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CHAPTER 11 Alkenes; Infrared Spectroscopy and Mass Spectroscopy

Vollhardt 6e Lecture PowerPoints - Chapter 11

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Page 1: Vollhardt 6e Lecture PowerPoints - Chapter 11

CHAPTER 11

Alkenes; Infrared Spectroscopy and Mass Spectroscopy

Page 2: Vollhardt 6e Lecture PowerPoints - Chapter 11

Naming the Alkenes11-1

Alkenes are characterized by the presence of a double bond.

The general formula of an alkene is CnH2n, the same as for a cycloalkane.

Common nomenclature for alkenes replaces the corresponding alkane suffix –ane with -ylene.

Page 3: Vollhardt 6e Lecture PowerPoints - Chapter 11

IUPAC nomenclature replaces the alkane suffix –ane with –ene (ethene, propene, etc.).

Rules for naming alkenes:

Rule 1: Find the longest chain that includes both carbons of the double bond.

Page 4: Vollhardt 6e Lecture PowerPoints - Chapter 11

Rule 2: Indicate the location of the double bond in the main chain by number starting at the end of the chain closest to the double bond.

The two double bond carbons in cycloalkenes are numbered 1 and 2.

Alkenes with the same formula but differing in the location of the double bond are called double-bond isomers.

A 1-alkene is referred to as a terminal alkene; the others are called internal.

Page 5: Vollhardt 6e Lecture PowerPoints - Chapter 11

Rule 3: Add substituents and their positions as prefixes to the alkene stem.

If the stem is symmetric, begin from the end giving the first substituent the lowest possible number.

Page 6: Vollhardt 6e Lecture PowerPoints - Chapter 11

Rule 4: Identify any cis/trans stereoisomers. These are examples of diastereomers, or stereoisomers that are not mirror images of each other.

In cycloalkenes, trans isomers are stable only for the larger ring sizes.

Page 7: Vollhardt 6e Lecture PowerPoints - Chapter 11

Rule 5: Use the IUPAC E,Z system when cis/trans labels are not applicable (3 or 4 different substituents attached to the double-bond carbons).

Apply the sequence rules devised for R,S substituent priorities to the two groups on each double-bond carbon.

If the two groups of highest priority are on opposite sides of the double bond, the molecule is an E isomer. If they are on the same side of the double bond, the molecule is a Z isomer.

Page 8: Vollhardt 6e Lecture PowerPoints - Chapter 11

Rule 6: Give the hydroxy functional group precedence over the double bond in numbering a chain.

Alcohols containing double bonds are named alkenols. The stem incorporating both functions is numbered to give the OH carbon the lowest possible assignment.

The last “e” and “alkene” is dropped in naming alkenols.

Page 9: Vollhardt 6e Lecture PowerPoints - Chapter 11

Rule 7: Substituents containing a double bond are named alkenyl.

The numbering of a substituent chain containing a double bond begins at the point of attachment to the basic stem.

Page 10: Vollhardt 6e Lecture PowerPoints - Chapter 11

Structure and Bonding in Ethene: The Pi Bond11-2

The double bond consists of sigma and pi components.

Ethene is planar. It contains two trigonal carbon atoms having bond angles close to 120o.

The hybridization of the carbon atoms is best described as sp2.

The three sp2 orbital on each carbon form bonds to two hydrogen atoms and to the other carbon atom.

The remaining unhybridized p orbital on each carbon overlap to form a bond. The electron density of the bond is equally distributed above and below the plane of the molecule.

Page 11: Vollhardt 6e Lecture PowerPoints - Chapter 11

The pi bond in ethene is relatively weak.

The overlap of the two sp2 orbitals to form the bond connecting the two carbon atoms is much greater than the overlap of the two p orbitals to form the bond.

As a consequence, the bond contributes more to the double bond strength than does the bond.

Page 12: Vollhardt 6e Lecture PowerPoints - Chapter 11

The relative energies of the bonding and antibonding and orbitals can be summarized:

Page 13: Vollhardt 6e Lecture PowerPoints - Chapter 11

Thermal isomerization allows us to measure the strength of the pi bond.

Thermal isomerization involves the interconversion of the cis form and the trans form of a double bond at high temperature.

During the isomerization process, the bond between the two carbon atoms is broken and the p orbitals on the two carbon atoms become perpendicular to each other (transition state).

The activation energy for this process is roughly the same as the contribution to the double-bond energy.

