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Why is the molecular peak not observed in the mass spectrum of 2,2-dimethylbutane?
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Why is the molecular peak not observed in the mass spectrum of 2,2-dimethylbutane?
The answer probably has to do with how the mass spectrum was obtained.
Most mass spectra that are provided online, including the ones on the NIST WebBook, are electron ionisation mass spectra. That means that you put the sample in a chamber, accelerate electrons to a really high speed (giving them approximately $70~\mathrm{eV}$ of energy), and then bombard the sample with those high-energy electrons in order to ionise the sample.
The ionisation of these organic molecules doesn't take that much energy - $15~\mathrm{eV}$ is a good approximation. So you have an extra $55~\mathrm{eV}$ of energy in each molecule. How can the molecule get rid of this excess energy? The simplest way is to break chemical bonds; a typical bond strength lies between $3$ and $10~\mathrm{eV}$.
Branched compounds in general do tend to fragment more easily than straight-chain compounds. The reason is simple. If you think about n-hexane, no matter which bond you cleave, you're going to end up with a primary cation (or radical). However, if you cleave at one of the branching points of 2,2-dimethylbutane, you'll get a tertiary cation (or radical). A more stable product means that fragmentation occurs more readily.
However, the fact that the molecular ion is not observed is not really unique to 2,2-dimethylbutane. Lots of EI mass spectra don't have the $\ce{M+}$ peak. If you want to observe the $\ce{M+}$ peak, it's a better idea to use a gentler ionisation technique, such as chemical ionisation or electrospray ionisation.
The answer probably has to do with how the mass spectrum was obtained.
Most mass spectra that are provided online, including the ones on the NIST WebBook, are electron ionisation mass spectra. That means that you put the sample in a chamber, accelerate electrons to a really high speed (giving them approximately $70~\mathrm{eV}$ of energy), and then bombard the sample with those high-energy electrons in order to ionise the sample.
The ionisation of these organic molecules doesn't take that much energy - $15~\mathrm{eV}$ is a good approximation. So you have an extra $55~\mathrm{eV}$ of energy in each molecule. How can the molecule get rid of this excess energy? The simplest way is to break chemical bonds; a typical bond strength lies between $3$ and $10~\mathrm{eV}$.
Branched compounds in general do tend to fragment more easily than straight-chain compounds. The reason is simple. If you think about n-hexane, no matter which bond you cleave, you're going to end up with a primary cation (or radical). However, if you cleave at one of the branching points of 2,2-dimethylbutane, you'll get a tertiary cation (or radical). A more stable product means that fragmentation occurs more readily.
However, the fact that the molecular ion is not observed is not really unique to 2,2-dimethylbutane. Lots of EI mass spectra don't have the $\ce{M+}$ peak. If you want to observe the $\ce{M+}$ peak, it's a better idea to use a gentler ionisation technique, such as chemical ionisation or electrospray ionisation.
Ill add that once ionized, molecules can also relocate the charge to form a more stable ion. It just depends on the molecule and where it was ionized.More
The answer probably has to do with how the mass spectrum was obtained.
Most mass spectra that are provided online, including the ones on the NIST WebBook, are electron ionisation mass spectra. That means that you put the sample in a chamber, accelerate electrons to a really high speed (giving them approximately $70~\mathrm{eV}$ of energy), and then bombard the sample with those high-energy electrons in order to ionise the sample.
The ionisation of these organic molecules doesn't take that much energy - $15~\mathrm{eV}$ is a good approximation. So you have an extra $55~\mathrm{eV}$ of energy in each molecule. How can the molecule get rid of this excess energy? The simplest way is to break chemical bonds; a typical bond strength lies between $3$ and $10~\mathrm{eV}$.
Branched compounds in general do tend to fragment more easily than straight-chain compounds. The reason is simple. If you think about n-hexane, no matter which bond you cleave, you're going to end up with a primary cation (or radical). However, if you cleave at one of the branching points of 2,2-dimethylbutane, you'll get a tertiary cation (or radical). A more stable product means that fragmentation occurs more readily.
However, the fact that the molecular ion is not observed is not really unique to 2,2-dimethylbutane. Lots of EI mass spectra don't have the $\ce{M+}$ peak. If you want to observe the $\ce{M+}$ peak, it's a better idea to use a gentler ionisation technique, such as chemical ionisation or electrospray ionisation.
The answer probably has to do with how the mass spectrum was obtained.
Most mass spectra that are provided online, including the ones on the NIST WebBook, are electron ionisation mass spectra. That means that you put the sample in a chamber, accelerate electrons to a really high speed (giving them approximately $70~\mathrm{eV}$ of energy), and then bombard the sample with those high-energy electrons in order to ionise the sample.
The ionisation of these organic molecules doesn't take that much energy - $15~\mathrm{eV}$ is a good approximation. So you have an extra $55~\mathrm{eV}$ of energy in each molecule. How can the molecule get rid of this excess energy? The simplest way is to break chemical bonds; a typical bond strength lies between $3$ and $10~\mathrm{eV}$.
Branched compounds in general do tend to fragment more easily than straight-chain compounds. The reason is simple. If you think about n-hexane, no matter which bond you cleave, you're going to end up with a primary cation (or radical). However, if you cleave at one of the branching points of 2,2-dimethylbutane, you'll get a tertiary cation (or radical). A more stable product means that fragmentation occurs more readily.
However, the fact that the molecular ion is not observed is not really unique to 2,2-dimethylbutane. Lots of EI mass spectra don't have the $\ce{M+}$ peak. If you want to observe the $\ce{M+}$ peak, it's a better idea to use a gentler ionisation technique, such as chemical ionisation or electrospray ionisation.
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