CHEM 2311L/2411LLab Report Cover Sheet

Name: _________________________

TA: Doug Jackson/ Matt McKinley/ Tom Irvin

Date: _________________________

Experiment: _____________________________________________________

Pre-lab (25 points)Title ____________ (2 points)Introduction ____________ (13 points)Table of Reagents ____________ (5 points)Safety Information ____________ (5 points)

In-lab (20 points)Procedure/Data/Obs. ____________ (5 points)Data Sheet ____________ (5 points)Quality Data/Spectra ____________ (10 points)Recorded

Post-lab (50 points)Results/Spectra, Discussion ____________ (35 points)and ConclusionsCalculations ____________ (10 points)Answers to Questions ____________ (5 points)

Discretionary Points (5 points) ____________` (5 points)

Total ____________ (100 points)

TA Comments:

Dehydration of Alcohols

Introduction

One important way to form alkenes in organic synthesis is by dehydration of an

alcohol precursor. The dehydration reaction proceeds under thermal conditions in

refluxing acid through a carbocation intermediate.

The thermal conditions for this reaction are high temperature and a long reaction

time. Under these conditions, if multiple products are possible, then the more stable

product should be the major product and predominate. The more stable product is known

as the “Zaitsev Product.”

It is not known to what degree the Zaitsev Product will predominate in the

reaction. In other words, it is not able to predict what the product percentage will be in

the final product of a reaction.

The goal of this experiment is to dehydrate 3-methyl-3-pentanol to form alkene

products in some unknown ratio. Proton nuclear magnetic resonance spectroscopy will

be used to determine the ratio by looking at the relative proton integration. Using the

integrated spectra of the product mixture, the product distribution (% of each product)

will be determined for the final sample.

NMR instruments rely on large magnetic fields to align nuclear spins in a way to

make them detectable to a probe. Every proton of a compound in a different chemical

environment gives a different H NMR signal because every position in a compound

experiences a shielding effect due to its electron cloud. The chemical shift of the

spectrum is affected by several things, the close a proton is to an electron withdrawing

atom, the greater the shift, the proximity to a double bond or aromatic ring.

Table of Reagents

Name Structure Molecular Weight

Boiling Point

Melting Point

3-methyl-3-pentanol

102.174 g/mol 122.4 °C -23.6 °C

Phosphoric Acid 98 g/mol 158 °C 42.35 °C

Safety

Always lab goggles during the experiment. Remember to never heat a sealed

apparatus. Phosphoric acid is corrosive, so in addition to goggles, gloves are required.

The starting material and products of the experiment are flammable so avoid any sources

of ignition.

Sarah Merkel

Experiment 6

Lab Partners: Meg Adams and Jesse Cann

Lab Performed on October 2nd, 2012

Data and Calculations

Amount of 3-methyl-3-pentanol added 2.01 gramsAmount of Phosphoric Acid 0.25 mLMass of collection vial 1.832 gramsMass of collection vial + sample 3.028 gramsMass of Sample 1.196 grams

% Of original recovered = 1.196grams2.01grams

x100=59.5 %

Shift Integration6.5 ppm 12 mm5.9 ppm 4 mm

Ratio of Integration = 12mm4mm

=31

Ratio of Hydrogens = 21

% Distribution of 3-methyl-2-pentene = 35x100=60 %

% Distribution of 2-ethyl-1-butene = 25x100=40 %

Mechanism of Dehydration of Alcohol

Discussion

From the proposed mechanism above, the dehydration of the 3-methyl-3-pentanol

will produce two different products. From the Zaitsev’s Rule, the most stable product

will predominate. The most stable product according to the Zaitsev’s Rule is the one that

is typically the most substituted one.

The percent of original recovered in this experiment is 59.5 %. This Experiment

produces two different products, 3-methyl-2-pentene and 2-ethyl-1-butene and also gives

off a water molecule and a phosphoric acid. From the NMR spectrum, a H2O impurity is

found in the region around 4.5 ppm. The existence of this peak proves that a small water

impurity is present in the final sample.

To determine the product distribution in the final sample, an HNMR spectrum

was taken. From the spectrum, the ratio of integration can be determined. In the

spectrum taken, the ratio of integration is 3:1. One product contains 2 hydrogens, while

the other product only contains 1 hydrogen. To account for the difference in protons, the

integration number for the product that only contains 1 hydrogen is multiplied by 2. The

number integration ratio is 3:2. To determine the product distribution of the product with

the integration value of 3, a simple calculation of 3/5 leads to the percent distribution of

60%. To determine the product distribution of the product with the integration value of

2, a simple calculation of 2/5 leads to the percent distribution of 40%.

The final step in the NMR analysis is to determine which peak belongs to which

product. This analysis is done by looking at the multiplicity splitting of the peaks. The

peak located at 6.5 ppm has what appears to be a quartet. The peak located at 5.9 ppm

has what appears to be a singlet. From this information, it is determined that the peak

located at 6.5 ppm belongs to 3-methyl-2-pentene due to hydrogen being split into a

quartet by 3 hydrogens. The peak located at 5.9 ppm belongs to 2-ethyl-1-butene because

of the 2 hydrogens that are not split by neighboring hydrogens.

Now that the peaks have been labeled, it is evident that the 3-methyl-2-pentene is

the major product and is 60% of the product of the dehydration of the alcohol. The 2-

ethyl-1-butene is the minor product and is 40% of the product of the dehydration of the

alcohol.

These findings agree with Zaitsev’s Rule because the major product is the one

that is the most substituted and the minor product is less substituted.

Post-Lab Questions

1. The proposed product distribution does agree with Zaitsev’s rule. The product

found 60%