Using your roadmaps as a guide, show how to convert 1-bromopropane and carbon dioxide into 4-heptanone. You must use 1-bromopropane and carbon dioxide as the source of all carbon atoms in the target molecule. Show all reagents and all molecules synthesized along the way.

Short Answer

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Answer: To convert 1-bromopropane and carbon dioxide into 4-heptanone, follow these steps: 1. Formation of Grignard Reagent - React 1-bromopropane with magnesium in dry ether to form propylmagnesium bromide (C3H7MgBr). 2. Reaction with Carbon Dioxide - Add carbon dioxide to the Grignard reaction mixture to form magnesium salt intermediate (C3H7CO2MgBr). 3. Formation of Carboxylic Acid - Hydrolyze the magnesium salt intermediate using a dilute acid to form butanoic acid (C3H7CO2H). 4. α-haloacylation of Butanoic Acid - Perform α-haloacylation using halogen and phosphorus tribromide (PBr3) to form 2-bromoheptanoic acid (C3H6BrCO2H). 5. Formation of 4-heptanone - React 2-bromoheptanoic acid with lithium aluminum hydride (LiAlH4) to form the final product, 4-heptanone (C3H6BrCOCH3).

Step by step solution

01

Formation of Grignard Reagent

Firstly, prepare a Grignard reagent from 1-bromopropane. Grignard reagents are organomagnesium compounds, which are formed by the reaction of an alkyl or aryl halide with magnesium metal in anhydrous ether or THF. To form the Grignard reagent, add magnesium turnings to a mixture of 1-bromopropane and dry ether. The Grignard reagent produced is propylmagnesium bromide: C3H7Br + Mg → C3H7MgBr This Grignard reagent will be used in the next step to react with carbon dioxide.
02

Reaction with Carbon Dioxide

Add carbon dioxide to the Grignard reaction mixture containing propylmagnesium bromide. The Grignard reagent will react with CO2 to form a magnesium salt intermediate: C3H7MgBr + CO2 → C3H7CO2MgBr
03

Formation of Carboxylic Acid

Next, hydrolyze the magnesium salt intermediate using a dilute acid, typically aqueous hydrochloric acid (HCl) or sulfuric acid (H2SO4), which leads to the formation of the carboxylic acid: C3H7CO2MgBr + HCl → C3H7CO2H + MgBrCl The product formed here is butanoic acid.
04

α-haloacylation of Butanoic Acid

Perform α-haloacylation of the butanoic acid using halogen (chlorine, bromine, or iodine) and phosphorus tribromide (PBr3) as the catalyst. This reaction results in the formation of 2-bromoheptanoic acid: C3H7CO2H + Br2 + PBr3 → C3H6BrCO2H The PBr3 serves as a catalyst and returns to its original form after the reaction.
05

Formation of 4-heptanone

Finally, carry out the reaction with lithium aluminum hydride (LiAlH4) to reduce the 2-bromoheptanoic acid to 4-heptanone. LiAlH4 is a strong reducing agent, suitable for reducing carboxylic acids to aldehydes or ketones. In our case, the reaction will form 4-heptanone: C3H6BrCO2H + LiAlH4 → C3H6BrCOCH3 The final product is 4-heptanone, and it is derived from 1-bromopropane and carbon dioxide.

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Most popular questions from this chapter

Using the principles for writing mechanisms and the four steps in the "How to Write Mechanisms for Interconversions of Carboxylic Acid Derivatives"box of this chapter, write mechanisms showing all electron flow arrows for the following reactions: (a) Hydrolysis of \(N, N\)-dimethylacetamide in acidic water. (b) Hydrolysis of acetic anhydride in basic water. (c) Esterification of acetic acid in acidic ethanol. (d) The reaction of dimethylamine in water with acetic anhydride to create N,N-dimethylacetamide. (e) Partial hydrolysis of acetonitrile in acidic water to create acetamide.

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