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  1. Problems
  2. /Four compounds are each treated first with dilute and then with concentrated sodium hydroxide…
HardChemistryHigh School (11–12)Organic Compounds with Functional Groups › Aldehydes, Ketones and Carboxylic AcidsNumericalConceptual

Four compounds are each treated first with dilute and then with concentrated sodium hydroxide…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Four compounds are each treated first with dilute and then with concentrated sodium hydroxide: methanal, 2,2-dimethylpropanal, butanal and cyclohexanone. How many of them undergo the Cannizzaro reaction in concentrated base? Butanal (72.11 g/mol72.11\ \text{g/mol}72.11 g/mol) gives an aldol in dilute base, and the aldol loses water on heating to give 2-ethylhex-2-enal (126.20 g/mol126.20\ \text{g/mol}126.20 g/mol): 2 C3H7CHO→C8H14O+H2O2\,\text{C}_3\text{H}_7\text{CHO} \to \text{C}_8\text{H}_{14}\text{O} + \text{H}_2\text{O}2C3​H7​CHO→C8​H14​O+H2​O. Find the mass of 2-ethylhex-2-enal from 7.21 g7.21\ \text{g}7.21 g of butanal, in a 100 %100\ \%100 % yield.

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Need a reminder? The key ideaShowHide

An alpha hydrogen is needed for the aldol reaction (dilute base). With no alpha hydrogen, concentrated base gives the Cannizzaro reaction. A crossed aldol of two enolisable compounds gives four products. The carbonyl group is meta-directing. For the masses, use moles = mass / molar mass and the equation ratio.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. Cannizzaro. It needs an aldehyde with no alpha hydrogen. Methanal and 2,2-dimethylpropanal have none. Butanal and cyclohexanone have alpha hydrogens, and they do aldol reactions. So 2 compounds.

Step 2. Moles of butanal. 7.2172.11=0.100 mol\dfrac{7.21}{72.11} = 0.100\ \text{mol}72.117.21​=0.100 mol.

Step 3. Product. Two moles of butanal give one mole of the product, so 0.0500 mol0.0500\ \text{mol}0.0500 mol.

Step 4. Mass. 0.0500 mol×126.20 g/mol=6.31 g0.0500\ \text{mol} \times 126.20\ \text{g/mol} = 6.31\ \text{g}0.0500 mol×126.20 g/mol=6.31 g.

Answer: 2 compounds, and 6.31 g6.31\ \text{g}6.31 g.

Check. The product mass is less than the mass of butanal because water is lost: 7.21−0.90=6.317.21 - 0.90 = 6.317.21−0.90=6.31, with 0.0500 mol×18.02 g/mol=0.90 g0.0500\ \text{mol} \times 18.02\ \text{g/mol} = 0.90\ \text{g}0.0500 mol×18.02 g/mol=0.90 g of water.

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  1. Aldol and Cannizzaro reactions, and the benzene ringReactions of the Carbonyl Group: Addition, Redox and Aldol

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