Question

Show how you might bring about the following conversions. For any conversion involving more than one step, show each intermediate compound.

Short answer

Question: Explain the three conversions mentioned in the given solution. Answer: The three conversions mentioned are: 1. Ethanol to ethene through dehydration: Ethanol is converted into ethene by removing a water molecule using a strong acid catalyst like sulfuric acid. 2. Cyclohexene to cyclohexanol through hydration: Cyclohexene is converted into cyclohexanol by adding a water molecule across the double bond in the presence of a strong acid catalyst like sulfuric acid or phosphoric acid. 3. Propane to propanoic acid through a two-step process (halogenation and hydrolysis): Propane is first converted to 1-chloropropane via halogenation in the presence of sunlight (UV light). Then, 1-chloropropane is converted to propanoic acid through hydrolysis using a strong base like sodium hydroxide followed by acidification.

Step-by-step solution

Conversion 1: Ethanol to Ethene (Dehydration)

Starting from ethanol (C2H5OH), we can convert it into ethene (C2H4) by removing a water molecule (HOH). The reaction can be facilitated with a strong acid catalyst, such as sulfuric acid. The overall reaction is shown below in a balanced form: C2H5OH ->[-H_2O] C2H4 + H2O

Conversion 2: Cyclohexene to Cyclohexanol (Hydration)

Starting from cyclohexene (C6H10), we can convert it into cyclohexanol (C6H11OH) by adding a water molecule across the double bond present in cyclohexene. The reaction can be done in the presence of a strong acid catalyst like sulfuric acid or phosphoric acid. The overall reaction is shown below in a balanced form: C6H10 + H2O ->[H^+] C6H11OH

Conversion 3: Propane to Propanoic Acid (Halogenation followed by Hydrolysis)

This conversion needs two steps: Step 1- Halogenation: Starting from propane (C3H8), we will replace one of the hydrogen atoms with a halogen atom, say chlorine (Cl) to form 1-chloropropane (C3H7Cl). The reaction can be done in the presence of sunlight (UV light) for the free radical halogenation. The overall reaction will be: C3H8 + Cl2 ->[\text{UV light}] C3H7Cl + HCl Step 2- Hydrolysis: For the second step, we will convert 1-chloropropane (C3H7Cl) into propanoic acid (C3H6O2) through hydrolysis. The hydrolysis can be done by treating the compound with a strong base like sodium hydroxide (NaOH) followed by acidification. The overall reactions for this step will be: C3H7Cl + NaOH -> C3H7ONa + NaCl C3H7ONa + HCl -> C3H6O2 + NaCl Now we have converted propane into propanoic acid following a two-step process.