IR NMR Conversion of Alcohols to Alkyl Halides PDF

Preview

Summary

IR NMR Conversion of Alcohols to Alkyl Halides...

Description

Conversion of Alcohols to Alkyl Halides and Analysis by NMR and IR Individual Lab Report CH 236 – M5 12/04/2015 Kayla Hazelwood Table 1: Table of Reagents Compound

Molecular Weight (g/mol) 60.10 88.15 122.99 122.99 151.04 151.04 102.89 18.02 98.08

1-propanol 2-pentanol 1-bromopropane 2-bromopropane 2-bromopentane 3-bromopentane Sodium Bromide Water Sulfuric Acid

Density (g/mL)

Boiling Point (°C) 97 118-119 71 59 116-117 118-119 1396 100 337

.804 .812 1.35 1.31 1.22 1.22 3.21 1.00 1.84

Melting Point (°C) -194.8 -99.4 -112.5 -128.2 -95.6 -126.2 747 0.00 10.0

Results: Analysis by IR and NMR indicate that 1-bromopropane was formed from 1-propanol, and 2-bromopentane and 3-bromopentane were formed from 2-pentanol. The corresponding graphs and data are shown below.

Table 2: Products Formed from the Alcohols Product 1-bromopropane

Mass (grams) 1.363

Percent Yield (%) 33

2-bromopentane and 3-bromopentane

1.532

44

Calculations 1: Theoretical and Percent Yield of Products 1𝑚𝑜𝑙

𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑌𝑖𝑒𝑙𝑑: 2.5𝑚𝐿 𝑜𝑓 1 − 𝑝𝑟𝑜𝑝𝑎𝑛𝑜𝑙 × 0.804𝑚𝐿𝑔 ×

60.10𝑔

𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑌𝑖𝑒𝑙𝑑: 2.5𝑚𝐿 𝑜𝑓 2 − 𝑝𝑒𝑛𝑡𝑎𝑛𝑜𝑙 × 0.812𝑚𝐿𝑔 ×

88.15𝑔

×

𝑔 = 4.113 𝑔 1-bromopropane 122.99𝑚𝑜𝑙

151.04

𝑔 𝑚𝑜𝑙

1𝑚𝑜𝑙

×

1 𝑚𝑜𝑙 1 𝑜𝑟 2−𝑏𝑟𝑜𝑚𝑜𝑝𝑟𝑜𝑝𝑎𝑛𝑒 1𝑚𝑜𝑙 1−𝑝𝑟𝑜𝑝𝑎𝑛𝑜𝑙

×

1 𝑚𝑜𝑙 2 𝑜𝑟 3−𝑏𝑟𝑜𝑚𝑜𝑝𝑒𝑛𝑡𝑎𝑛𝑒

×

1𝑚𝑜𝑙 2−𝑝𝑒𝑛𝑡𝑎𝑛𝑜𝑙

= 3.478 𝑔 2-bromopentane or 3-bromopentane

1-bromopropane product from 1-propanol Percent Yield Mass of product = (Mass of vial w/product – Mass of vial)

Eq. 1

17.749 grams - 16.386 grams = 1.363 grams 𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑌𝑖𝑒𝑙𝑑 = (

𝐴𝑐𝑡𝑢𝑎𝑙 𝑌𝑖𝑒𝑙𝑑

𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑌𝑖𝑒𝑙𝑑

) × 100

Eq. 2

Percent Yield = (1.363 grams/ 4.113 grams) x 100% = 33% 2-bromopentane or 3-bromopentane from 2-pentanol Percent Yield Eq. 1

Mass of product =1.532 grams Percent Yield = (Actual Yield/Theoretical Yield) x 100%

Eq. 2

Percent Yield = (1.532 grams/ 3.478 grams) x 100% = 44%

Nuclear Magnetic Resonance (NMR) Spectra Graphs: Figure 1: NMR Spectra for 1-bromopropane

A C

B

In Figure 1 the NMR spectrum above shows the product of 1-propanol. The letters correlate to the structural compound in Figure 2 of specific hydrogens. There are three groups of non-equivalent hydrogens shown. With this data, the three signals are matched to the formed product of 1-bromopropane.

