I\'m trying to calculate the % yield of the reaction in the picture. I started w

Question

I'm trying to calculate the % yield of the reaction in the picture.

I started with 15g NaBr and 10ml n butyl alcohol. my final product is 3.03 grams. I've done some calculations and got a percent yield of 20.2% but I'm not sure if it's right. can I see your way of calculating the %yield so I can check my work? Thanks.

o02 Te5t Corvement reagents to convert tertiary alcohols to alkyl bromides or alkyl chlorides are hydrobromic acid and hydrochloric acid, respectively EXPERIMENT 8A Who 19. PREPARATION OF n-BUTYL BROMIDE This experiment illustrates method (c) above, the most economical method. n-Butyl alcohol (1- butanol) is converted to n-butyl bromide (1-bromobutane) by refluxing with aqueous sodium bromide and concentrated sulfuric acid NaBr CH2-CH2-CH2-CH2-Br CH3-CH2-CH2-CH2-OH H2SO4 Discussion This reaction presents an example of a fundamental reaction type known as a nucleophilic substitution. A leaving group in the starting material (in this case, the hydroxyl group, OH) is displaced by a nucleophile, or Lewis base (in this case, Br)) The exact mechanism of the substitution process

Explanation / Answer

One mole NaBr reacts with one mole n-butanol gives one mole 1-bromobutane (from equation).

Moles of NaBr = mass of NaBr/ molar mass of NaBr = 15 g/(102.89 g/mol) = 0.146 mol

So, moles of 1-bromobutane produced from 0.146 moles of NaBr is 0.146 moles.

Mass of n-butanol = volume of n-butanol x density of n-butanol = 10 mL x (0.81 g/mL) = 8.1 g

Moles of n-butanol = mass of n-butanol/molar mass n-butanol = 8.1 g/(74.12 g/mol) = 0.109 mol

So, the moles of 1-bromobutane produced from 0.109 moles of n-butanol is 0.109 mol.

Since n-butanol produced less amount of product, it is the limiting reactant and the amount of product corrosponding to the limiting reactant.

Mass of 1-bromobutane = moles 1-bromobutane x molar mass of 1-bromobutane

                                       = 0.109 mol x (137.02 g/mol)

                                       = 14.9 g

So, the theoretical yield is 14.9 g.

Actual yield is 3.03 g

% yield = (actual yield / theoretical yield)x100

            = (3.03 g/14.9 g)x100

            = 20.3%

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