Is it possible to do a Friedel-Crafts alkylation on a plain alkane such as the one I've attached below, or is it only reserved for EArS mechanisms?

Also, I was wondering what would happen if I tried this reaction with another halogen on the starting substrate (if I had 1-bromoethane for example)...

Substrate is an ethane, reactant is 1-chloropropane (reacts with AlCl3), product is 2-methylpentane.

Edit 1: I realize now that the top reaction doesn't make sense. Rather than a Friedel-Crafts Alkylation, I think I found a mechanism to make this type of reaction work (SN2) with Br as a leaving group and CN as a strong nucleophile, but I'm not sure...

1-bromoethane as the substrate, reacts with butyronitrile; the reagent, to form (S)-2-ethyl-3-methylbutanenitrile as the product

Edit 2: I miscounted the carbons. 1-bromoethane as the substrate, reacts with pentanenitrile; the reagent, to form (S)-2-ethyl-3-methylbutanenitrile as the product

Edit 3: This is what I understand, from the suggestions given...

1-bromoethane as the substrate, reacts with isobutyronitrile; the reagent to form (S)-2-ethyl-3-methylbutanenitrile as the product + an enantiomer

  • $\begingroup$ I assume you've seen the FC alkylation mechanism before. How do you think it could be extended to this? Can you draw out a proposed mechanism? $\endgroup$
    – orthocresol
    Dec 10 '20 at 15:22
  • $\begingroup$ For a moment, I thought it were an example of ~Wurtz-Fittig reaction~, but your reaction does not include elemental sodium. $\endgroup$
    – Buttonwood
    Dec 10 '20 at 16:09
  • $\begingroup$ yes, I don't think that my first attempt makes sense. I tried another mechanism in my edit to form a different product. $\endgroup$ Dec 10 '20 at 17:33
  • $\begingroup$ The edit also does not make sense. Count the number of carbons. $\endgroup$
    – Zhe
    Dec 10 '20 at 17:49
  • $\begingroup$ oops I missed that. but can that mechanism still work? $\endgroup$ Dec 10 '20 at 17:54

It appears that you are having some difficulty with this synthesis. The suggestion in the Comments that you employ lithium diisopropylamide (LDA) to generate the anion of a nitrile is a wise one. The anion is usually formed a -78oC in tetrahydrofuran (THF). The bromide is added and the reaction mixture is allowed to warm to ambient temperature. There are two possible bonds that can be made in what is termed a retrosynthetic analysis. One by method A and another by method B. The bonds formed are shown in red in structures 2. Method A is better than method B because an SN2 reaction is being conducted on a primary halide, ethyl bromide. Method B requires an SN2 reaction on a secondary halide, 2-bromopropane, a process that is hindered and in competition with an E2 elimination that would produce propene 4 and the starting nitrile 3. Nitrile 2 must be a racemate given its method of formation. It is comprised of equal amounts of the two enantiomers, (R)-2 and (S)-2. By the methods employed here, a single enantiomer cannot be formed. Thus, a "normal bond" (not dashed nor bold) is shown in nitriles 2.

  • $\begingroup$ yes, I am struggling with this, thank you so much. so in order to get the desired product I absolutely have to start with the nitrile on my substrate? $\endgroup$ Dec 11 '20 at 0:39
  • $\begingroup$ If you must start with nitrile 3, add the anion to acetone and dehydrate to the conjugated nitrile. Then the double bond has to be reduced without affecting the nitrile. Perhaps Pd/C and H2 in ETOAc at atmospheric pressure. But that's another question. $\endgroup$
    – user55119
    Dec 11 '20 at 1:05
  • $\begingroup$ thank you!! I really appreciate all your help! $\endgroup$ Dec 11 '20 at 1:54

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