Conversion Reactions (III) Problem 5

Description:

Problem

What reactions will bring about the following conversions?

1. 1 − Bromopropane to 2 − Bromopropane

2. Benzyl alcohol to 2 − phenylethanoic acid

3. Aniline to chlorobenzene

4. 2 − Chloropropane to 1 − propanol

Solution

1 − Bromopropane to 2 − Bromopropane

CH₃CH₂CH₂Br → CH₃ − CH(Br) − CH₃

• The reactant has bromine at C1 carbon while the product has bromine at C2 carbon.

• Dehydrohalogenation of 1 − Bromopropane gives propene.

• Markovnikov addition of HBr on propene gives 2 − Bromopropane where bromine is at C2 carbon.

2. Benzyl alcohol to 2 − phenylethanoic acid

• The product has 1C atom more than the reactant.

• Thus, we need to convert –OH to –CN using KCN, followed by complete hydrolysis to give acid.

• However, we know that direct substitution of alcohol cannot take place as –OH is a poor leaving group.

• Hence, we convert it to benzyl chloride followed by nucleophilic substitution using KCN.

3. Aniline to chlorobenzene

• Direct substitution of −NH₂ by chlorine cannot take place as NH₂ is a very poor leaving group.

• Thus, it is first converted to diazonium salt followed by substitution with CuCl/HCl.

4. 2 − Chloropropane to 1 − propanol

CH₃CH(Cl)CH₃ → CH₃CH₂CH₂OH

• The reactant has –Cl at C2 while product has –OH at C1.

• First, convert it into an alkene followed by Antimarkovnikov addition of HBr.

• Lastly, substitution with aqueous KOH is carried out to get 1 − propanol as the final product.

• The steps involved are dehydrohalogenation, using alc.KOH/Δ, reaction with HBr/peroxide to get Antimarkovnikov’s product CH₃CH₂CH₂Br, substitution using aq.KOH/Δ.