please determine and show/explain how much aldehyde is added. For both of the al

Question

please determine and show/explain how much aldehyde is added. For both of the aldehydes listed in step 2.

Scheme 21.2 provides the mechanism for the reaction In this experiment you will compare the time of reac- ction with that of your partner and comment on what most likely leads to the observed dif- o form a hypothesis for this difference, a solid understanding of the mechanism is necessary In order t ference. It is important to note that the equilibrium is shifted to the product because the p solution and the extended conjugation makes the product considerably more stable EXPERIMENTAL PROCEDURE Place 300 of acetophenone (2.58 mmol) into the IOrnL round bot torn flask and add 3 mL of cthanol and a spinvane Oblain an equal number of moles of your assigned aldehyde and add it to the round bottom flask If you were assigned a liquid aldehyde, you will need to calculate the volume required. Sir the con- tents of the flask until everything is homogenous. The aldehydes are 4-chlorobenzaldehyde and 1. 2. You need to look up the physical properties of these compounds before coming to lab (www.sigmaaldrich.com) Obtain a single pellet of sodium hydroxide. Place the pellet on a piece of weigh paper and fold the weigh paper in half. Gently crush the pellet with your spatula into a powder. Transfer the powder to 3. the flask and record the time of addition

Explanation / Answer

As 4-chlorobenzaldehyde is a colorless to yellow powder or white crystalline solidhaving molecular weight of 140.566 g/mol

i.e.1 mol contains 140.566 g of 4-chlorobenzaldehyde

or 1000 mmol

contains 140.566 g of 4-chlorobenzaldehyde

so, 2.58 mmol contains (140.566/1000)X 2.58 =0.36266 grams of 4-chlorobenzaldehyde

similarly, for p-tolualdehyde, as it is acolorless liquid having molecular weight 120.151 g/mol

i.e. 1 mol contains 120.151 g of p-tolualdehyde

or 1000 mmol contains 120.151 g of p-tolualdehyde

so, 2.58 mmol contains (120.151/1000)X 2.58 =0.3099 grams of p-tolualdehyde

As it is liquid, so to calculate in volume, we have to divide mass by density of p-tolualdehyde

so, volume=0.3099 g /1.019 g/ml

=0.3041 ml of p-tolualdehyde is required

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