Page 14: Vollhardt 6e Lecture PowerPoints - Chapter 11

The measured activation energy for this process is about 65 kcal mol-1. The total energy of the ethene double bond is 173 kcal mol-1, which means the bond energy must be about 108 kcal mol-1.

The alkenyl hydrogens are more tightly held in alkenes than the C-H bonds in the corresponding alkanes. As a result, addition to the weaker bond characterizes the reactivity of alkenes in radical reactions, rather than hydrogen abstraction.

Page 15: Vollhardt 6e Lecture PowerPoints - Chapter 11

Physical Properties of Alkenes11-3

The boiling points of alkenes are very similar to the corresponding alkanes.

The melting points of alkenes are lower than those of the corresponding alkanes.

The presence of a trans double bond lowers the melting point slightly, while the presence of a cis double bond lowers the melting point significantly more.

The effect of a double bond on melting point is due to the disruption of packing of molecules in the crystal lattice compared to the packing of saturated molecules.

Page 16: Vollhardt 6e Lecture PowerPoints - Chapter 11
Page 17: Vollhardt 6e Lecture PowerPoints - Chapter 11

Cis double bonds often exhibit weak dipolar character. The degree of s orbital character in a sp2 carbon is larger than in an sp3 carbon (alkane) which makes the sp2 carbon a weak electron withdrawing group.

Trans double bonds, on the other hand, generally have little dipolar nature since the dipoles involved oppose each other.

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The electron-attracting character of the sp2 carbon also accounts for the increased acidity of the alkenyl hydrogen, compared to its saturated counterpart.

Ethene is still a very poor source of protons compared to alcohols or carboxylic acids.

Page 19: Vollhardt 6e Lecture PowerPoints - Chapter 11

Nuclear Magnetic Resonance of Alkenes11-4

The pi electrons exert a deshielding effect on alkenyl hydrogens.

The proton NMR spectra of trans-2,2,5,5-tetramethyl-3-hexene shows only two peaks. The methyl protons and alkenyl protons are too far from each other to produce detectable coupling.

The resonance of the allenyl protons at 5.30 ppm is typical of hydrogens bound to alkenyl carbons.

Terminal alkenyl hydrogens (RR’C=CH2) resonate at 4.6–5.0 ppm. Internal alkenyl hydrogens (RCH=CHR) resonate at 5.2–5.7 ppm

Page 20: Vollhardt 6e Lecture PowerPoints - Chapter 11

The deshielding for alkenyl hydrogens has two causes.

Less important is the electron withdrawing effect of the sp2 hybridized carbon.

More important is the effect of the external magnetic field on the cloud of electrons. The electrons are forced to assume a circular motion when the magnetic field is perpendicular to the double bond axis.

The circular motion of the electrons induces a second magnetic field which reinforces the external field.

Page 21: Vollhardt 6e Lecture PowerPoints - Chapter 11

Cis coupling through the double bond is different from trans.

Unsymmetrically substituted double bonds lead to non-equivalent alkenyl hydrogens which leads to spin-spin coupling.

Within a set of cis/trans isomers, the coupling constant for the trans isomer, J, is always larger than for the cis isomer.

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Coupling between hydrogens on adjacent carbons is called vicinal.

Coupling between hydrogens on the same carbon is called geminal and is usually small in alkenes.

Coupling to neighboring alkyl hydrogens (allylic) and 1,4- or long-range coupling is also possible, which may produce complicated spectral patterns.

Page 23: Vollhardt 6e Lecture PowerPoints - Chapter 11

Further coupling leads to more complex spectra.

In 3,3-dimethyl-1-butene Ha resonates at 5.86 ppm in the form of a doublet with two relatively large coupling constants (Jab=18 Hz, Jac=10.5 Hz). Hb and Hc also absorb as doubles due to their coupling to Ha and their mutual coupling (Jbc= 1.5 Hz).

Page 24: Vollhardt 6e Lecture PowerPoints - Chapter 11

In 1-pentene, there is additional coupling to the attached alkyl group. In addition, the double bond causes a slight deshielding of the allylic CH2 group.

The coupling between the allylic hydrogens and the neighboring alkenyl hydrogen is about the same as the coupling with the two CH2 hydrogens on the other side.

As a result, the multiplet for the allylic CH2 group appears as a quartet.

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Alkenyl carbons are deshielded in 13C NMR.

Relative to alkanes, corresponding alkene carbons absorb at about 100 ppm lower field.

Page 26: Vollhardt 6e Lecture PowerPoints - Chapter 11

Infrared Spectroscopy11-5

IR spectroscopy measures the vibrational excitation of atoms around the bonds that connect them.

