Topic 20 · page 6 of 15

Grignard reagents and carbon dioxide

A Grignard reagent is a halogenoalkane that has been turned inside out. Its carbon atom is now the electron-rich end, so it forms a new carbon-carbon bond with carbon dioxide and gives a carboxylic acid with one more carbon.

13 min read20.2

Why lengthen the chain? 20.2

The reactions you know so far change a functional group and leave the carbon chain alone. To build a bigger molecule from a smaller one you must make a new carbon–carbon bond. You have met one way already: halogenoalkane with cyanide ions gives a nitrile, adding one carbon. Grignard reagents are a second way, and they can add more than one carbon.

Making a Grignard reagent 20.2

CHX3CHX2Br+Mg→dry etherCHX3CHX2MgBr \ce{CH3CH2Br + Mg ->[\text{dry ether}] CH3CH2MgBr}

The reagent in this example is ethylmagnesium bromide. The magnesium metal is used as turnings, and the mixture is warmed gently under reflux.

In the halogenoalkane the carbon next to the halogen is δ+, because halogens are more electronegative than carbon. In the Grignard reagent the carbon is bonded to magnesium, which is less electronegative than carbon, so that carbon is now δ−.

CHX3CHX2δ− ⁣− ⁣MgBrδ+ \overset{\delta-}{\ce{CH3CH2}}\!-\!\overset{\delta+}{\ce{MgBr}}

The δ− carbon acts as a nucleophile. It attacks a δ+ carbon, such as the carbon in CO₂ or in a C=O group, and makes a new C–C bond.

Everything must be dry

Grignard reagents react with water, and the reagent is destroyed.

CHX3CHX2MgBr+HX2O→CHX3CHX3+Mg(OH)Br \ce{CH3CH2MgBr + H2O -> CH3CH3 + Mg(OH)Br}

The product is an alkane, so no carbon chain is lengthened. For this reason the halogenoalkane, the magnesium, the apparatus and the ether solvent must all be dry. Ether is used because it has no O–H or other acidic hydrogen, so it does not react with the reagent, and because its oxygen atoms help to keep the reagent in solution.

Grignard reagents and carbon dioxide 20.2

Carbon dioxide has a δ+ carbon. The Grignard reagent attacks it and forms a salt of a carboxylic acid. Adding dilute acid then releases the acid itself.

MethodFrom halogenoalkane to carboxylic acid

  1. Mg in dry ether turns the halogenoalkane R–X into R–MgX.
  2. Carbon dioxide (solid, called dry ice, or the gas) is added in dry ether.
  3. Dilute acid (for example HCl(aq)) is added to release R–COOH.
CHX3CHX2MgBr+COX2→CHX3CHX2COOMgBr \ce{CH3CH2MgBr + CO2 -> CH3CH2COOMgBr}
CHX3CHX2COOMgBr+HX+→CHX3CHX2COOH+MgX2++BrX− \ce{CH3CH2COOMgBr + H+ -> CH3CH2COOH + Mg^2+ + Br^-}
A scheme in three stages. Bromoethane, with two carbon atoms, reacts with magnesium in dry ether to give ethylmagnesium bromide. This reacts with carbon dioxide in dry ether to give a magnesium salt. Dilute acid then gives propanoic acid with three carbon atoms. The carbon and oxygen atoms that came from carbon dioxide are highlighted.
Bromoethane has two carbons and propanoic acid has three. The extra carbon, with its two oxygen atoms, comes from carbon dioxide.

The carbon of the carboxyl group, R–COOH, comes from CO₂. The acid has one more carbon than the halogenoalkane you started with.

Worked exampleWhich halogenoalkane makes butanoic acid?

  1. Butanoic acid, CH₃CH₂CH₂COOH, has four carbons. The COOH carbon came from CO₂, so the halogenoalkane has three carbons.
  2. The halogenoalkane is 1-bromopropane, CH₃CH₂CH₂Br.
  3. Route: CH₃CH₂CH₂Br → CH₃CH₂CH₂MgBr (Mg, dry ether) → CH₃CH₂CH₂COOMgBr (CO₂, dry ether) → CH₃CH₂CH₂COOH (dilute acid).

Two ways to add one carbon

Cyanide route Grignard route
Steps R–X to R–CN, then hydrolyse R–X to R–MgX, then CO₂, then acid
Reagents KCN in ethanol, heat; then dilute HCl, heat Mg, dry ether; CO₂; dilute acid
Product R–COOH, one more carbon R–COOH, one more carbon
Hazard cyanide is very toxic ether is highly flammable; strictly dry conditions

Check yourself

  1. Write an equation, with a name for the product, for the formation of a Grignard reagent from 1-bromobutane.
  2. Give the reagents and conditions for the conversion of 1-bromobutane into pentanoic acid.
  3. Write an equation for the reaction of CH₃CH₂CH₂MgBr with water. Why does this matter in the laboratory?
  4. Which halogenoalkane, with Mg and then CO₂, gives 2-methylpropanoic acid, (CH₃)₂CHCOOH?
  5. A student made propanoic acid from bromoethane in 60% yield. The theoretical yield from 5.45 g of bromoethane was 3.70 g of acid. Calculate the mass of acid obtained.
  6. Compare the carbon chain length when bromoethane reacts (a) with KCN, then dilute acid, and (b) with Mg, CO₂, then dilute acid.

Answers cover these until you have tried the questions

  1. CHX3CHX2CHX2CHX2Br+Mg→CHX3CHX2CHX2CHX2MgBr\ce{CH3CH2CH2CH2Br + Mg -> CH3CH2CH2CH2MgBr} in dry ether. The product is butylmagnesium bromide.
  2. Mg in dry ether to give CH₃(CH₂)₃MgBr; add CO₂ (dry ice) in dry ether; then add dilute acid, for example HCl(aq).
  3. CHX3CHX2CHX2MgBr+HX2O→CHX3CHX2CHX3+Mg(OH)Br\ce{CH3CH2CH2MgBr + H2O -> CH3CH2CH3 + Mg(OH)Br}. Water destroys the Grignard reagent and gives an alkane, so every reagent, the apparatus and the ether must be dry.
  4. The COOH carbon comes from CO₂, so the halogenoalkane is (CH₃)₂CHBr: 2-bromopropane. Mg in dry ether; CO₂; dilute acid.
  5. Mass obtained = 60% of 3.70 g = 0.60 × 3.70 = 2.22 g. (Check: n(CH₃CH₂Br) = 5.45 ÷ 108.9 = 0.0500 mol, giving 0.0500 × 74.0 = 3.70 g.)
  6. Both give propanoic acid, three carbons, which is one more than bromoethane. The cyanide route adds the carbon of CN, and the Grignard route adds the carbon of CO₂.