Preparation of Cyclohexene From Cyclohexanol

About this essay

A 42.89% yield cyclohexene was successfully synthesized from 10.0 mL cyclohexanol by unimolecular elimination (E1) through the dehydration of cyclohexanol and confirmed via a bromine test and the IR spectra. Introduction:

Alkenes are hydrocarbons that have carbon–carbon double bonds and are one of the many functional groups in organic molecules. Alkenes are sp2 hybridized and are unsaturated because two of their hydrogen’s are missing from the saturated alkane formula (CnH2n+2). Typically alkenes are synthesized by elimination reactions, however, organic molecules can also undergo substitution reactions amongst many other reactions like oxidation and reduction based on the structure of the organic molecule and the conditions under which the reaction is performed.

These reactions are possible because the carbon atom in organic molecules is electron deficient due to induced dipoles created by the presence of functional groups or more electronegative atoms/groups.

The organic molecule as a whole is referred to as the substrate. The most likely to be attacked carbon atom in the organic molecule is referred to as the electrophile and is and it is usually bonded a more electronegative atom/group such as a halide, oxygen or pseudo-halide, called the leaving group.

Get quality help now
checked Verified writer

Proficient in: Chemistry

star star star star 4.8 (309)

“ Writer-marian did a very good job with my paper, she got straight to the point, she made it clear and organized ”

avatar avatar avatar
+84 relevant experts are online
Hire writer

In a reaction, a leaving group leaves the substrate by heterolysis (a reaction in which the breaking of bonds leads to the formation of ion pairs) and it is replaced by a nucleophile (usually a base) which is attracted to the partial positive carbon atom because the nucleophile has excess electron pairs or a negative charge.

Get to Know The Price Estimate For Your Paper
Number of pages
Email Invalid email

By clicking “Check Writers’ Offers”, you agree to our terms of service and privacy policy. We’ll occasionally send you promo and account related email

"You must agree to out terms of services and privacy policy"
Write my paper

You won’t be charged yet!

The relative strength of the nucleophile determines its nucleophilicity. Nucleophilicity depends on many factors, including charge, basicity, solvent, polarizability, and the nature of the substituents present on the organic molecule. In general, a nucleophile containing a negatively charged reactive atom is better than a nucleophile containing a reactive atom that is neutral and basicity parallels nucleophilicity meaning a strong base is a good nucleophile and a week base is a weak nucleophile. Also, there are several factors that contribute to the ability of a group/ atom to function as a good leaving group, which includes the property of the carbon-leaving group bond (polarizability and strength) and the stability of the leaving group.

Weak based are good leaving groups and strong based are bad leaving groups. The substrate organic molecule containing the electrophile is characterized in terms of steric effect that entails the bulkiest of the substrate and it’s shape/size in 3D and whether the electrophilic carbon is a primary, secondary or tertiary carbon. The nature of the substrate, nucleophile, electrophile and solvent (polar protic or polar aprotic) used will determine whether an organic molecule will undergo a substitution or elimination reaction.

An elimination reaction is one in which an alpha- hydrogen and a LG are removed from an inorganic molecule in either a one or two step mechanism to form an alkene. The one and two-step mechanisms are bimolecular (E2) and unimolecular (E1) elimination reactions respectively, defined based on the rates of the reactions. The rates of the reactions are based on the kinematics of each reaction and not on the number of steps in the reaction. As implied by the names, E2 reactions have a rate factor of two (second-order) because the rate of the reaction is based on the concentration of the substrate and the nucleophile while E1 reactions have a rate factor of one (first-order) because the rate of the reaction is based solely on the concentration of the substrate. E2 is a one-step concerted process with individual transition states typically undergone by primary and secondary substrates, while E1 is a two-step process of elimination (carbocation formation and deprotonation) undergone by secondary and tertiary substrates only.

E2 requires a reasonably good nucleophile (strong base) and a polar protic solvent. E1 is favored by tertiary and secondary substrates structures with a bad nucleophile in the presence of a polar protic solvent in which the tertiary substrate is more reactive than the secondary due to greater stabilization of the carbocation and lower activation energy. The polar protic solvent also helps stabilize the carbocation. In an E1 reaction, the most substituted alkene (most stable) is preferred over the least substituted one according to Zaitsev’s rule. This is due to the stabilization of the carbocation intermediate formed by hyperconjugation which leads to rearrangements involving hydride shifts or alkyl shifts in which a hydrogen or an alkyl group with its pair of electrons relocate to the carbocation to form a more stable carbocation.

A substitution reaction is one in which a leaving group on a substrate is replaced by a nucleophile in a one or two step process called bimolecular substitution (Sn2) or unimolecular substitution (Sn1) respectively. Sn2 is undergone by primary and secondary substrates only preferable in a polar protic solvent with a relatively good nucleophile while Sn1 is undergone by secondary and tertiary substrates only preferably in a polar protic solvent with a poor nucleophile. Additionally, tertiary substrates are better than secondary substrates for Sn1 because of the stabilization of the carbocation with the polar protic solvent increasing the stability of the carbocation. There is always competition between substitution and elimination reactions depending on the conditions of the reaction. Under normal circumstances Sn1 is always accompanied by E1 as the minor product but E1 products can be favored by changing the conditions of the reaction such as higher temperatures, which always favor E1 products.

