When the following compound is treated with the base sodium hydride (NaH), an intramolecular S2 reaction occurs. Please draw both the organic intermediate and the major organic product obtained. Be sure to include the correct stereochemistry and charges. Cl OH
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- Enamines can serve as enolate surrogates in reactions at the a-carbon. In the reaction sequence, the structures of the enamine addition product – the initial zwitterion and its neutral tautomer – are shown. Draw the structures of the two reactants forming these intermediates, and draw the structure of the final product, obtained via hydrolysis of the neutral intermediate. initial zwitterionic intermediate neutral intermediate tautomerization Reactants H,0 hydrolysis product Draw the two reactants. Draw the hydrolysis product. Select Draw Rings More Erase Select Draw Rings More Erase H HPart 2 of 2 When the following compound is treated with the base sodium methoxide (NaOMe), an intramolecular S 2 reaction occurs. Please draw both the organic intermediate and the major organic product obtained. Be sure to include the correct stereochemistry and charges. O OH hyb) Listed below are several hypothetical nucleophilic substitution reactions. None is synthetically useful because the product indicated is not formed at an appreciable rate. In each case provide an explanation for the failure of the reaction to take place as indicated. OMe HO + OMe + OH HO + CH; OH
- Step 4: In an acid-catalyzed epoxide ring opening, the epoxide is first protonated and can then be attacked by weak nucleophiles. H HO: :Nuc Nuc Weak nucleophiles such as water and alcohols can react with the protonated epoxide. One of the main differences between base-catalyzed and acid-catalyzed epoxide ring openings is the regiochemical product expected. In an acid-catalyzed reaction, if one of the epoxide carbons is tertiary and the other primary, the nucleophile will attack at the more substituted carbon. If the epoxide is monosubstituted, the nucleophile will attack preferentially at the primary carbon vs. the secondary carbon. Which reaction will not give the indicated product? N 1. CH,CH,O 2. H₂O OH CH₂CH₂OH, H* N CH₂CH₂OH, H 1 CHICH,O 2. H₂O HO HO H- HThe following reaction involves two sequential Heck reactions. Draw structural formu- las for each organopalladium intermediate formed in the sequence and show how the final product is formed. Note from the molecular formula given under each structural formula that this conversion corresponds to a loss of H and I from the starting material. Acetonitrile, CH,CN, is the solvent. 1% mol Pd(OAc), 4% mol Ph,P CH,CN C4H171 C4H16Acyl transfer (nucleophilic substitution at carbonyl) reactions proceed in two stages via a "tetrahedral intermediate." Draw the tetrahedral intermediate as it is first formed in the following reaction. CI H₂N OH • You do not have to consider stereochemistry. • Include all valence lone pairs in your answer. • Do not include counter-ions, e.g., Na+, I, in your answer. • In cases where there is more than one answer, just draw one. Sn [F
- Please helpAlcohols are acidic in nature. Therefore, a strong base can abstract the acidic hydrogen atom of the alcohol in a process known as deprotonation. The alcohol forms an alkoxide ion by losing the proton attached to the oxygen atom of the hydroxyl ( -OH) group. The alkoxide formed can act as a base or a nucleophile depending on the substrate and reaction conditions. However, not all bases can abstract the acidic proton of alcohols and not all alcohols easily lose the proton. Deprotonation depends on the strength of the base and the acidity of the alcohol. Strong bases, such as NaNH2, can easily abstract a proton from almost all alcohols. Likewise, more acidic alcohols lose a proton more easily. Determine which of the following reactions would undergo deprotonation based on the strength of the base and the acidity of the alcohol. Check all that apply. ► View Available Hint(s) CH3CH,OH + NH3 →CH,CH,O-NH CH3 CH3 H3C-C-H+NH3 → H3 C-C-H OH O-NH CH3CH2OH + NaNH, → CH3CH,O-Na* + NH3 CHC12 Cl₂…Give the major organic product of the following reaction. H2 Ni
- Acyl transfer (nucleophilic substitution at carbonyl) reactions proceed in two stages via a "tetrahedral intermediate." Draw the tetrahedral intermediate as it is first formed in the following reaction. CI + H₂N • You do not have to consider stereochemistry. • Include all valence lone pairs in your answer. OH • Do not include counter-ions, e.g., Na+, I¯, in your answer. • In cases where there is more than one answer, just draw one.Enamines can serve as enolate surrogates in reactions at the a-carbon. In the following reaction sequence, the structures of the enamine addition product – the initial zwitterion and its neutral tautomer – are shown. Draw the structures of the two reactants forming these intermediates, and draw the structure of the final product, obtained via hydrolysis the neutral intermediate. reactants initial zwitterionic intermediate tautomerization hydrolysis product neutral intermediate NH н,о Н,о ↑Acid-catalyzed hydrolysis of the following epoxide gives a trans diol. CH3 CH3 CH3 НО, НО, H,SO, + H,O HO HO Only this glycol This glycol is is formed. not formed. Of the two possible trans diols, only one is formed. How do you account for this stereoselectivity?