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SN1 reaction - Wikipedia, the free encyclopedia

The SN1 reaction is a substitution reaction in organic chemistry. ...

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Question:what happens to the rate of an Sn1 reaction if you add in acetone

Answers:The effect depends on a lot of things like the charges of the reactants as well as what the original solvent was. If the reaction was originally done in a polar protic solvent, then adding acetone will drop the solvent polarity and will increase the transition energy barrier so the reaction will slow down.

Question:The reaction of 2-methyl-2-butanol with concentrated hydrochloric acid to produce 2-chloro-2-methylbutane is an example of Sn1 mechanism, right? It cannot undergo Sn2 mechanism because it is a tertiary structure. But then again, it might not react with Sn1 mechanism because chloride ion is a good nucleophile and Sn1 mechanism usually works with a weak nucleophile. Can this reaction occur via E1 mechanism? Since chloride ion is a weak base. I don't think it can occur through E2 mechanism because a strong base is needed. Please help!

Answers:The distinction of SN1 versus SN2 is which step is the rate limiting step. If bond cleavage precedes bond formation, then the rate will depend solely on the concentration of the substrate, this is SN1. The fact that tertiary carbons do not undergo an SN2 reaction is a pattern that I can think of examples that would make you question some aspects of that pattern. So, I would prefer tertiary carbons do no undergo SN2 reactions to cannot. This will be an SN1 reaction. Chloride is not necessarily a good nucleophile. It is when all other nucleophiles are excluded. If this reaction were attempted with t-butyl chloride and a good nucleophile, cyanide for example, the reaction would give elimination. An E2 elimination would compete with SN2 displacement. Since the substrate would not react at all in the presence of chloride (NaCl) as hydroxide is a stronger base, we need to make the leaving group a weaker base, water, by protonation. So the reaction conditions insure the reaction will take place by an SN1 (or E1) mechanism. Lets look further at the details of this reaction. It is difficult to control SN1 versus E1 reactions from taking place. If chloride, alcohol, or water were to deprotonate the carbocation to give the alkene. This would give the E1 elimination product. However, because it could reform the t-carbocation, it would do so and evidence of the E1 elimination would be hard to find. It is also important that the reaction be carried out in sufficient acid. Because an alcohol or water are stronger bases than chloride, they will be protonated more easily (as well as better nucleophiles). However, if the product forms, it will be less basic, will not protonate, and swing the equilibrium toward the chloride. So, this reaction is also controlled by Le Chatelier's principle. If you were to run this reaction with 1-butanol, the reaction would be an SN2 reaction. Some 1-butene would form as a by-product. This would form from competing E2 elimination. It would be difficult to know whether it formed from chloride or water as base, but neither are strong bases and the reaction would have been an E2 elimination.

Question:can a primary alkyl halide undergo sn1 reactions? If so under which conditions? thanks

Answers:usually Sn1 rxn undergoes only with secondary and tertiary alkyl halide and with primary the most favourable rxn would be sn2. So I dont think that alkyl halide would actually undergo such a unfavourablle rxn mechs.

Question:Which of the following 2 compounds will undergo Sn1 faster? a) A benzene ring with OCH3 in position 1, Br in position 4, and two methyl groups in postion five. b) on the second molecule, the methoxy is on carbon six, no substituent on carbon 1, and the Br and dimethyls are the same. Why?

Answers:There's something wrong with your question: "and two methyl groups in postion five" Each carbon in the benzene ring, BEFORE substitution, has at least two bonds due to adjacent carbons, and then a THIRD bond due to the double bonds that are typical of benzene rings. This means that, BEFORE substitution, each carbon already has three bonds. If you have two methyl groups on a single carbon, that means that the carbon will have 5 bonds, which is impossible. *** As an answer to your question, none of these compounds will undergo Sn1 at all, because neither of the two even exists!!! *** If carbon could somehow have 5 bonds, then my answer would be compound B. Reason: When the methoxy group comes off, you'll have a 2 carbocation. One of the two methyl groups could shift to that carbon, forming a more stable, 3 carbocation on the carbon that had two methyl groups. Of course, this isn't possible because the compounds don't exist.

From Youtube

E2 E1 Sn2 Sn1 Reactions Example 3 :E2 E1 Sn2 Sn1 Reactions Example 3

Sn1 Mechanism and Examples :Overview of the Sn1 substitution reaction and the different products we can form.