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# nucleic acids and carbohydrates

. Now we know that our molecular structure is composed of three implications that make up 2 CH3CH2CH2 structures.  But if you look at the circled implication from the First Signal, there  is an attachment on the first CH2 (‐CH2CH2).  This means that something attaches to this  carbon.  If we look at the original formula given to us, we see that we have accounted for everything except the oxygen [there are 7 hydrogen atoms x 2 = 14 hydrogen atoms total; there are 3  carbon atoms x 2 = 6 carbons total  this is how many carbons and hydrogens were in the  given molecular formula].  Now we have to account for the oxygen atom.  Because there is  an attachment to the CH2, we can attach the oxygen to this carbon.  Our structure will  therefore look like this: CH3CH2CH2‐O‐CH2CH2CH3 And there’s your molecular structure for the molecule! Note: Make sure to check your work!  Check the formula, integrals, number of signals,  splitting, etc. and make sure they are all correct.  Always perform an atom check and  compare it to the formula given.  Additionally, check to see if your structure agrees with the  DBE for the molecule.   Mass spectrum: m/z = 154 (M; 100%), m/z = 155 (11.23%), and m/z = 156 (0.26%) IR spectrum:  1 H‐NMR: 2.1 ppm (singlet; integral = 2), 1.9 ppm (singlet; integral = 1), and 1.1 ppm (singlet,  integral = 6) Now let’s do another example involving mass spectroscopy and infrared spectroscopy. 1. We first need analyze what the information the mass spectroscopy is telling us.  The first  m/z value is the M peak, which tells us how many nitrogen atoms are present in the  molecular formula.  If the m/z value is even, then there are zero or an even amount of  nitrogen atoms.  If the m/z value is odd, then there are an odd amount of nitrogen atoms. In this example, the M peak has an intensity of 154.  This is an even number, so there  are 0 or an even amount of nitrogens in the structure.