help with #2 Supply the requested information 24 a. Moles of t-butyl alcohol int

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help with #2

Supply the requested information 24 a. Moles of t-butyl alcohol introduced. b. Moles of HCL introduced c. The limiting reagent is d. Theoretical yield-moles-of t-butyl chloride e. Theoretical yield-grams-of t-butyl chloride. f. Actual yield-grams g. Percent yield of t-butyl chloride. Why was the excess hydrochloric acid neutralized by the addition of 5% aqueous sodium bicarbonate rather than an aqueous solution of sodium hydroxide? 2. oentlyf unnd iastood of e weak bise 3. s-Butyl chloride cannot be prepared from s-butyl alcohol by the method used for t- butyl chloride. Outline a procedure for the preparation of s-butyl chloride from s- butyl alcohol; accompany with a suitable balanced equation. Hint, consult your text book's index. Ort HC 5-butyl ehlore

Explanation / Answer

HCl + NaHCO3 ----> NaCl + H2O +CO2   enthalpy of reaction = -34 kJ/mol

HCl + NaOH   -----> NaCl + H2O             enthalpy of reaction = -58 kJ/mol

So, both reactions (HCl with NaOH and NaHCO3 ) are exothermic reaction. But the amount of heat released with NaOH is almost double of NaHCO3 . In simple words NaOH will react violently while NaHCO3 will react less violently with HCl.

Also, the product formed (2-Chlorobutane) will react with OH- of NaOH to form the reactant (butan-2-ol). But NaHCO3 will not undergo above stated reaction, as it is not a direct source of OH-. So to avoid the high concentration of OH- ion one uses NaHCO3. In other words, to avoid the reverse reaction one does not use NaOH.

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