Reaction Quenching: Our Fisher Esterification reaction is being quenched using sodium bicarbonate Why do we use this reagent? Explain, in words and by drawing out acid/base chemistry reactions, wha the quench would accomplish. Be sure to use pKą values in your explanation. HINT: there are 2 different quench reactions that occur. Look at the reagents and the mechanism for what species exis in the reaction mixture that can react with sodium bicarbonate. OH Isobutyric acid C4H8 02 MW: 88.11 g/mol + HO Isobutyl alcohol C4H10O MW: 74.12 g/mol H₂SO4 (cat.) Reflux O Isobutyl isobutyrate C8H1602 MW: 144.20 g/mol + H₂O

Reaction Quenching: Our Fisher Esterification reaction is being quenched using sodium bicarbonate Why do we use this reagent? Explain, in words and by drawing out acid/base chemistry reactions, wha the quench would accomplish. Be sure to use pKą values in your explanation. HINT: there are 2 different quench reactions that occur. Look at the reagents and the mechanism for what species exis in the reaction mixture that can react with sodium bicarbonate. OH Isobutyric acid C4H8 02 MW: 88.11 g/mol + HO Isobutyl alcohol C4H10O MW: 74.12 g/mol H₂SO4 (cat.) Reflux O Isobutyl isobutyrate C8H1602 MW: 144.20 g/mol + H₂O

Organic Chemistry: A Guided Inquiry

2nd Edition

ISBN:9780618974122

Author:Andrei Straumanis

Publisher:Andrei Straumanis

Chapter13: Substitution

Section: Chapter Questions

Problem 4E

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Reaction Quenching. Our cinnamaldehyde to cinnamyl alcohol reduction reaction is being quenched using HCl. Why do we use this reagent? Explain, in words and by drawing out acid/base chemistry reactions, what the quench would accomplish. Be sure to use pKa values in your explanation. HINT: there are 3 different quench reactions that occur. Look at the reagents and the mechanism for what species exist in the reaction mixture that can react with HCl. Cinnamaldehyde MW: 132.16 g/mol b.p. 248°C NaBH4, CH₂CH₂OH Diethyl Ether, 0°C OH Cinnamyl alcohol MW: 134.18 g/mol m.p. 30-33°C
Complete the mechanism for these Lewis acid-base reactions. Draw mechanistic arrows and partial charges where appropriate. Reaction A: CH₂NH₂ + HCl = 3 Reaction B: CH₂NH₂ + H₂O = 3 Reaction C: CH₂NH¯ + H₂O = 3 + Reaction D: CH₂NH 3 + H₂O 2 When each of these reactions has come to an equilibrium (where the relative amounts of each chemical species remains unchanged but the forward and reverse reactions are still occurring), predict whether each reaction will have more reactant or more product. Explain your predictions using your knowledge of atomic and molecular structures and electronegativity.
Give a complete arrow-pushing mechanism for the following reactions. Show all important resonance structures of the intermediates and show exactly where each proton comes from and goes to. Indicate the Lewis acid and Lewis base (LA or LB) and whether they are also Bronsted acids and bases (BA or BB) for each intermolecular reaction. To preview image Click Here EtOH HCI (cat.) 0 LOEt OEt + H₂O
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Alcohols 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…
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The pKa of our product is ~ 2-3 units lower than the pKa of phenol. Draw all possible resonance structures of the conjugate base of the product, and use these to explain why our product is so much more acidic than phenol.
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