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Factors influencing Sn1 and Sn2 mechanisms in sp3 carbon substitutions, Exams of Chemistry

Organic Chemistry IIPhysical Organic ChemistryOrganic Chemistry I

An overview of nucleophilic substitutions at sp3 carbon, focusing on Sn1 and Sn2 mechanisms and the factors that influence their course. Sn2 reactions are favored by sterically accessible substrates, strong and small nucleophiles, and low to moderate polarity solvents, while Sn1 reactions are favored by sterically hindered substrates, weak and small nucleophiles, and moderate to high polarity solvents. The document also discusses the effects of steric nature of the substrate, nature of the nucleophile, solvent used, and leaving group on the reaction. Examples of Sn1 and Sn2 substrates and competing reactions are provided.

What you will learn

  • What are some examples of Sn1 and Sn2 substrates and their corresponding products?
  • What are the differences between Sn1 and Sn2 mechanisms in nucleophilic substitutions at sp3 carbon?
  • What are the factors that influence the course of nucleophilic substitutions at sp3 carbon?

Typology: Exams

2021/2022

Uploaded on 09/12/2022

anarghya
anarghya 🇺🇸

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Download Factors influencing Sn1 and Sn2 mechanisms in sp3 carbon substitutions and more Exams Chemistry in PDF only on Docsity! HIGHLIGHTS OF NUCLEOPHILIC SUBSTITUTION REACTIONS INVOLVING sp3 CARBON Sn2 REACTIONS From a synthetic point of view, this is the most useful reaction. It provides a means to prepare many functional groups from alkyl halides, and therefore from alkanes through the free radical halogenation reaction. Nu + C L C Nu + L Nucleophile Electrophile, or substrate L= leaving group Nucleophilic substitution product + Alkane Cl2 or Br2 R H X2 R X Alkyl halide (chloride or bromide) light +R X OH R OH AlcoholAlkyl halide +R X CN R CN NitrileAlkyl halide etc. + X + X The Sn2 mechanism: a) is a single step process b) involves no intermediates c) involves only one transition state, which is of low polarity d) follows second order (bimolecular) kinetics. That is, rate=k[substrate][nucleophile] In nucleophilic substitutions at sp3 carbon, Sn2 mechanisms are favored by using: a) sterically accessible substrates b) strong (negatively charged), small nucleophiles c) low to moderate polarity solvents Stereochemically, if the electrophilic center in the substrate is chiral, the Sn2 reaction produces a product with inverted configuration. Sn1 REACTIONS From a synthetic point of view, the Sn1 reaction is less useful. It is prone to side reactions such as eliminations and carbocation rearrangements. The Sn1 mechanism: a) is a multistep process b) occurs with formation of carbocation intermediates in the rate determining step c) involves one transition state per step. The rate-determining step involves a high polarity transition state d) follows first order (unimolecular) kinetics. That is, rate=k[substrate] In nucleophilic substitutions at sp3 carbon, Sn1 mechanisms are favored by using: a) sterically hindered substrates b) weak (neutral), small nucleophiles c) moderate to high polarity solvents that can stabilize the transition state and the carbocation intermediate Stereochemically, if the electrophilic center in the substrate is chiral, the Sn1 reaction produces a racemic product. The relative proportions of the enantiomers depend on the specific reaction, but will typically be close to 50/50. FACTORS THAT AFFECT THE COURSE OF NUCLEOPHILIC SUBSTITUTIONS AT sp3 CARBON 1. STERIC NATURE OF THE SUBSTRATE. Steric accessibility of the electrophilic center in the substrate is probably the most important factor that determines if a nucleophilic substitution will follow an Sn1 or an Sn2 mechanism. EXAMPLES OF Sn2 (sterically accessible) SUBSTRATES CH3 Br CH3 CH2 Cl primary substrates H3C CH CH3 Cl Br unhindered secondary substrates Br primary allylic halides EXAMPLES OF Sn1 (sterically hindered) SUBSTRATES C CH3 CH3 H3C Br H3C Br tertiary halides Cl hindered secondary halides C CH3 CH3 H3C CH2 Cl hindered primary halides The above example also shows the reason why, when the nucleophile is water or an alcohol, the group that replaces the leaving group in the product is the conjugate base of water (OH) or the alcohol (RO respectively). Another example illustrates a similar point. Can you provide a step by step mechanism (it might be in the test, you never know)? Br The leaving group (Br) is on a secondary carbon, but this carbon is next to a tertiary carbon. The nucleophile is water, therefore the expected product is an alcohol. The product will consist of a mixture of the expected secondary alcohol (minor) and a tertiary alcohol (major) due to the rearrangement to a more stable cation shown below. H2O acetone (polar solvent) OH expected product (minor) + OH rearranged product (major) CH3 Br CH3OH (solvent and nucleophile) CH3 OCH3 minor + H3C OCH3 major notice that the conjugate base of the nucleophile (in red) has replaced the leaving group Br (1) Sn1 secondary cation hydride r.d.s. (2) tertiary alcohol conjugat acid (4) H2O + H3O H H shift tertiary cation, more stable H H2O O HH (3) O HH OH a proton transfer (acid-base reaction) yields the free alcohol Sn1 Sn1 REACTIONS AND REARRANGEMENTS INVOLVING PRIMARY SUBSTRATES Primary substrates normally do not follow Sn1 mechanisms because they do not form stable cations. However, a hindered primary substrate can be forced into an Sn1 mechanism if sufficient energy and time are allowed, for example boiling the substrate in a nucleophilic solvent such as ethanol. The nucleophile cannot do a backside attack, and the substrate cannot form stable cations. In this case the substrate will begin to rearrange as the leaving group departs. This avoids formation of a primary cation. As the leaving group departs, a positive charge begins to develop on the carbon bearing the leaving group, and the rearrangement process starts, all in unison. The reaction of neopentyl bromide with ethanol illustrates this point. As the positive charge develops on the primary carbon while bromine leaves, the methyl group is migrating to an adjacent position to forma more stable cation. H3C C CH2 CH3 (2) CH3 + H O O H H2O O Substitution product (an ether) H3C C CH3 CH3 CH2 Br CH3CH2OH heat H3C C CH3 C H3 C Br H H δ + δ − H3C C C H2 CH3 CH3 (1) δ+
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