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  1. Problems
  2. /A hexose (molar mass 180.16 g/mol) is treated with excess acetic anhydride…
MediumChemistryHigh School (11–12)Organic Compounds with Functional Groups › BiomoleculesNumerical

A hexose (molar mass 180.16 g/mol) is treated with excess acetic anhydride…

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.

A hexose (molar mass 180.16 g/mol180.16\ \text{g/mol}180.16 g/mol) is treated with excess acetic anhydride. A sample of 1.80 g1.80\ \text{g}1.80 g of the sugar gives 3.90 g3.90\ \text{g}3.90 g of the acetylated product, with every reactive -OH\text{-}\text{OH}-OH group changed to -OCOCH3\text{-}\text{OCOCH}_{3}-OCOCH3​. Each change replaces an H atom by a COCH3\text{COCH}_{3}COCH3​ group and adds 42.04 g/mol42.04\ \text{g/mol}42.04 g/mol. Find how many -OH\text{-}\text{OH}-OH groups the sugar has.

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mmol
g/mol
Need a reminder? The key ideaShowHide

A carbohydrate is a polyhydroxy aldehyde or ketone. An aldose has n−2n - 2n−2 stereocentres and a ketose has n−3n - 3n−3, where nnn is the number of carbons in the chain. The number of stereoisomers is 2centres2^{\text{centres}}2centres, and half of them belong to the D series. An empirical formula comes from the moles of each element in 100 g.

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.

Plan. Each acetylated -OH\text{-}\text{OH}-OH adds a fixed mass. Divide the mass gained per mole by it.

Step 1. Amount of the sugar. 1.80 g/180.16 g/mol=9.99 mmol1.80\ \text{g} / 180.16\ \text{g/mol} = 9.99\ \text{mmol}1.80 g/180.16 g/mol=9.99 mmol.

Step 2. Mass gained. 3.90−1.80=2.10 g3.90 - 1.80 = 2.10\ \text{g}3.90−1.80=2.10 g. For one mole: 2.10 g/0.00999 mol=210.2 g/mol2.10\ \text{g} / 0.00999\ \text{mol} = 210.2\ \text{g/mol}2.10 g/0.00999 mol=210.2 g/mol.

Step 3. Groups. 210.2/42.04=5.0210.2 / 42.04 = 5.0210.2/42.04=5.0.

Answer: the sugar has 5 hydroxyl groups, so it makes a pentaacetate.

Check. The pentaacetate has a molar mass of 180.16+5×42.04=390.4 g/mol180.16 + 5 \times 42.04 = 390.4\ \text{g/mol}180.16+5×42.04=390.4 g/mol, and glucose pentaacetate is C16H22O11\text{C}_{16}\text{H}_{22}\text{O}_{11}C16​H22​O11​ with 390.3 g/mol390.3\ \text{g/mol}390.3 g/mol.

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