
Interpretation:
The reason for the statement “The mass spectrum of 3-methylpentane has a peak of very low relative abundance at
Concept introduction:
Mass spectrometry involves formation of ions in a mass spectrometer followed by separation and detection of the ions according to mass and charge.
The mass spectrum is a graph with mass (
One of the highest value m/z may or may not represent the molecular ion, that is, the ion with the formula weight of the original compound. When present, the molecular ion is usually not the base peak because the ions from the original compound molecule tend to fragment, thereby resulting in the other m/z peaks in the spectrum.
Small peaks having m/z values 1 or 2 higher than the formula weight of the compound are because of 13C and other isotopes.
Molecular ions formed by EI mass spectrometry are high-energy species and these are called radical cations because it contains both an unshared electron and a positive charge. Fragmentation of molecular ion means that a complex molecule is broken into smaller molecules and these fragments can undergo more breaking and so on.
Cleavage of a single bond produces a cation and a radical. The cation can be detected using positive ion mass spectrometry and because the radical is not charged, it will be undetected.
Chain branching or cleavage of a single bond at branch points will occur such that the carbocation formed is more stable. In simple words, the stability of carbocation determines the fragmentation pattern.

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Chapter 9 Solutions
Organic Chemistry
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- For the reaction below: 1. Draw all reasonable elimination products to the right of the arrow. 2. In the box below the reaction, redraw any product you expect to be a major product. 田 Major Product: Check ☐ + I Na OH esc F1 F2 2 1 @ 2 Q W tab A caps lock S #3 80 F3 69 4 σ F4 % 95 S Click and drag to sta drawing a structure mm Save For Later 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use GO DII F5 F6 F7 F8 F9 F10 6 CO 89 & 7 LU E R T Y U 8* 9 0 D F G H J K L Z X C V B N M 36arrow_forwardProblem 7 of 10 Draw the major product of this reaction. Ignore inorganic byproducts. S' S 1. BuLi 2. ethylene oxide (C2H4O) Select to Draw a Submitarrow_forwardFeedback (4/10) 30% Retry Curved arrows are used to illustrate the flow of electrons. Use the reaction conditions provided and follow the arrows to draw the reactant and missing intermediates involved in this reaction. Include all lone pairs and charges as appropriate. Ignore inorganic byproducts. Incorrect, 6 attempts remaining :0: Draw the Reactant H H3CO H- HIO: Ö-CH3 CH3OH2* protonation H. a H (+) H Ο CH3OH2 O: H3C protonation CH3OH deprotonation > CH3OH nucleophilic addition H. HO 0:0 Draw Intermediate a Xarrow_forward
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