1
$\begingroup$

I know that "R-X + Mg" gives RMgX.

I want to ask if the Halide is Poly, then How's the formation of Grignard take place?

For example. If 1,4 - Dibromobutane react with Mg, then what will be the product?

$\endgroup$
1
  • $\begingroup$ As far as I know, side products such as MgBr2 form. Grignard Reagents contain only one halogen atom in their 'molecules'. $\endgroup$
    – Kartik
    Apr 28 '19 at 17:32
2
$\begingroup$

1,4-dibromobutane reacts with magnesium to form cyclobutane as main product. The the side products are buta-1,3-diene, ethene, butane, (E) and (Z) but-2-ene. The reaction has been performed in xylene solvent to give a mixture of hydrocarbon(81%) and cyclobutane(13%). The best yield for cyclobutane from 1,4-dibromobutane can be obtained by using lithium amalgam in dioxane for 3 hours.

enter image description here


In general, for any dibromoalkane reacting with magnesium will always lead to formation of alkene or cycloalkane and magnesium bromide as side product. For e.g.- 1,2 dibromoethane reacts with magnesium to form ethene and magnesium bromide. In fact, it is being used as activating agents as its action can be monitored by the observation of bubbles of ethylene and also the side-products($\ce{MgBr2}$) is almost negligible and does not affect the reaction. (Wikipedia)

$$\ce{Mg + BrC2H4Br → C2H4 + MgBr2}$$

The reaction mechanism is that any dibromoalkane reacting with magnesium will first lead to magnesium bromine adduct of the alkane chain which will then decompose to form alkene or cycloalkane and magnesium bromide as side product.

$$\ce{Br-(CH2)_n-Br + 2 Mg ->[ether] BrMg-(CH2)_n-MgBr}$$

1,3-dibrompropane reacting with magnesium will form cyclopropane and magnesium bromide as side product.

$$\ce{BrCH2CH2CH2Br + Mg -> (\Delta) + MgBr2}$$


Scientists have discovered more exotic species called magnesacycloalkanes which is said to maintain a equilibrium between magnesium bromine adduct of the alkane chain and the final product. Some of the species include:

enter image description here

enter image description here

Now, the species formation depends on position of bromine atoms in alkane chain:

  1. 1,3 DiGrignard reagent species: formed by reacting 1,3-Dibromopropane and magnesium to form 3-Bis(bromomagnesio)propane (4a) and 1,6—bis(bromomagnesio)hexane (15%) as side product.

enter image description here

  1. 1,2-DiGrignard reagent species: Formed when aromatic 1,2 Dihalides are used

enter image description here

  1. 1,1-DiGrignard reagent species: Also called methylene—di—Grignard reagent It is prepared from dibromo- or diiodomethane.

$$\ce{CH2Br2 + Mg ->[(i-Pr)2O] CH2(MgBr)2 }$$

References

  1. Bickelhaupt, F.. “Di-Grignard reagents and metallacycles.” Pure and Applied Chemistry 58 (1986): 537 -542, DOI: 10.1351/pac198658040537
  2. https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/alhalrx4.htm
  3. Science of Synthesis: Houben-Weyl Methods of Molecular Transformations Vol. 48: Alkanes, Georg Thieme Verlag, May 2014.
$\endgroup$
3
  • $\begingroup$ I asked for 1,4 - Dibromobutane, but you took 1,2 - Dibromoethane, I can understand that both bromine gone with Mg and two adjacent C free radical made a bond. But if the chain is long then Will it form cyclic? Please give the answer for the asked reaction.... $\endgroup$ Apr 29 '19 at 1:59
  • $\begingroup$ @HelenaWilliams I have updated my answer. Hope, it will answer your question. $\endgroup$ Apr 29 '19 at 4:51
  • $\begingroup$ See also this question. When the halogen atoms are conjugated to opposite ends of a double or triple bond, a 1,4 elimination occurs leading to a magnesium-free, reduced product. $\endgroup$ May 19 '19 at 12:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.