The positions of the absorption lines are related to the types of functional groups present.

The IR spectrum as a whole is unique for each individual substance.

Page 27: Vollhardt 6e Lecture PowerPoints - Chapter 11

Absorption of infrared light causes molecular vibrations.

The infrared region is range of the electromagnetic spectrum just below visible light. Absorption of light of this wavelength causes vibrational excitation of the bonds in a molecule.

Middle infrared light (λ~2.5-16.7 μm, or 600-4000 cm-1) has energies from 1 to 10 kcal mol-1 and is most useful to the chemist.

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Hooke’s law relates the parameters affecting the vibrational frequency of two weights connected by a spring.

The vibrational frequency of two atoms connected by a bond is also accurately described by Hooke’s law:

Page 29: Vollhardt 6e Lecture PowerPoints - Chapter 11

However, the infrared spectrum of a molecule is significantly more complex than the vibrational frequencies of all of the bonds present.

Various bending motions, and combinations of stretching and bending are also excited by IR radiation, which leads to complicated patterns.

Page 30: Vollhardt 6e Lecture PowerPoints - Chapter 11

Fortunately, the vibrational bands of many functional groups appear at characteristic wavenumbers, and the entire IR spectrum of a given compound is unique and can be distinguished from that of any other substance.

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Functional groups have typical infrared absorptions.

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Compare the IR spectra of pentane and hexane:

Above 1500 cm-1 the C-H stretching absorptions typical of alkanes can be seen.

Since no function groups are present, no absorptions are seen in the region from 2840–3000 cm-1.

Below 1500 cm-1, the fingerprint region, C-C stretching and C-C and C-H bending motions absorb to give complicated patterns.

All saturated hydrocarbons show peaks at 1460, 1380, and 730 cm-1.

Page 33: Vollhardt 6e Lecture PowerPoints - Chapter 11

Now compare hexane to 1-hexene:

An additional peak at 3080 cm-1 can be seen which is due to the stronger Csp2-H bond.

The C=C stretching band should appear between 1620 and 1680 cm-1 and is seen at 1640 cm-1.

The two signals at 915 and 995 cm-1 are characteristic of a terminal alkene.

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Several other strong bending modes are characteristic for the substitution patterns in alkenes:

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The O-H stretching absorption is the most characteristic band in the IR spectra of alcohols. This appears as a broad band over the range 3200–3650 cm-1. This is due to hydrogen bonding.

Dry, dilute alcohols show a sharp narrow band in the range 3620–3650 cm-1.

Haloalkane C-X stretching frequencies are too low (<800 cm-1) to be useful for characterization.

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Degree of Unsaturation: Another Aid to Identifying Molecular Structure

11-6

Knowledge of the degree of unsaturation, defined as the numbers of rings and bonds present in a molecule, is useful information when determining the structure of a compound.

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A fully saturated hydrocarbon will have 2n+2 hydrogen atoms for every n carbon atom.

Consider the compounds in the class C5H8. This compound is 4 hydrogens short of being saturated, so its degree of unsaturation is 4/2 = 2.

All molecules having this formula must have a combination of rings and bonds adding up to 2.

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The presence of heteroatoms may affect the calculation.

The presence of a halogen atom decreases the number of hydrogens by one.

The presence of a nitrogen atom increases the number of hydrogens by one.

The presence of oxygen or sulfur does not affect the number of hydrogens.

To determine the degree of unsaturation:

Step 1: Hsat = 2nC + 2 – nX + nN

Step 2: Degree of unsaturation = (Hsat – Hactual)/2

Page 39: Vollhardt 6e Lecture PowerPoints - Chapter 11

Catalytic Hydrogenation of Alkenes: Relative Stability of Double Bonds

11-7

Hydrogen gas and an alkene will react when mixed in the presence of a catalyst such as platinum or palladium.

Two hydrogen atoms are added to the alkene in a reaction called hydrogenation, which is very exothermic.

The heat released is called the “heat of hydrogenation” and has a typical value of about -30 kcal mol-1 per double bond.

Page 40: Vollhardt 6e Lecture PowerPoints - Chapter 11

The heat of hydrogenation is a measure of stability.

The relative stabilities of related alkenes can be determined by measuring their heats of combustion.

The thermodynamic stability of the butenes increases in the order: 1-butene < cis-2-butene < trans-2-butene.

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Highly substituted alkenes are most stable; trans isomers are more stable than cis.

The relative stability of the alkenes increases with increasing substitution (hyperconjugation), and trans isomers are usually more stable than cis isomers (crowding).