Figure 2: Structure of 1-bromopropane

Table 3: 1-bromopropane NMR Data Characteristics Group Splitting Pattern (n+1) Integration ppm Triplet (3) 3 ~1 A Sextet (6) 2 ~2 B Triplet (3) 2 3~4 C These groups in Table 3 correspond to 1-bromopropane in Figure 2 shown above. Each group is color coded in the structure and table to compare and match non-equivalent hydrogens to the NMR spectra.

Figure 3: NMR Spectra for 2-bromopentane and 3-bromopentane

2

A B

3

E

C D

1

In Figure 3 the NMR spectrum above shows the products of 2-pentanol. The letters and numbers correlate to the structural compounds shown in Figures 4 and 5. There are eight groups of non-equivalent hydrogens shown. Below shows matching letters and numbers to integration numbers in the NMR. Letter (ppm)

A 3.42

B 0.86

C 2.48

D 0.31

E 3.00

1 2.24

2 1.80

3 0.34

Figure 4: Structure of 2-bromopentane

Table 4: 2-bromopentane NMR Data Characteristics Group Splitting Pattern (n+1) Integration ppm Triplet (3) 3 3~4 A Sextet (6) 1 3~4 B Quartet (4) 2 1~2 C Sextet (6) 2 1~2 D Triplet (3) 3 1~2 E These groups in Table 4 correspond to 2-bromopentane in Figure 4 shown above. Each group is color coded in the structure and table to compare and match non-equivalent hydrogens to the NMR spectra. Figure 5: Structure of 3-bromopentane 3 2 1

Table 5: 3-bromopentane NMR Data Characteristics Group Splitting Pattern (n+1) Integration ppm Triplet (3) 3 3~4 A Quintet (5) 2 ~2 B Quintet (5) 1 ~1 C These groups in Table 5 correspond to 1-bromopropane in Figure 5 shown above. Each group is color coded in the structure and table to compare and match non-equivalent hydrogens to the NMR spectra. Calculations 2: Product Distribution of 2-bromopentane and 3-bromopentane 𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 =

# 𝑜𝑓 ℎ𝑦𝑑𝑟𝑜𝑔𝑒𝑛𝑠 𝑡ℎ𝑎𝑡 𝑐𝑎𝑛 𝑏𝑒 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒𝑑 𝑜𝑛 𝑎 𝑔𝑖𝑣𝑒𝑛 𝑐𝑎𝑟𝑏𝑜𝑛 # 𝑜𝑓 𝑡𝑜𝑡𝑎𝑙 ℎ𝑦𝑑𝑟𝑜𝑔𝑒𝑛𝑠 𝑡ℎ𝑎𝑡 𝑐𝑎𝑛 𝑏𝑒 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒𝑑

2 − 𝑏𝑟𝑜𝑚𝑜𝑝𝑒𝑛𝑡𝑎𝑛𝑒: 3 − 𝑏𝑟𝑜𝑚𝑜𝑝𝑒𝑛𝑡𝑎𝑛𝑒:

Eq. 3

1 = 0.091 11

2 = 0.182 11

The integration ratio shows that the likely hood of 3-bromopentane forming is 0.182 to 0.091. This is a 2 to 1 ratio in determining the major product.

Infrared Spectra Graphs: Figure 6: IR of 1-bromopropane

Csp3-H Stretch

Finger print region

In the IR above, the Csp3-H functional group is shown with the wavenumber being less than 3000 cm-1. The peak is not intense.

Figure 7: IR of the 2-bromopentane and 3-bromopentane

Csp3-H Stretch

Finger print region

In the IR above, the Csp3-H functional group is shown with the wavenumber being less than 3000 cm-1. The peak is not intense.

Table 6: Comparison of Functional Groups in IR Starting Material 1-propanol 2-pentanol Products 1-bromopropane 3-bromopropane

Key Peaks -OH stretch at 3300cm-1, C-C stretch at 1200cm-1, Csp3-H stretch at 2900cm-1 Lacks -OH stretch at 3300cm-1, C-C stretch at 1200cm-1, Csp3-H stretch at 2900cm-1 Lacks -OH stretch at 3300cm-1, C-C stretch at 1200cm-1, Csp3-H stretch at 2900cm-1 Lacks -OH stretch at 3300cm-1, C-C stretch at 1200cm-1, Csp3-H stretch at 2900cm-1

