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- 1) Hydrolysis of Methyl Salicylate: Methyl Salicylate (C8H8O3) + sodium Hydroxide (NaOH) -> sodium Salicylate ( NaC7H5O3) + Methanol (CH3OH) 2) Acidification of Sodium Salicylate: Sodium salicylate (NaC7H5O3) + HCL -> Salicylic acid (C7H603) + sodium chloride (NaCL) Could you find the theoretical yield for me? 2.15g of methyl salicylate was used. 25ml of sodium hydroxide (2m solution) was used for hydrolysis and 2ml of hydrochloric acid was used for the acidification. This produced a yield of 1.53g of salicylic acid.Calcium carbonate and lactic acid used to prepare Calcium lactate mixture. Whu?Complete the reaction of propylene oxide with methylamine
- Use the following reagent table for the reaction of : cyclohexanol + H2 SO4 → cyclohexene Chemical MW (g/mol) Density (g/mL) mmols used Amount used cyclohexanol 100.16 0.9624 1.50 mL Sulfuric Acid 98.08 1.83 1.0 mL cyclohexene 82.14 How many mmols of sulfuric acid are used? (answer with 2 decimal places, no units)An important step in one synthesis of carboxylic acids is the deprotonation of diethyl malonate and its alkyl-substituted derivative: Base CH;CH2O OCH,CH3 CH;CH,0 OCH2CH3 H2 Diethyl malonate Base CH;CH,0 °C `OCH,CH3 CH;CH,O OCH,CH3 R Alkyl substituted diethyl malonate NaOH can deprotonate diethyl malonate effectively, but NaOC(CH3)3 is typically used to deprotonate the alkyl-substituted derivative. Explain why.Give the product for the reaction.
- 78) Why are primary alcohols more acidic than tertiary alcohols, in general? Because tertiary alcohols have less acidic hydrogens. Because primary alcohols have less electron-donating groups which create more effective charge separation between oxygen and hydrogen atom, making it more acidic. Because primary alcohols are more polar than tertiary alcohols. Because primary alcohols have less electron-donating groups which decrease electron density on oxygen, making it more susceptible for H+ to depart from the structure.give me some example of drug synthesis that are obtained from name reactionNaOH, 20, 5°C Determine the predominant product(s) from the reaction shown below, using the given compounds. O ОН O он
