# How can I draw possible structures for an ester that has a molecular ion with an m/z value of 74?

##### 2 Answers
Jun 19, 2015

You subtract the masses of the atoms you know are present, and then you work out all possible combinations of atoms that could give you the remaining mass.

#### Explanation:

You know that the compound is an ester, $\text{RCOOR}$'.

The $\text{COO}$ has a mass of 44, so the mass of $\text{R + R}$' must equal $74 - 44 = 30$.

If both $\text{R}$ groups are saturated, their total formula is ${C}_{n} {H}_{2 n + 2}$.

$12 n + 1 \left(2 n + 2\right) = 14 n + 2 = 30$

n = (30 – 2)/14 = 2

The possibilities are: $n = 0 + 2$ and $n = 1 + 1$.

The corresponding formulas are

${\text{HCOOC"_2"H}}_{5}$ (ethyl formate)

and

${\text{CH"_3"COOCH}}_{3}$ (methyl acetate)

Sep 14, 2015

This seems to be in the context of mass spectroscopy, so I'm assuming that you're trying to identify an unknown compound by looking at its spectrum and knowing that it's an ester ahead of time from, say, IR spectroscopy.

A molecular mass (MM) of $14$ would likely be ${\text{CH}}_{2}$, while ${\text{CH}}_{3}$ is about $15$. Oxygen is $16$ per atom. You know that you have at least ${\text{RCOOCH}}_{3}$ or $\text{RCOOCH"_2"R'}$. Having a MM of $74$ makes me think:

$74 = {\overbrace{15}}^{\text{remaining methyl" + overbrace(12 + 16*2 + 15)^"ester group}}$

So this suggests that your compound could be methyl acetate.

Another isomeric possibility, if you move the left methyl group to the rightmost connection:

$74 = {\overbrace{1 + 12 + 16 \cdot 2 + 14}}^{\text{ester+aldehyde group" + overbrace(15)^"remaining methyl}}$

which is ethyl formate.

It is unclear which one is more likely, but if you also have the IR spectrum, that should help. An aldehyde bending vibration peak on there is somewhere near ${\text{1740~1720 cm}}^{-} 1$. Unfortunately, an ester bending vibration peak is around ${\text{1750~1735 cm}}^{-} 1$.

Fortunately, methyl acetate would have one well-defined peak, whereas ethyl formate would give you two indistinct, overlapping peaks in that vicinity---the same carbonyl is shared in an aldehyde and ester group. That aldehyde+ester peak would be wider and less well-defined.

Also, ethyl formate would give a very obvious wide peak near ${\text{3000 cm}}^{-} 1$ for the ester stretching vibration.

Here are the IR spectra of each one for you to compare.

Methyl acetate:

Ethyl formate: