Haloalkanes and Haloarenes · Lesson 11 of 16
Organometallic Reagents and Carbon–Carbon Coupling
“Use carbon–metal polarity and coupling reactions to track carbon skeletons and recognise practical limitations.”
• Explain formation and proton sensitivity of Grignard reagents in dry ether. • Predict hydrocarbon and deuterium-quench products with correct carbon counts. • Distinguish Wurtz, Fittig and Wurtz–Fittig coupling equations. • Evaluate mixed products and elimination limitations instead of treating every formal coupling as a clean synthesis.
The same carbon changes from electrophilic to nucleophilic
In an alkyl halide the C–X bond makes carbon relatively electron-poor. When magnesium enters the bond under appropriate anhydrous ether conditions, RMgX is formed and the C–Mg bond is strongly polar toward carbon. Carbon now has carbanion-like, basic and nucleophilic character. This reversal of polarity explains why an alkyl halide and its organomagnesium derivative have very different reactions even though the carbon skeleton can be unchanged.
The formal formation equation is RX+Mg→RMgX in dry ether. Ether coordinates to magnesium and helps stabilise the organometallic system; it is not merely an inert word written on an arrow. Real solutions contain solvated and associated species, so a simple R–Mg–X formula is a useful representation rather than a complete single-molecule description of every solution. Appropriate chlorides, bromides and iodides can be used, while reaction initiation and competing functional groups matter.
Problem
What happens to ethylmagnesium bromide on adding water?
- 1.The ethyl group behaves as a strong base and takes a proton from water.
- 2.A balanced formal equation is C₂H₅MgBr+H₂O→C₂H₆+Mg(OH)Br.
- 3.The two-carbon group becomes ethane; it does not become ethanol by attaching OH to carbon. Proton transfer destroys the organomagnesium reagent before its intended carbon–carbon bond-forming use.
Alcohols and compounds with sufficiently acidic O–H or N–H bonds can similarly quench a Grignard reagent. “All amines” is too broad: an N–H proton is needed for that particular proton-transfer statement, and tertiary amines have no N–H. Moisture exclusion therefore concerns both solvent and substrate functional groups. It follows from the reagent’s basicity, not from a belief that water always breaks every metal bond identically.
Problem
Infer R when RMgBr gives (CH₃)₂CHD on quenching with D₂O.
- 1.The newly transferred atom is D, so replace D conceptually by the original C–Mg attachment to identify the carbon fragment.
- 2.R=(CH₃)₂CH– and the precursor is isopropyl bromide. Formation gives (CH₃)₂CHMgBr; D₂O quench gives (CH₃)₂CHD+Mg(OD)Br.
- 3.Carbon count stays three. Heavy water labels the protonation position; it does not add a carbon or change the skeleton into butane.
Bromobenzene can likewise form phenylmagnesium bromide with Mg in dry ether, and protonation gives benzene. Its formation is not an ordinary aryl SN1 or SN2 example; metal insertion follows different chemistry. A cyclohexyl bromide gives cyclohexylmagnesium bromide and then cyclohexane on hydrolysis, preserving the six-membered saturated ring.
Wurtz joins two alkyl fragments
The formal Wurtz equation is 2RX+2Na→R–R+2NaX in dry ether. Two equivalent ethyl fragments give butane; two n-butyl fragments give n-octane. A two-carbon starting alkyl halide therefore makes a four-carbon coupled skeleton, not an alkene of the same carbon count. The method is most useful as a simple pattern for symmetrical products from suitable primary halides.
Problem
Compare Wurtz products of n-butyl bromide and isobutyl bromide.
- 1.n-Butyl bromide has fragment CH₃CH₂CH₂CH₂–. Joining two reacting terminal carbons gives unbranched n-octane.
- 2.Isobutyl bromide has fragment (CH₃)₂CHCH₂–. Joining two such CH₂ groups gives (CH₃)₂CHCH₂CH₂CH(CH₃)₂, 2,5-dimethylhexane.
- 3.Both products are C₈H₁₈, so a molecular formula cannot distinguish them. The Wurtz product can help infer which primary C₄H₉Br skeleton was present.