Discussion: Alkyl halide products were prepared from the original alcohols in lab by a series of techniques. First reflex was completed to act as acid-catalyzed dehydration to start the reaction. Reflux reduces product loss by using a small amount of solvent. In this procedure sulfuric acid, sodium bromide, and alcohol were placed in a round bottom flask and heated until condensed in the apparatus. Reflux is used to allow the newly formed product to re-condense multiple times; it will not evaporate out. Next, distillation was used to separate compounds based on boiling points. The alkyl halide products were boiled and re-condensed into the receiving flask. The purpose of distillation was to collect pure products in the distillate and boil off the impurities. In distillation the temperature range of the distillate is similar to the products, because of the high yield from the reaction. By reflux, the reaction was allowed time for the products to form before they were separated out by distillation. Therefore, the temperature would then coincide with the correct range because the products were formed. Further, the product was then separated and dried. Separation was done with a separatory funnel to filter out the product. The products were denser than water, and this caused water to rise to the top and the organic layer separated to the bottom. This made the product easy to be collected, pouring out the bottom layer into a conical vial. Finally, the product was dried with anhydrous Na2SO4. The purpose of this was to remove excess water, it was absorbed because it was polar and like dissolves like; this separates impurities from the product. The reason reflux was used before simple distillation is because reflux allowed the reaction to run to completion. If distillation was performed first, then only the reactants would have been distilled. Reflux is heating the products (acting as a catalyst) in a closed system, and it will not evaporate out so more product is collected when condensing. Reflux helped make the reaction more efficient. The products produced for both substrates were of decent percent yields. By analysis of IR and NMR it was determined that the respective products and their yields were 33% 1bromopropane produced and 44% 3-bromopentane. When accounting for errors, many can be considered when setting up the apparatuses for reflux and simple distillation. Heating too slowly could make the reaction happen slower, or not happen at all if at too low of a temperature. Other errors include the separation and drying techniques. If the reaction did not happen correctly to form organic products, the separation of the alkyl halide products would have been impure. In

drying, if the anhydrous Na2SO4 did not completely dry the product, then water would have been left as an impurity and this would have skewed data showing an OH stretch in IR and more signals in NMR. All errors effect the efficiency of the reaction of the conversion of alcohols to alkyl halides. This means errors must be accounted for when considering percent yield. Determining if and what products are formed from alcohols can be easily seen with infrared spectrums (IR). In both reactions of the products, the OH group was hydrated to H2O and then functioned as a leaving group. Once water left, bromine came in through a substitution mechanism. If this reaction happened, it will show up in the IR graphs. Comparing the reactants to the products, the reactants contain an OH functional group while the products do not. In IR OH stretches show up in a broad spectrum, sometimes wiping out the C-H stretch. Both IR graphs in Figure 6 and 7 do not show any OH stretch, and therefore confirms that the product was formed because the OH was substituted. Table 6 compares the groups of the reactants versus the products, and shows the products in the IR showing a Csp3-H stretch. Although IR is effective for determine the functional groups present it cannot determine the structures of the compounds. IR will show what type of product is present, but not how the structure is connected, and it won’t show how many products are formed. The formation of alkyl halide products and their structures can be determined by nuclear magnetic resonance (NMR) and identified by non-equivalent hydrogen groups shown by differing signals. Other characteristics include splitting pattern, integration, and the position of signals on the x-axis measured by parts per million (ppm.) The closer the hydrogen group is to the more substituted group, the more downfield it is pushed and it has a higher ppm, whereas the further away the group is from the substituted group the signal is more up field and closer to 0 ppm. The splitting pattern is based on the n+1 rule, where the number of hydrogens on the neighboring carbons are counted up and then and addition +1 is added. The splitting pattern gives the numbers of peaks in the signal. The NMR can also help identify how many hydrogens are on each carbon for their specific hydrogen connectivity, and the structure of the compound. Next, the integration number is given by the NMR machine by a series of calculations. However, integration numbers can reflect the number of hydrogens present on specific carbons. The NMR spectrums and the compound displayed in figures 1 and 3 are determined by tables 3, 4, and 5. In the tables each compound is divided up by letters and numbers and then analyzed for NMR characteristics such as splitting pattern, integration, and position on the spectrum in. The tables display specific product characteristics and then are matched to the NMR signals to reveal that 1bromopropane, and 3-bromopentane were formed. NMR is useful for determining hydrogen connectivity from signals on the graph. However, other signals can indicate the presence of other products or solvents left in the solution. For example both NMR spectrums in figures 1 and 3 show a signal at 0 ppm. These peaks are not hydrogen groups, they represent the marker (TMS-tetramethylsilane) to measure as a standard peak. Although there is no peak shown at 7.26 in the figures, this can appear as CHCl3. This is not shown because the graphs are cropped, but the compound is used as a solvent to aid the process of NMR. These peaks at these ppm values of 0 and 7.26 can skew the data, but not drastically. Because two other compounds are involved in the spectra reading, they could act

