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Oraganic Chemistry I: Oxidizing Methoxybenzyl Alcohol 1) Explain why a phase-tra

ID: 911068 • Letter: O

Question

Oraganic Chemistry I: Oxidizing Methoxybenzyl Alcohol

1) Explain why a phase-transfer catalyst is used when oxidizing methoxybenzyl alcohol using NaOCl.

2)

substance                             quantity          molar mass (g/mole)                 bp (°C)

bleach (NaOCl)                       20 mL

dichloromethane                     18 mL              84.93                                        40°C

ethyl acetate                           40 mL              88.11                                         76—77°C

magnesium sulfate,

anhydrous                               1 g                 120.37

methoxybenzaldehyde

(reaction product)                                           136.15                                        248°C

p-methoxybenzyl alcohol         1.5 g             138.15

sodium chloride,

saturated solution                    20—30 mL

sodium hydroxide,5%              20 mL

tetrabutylammonium

hydrogen sulfate                     0.15—0.2 g     339.54

Using the data in the Reagents and Properties table above, calculate the theoretical yield of methoxybenzaldehyde in grams, assuming you start with 1.50 g of the alcohol. (Assume the alcohol is the limiting reactant). Please, show your work.

Explanation / Answer

1) methoxybenzyl alcohol will be soluble in oreganic solvent, whereas NaOCl will be soluble in water. Hence to bring these two reactants together  a phase-transfer catalyst shall be used.

In this reaction, Tetrabultyleammonium hydrogen sulphate (TBAS) shall be used as a phase-transfer catal.

2) The Balance euation for the above reation is

     C6H4Cl-CH2-OH + NaOCl --> C6H4Cl-CH=O

    That means 1 mole of methoxybenzyl alcohol will react with 1 mole of NaOCl

    From given data No. of moles of methoxybenzyl alcohol = 1.5 g/ 138.15 g/mole = 0.01086 moles

    Hence no. of moles of methoxybenzaldehyde = 0.01086 moles

   the theoretical yield = 0.01086 moles x 136.15 g/mole =1 .478 g

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