A common way to synthesize alkenes is by the dehydration of alcohol (scheme 5). The reaction is acid catalyzed to transform the hydroxide (OH-) bad leaving group into a good leaving group (H2O) and so a basic nucleophile cannot be used because it cannot provide the proton needed to transform OH- to H2O. The acid used has to be non-nucleophilic such as phosphoric acid or sulfuric acid in a high concentration for an E1 product to be favored over an Sn1 product. High temperatures will also favor E1 over Sn1 due to the energies of the reaction as Gibbs free energy (ΔG) becomes significantly negative. The entropy of a Sn1 reaction is zero while the entropy of an E1 reaction is greater than zero, so increasing the temperature make ΔG even more negative increasing the spontaneity of E1.

To test the effectiveness of a reaction producing an alkene, the product of the reaction can be tested for the presence of an alkene using the bromine test or infrared spectroscopy. The bromine test occurs by halogenation at the double bond in which two bromine radicals break the double bond by binding to the two carbons in the double bond. The reaction occurs as a bromine radical attacks one of the sides of the p-orbital in a pi-bond forcing the bond to cleave homolytically and results with a bromine-carbon bond being formed at the point where the p-orbital of the carbon was attacked. The presence of an alkene is indicated by the disappearance of the deep brown coloration of bromine, which happens because the bromine has been consumed by the reaction of the unknown sample. Infrared spectroscopy (IR) is a technique by which a molecule is analyzed based on its absorbance of infrared light due to the functional groups present within the molecule in a single non-destructive test.

IR is possible because every bond has a natural frequency and the amount of energy applied to a bond affects the amplitude of the vibration of a bond. Therefore it is possible to detect functional groups on an IR spectrum, as each functional group will have a specific natural frequency that is characteristic of that group, which corresponds to a specific absorption region of the spectrum. The natural vibration frequency of any bond relies on its bond order (single, double, triple) and the bonded atoms such as hydrogen, oxygen or nitrogen as determined by Hooke’s law (scheme7). Hooke’s law gives a basic idea of how the bond between atoms will act, in comparison to a spring, with the two atoms acting like weights attached to the spring.

By performing an IR spectroscopy on the product of a reaction, it is possible to discern whether a reaction took place, or if the right reaction took place, by analyzing the IR spectrum for that product, comparing the peaks to standard peaks. Alkenyl alkenes have a standard absorption frequency of 1620 cm-1-1680 cm-1 and aromatic alkenes have a standard absorption frequency of 1500 cm-1&1600 cm-1 and so these peaks (aromatic or alkenyl)1 have to be present in the IR spectrum of the product to indicate the presence of an alkene. Reagent Table:

Cyclohexene was synthesized from cyclohexanol by unimolecular elimination (E1) through the dehydration of cyclohexanol. Phosphoric acid was used to catalyze the reaction and the unimolecular elimination was favored by heating the reaction at a high temperature and also by the use of the non-nucleophilic phosphoric acid. Both a bromine test and Infrared spectroscopy were performed to test the presence of an alkene in the dehydration product formed. Results:

Temperature at which stable distillate was collected = 80.0 °C Volume cyclohexanol collected = 10.0 mL = 9.62 g = 0.0960 mol Theoretically: 1 mol cyclohexanol = 1 mol of cyclohexene
Theoretical moles cyclohexene = 0.096 mol
Theoretical Yield cyclohexane: = 7.88 g
Actual mass of cyclohexene collected = 3.38 g
% Yield cyclohexene: 42.89%
IR Spectrum Anomalies:
Group (assumed)
Wavenumber (cm-1)
Ketone (reactant IR)
Alcohol (product IR)
Carbon Fingerprints

The phosphoric acid catalyzed dehydration of cyclohexanol to cyclohexene proceeds by an E1 mechanism. After protonation of the alcohol, formation of a carbocation occurs. This is the slow step or rate-determining step of the reaction. The formation of the alkene product occurs by elimination of a hydrogen on a β-carbon to the carbocation. The mechanism is shown below: In the presence of a strong acid, an alcohol can be dehydrated to form an alkene. The acid used in this experiment was 85% phosphoric acid and the alcohol was cyclohexanol. The phosphoric acid is a catalyst and as such increased the rate of reaction but did not affect the overall stoichiometry.

In this reaction, as the alcohol and acid were heated, alkene and water were produced and co-distilled into a collection vial. As in any distillation, unless precautions are taken, some of the product would have been lost as hold-up in the apparatus. Hold-up would result in a reduced yield of product. To overcome this problem and to ensure that a maximum amount of product is distilled, a higher boiling “chaser” solvent was added to the distillation flask and the distillation is continued until the temperature rose well over the BP of cyclohexene. At this point it was assumed that all product has distilled into the collection flask.

