Question

How do the spectra for electron-impact, field ionization, and chemical ionization sources differ from one another?

Short answer

In filed ionization, the electrons are formed under the influence of a large electrical field. The large electric filed is ${10}^8\mathrm{V}/\mathrm{cm}$. Field ionization also produces very less fragmentation. This spectrum contains $(\mathrm{M}+1{)}^{+}$peaks. These peaks correspond to smaller mass fragments arise in electron-impact ionization spectrum are absent.

Step-by-step solution

Given information

In electron-impact ionization, an electron beam generated from a tungsten or rhenium filament, is used to ionize gas phase atoms or molecules. Interaction with an $70\mathrm{V}$electron most likely removes the electron with lowest ionization energy. The resulting molecular ion $\left({\mathrm{M}}^{*+}\right);$can have so much extra energy that it breaks into fragments.

$\underset{\begin{array}{l}\text{Analyte}\\ \text{molocule}\end{array}}{\mathrm{M}}+{\mathrm{e}}^{-}\to {\mathrm{M}}^{*+}+2{\mathrm{e}}^{-}$

Therefore, the molecular ion peak is small or absent in electron-impact ionization spectrum.

Calculating chemical ionization

In chemical ionization, a small variety of gaseous atoms are ionized with the aid of using collision with ions produced with the aid of using electron bombardment of the reagent gas. Chemical ionization produces much less fragmentation than electron ionization.

The following ions result from the addition or subtraction of a proton in the presence of the reagent ion:

$\underset{\begin{array}{l}\text{Analyte}\\ \text{molecule}\end{array}}{\mathrm{MH}}\underset{\text{reagent ion}}{{\mathrm{CH}}_5^{+}}\to \underset{(\mathrm{M}+1{)}^{*}\text{ion}}{{\mathrm{MH}}_2^{+}}+{\mathrm{CH}}_4$

$\underset{}{\begin{array}{c}\mathrm{MH}\\ \text{Analyte}\\ \text{molecule}\end{array}+\underset{\text{reagent ion}}{{\mathrm{C}}_2{\mathrm{H}}_5^{+}}\to }\underset{(\mathrm{M}-1{)}^2\text{ion}}{{\mathrm{M}}^{+}}+{\mathrm{C}}_2{\mathrm{H}}_6$

Therefore, the chemical ionization spectrum contains $(\mathrm{M}+1{)}^{+}$or $(\mathrm{M}-1{)}^{+}$peaks.