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The so-called nitrogen rule states that if a compound has an odd number of nitrogen atoms, the value of \(m / z\) for its molecular ion will be an odd number. Why?

Short Answer

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Answer: The nitrogen rule states that if a compound has an odd number of nitrogen atoms, the value of \(m/z\) for its molecular ion will be an odd number. This relationship occurs because nitrogen is the only element with an odd atomic mass among the common elements in organic compounds. When a compound contains an odd number of nitrogen atoms, it results in an odd contribution to the mass-to-charge ratio, leading to an odd \(m/z\) value according to the nitrogen rule.

Step by step solution

01

Understand the nitrogen rule and its relation to the mass-to-charge ratio

The nitrogen rule states that if a compound contains an odd number of nitrogen atoms, then its mass-to-charge ratio (\(m/z\)) of the molecular ion will be an odd number. The mass-to-charge ratio (\(m/z\)) is determined by the sum of atomic masses of all the elements in the compound.
02

Analyze the structure of a compound with respect to nitrogen atoms

Let's consider how nitrogen atoms affect the mass-to-charge ratio (\(m/z\)) of a compound. Nitrogen has an atomic mass of 14, which is an even number. If we add nitrogen atoms to a compound, adding an even (odd) number of nitrogen atoms will maintain the overall parity (odd or even) of the compound's mass-to-charge ratio, while adding an odd (even) number of nitrogen atoms will change the overall parity of the compound's mass-to-charge ratio.
03

Determine the effect of other elements on the mass-to-charge ratio

We should also analyze how other elements in a compound affect its mass-to-charge ratio. Most organic compounds contain carbon (C), hydrogen (H), oxygen (O), and sometimes sulfur (S). All these elements have even atomic masses: C = 12, H = 1, O = 16, S = 32. Therefore, adding or removing a combination of these elements will not change the overall parity of the mass-to-charge ratio due to them having even atomic masses.
04

Connect the number of nitrogen atoms to the overall mass-to-charge ratio

Since a compound's atomic mass is the sum of all its constituent elements' atomic masses, and nitrogen is the only element with an odd atomic mass, a compound will have an odd \(m/z\) value when it contains an odd number of nitrogen atoms. This is because any combination of other elements, all with even atomic masses, will result in an even contribution to the overall atomic mass, whereas the odd contribution from nitrogen atoms induces the odd \(m/z\) value.
05

Explain the nitrogen rule

In conclusion, the nitrogen rule states that if a compound has an odd number of nitrogen atoms, the value of \(m/z\) for its molecular ion will be an odd number. This is because nitrogen (N=14) is the only element with an odd atomic mass among the common elements in organic compounds. When a compound contains an odd number of nitrogen atoms, it results in an odd contribution to the mass-to-charge ratio, leading to an odd \(m/z\) value according to the nitrogen rule.

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Most popular questions from this chapter

The molecular ion for compounds containing only \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{O}\) always has an even mass-to-charge value. Why? What can you say about the mass-to-charge ratio of ions that arise from fragmentation of one bond in the molecular ion? from fragmentation of two bonds in the molecular ion?

For primary amines with no branching on the carbon bearing the nitrogen, the base peak occurs at \(m / z\) 30. What cation does this peak represent? How is it formed? Show by drawing contributing structures that this cation is stabilized by resonance.

The base peak in the mass spectrum of propanone (acetone) occurs at \(m / z 43\). What cation does this peak represent?

For which compounds containing a heteroatom (an atom other than carbon or hydrogen) does the molecular ion have an even-numbered mass? For which does it have an odd-numbered mass? (a) A chloroalkane with the molecular formula \(\mathrm{C}_{n} \mathrm{H}_{2 n+1} \mathrm{Cl}\) (b) A bromoalkane with the molecular formula \(\mathrm{C}_{n} \mathrm{H}_{2 n+1} \mathrm{Br}\) (c) An alcohol with the molecular formula \(\mathrm{C}_{n} \mathrm{H}_{2 n+1} \mathrm{OH}\) (d) A primary amine with the molecular formula \(\mathrm{C}_{n} \mathrm{H}_{2 n-1} \mathrm{NH}_{2}\) (e) A thiol with the molecular formula \(\mathrm{C}_{n} \mathrm{H}_{2 n+1} \mathrm{SH}\)

Predict the relative intensities of the \(M\) and \(M+2\) peaks for the following. (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}\) (c) \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{Br}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{SH}\)

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