The collection flask contained cyclohexene, water, toluene, and small amounts of other impurities. Because any water present would interfere with the distillation (water will co-distill and will not separate), prior to a final distillation, to obtain pure product, all water was removed. This was done in two steps. First, the sample was mixed well with an aqueous saturated sodium chloride solution (saturated salt) and the lower aqueous layer was removed and discarded.

Next, anhydrous Na2SO4, an inorganic drying agent that binds strongly with water and thus removes any traces of water from the solution, was added. After about five minutes, the solution was separated from the pellets and transferred to the clean and dry fractional distillation apparatus. The dried solution was then fractionally distilled to produce purified cyclohexene. In order achieve optimal separation the distillation was performed at a slow and steady rate. Also, to ensure that the fraction collected as product is relatively pure cyclohexene, this fraction was collected over a narrow range at the boiling point of cyclohexene. Calculating the Percent Yield:

The actual yield and theoretical yield of cyclohexanol were calculated in order to determine the percentage yield of the compound. Based on the calculation, the percentage yield obtained was 42.819% First, one must write out the balanced equation the preparation of cyclohexene from cyclohexanol:

The equation shows that one reactant produces one product (water is also a product but we are only interested in the cyclohexene here) in a 1:1 ratio.
Note that the phosphoric acid is a catalyst and is not involved in the yield calculation. Therefore, one mole of cyclohexanol should produce one mole of cyclohexene. Since 10.0 mL of cyclohexanol is used, the mass cyclohexanol (g) can be represented by:

Density cyclohexanol (g/mL) * Volume cyclohexanol (mL)
0.962 g/mL * 10.0 mL = 9.62 g cyclohexanol
Now, this can be converted to moles by dividing by the molecular weight of cyclohexanol (MW = 100.2 g/mol):
9.62 g / 100.2 g/mol = 0.096 mol cyclohexanol
Because 1 mol of cyclohexanol should produce 1 mol of cyclohexene, 0.096 mol cyclohexanol should theoretically produce 0.096 mol cyclohexene. Convert this number of moles of cyclohexene to grams of cyclohexene by multiplying by the MW of cyclohexene (82.1 g/mol):

0.096 mol * 82.1 g/mol = 7.88 g cyclohexene
In other words, 9.62 g of cyclohexanol should produce 7.88 g of cyclohexene. This is the best-case yield also known as the theoretical yield. The theoretical yield is what would be obtained in an ideal world, if every molecule of cyclohexanol were converted to a molecule of cyclohexene. The percent yield is the percentage of the theoretical yield that you actually obtain after isolating product at the end of the procedure. After the final fractional distillation of the cyclohexene, 3.38 g was collected. Assuming that the 3.38 g that was obtained was 100% pure, the percent yield then would be:

% yield = (actual / theoretical) x 100 = (3.38 g / 7.88 g) * 100% = 42.89%

The yield of cyclohexene was so low because the reaction is in equilibrium between the SN1 and E1 products. The desired product was cyclohexene, however, in order to be sure that it is the product obtained, further analysis must be done. One way to access the purity is by looking at the distillation temperature. The temperature of the distilled vapors corresponds to the boiling point of the distillate. By monitoring the temperature of the distillation vapors and comparing them with the known boiling point of the desired product, it can be determined if the product is impure. . On the final distillation of cyclohexene, the product was collected at a boiling range of 80-84 °C which is very close to the boiling point of pure cyclohexene at 82.8 °C. This supports the belief that the distillate collected was pure cyclohexene. After the distillate was collected Infrared Spectroscopy was used to analyze the product.

This is an important test because IR identifies functional groups of the structure. Attached is the print out showing the infrared spectrum cyclohexene. Of note, the C=C absorption, characteristic of alkenes at about 1620 cm-1, was absent. The reason for this can be attributed to the symmetrical nature of the cyclohexene, and there is no significant dipole change on IR excitation of the molecule. Evidence to support the isolation of cyclohexene was provided by the absence of an alcohol absorption at 3400-3650 cm-1 and the presence of C-H absorption characteristic of alkenes at 3432.28cm-1. According to the IR graph obtained for the product, both sp2 and sp3 hybridized carbon hydrogen stretch was identified. This indicated that there is the presence of a pi bond. Also the IR graph lacked a hydroxyl group peak. This strong, broad peak is an indicator that an alcohol is present, and its absence indicated that the product did not contain any OH groups.

Works Cited:
( Padias

Cite this page

Preparation of Cyclohexene From Cyclohexanol. (2016, Mar 28). Retrieved from

Preparation of Cyclohexene From Cyclohexanol
Live chat  with support 24/7

👋 Hi! I’m your smart assistant Amy!

Don’t know where to start? Type your requirements and I’ll connect you to an academic expert within 3 minutes.

get help with your assignment