Question

Show how to prepare each compound from 1-phenyl-1-propanone. CCC(=O)c1ccccc1 CC(Br)C(=O)c1ccccc1 (a) CCC(=O)c1cccc(Br)c1 (b) 1-Phenyl-1-propanone (racemic)

Short answer

Question: Show how to prepare the following compounds starting from 1-phenyl-1-propanone: (a) 2-bromo-1-phenyl-1-propanone (b) 3-bromo-1-phenylpropan-1-one. Answer: (a) To synthesize 2-bromo-1-phenyl-1-propanone, first, generate the enolate ion by treating 1-phenyl-1-propanone with sodium ethoxide (NaOEt) in ethanol. Then, perform an alpha-halogenation reaction by reacting the enolate ion with bromine (Br2) to obtain 2-bromo-1-phenyl-1-propanone. (b) To synthesize 3-bromo-1-phenylpropan-1-one, first, convert 1-phenyl-1-propanone to its enol form using acid-catalyzed keto-enol tautomerization with H2SO4 in water. Then, perform electrophilic aromatic substitution using bromine (Br2) and a Lewis acid catalyst like aluminum(III) bromide (AlBr3). Hydrolyze the resulting enol to obtain 3-bromo-1-phenylpropan-1-one.

Step-by-step solution

Target compound (a)

The first compound to be synthesized is given by the SMILES notation: CC(Br)C(=O)c1ccccc1. This molecule is 2-bromo-1-phenyl-1-propanone.

Convert 1-phenyl-1-propanone to its enolate ion

To introduce the bromine atom at the alpha carbon, we need to start by converting the carbonyl compound into its enolate ion. This can be done by treating 1-phenyl-1-propanone with a base like sodium ethoxide (NaOEt) in ethanol. This deprotonates the alpha hydrogen, generating the enolate ion as a nucleophilic species. 1-Phenyl-1-propanone + NaOEt -> Enolate ion

Perform an alpha-halogenation reaction with bromine

To introduce the bromine atom at the alpha position, treat the enolate ion with a halogenating agent like bromine (Br2). The enolate ion reacts with the bromine, resulting in an alpha-brominated product, which is 2-bromo-1-phenyl-1-propanone. Enolate ion + Br2 -> 2-bromo-1-phenyl-1-propanone Now let's move on to target compound (b).

Target compound (b)

The second compound to be synthesized is given by the SMILES notation: CCC(=O)c1cccc(Br)c1. This molecule is 3-bromo-1-phenylpropan-1-one.

Convert 1-phenyl-1-propanone to its enol form

To perform the bromination on the aromatic ring, we should first convert the carbonyl group of 1-phenyl-1-propanone into an enol form. This can be done by using acid-catalyzed keto-enol tautomerization. Adding a small amount of a strong acid like H2SO4 in water will facilitate this transformation. 1-Phenyl-1-propanone + H2SO4 -> Enol form

Perform electrophilic aromatic substitution with bromine

Next, treat the enol form with the halogenating agent bromine (Br2) in the presence of a Lewis acid catalyst like aluminum(III) bromide (AlBr3) to facilitate electrophilic aromatic substitution. The bromine molecule is activated by the Lewis acid, generating a positively charged bromonium ion, which can react with the aromatic ring. This reaction introduces the bromine atom in the 3-position of the benzene ring, an ortho position relative to the -CO- group. Finally, hydrolyze the resulting enol to obtain 3-bromo-1-phenylpropan-1-one. Enol form + Br2 + AlBr3 -> Enol form of 3-bromo-1-phenylpropan-1-one Enol form of 3-bromo-1-phenylpropan-1-one + H2O -> 3-bromo-1-phenylpropan-1-one