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- For the following reactions, predict the likely product(s) of the reaction or write 'no reaction." Me H Me °x Me Me Me Br Me Me Me Nal, NaOCI EtOH, H₂O HNO3 ACOH AICI 3Given each value, determine whether the starting material or product is favored at equilibrium. a. Keq = 0.5 b. ΔG° = −100 kJ/mol c. ΔH° = 8.0 kJ/mol d. Keq = 16 e. ΔG° = 2.0 kJ/mol f. ΔH° = 200 kJ/mol g. ΔS° = 8 J/(K · mol) h. ΔS° = −8 J/(K · mol)Given these reactions, answer the questions (one answer can be used twice) NaOEt 1 YBr EtOH YBr Br KOtBu || ||| Br IV Reaction I is most likely Reaction II is most likely Reaction III is most likely Reaction IV is most likely NaCN ✓ [Choose] E2 E1 SN1 SN2 [Choose] [Choose ]
- Consider the reaction between aqueous solutions of ammonia and sodium cyanide. Select the products of the net ionic equation and whether products or reactants are favoured in the reaction. HCN(aq) NH4*(aq) NH2 (aq) OH (aq) | H20(?)Predict the product for the following reaction. ✔Saved CI || |||| IV NaCN CN ∞∞ m 111 CN CN IV CN+ F₂ catalyst The above reactants would undergo a catalyst .ل= + H H The above reaction would be an example of HHHH |||| H-C=C-C-C-C-C-H+2H-H | | | | HHHH The above reactants would undergo a(n) catalyst reaction ◆ reaction Isopropyl alcohol (propan-2-ol) is used as a solvent for industrial processes, a disinfectant solution, and a de-icer solution for vehicles. One method of producing isopropyl alcohol is to react propene with water in the presence of an acid catalyst. This reaction is classified as a(n) ◆ reaction.
- HO - CH, НО - ОН I CH_ - :0: solute CH ОН - NaCl ОН I CH - CH ОН - :0: || с :0: || - · C — CH2 — CH2 — C — ОН н Which is the better solvent? Н н C | H 1 C C 1 Н CH3- H2O CH₂OH C ||| C CH3(CH2), ОН :0: || S о- CH3 CH₂CH₂OH Н НQurston 18 Consuder the reatons shown. CI (8) t HCI cl + H2 What type of resction in it? Which is the corred product A,B or C? Name the first reactant the second reactont Name theGiven each value, determine whether the starting material or product is favored at equilibrium. a.Keq = 0.5 b.ΔGo = −100 kJ/mol c.ΔHo = 8.0 kJ/mol d.Keq = 16 e. ΔGo = 2.0 kJ/mol f.ΔHo = 200 kJ/mol g. ΔSo = 8 J/(K•mol) h.ΔSo = −8 J/(K•mol)