₃-CH₂-CH₂-CH₂-CH₂-CH₂-CH₂-CH₂-CH(Br)-CH(Br)-CH₂-CH₂-CH₂-CH₂-CH₂-CH₂-CH₂-COOH
или
C₁₇H₃₃COOH + Br₂ --> C₁₇H₃₃(Br)₂COOH
в результате присоединения брома по месту двойной связи в олеиновой кислоте происходит обесцвечивание бромной жидкости.

Answer 3

Reaction of Bromination of Oleic Acid

When oleic acid is reacted with a solution of bromine in tetrachloromethane, the bromine solution becomes colorless. This is due to the addition of bromine to the double bond in the oleic acid molecule, resulting in the formation of a colorless dibromo compound.

The reaction can be represented as follows: C17H33COOH + Br2 → C17H33COBr + HBr

This reaction involves the addition of bromine to the double bond of the oleic acid molecule, resulting in the formation of dibromo compound and hydrogen bromide as a byproduct.

This reaction is a typical example of the bromination of unsaturated fatty acids, where the double bond in the fatty acid molecule undergoes addition reaction with bromine.