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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?

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  • $\begingroup$ 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
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In general, for any dibromoalkane reacting with magnesium will always lead to forming alkene or cycloalkane and magnesium bromide as side product. For ex- 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($\ce{MgBr2}$) in almost negligible and does not affect the reaction. (Wikipedia)

$$\ce{Mg + BrC2H4Br → C2H4 + MgBr2}$$

Now in general the reaction mechanism is that for 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.

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

If 1,3 dibrompropane reacts with magnesium, it will form cyclopropane and magnesium bromide.

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

Now on researching more on this topic, scientists have discovered a more exotic species i.e magnesacycloalkanes which maintains a equilibrium between magnesium bromine adduct of the alkane chain.

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and from that more number of exotic species has been discovered.

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Now, products vary depending on position of bromine atoms in alkane chain:

  1. 1,3 DiGrignard reagents 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

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  1. 1,2-DiGrignard reagents Here, aromatic 1,2 Dihalides is used and no side product is observed

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  1. 1,1-DiGrignard reagent The methylene—di—Grignard reagent can be prepared from dibromo- or diiodomethane.

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


As for your reaction i.e 1,4-dibromobutane and magnesium, it has been observed that the reaction various side products i.e buta-1,3-diene, cyclobutane, ethene, butane, (E) and (Z) but-2-ene. The reaction has been performed in xylene solvent to give a mixture of hydrocarbond(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.(Ref. 3)

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References

  1. Pure & AppL Chem., Vol. 58, No.4, pp. 537—542, 1986. DOI - https://pdfs.semanticscholar.org/8ca4/6e6b43bcb1c336fb47d33a372a144ca31bc0.pdf
  2. https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/alhalrx4.htm
  3. https://books.google.co.in/books?id=DiKGAwAAQBAJ&pg=PA619&lpg=PA619&dq=1-2+dibromobutane+%2B+magnesium&source=bl&ots=JYxN8jnYva&sig=ACfU3U1F56uq_adwsyXVIqO5afuxPnl9XA&hl=en&sa=X&ved=2ahUKEwiwis2AtPThAhUBOI8KHZoKAHwQ6AEwDnoECAgQAQ#v=onepage&q=1-2%20dibromobutane%20%2B%20magnesium&f=false
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  • $\begingroup$ 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.... $\endgroup$ – HelenaWilliams Apr 29 '19 at 1:59
  • $\begingroup$ @HelenaWilliams I have updated my answer. Hope, it will answer your question. $\endgroup$ – Nilay Ghosh 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. $\endgroup$ – Oscar Lanzi May 19 '19 at 12:07

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