Experimental [Notes 4, 5, and 6| OH NaBH 4. OH OH Step I: Procedure for the Sodi

Question

Experimental [Notes 4, 5, and 6| OH NaBH 4. OH OH Step I: Procedure for the Sodium Borohydride Reduction of 3-Ethoxy-4-hydroxybenzaldehyde: Dissolve 1.62 g of 3-ethoxy-4-hydroxybenzaldehyde in 10 l of 10 M NaOH in a 125 mL Erlenmeyer flask with a magnctic stir bar. Cool the flask in a cold-water bath, and then add, in small portions, 0.30 g of NaBH4 over five minutes. Remove the cold-water bath and allow the reaction flask to warm to room temperature over twenty minutes. Cool the reaction flask again in the ice-water bath, and slowly add 2.5 M HCI (Add the hydrochloric acid solution SLOWLY; there is a lot of foaming. Caution: Hydrogen gas evolution!). Check the acidity of the reaction mixture once the gas evolution begins to be less pronounced, somewhere around 7-8 mL. Keep adding the hydrochloric acid until the reaction mixture is acidic. Let the flask sit in the ice-water bath for an additional five minutes, then vacuum filter the precipitate, washing with cold water (4 x 10 mL). Allow the solid to dry with the vacuum on for a good fifteen minutes; water is hard to get rid of. This material should be pure enough for characterization (mass, 'H NMR, IR and mp) and for the next reaction.

Explanation / Answer

First part/ Step 1:

Given reaction is

3-ethoxy-4-hydroxybenzaldehyde   + NaBH4,NaOH,H2O   --------------------> 3-ethoxy-4-hydroxy benzyl alcohol

1 mol                                                                                                         1 mol

166.17 g                                                                                                     168.19 g

1.62 g                                                                                                              ?

Hence,

                                    ? = ( 1.62 g/ 166.17 g) x 168.19 g product

                                        = 1.64 g product

This is the theoretical yield of the product i.e 1.64 g.

But, given that actual yield of the product = 1.36 g

Therefore,

percent yield of the reaction = ( actual yield /theoretical yield ) x 100

                                          = ( 1.36 g /1.64 g) x 100

                                        = 82.9 %

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Therefore,

theoretical yield of the product = 1.64 g

actual yield of the product = 1.36 g

percent yield of the reaction = 82.9 %

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