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  1. Problems
  2. /4.40 g of ethanal (44.05 g/mol) is warmed with an excess of methanol (32.04 g/mol) and a little dry…
HardChemistryHigh School (11–12)Organic Compounds with Functional Groups › Aldehydes, Ketones and Carboxylic AcidsNumericalConceptual

4.40 g of ethanal (44.05 g/mol) is warmed with an excess of methanol (32.04 g/mol) and a little dry…

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.

4.40 g4.40\ \text{g}4.40 g of ethanal (44.05 g/mol44.05\ \text{g/mol}44.05 g/mol) is warmed with an excess of methanol (32.04 g/mol32.04\ \text{g/mol}32.04 g/mol) and a little dry HCl. The acetal 1,1-dimethoxyethane (90.12 g/mol90.12\ \text{g/mol}90.12 g/mol) forms in a yield of 70 %70\ \%70 %, with one mole of water for each mole of acetal. Write the equation. What is the least mass of methanol that the reaction needs? What mass of acetal is obtained? Say why dry HCl is used.

Check your answer

Work it out on paper, then type your number. The answer itself stays hidden.

g
g
Need a reminder? The key ideaShowHide

A nucleophile attacks the carbonyl carbon, the carbon changes from sp2sp^2sp2 to sp3sp^3sp3, and the oxygen then takes a proton. Alkyl groups lower the positive charge and crowd the carbon, so methanal > other aldehydes > ketones. HCN gives a cyanohydrin, NaHSO3_33​ a bisulphite compound, alcohols with dry HCl an acetal, and H2N-Z\text{H}_2\text{N-Z}H2​N-Z compounds a C=N compound and water.

The worked solution

Have you tried it?

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. Equation. CH3CHO+2CH3OH⟶CH3CH(OCH3)2+H2O\text{CH}_{3}\text{CHO} + _{2}\text{CH}_{3}\text{OH} \longrightarrow \text{CH}_{3}\text{CH}(\text{OCH}_{3})_{2} + \text{H}_{2}\text{O}CH3​CHO+2​CH3​OH⟶CH3​CH(OCH3​)2​+H2​O

Step 2. Amount. 4.4044.05=0.0999 mol\dfrac{4.40}{44.05} = 0.0999\ \text{mol}44.054.40​=0.0999 mol.

Step 3. Methanol. Two moles for each mole of ethanal: 2×0.0999=0.200 mol2 \times 0.0999 = 0.200\ \text{mol}2×0.0999=0.200 mol, which is 0.200×32.04=6.40 g0.200 \times 32.04 = 6.40\ \text{g}0.200×32.04=6.40 g.

Step 4. Acetal. 0.0999 mol×90.12 g/mol×0.70=6.30 g0.0999\ \text{mol} \times 90.12\ \text{g/mol} \times 0.70 = 6.30\ \text{g}0.0999 mol×90.12 g/mol×0.70=6.30 g.

Step 5. Dry HCl. The reaction is an equilibrium that makes water. Water would push it back, and it would also hydrolyse the acetal. Dry conditions keep it on the acetal side.

Answer: 6.40 g6.40\ \text{g}6.40 g and 6.30 g6.30\ \text{g}6.30 g.

What if? Dilute aqueous acid reverses the reaction and gives back ethanal and methanol.

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