Williamson Ether Synthesis postlab PDF

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Williamson Ether Synthesis postlab Wes James Percent yield and limiting reagent Limiting reagent: 3.46 g of 4-bromophenol / 173.01 (g/mol) = 0.02 mol 0.02 mol * (1 mol alkyl bromide / 1 mol 4-bromophenol) = 0.02 mol alkyl bromide required 3.019 g alkyl bromide / 151.04 (g/mol) = 0.02 mol used Equal moles of each used. Theoretical Yield: 3.46 g / 173.01 (g/mol) = 0.02 mol 4-bromophenol 0.02 mol * (1 mol product / 1 mol 4-bromophenol) = 0.02 mol product 0.02 mol product * 243.144 (g/mol) = 4.863 g product Percent Yield: Percent yield = (experimental yield/theoretical yield) * 100 (0.944 g / 4.863 g) * 100 = 19.4 %

Results/Conclusions To find the initial alkyl halide we need to analyze the H NMR and the IR. In the NMR there are two doublets at 6.7 and 7.8, these would be indicative of the two types of hydrogens on the benzene rings. The one farther left are the hydrogens closer to the Br. The next peak we see is at 4.0 and is a triplet with an integration of 2. This would show that there are two neighbors and that the peak itself contains two hydrogen. This would be indicative of the hydrogens that are next to the oxygen atom. The fact that there are neighbors means that it is next to 2 more hydrogen. The next peak at 1.6 is hard to decipher but it could be a doublet and has an integration of 2.6 which should round to 3 but I think that it should actually be 2 which will be discussed farther into the explanation. The last peak at 1.0 is also a doublet like the last one and has an integration of 6.43 which is high but would round down to 6. This peak is the largest reason that I think the final product is 1-bromo-4(isopentyloxy)benzene. These hydrogens would correspond with the six hydrogen on the end of the molecule. Now there is one peak missing that I would expect between the peaks at 1.0 and 1.6. This peak would be the one hydrogen that is on the carbon where the split occurs between the two carbons that contain the 6 hydrogens in the peak at 1.0. This peak not existing could be a reason that the peak at 1.6 is closer to an integration of three than two. This hydrogen could also be partly represented in the first peak at 1.0 with an integration that is higher than just 6 at 6.43. The H NMR has multiple signs that

point to the product being 1-bromo-4-(isopentyloxy)benzene. This would make the unknown alkyl halide 1-bromo-3-methylbutane.

The IR is a little more difficult to decipher. When looking at the spectrum, the peaks around 1450-1500 cm-1 indicate the presence of a benzene ring. The peaks around 3000-3200 cm-1 indicate the presence of sp2 hybridized carbons which are the carbons of the ring which confirms the presence of the ring. The small peaks around 2900 cm-1 represent the sp3 hybridized carbons present which come from the alkyl halide. The peak around 1250 cm-1 represents the alkyl ether. When looking at this IR spectrum, it can be said that there is no remaining starting material because there is no broad, distinct alcohol peak around 3200-3000 cm-1. If there was starting material remaining, there would have been an alcohol peak since one of the starting materials was 4-bromophenol....