Question

помогите плиз, очень надоКакой углеводород получится при действии спиртового раствора щелочи на 2-бром-2-метилбутан? Назовите его. Напишите для него реакции с HCl и H2 SO4.

Answer 1

Согласно правилу Зайцева тут образуется 2-метилбутен-2 СН3-СH=C(CH3)-CH3

 

СН3-СH2-CBr(CH3)-CH3 +KOH = СН3-СH=C(CH3)-CH3 +KBr+H2O

 

СН3-СH=C(CH3)-CH3 +  HCl = СН3-СH2-CCl(CH3)-CH3 ( 2-хлор-2-метилбутан) 

 

СН3-СH=C(CH3)-CH3 +  H2SO4 = СН3-СH2-C(SO3H)(CH3)-CH3 

( 2-метил-2-бутансульфокислота)

Answer 2

Formation of Hydrocarbon:

When a 2-bromo-2-methylbutane reacts with a solution of alcohol and alkali, it undergoes an elimination reaction to form an alkene. In this case, the reaction would result in the formation of 2-methylpropene.

The reaction can be represented as follows: ``` 2-bromo-2-methylbutane + Alcoholic KOH → 2-methylpropene + KBr + H2O ```

Reactions with HCl and H2SO4:

2-methylpropene can undergo addition reactions with hydrochloric acid (HCl) and sulfuric acid (H2SO4).

1. Reaction with HCl: - The addition of HCl to 2-methylpropene results in the formation of 2-chloro-2-methylpropane. - The reaction can be represented as: ``` 2-methylpropene + HCl → 2-chloro-2-methylpropane ```

2. Reaction with H2SO4: - When 2-methylpropene reacts with concentrated sulfuric acid (H2SO4), it undergoes an electrophilic addition reaction to form 2-methylpropan-2-ol (2-methyl-2-propanol). - The reaction can be represented as: ``` 2-methylpropene + H2SO4 → 2-methylpropan-2-ol ```

These reactions demonstrate the chemical transformations that occur when 2-bromo-2-methylbutane reacts with an alcoholic solution of alkali, and the subsequent reactions of the resulting hydrocarbon with HCl and H2SO4.