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In a book I refer, it is said that:

Both Ionic [carbanion] mechanism as well as free radical mechanism are widely accepted for Wurtz reaction.

When I read this statement, I was thinking, which of these mechanisms would be able to explain the order of reactivity of different substrates towards Wurtz reaction. My conclusion was that the rate of the reaction should follow the below mentioned order:

$1^\circ > 2^\circ > 3^\circ$ if it takes place through carbanion mechanism; and

$3^\circ > 2^\circ > 1^\circ$ if it takes place through free radical mechanism

where $1^\circ$ refers to primary alkyl halide, $2^\circ$ refers to secondary alkyl halide, and $3^\circ$ refers to tertiary alkyl halide.

So, my doubt is what is the correct consolidated order of reactivity of alkyl halides towards Wurtz reaction?

I know my question could be a possible duplicate of this.

However, I did not get a satisfactory explanation over there and hence have asked it as a new question.

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  • $\begingroup$ I'd say free radical mechanism $\endgroup$ – user93057 May 10 at 6:20
  • $\begingroup$ @user93057 Eventhough 3° free radical is more stable,it does not give the expected product in the wurtz reaction.It is also reported that 3° alkyl halides undergo disproportionation reaction under conditions maintained for wurtz reaction. So,will the order of reactivity be 2°>1°>3° (where this has the same meaning as what I have explained in my question)? $\endgroup$ – Chem-Learner May 10 at 7:39
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The Wurtz reaction is an organic reaction, in which two molecules of an alkyl halide (e.g., $\ce{R-Cl}$) are coupled in the presence of $\ce{Na, Mg}$, or other reactive metals to form an alkane (original classical Wurtz reaction used $\ce{Na}$ metal). The mechanism begins with a single electron transfer (SET) from sodium metal to the alkyl halide, which dissociates to form an alkyl radical ($\ce{R^.}$) in a similar fashion to the formation of a Grignard reagent and sodium halide salt (e.g., $\ce{NaCl}$):

$$\ce{R-X + Na^\circ -> R^. + NaX} \; \text{ where } \ce{X = Cl, Br,} \text{ or } \ce{I}$$

A second molecule of sodium undergoes another SET to the alkyl radical to form a nucleophilic carbanion (e.g., $\ce{R-}$):

$$\ce{R^. + Na^\circ -> R^- + Na+} $$

This alkylmetal intermediate can be isolated (e.g., alkyllithium) and has been isolated in a several cases. The carbanion then attacks another molecule of alkyl halide in a nucleophilic substitution reaction ($\mathrm{S_N2}$) to form the final coupled product (e.g., $\ce{R-R}$) and another molecule of sodium halide salt:

$$\ce{R^- + R-X -> R-R + X-} $$

Thus, the mechanism is believed to be undergoing in both free radical and ionic pathways in particular steps. The depicted diagram of the mechanism is illustrated in below:

Wurtz reaction

In a review of the Wurtz reaction (Ref.1) revealed that the reaction is severely limited in scope as demonstrated by the following observations:

  1. In the synthesis of noncyclic systems (dimerization or cross-coupling) reasonable yields require the use of primary alkyl halides, with iodides giving the best results. Secondary halides give very poor results;
  2. Cross-coupling reactions (using a mixture of two different alkyl halides) give approximately statistical ratios of cross-coupled : dimeric products, due to the high reactivity of the organosodium intermediates involved. Thus, in the synthesis of 3-methyluntriacontane (4 ($\ce{R-R'}$); see above diagram), coupling of the iodide (1; $\ce{R-I}$) with 1-iodooctadecane (2; $\ce{R'-I}$) under standard Wurtz conditions gave a mixture of 3 ($\ce{R-R}$) from dimerization of 1, 4 ($\ce{R-R'}$) from the desired cross-coupling, and 5 ($\ce{R'-R'}$) from dimerization of 2, in the ratio $21:50:29$.

However, This classical reaction is of value in the preparation of cyclic systems, notably [2.2]phanes and bicyclobutanes (e.g., bicyclo[1.1.0]butane), as well as in the preparation of symmetric dimers. The Wurtz reaction provides one of the most successful approaches to these class of compounds, by ring-closure reactions. The reaction may proceed either intra- or inter-molecularly, with intramolecular couplings giving better yields (Ref.1):

Bicyclo systems from Wurtz reaction

The modest yields of [2.2]phane products are due to competing oligomerization/cyclization reactions, which is demonstrated by the product distribution obtained on treatment of the dibromide (6) with sodium/tetraphenylethane (TPE) in THF at $\pu{-80 ^\circ C}$ (See above diagram). A homologous series of products has been obtained under these conditions, the approximate yields of the metacyclophanes being obtained were given in the scheme.

In conclusion, the Wurtz reaction is seldom used because of side reactions and it is intolerant of a range of functional groups. It has limited use to the synthesis of symmetric alkanes and cycloalkanes. If two dissimilar alkyl halides are taken as reactants in cross coupling reactions, the resulting product is a mixture of alkanes that is often difficult to separate. In the case of (1,3), (1,4), (1,5), (1,6) dihalides (when used as a reactant), the reaction leads to formation of cyclic products instead of intended dimerization. The Wurtz reaction fails in case of tertiary halides. In addition, since the reaction involves free radical species in one part of the mechanism, a side reaction occurs to produce an alkene (specially with 1,2-dihalides). This side reaction becomes more significant when the alkyl halides are bulky at the halogen-attached carbon atom.

There is an organometalic compound in the mechanism at one stage, which undergoes a $\mathrm{S_N2}$ replacement of halide ion of available second $\ce{R-X}$ molecule. Whether these types of organometalic compounds are ionic (cabanionic nature of $\ce{R-}$) has been discussed and reviewed using quantum chemical calculations recently (Ref.2).

Thus, it is safe to say that the reaction is better for primary ($1^\circ$) halides, regardless of whether the mechanism is free radical oriented or carboanion oriented. Even more than 150 years after Charles Adolphe Wurtz introduced the reaction in 1855, the mechanism of the reaction is still up in air, regardless of involvement of both in the mechanism.

References:

  1. David C. Billington, "Chapter 2.1: Coupling Reactions Between $\mathrm{sp^3}$ Carbon Centers (https://doi.org/10.1016/B978-0-08-052349-1.00068-8)," In Comprehensive Organic Synthesis, Volume 3: Carbon–Carbon $\sigma$-Bond Formation; Barry M. Trost, Ian Fleming, Editors-in-Chief; Elsevier Science Ltd.: New York, NY, 1991, pp. 413-434 (ISBN: 978-0-08-052349-1).
  2. Christoph Lambert, Paul von Ragué Schleyer, “Are Polar Organometallic Compounds “Carbanions”? The Gegenion Effect on Structure and Energies of Alkali‐Metal Compounds,” Angew. Chem., Intl. Ed. 1994, 33(11), 1129-1140 (https://doi.org/10.1002/anie.199411291).
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  • $\begingroup$ Should it be carbanion or free radical reaction $\endgroup$ – An enthusiast May 11 at 11:42
  • $\begingroup$ There are evidence for both. $\endgroup$ – Mathew Mahindaratne May 11 at 15:08
  • $\begingroup$ I know that carbanion had been trapped by using ${CO_2}$ was there a proof for free radical as well? $\endgroup$ – An enthusiast May 11 at 15:59
  • $\begingroup$ It was mention that formation of alkene can be discussed by only free radical mechanism. I disagreed because any carbanion acting as a base could do that. However, SET can be done only by metal acing as a free radical. $\endgroup$ – Mathew Mahindaratne May 11 at 16:12

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