# HW answers - 1)/z=77(40,112(100,114(33, Identifythecompound...

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1 ) A compound gives a mass spectrum with peaks at m/z = 77 (40%), 112 (100%), 114 (33%), and essentially no other peaks. Identify the compound. First, your molecular ion peak is 112 and you have a M+2 peak at 114. Therefore, you have a halogen. Now, your molecular ion peak and M+2 peak are in a 3 to 1 ratio. This means chlorine. So, 112-35=77 # C’s 77/12=6 carbons so C 6 H 5 Cl. DOUS (2(6)+2-5-1)/2=4 Cl

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2 ) While organizing the undergraduate stockroom, a new chemistry professor found a half-gallon jug containing a cloudy liquid (bp 100– 105 °C), marked only "STUDENT PREP". She ran a quick mass spectrum, which is shown below. As soon as she saw the spectrum (without even checking the actual mass numbers), she said, "I know what it is." What compound is the "student prep"?
So molecular ion peak at 136 and M+2 peak at 138, so halogen present. They are in a 1:1 ratio so Br. So 136-79=57/12 = 4x12=48 57-48=9, C 4 H 9 Br (2(4)+2-9-1)/2=0 Br The peaks at 107 (C 2 H 5 ) and 93(C 3 H 7 ) tell us it is a linear chain instead of a branched one.

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3 ) A laboratory student added 1-bromobutane to a flask containing dry ether and Mg turnings. An exothermic reaction resulted, and the ether boiled vigorously for several minutes. Then she added acetone to the reaction mixture, and the ether boiled even more vigorously. She added dilute acid to the mixture, and separated the layers. She evaporated the ether layer, and distilled a liquid that boiled at 143 °C. GC–MS analysis of the distillate showed one major product with a few minor impurities. The mass spectrum of the major product is shown below. Show the structure of this major product.
Br Mg/ether MgBr O O - H + OH The molecular ion peak should be at 116, but the loss of a carbon from the quatenary C forms a stable carbocation at 101.

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