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Be careful! Ethanol is not a strong acid (pKa~16). So its conversion to ethoxide anion is not so easy. Rather, it would attack the C center by its lone pair and then it would release proton to form the product. Actually here what I think is, the hetero atom O just near to the C undergoing substitution will assist by its lone pair to kick out chloride and ...

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When you were taught that we don't see $S_{\mathrm{N}}2$ reactions at $sp^{2}$ centers, what evidence or rationale was given? In organic chemistry, very few things are definite. When they are, it's usually because of being forbidden by quantum mechanics (e.g., symmetry in pericyclic reactions). Otherwise, they are just very unlikely, unless you somehow ...

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Because you are a beginner to organic chemistry, I'll tell you this: Both esterification (e.g., Fischer esterification) and de-esterification reactions are reversible. The question you had at the end is why the last step of base catalyzed de-esterification reaction is only forward reaction. As you have learned in general chemistry class, acid-base reaction ...

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As most of us chemists have understood and experienced, the Grignard reagent have undergone the many reactions with suitable substrates. Yet, the most familiar and probably the most used reaction is the addition to a carbonyl group to give $2^\circ$- or $3^\circ$-alcohol. The Grignard reagents' affinity for a corbonyl group in the presence of an acidic $\... -1 It's true that tertiary alcohols cannot react with PX3 where X is a halogen (usually Cl or Br). The reason lies in the mechanism. The attack of X- on the protonated alcohol is Sn² which is not possible on 3° carbon. PS : Refer to the mechanism attached! 0 This is what I found from your reference to Clayden. You just had to read the next paragraph ( I have highlighted the point ). You can see that March and Clayden have agreed on the same point, that electron withdrawl from the phenyl/allyl group increases the effective conjugation in the transition state. So I hope You can understand what will be the correct ... 0 It has to do with an (equimolar) Racemic mixture being formed! Note that a racemic mixture still has chiral atoms but cannot contribute to any net rotation of the plane polarized light making it optically inactive as a combined product 1 I think it's difficult to say for sure without actually making experiments. It should depend a lot on the conformation of the reactant, the substituents, the leaving group and also on the solvent. Nevertheless, I would imagine that it should be possible in some cases if you have a good leaving group. In the case of a substituted cis-decalin, for example (I ... 2 Further to my comment - the solubility of Sodium Fluoride in Acetone is 2.4 microgrammes per 100g of solvent i.e. essentially insoluble source. There will be no F- in solution so the reaction will not occur. 3 That's not the right view. The$S_{N}2$mechanism does not favor strong nucleophiles. That statement doesn't make sense at a high level. Mechanisms can't favor anything. The key here is to look at what controls reactivity in$S_{N}2$and$S_{N}1\$ reactions. The former is concerted where the bond breaking and bond forming take place at the same time. ...

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step-1: cleavage of C=O bond: the C=O bond undergoes heterolytic cleavage to give positive charge on carbon and negative charge on oxygen just like almost all other organometallic reagents react with a ketone, for example a grignard reagent (not exactly, but similar to it). step-2:(the real step which caused deviation from your answer)-resonance of ...

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In accordance with what @SteffX said, it really depends on the conditions and every reaction has its own story. However a couple of factors favour the formation of the given product above others. There include the fact the MeOH might help to generate the "activated Nucleophile" anion at amide, being a polar protic solvent, it provides higher solvation to the ...

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We do get meta-chlorotoluene when toluene is chlorinated by EAS, but only in minor amounts. This paper describes a process designed to selectively produce the ortho isomer over the para one, but in addition to some para-chlorotoluene still being formed the abstract reports 4.0% meta-chlorotoluene. There are also small amounts of dichlorinated products and ...

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This is simply wrong. The person setting this question has assumed that the hydrazine will be completely selective for the ketone over the alkyl bromide and that the strong basic conditions will also cause elimination of the Br- to give the styrene. Selectivity seems most unlikely give the typical forcing conditions used in W-K reductions Hydrazine is a ...

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