as impurities and alter product peaks. Though, this is not significant enough to consider in the final products shown. The NMR in Figure 1 was determined to be the single product of 1bromopropane formed from 1-propanol. This was confirmed by the structure of the compound in Figure 2 and its distinct non-equivalent hydrogen groups characteristics identified in Table 3. This reaction underwent an SN2 mechanism because of substitution at the primary carbon as a favorable condition for a concerted reaction.1 NMR is a useful tool in distinguishing between two compounds in a mixture.2 The spectra shows different types of hydrogen bonding based on neighboring carbons with the n+1 rule, integration, and splitting patterns. The NMR in Figure 3 shows a mixture of the two products formed by acid-catalyzed dehydration of 2-pentanol. By looking at the two different structures formed in figures 4 and 5 of 2-bromopentane and 3-bromopentane, the non-equivalent hydrogen groups can be matched to specific signals on the NMR by distinct characteristics shown in tables 4 and 5. The letters on the structures and the graphs are labeled by different hydrogen connectivity and identified as either a signal for 2-bromopentane or 3-bromopentane. The NMR in Figure 3 shows that both products were formed and are present, but the SN1 favorable product formed is 3-bromopentane. This is the major product because of Zaitsev’s rule and the inductive effect whereas the more substituted group is closer to the middle of the compound for stability of the carbocation and to spread the charge throughout the molecule.3 The more spread out the charge is, the more stable the compound or carbocation is. SN1 occurred because of rearrangement, whereas sn2 cannot rearrange. The results for the NMR graphs matched to their products make logical sense. Especially in this case of a mixed product, the product distortion can be calculated to also determine the likely hood of 3-bromopentane to form over 2-bromopentane. These calculations are outlined in Calculations 2, and Equation 3, showing 3-bromopentene ratio is twice as likely to form as the major product. Conclusion: The NMR and IR were used in this lab to identify and confirm the alkyl halide products formed from alcohols through substitution reactions. The reaction of 1-propanol only formed one product, while 2-pentanol formed both products. This happened because 2-pentanol could have underwent both SN1 and SN2 mechanisms, because of the secondary carbon substitution, and ability to rearrange. Both products were yield to completion shown in the NMR. The reason 3bromopentane is the major product over 2-bromopentane is because of the stability of the carbocation intermediate is favored in the center of the structure. On the other hand, 1-propanol could only form products through SN2 because of the primary carbon substitution. My partner and I performed the reaction with 2-pentanol and the percent yield was higher than that of other groups in the lab. The 1-propanol data was obtained from another lab group. Both percent yields were reasonable amounts of products formed. It was determined that 1-bromopropane formed a 33% yield from 1-propanol and that 3-bromopentane formed a 44% yield from 2-pentanol. To have a higher efficiency, improvement can be made by drying out the product more with the anhydrous Na2SO4 or heating slower in reflux and distillation. Drying will help purify the product, and heating slower will reduce the amount reagents and products evaporated. These improvements can lead to higher percent yields, and better lab techniques.

Works Cited 1. Comparing the SN1 and SN2 Reactions http://www.masterorganicchemistry.com/2012/08/08/comparing-the-sn1-and-sn2reactions/ (accessed 2015). 2. Williamson, K. L. (1994) Macroscale and Microscale Organic Experiments, 2nd Ed. p268; revise 10/9/00. Chem.psu.edu. Houghton Mifflin, Boston. 3. Utah Valley University. OChemPal, http://science.uvu.edu/ochem/, Zaitsev’s Rule (accessed October 23, 2015)....