Question #6d344

1 Answer
Aug 9, 2016

#3# hydroxyl groups.

Explanation:

The idea here is that when you're treating your unknown compound #"X"# with excess acetyl chloride, #"CH"_3"COCl"#, the hydrogen atoms that belong to a hydroxyl groups, #-"OH"#, will be replaced by an acetyl group, #"CH"_3"CO"-#.

You don't need to know how that happens, i.e. the mechanism, all you need to know is that it does happen.

This is exactly what takes place when you react acetyl chloride with an alcohol, #"R"-"OH"#

#color(red)("CH"_ 3"CO")"Cl" + "R"-"OH" -> color(red)("CH"_3 "CO")"OR" + "HCl"#

Now, notice what happens when one hydroxyl group is present. The molecule, which in this example is #"ROH"#, is losing a hydrogen atom, #"H"#, and gaining an acetyl group.

If you take its initial relative molecular mass to be #"M"#, you can say that after the reaction is complete, its molar mass will be

#M_"after" = M - 1 + 43 = M + 42#

Here #1# represents the relative atomic mass of hydrogen and #43# represents the relative molecular mass of the acetyl group. One hydrogen atom jumps off, one acetyl group takes its place.

This tells you that for every hydroxyl group present on compound #"X"#, its relative molecular mass will increase by #42# after its treatment with excess acetyl chloride.

The problem tells you that the relative molecular mass of compound #"X"# increased by #126# following the reaction. This means that it contained

#126 color(red)(cancel(color(black)("increase"))) * "1 hydroxyl group"/(42 color(red)(cancel(color(black)("increase")))) = "3 hydroxyl groups"#

Therefore, you can say that compound #"X"# contained #3# hydroxyl groups.