What is the rule of 13?

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What is the rule of 13?

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Answer 1 · 2015-03-27T18:22:04

The rule of 13 states that the formula of a compound is a multiple $n$ $n$ of 13 (the molar mass of $CH$ $"CH"$ ) plus a remainder $r$ $r$ .

Explanation:

The formula is used in interpreting mass spectra.

$Molecular formula = C_{n} H_{n + r}$ $"Molecular formula" = "C"_n"H"_(n+r)$

If you have heteroatoms, you adjust the formula. For example:

For $O$ $"O"$ , add $O$ $"O"$ and subtract ${CH}_{4}$ $"CH"_4$ .
For $N$ $"N"$ , add $N$ $"N"$ and subtract ${CH}_{2}$ $"CH"_2$ .
For $^{35} Cl$ $""^35"Cl"$ , add $Cl$ $"Cl"$ and subtract $C_{2} H_{11}$ $"C"_2"H"_11"$ .

For example, benzamide has a molecular ion at 121, which is an odd number.

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$\frac{121}{13} = 9 r 4$ $121/13 = 9r4$

The rule of 13 predicts a formula of $C_{9} H_{9 + 4} = C_{9} H_{13}$ $"C"_9"H"_(9+4) = "C"_9"H"_13$ .

The odd mass and the odd number of $H$ $"H"$ atoms make it reasonable to add an $N$ $"N"$ atom and subtract ${CH}_{2}$ $"CH"_2$ :

$N-15 = {N - (CH+H) = N-CH}_{2}$ $"N-15" = "N - (CH+H) = N-CH"_2$

This gives a formula $C_{8} H_{11} N$ $"C"_8"H"_11"N"$ .

The peak at 77 suggests $C_{6} H 5$ $"C"_6"H"5$ , so the peak at 44 is the other fragment.

$\frac{44}{13} = 3 r 5$ $44/13 = 3r5$ , which gives a formula of $C_{3} H_{3 + 5} = C_{3} H_{8}$ $"C"_3"H"_(3+5) = "C"_3"H"_8$ .

We know this contains $N$ $"N"$ , so the formula becomes $C_{2} H_{6} N$ $"C"_2"H"_6"N"$ .

The infrared spectrum may give further information.

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If it shows a strong peak at ${1640-1690 cm}^{-1}$ $"1640-1690 cm"^"-1"$ .

This shows the presence of an O atom.

$O - 16 = 16 - (CH + 2) = {16 - CH}_{4}$ $"O - 16" = 16 -("CH + 2") = "16 - CH"_4$

The formula becomes ${CH}_{2} NO$ $"CH"_2"NO"$ .

The two absorptions in the region ${3100-3500 cm}^{-1}$ $"3100-3500 cm"^"-1"$ suggest a primary amide ${CONH}_{2} (m / z = 44)$ $"CONH"_2 (m//z = 44)$ .

The structure becomes

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What is the rule of 13?

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