With two different alkyl halides, R–R, R–R′ and R′–R′ products can be possible, so an unsymmetrical alkane is not normally obtained as a single predictable pure product. Secondary and especially tertiary halides can also undergo elimination and other side reactions. A formal dimer equation does not establish an efficient practical route.
Problem
A drawn coupling product is (CH₃)₃C–C(CH₃)₃. What fragment is implied, and how should the route be qualified?
- 1.Cut the central joining bond: each fragment is tert-butyl, (CH₃)₃C–, so a formal R′X precursor is a tert-butyl halide.
- 2.The structural inference predicts R′MgX as tert-butylmagnesium halide and its protonated hydrocarbon as isobutane in this structural identification scheme.
- 3.It does not prove a clean Wurtz synthesis of this severely crowded dimer. A tertiary halide has strong elimination competition, so this is formal fragment identification rather than an unqualified recommended preparative yield.
Fittig and Wurtz–Fittig change the fragment classes
| Coupling family | Balanced representative equation | Organic skeleton |
|---|---|---|
| Wurtz | 2C₂H₅Br+2Na→C₄H₁₀+2NaBr | Alkyl–alkyl |
| Fittig | 2C₆H₅Br+2Na→C₆H₅–C₆H₅+2NaBr | Aryl–aryl: biphenyl |
| Wurtz–Fittig | C₆H₅Br+CH₃Br+2Na→C₆H₅CH₃+2NaBr | Aryl–alkyl: toluene |
All three compact examples use sodium and dry ether in their intended descriptions. A mixed aryl/alkyl coupling can also have competing homocouplings; product purification and substrate limitations matter. The Wurtz–Fittig equation requires two sodium atoms and two halide-salt equivalents for the two initial C–X bonds. Omitting a coefficient simply because an abbreviated scheme shows only one NaX would violate the full atom balance.
Problem
Plan the two-stage route.
- 1.Brominate benzene with Br₂/FeBr₃ under suitable ring-substitution conditions to give bromobenzene plus HBr.
- 2.Apply the formal Fittig coupling: 2C₆H₅Br+2Na→C₆H₅–C₆H₅+2NaBr in dry ether.
- 3.The product has two six-carbon aromatic rings joined by one C–C bond: twelve carbons. It is biphenyl, also historically called diphenyl, not a fused two-ring naphthalene skeleton.
Quick Quiz
Quiz
What character does carbon have in RMgX?
Explanation: The C–Mg polarisation is toward the more electronegative carbon.
What does water do to C₂H₅MgBr?
Explanation: The carbon fragment takes H from water and the Grignard reagent is quenched.
Which reaction couples two aryl halides to a biaryl?
Explanation: Fittig has the aryl–aryl coupling pattern.
Why is a mixture of two alkyl halides a poor general route to a pure unsymmetrical Wurtz product?
Explanation: More than one pair of alkyl fragments can join.
What is the Wurtz product of isobutyl bromide in the formal symmetric coupling?
Explanation: Joining the two CH₂Br reaction centres preserves both branched isobutyl fragments.
Practice Problems
- Write formation and D₂O quench of cyclohexylmagnesium bromide.
- Why must an alcohol group be considered before preparing a Grignard reagent?
- Write the formal Wurtz route from 1-chlorobutane to n-octane.
- Give the formal product from coupling two sec-butyl fragments and state a limitation.
- Write balanced Wurtz–Fittig production of ethylbenzene from suitable bromides.
- Explain why tertiary-amine nitrogen is not covered by “every amine protonates a Grignard reagent”.
C₆H₁₁Br+Mg→C₆H₁₁MgBr in dry ether; C₆H₁₁MgBr+D₂O→C₆H₁₁D+Mg(OD)Br. The deuterium occupies the original carbon–magnesium attachment position; the six-carbon ring remains intact.
Key Takeaways
• Magnesium insertion reverses the effective carbon polarity of a suitable halide. • Dry coordinating ether and compatible functional groups are essential to Grignard chemistry. • Proton or deuterium transfer converts the organomagnesium carbon into a hydrocarbon attachment. • Wurtz, Fittig and Wurtz–Fittig describe alkyl–alkyl, aryl–aryl and mixed coupling families. • Coupling preserves fragment connectivity while adding their carbon counts. • Mixed and hindered substrates can give competing products rather than a clean formal target. • Balanced sodium-coupling equations account for both initial halogens.