Question

Помогите пожалуйста решить .Срочно!!!1)C3H6СL2+KOH=X 2)X + K2Cr2O7 + H2SO4=CH3CH2COOH 3)CH3CH2COOH + Cl2= X2 4)X2 +NaHCO3= X3 5)X3 + CH3CH2Br=X4

Answer 1

1)C3H6CI2 + 2KOH = C3H6(OH)2 + 2KCI;

2)4C3H6(OH)2 + K2Cr2O7 + 3H2SO4 = 4CH3-CH2-COOH + K2SO4 + 2CrSO4 + 7H2O;

3)CH3-CH2-COOH + CI2 = CH3-CHCI-COOH + HCI;

4)CH-CHCI-COOH + NaHCO3 = CH3-CHCI-COONa + CO2 + H2O;

5)CH3-CHCI-COONa + CH3-CH2-Br = CH3-CHCI-COO-CH2-CH3 + NaBr.

Answer 2

Step 1: C3H6Cl2 + KOH → X

The reaction between 1-chloropropane (C3H6Cl2) and potassium hydroxide (KOH) is a nucleophilic substitution reaction where hydroxyl ion (OH−) acts as the nucleophile. The hydroxyl ion attacks the electrophilic carbon atom bonded to the chlorine, displacing the chlorine and forming an alkoxide. The product is propan-1-ol (C3H7OH), which I will denote as compound X.

Step 2: X + K2Cr2O7 + H2SO4 → CH3CH2COOH

Propan-1-ol (compound X from the previous step) undergoes an oxidation reaction with acidified potassium dichromate (K2Cr2O7 + H2SO4) to form propanoic acid (CH3CH2COOH). Dichromate acts as the oxidizing agent.

Step 3: CH3CH2COOH + Cl2 → X2

Propanoic acid reacts with chlorine gas (Cl2) in a halogenation reaction to form chloroethanoic acid (CH2ClCOOH), which I will denote as compound X2. Chlorine acts as the halogen.

Step 4: X2 + NaHCO3 → X3

Chloroethanoic acid (compound X2) undergoes a base-catalyzed elimination reaction with sodium bicarbonate (NaHCO3) to form chloroethene (C2H3Cl), which I will denote as compound X3. Sodium bicarbonate acts as the base.

Step 5: X3 + CH3CH2Br → X4

Chloroethene (compound X3) undergoes an addition reaction with bromoethane (CH3CH2Br) to form 1,2-dibromoethane (CHBr2CH2Br), which I will denote as compound X4. This is an example of an addition reaction across the double bond of the alkene.