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Chemistry (7th Edition)
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- 7. Our textbook says that the fragmentation that occurs in the mass spectrometry of alkanes can be understood by realizing that "the differences in energy among ... tertiary, secondary, primary and methyl carbocations in the gas phase are much greater than the differences among comparable radicals. Therefore, where alternative modes of fragmentation are possible, the more stable carbocation tends to form in preference to the more stable radical." Given this information, which one of the following hexane isomers (all C6H14) is most likely to have a strong M-15 peak (that is, a peak at m/z 71)? HINT: You're looking for a compound that forms a 3° carbocation after loss of an electron and a CH³· radical. A) n-hexane D) 2-methylpentane B) 2,2-dimethylbutane E) 3-methylpentane C) 2,3-dimethylbutanearrow_forwardPlease help graph these plots below:arrow_forwardDraw the major product formed for each reaction. Assume the reactions are irreversible. Include stereochemistry when products contain stereocenter(s). It may be helpful to first identify whether the reaction is a substitution (and SN1 or SN2), an elimination (and E1 or E2) or a carbonyl addition. (a) 1 equiv means for every molecule of substrate, there is one molecule of NaOCH3 Br Br NaOCH 3 (1 equiv) 0 °C (b) Draw only the substitution product. Both elimination and substitution occur here. (၁) CH3 Br Br CH3OH NaOCH3 80 °C (d) "Then" means CHзl is added after the Grignard reacts. H3C MgBr (1 equiv) then CH3larrow_forward
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