Page 42: Vollhardt 6e Lecture PowerPoints - Chapter 11

An exception to this stability rule is in medium-ring and smaller cycloalkenes. The trans isomers of cycloalkenes are much more strained than are the corresponding cis isomers.

The smallest isolated simple trans cycloalkene is trans-cyclooctene which is 9.2 kcal mol-1 less stable than the cis isomer is very twisted.

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Preparation of Alkenes from Haloalkenes and Alkyl Sulfonates: Bimolecular Elimination Revisited

11-8

Two approaches to the synthesis of alkenes are elimination reactions and the dehydration of alcohols.

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Regioselectivity in E2 reactions depends on the base.

Haloalkanes (or alkyl sulfonates) in the presence of strong base can undergo elimination of HX with the simultaneous formation of a C=C double bond.

In the cases where the hydrogen atom can be removed from more than one carbon atom in the structure, the regioselectivity of the reaction can be controlled to a limited extent.

Page 45: Vollhardt 6e Lecture PowerPoints - Chapter 11

Consider the dehydrobromination of 2-bromo-2-methylbutane.

Elimination of HBr proceeds through attack by the base on one of the neighboring hydrogens situated anti to the leaving group.

The transition state leading to 2-methyl-2-butene is slightly more stabilized than the one leading to 2-methyl-1-butene.

The more stable product is formed faster because the structure of the transition state resembles that of the products.

Page 46: Vollhardt 6e Lecture PowerPoints - Chapter 11

Elimination reactions that lead to the more highly substituted alkene are said to follow the Saytzev rule.

The double bond preferentially forms between the carbon that contained the leaving group and the most highly substituted adjacent carbon that bears a hydrogen.

Page 47: Vollhardt 6e Lecture PowerPoints - Chapter 11

When a more hindered base is used, more of the thermodynamically less favored terminal alkene is generated.

Removal of a secondary hydrogen (C3 in the starting bromide) is sterically more difficult than abstracting a more exposed methyl hydrogen when a hindered base is used.

The transition state leading to the more stable product is increased in energy by steric interference with the bulky base.

An E2 reaction that generates the thermodynamically less favored isomer is said to follow the Hofmann rule.

Page 48: Vollhardt 6e Lecture PowerPoints - Chapter 11
Page 49: Vollhardt 6e Lecture PowerPoints - Chapter 11

E2 reactions often favor trans over cis.

The E2 reaction can lead to cis/trans alkene mixtures, in some cases with selectivity.

This and related reactions appear to be controlled by the relative thermodynamic stabilities of the products. The more stable trans double bond is formed preferentially.

Complete selectivity is rare in E2 reactions, however.

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Some E2 processes are stereospecific.

The preferred transition state of elimination places the proton to be removed and the leaving group anti with respect to each other.

When Z or E isomers are possible, stereospecific reactions may occur.

Page 51: Vollhardt 6e Lecture PowerPoints - Chapter 11

Preparation of Alkenes by Dehydration of Alcohols11-9

When alcohols are treated with mineral acid at elevated temperatures, dehydration (E1 or E2) occurs, resulting in alkene formation.

As the hydroxy bearing carbon becomes more substituted, the ease of elimination of water increases.

Page 52: Vollhardt 6e Lecture PowerPoints - Chapter 11

Secondary and tertiary alcohols dehydrate by an E1 mechanism.

The protonated hydroxy forms an alkyloxonium ion providing a good leaving group: water.

Loss of water forms a secondary or tertiary carbocation.

Deprotonation forms the alkene.

Carbocation side reactions (hydrogen shifts, alkyl shifts, etc.) are possible.

Page 53: Vollhardt 6e Lecture PowerPoints - Chapter 11

The thermodynamically most stable alkene or alkene mixture usually results from unimolecular dehydration in the presence of acid.

Whenever possible, the most highly substituted system is generated.

Trans-substituted alkenes predominate if there is a choice.

Treatment of primary alcohols with mineral acids at high temperatures also leads to alkenes.

The reaction of propanol yields propene.

The reaction proceeds by protonation of the alcohol, followed by attack by HSO4

- or another alcohol molecule (E2 reaction) to remove a proton from one carbon atom and water from the other.

Page 54: Vollhardt 6e Lecture PowerPoints - Chapter 11

Important Concepts11

1. Alkenes – Unsaturated molecules• IUPAC names are derived from the longest chain

containing the double bond as the stem.• Double bond isomers: terminal, internal, cis, and

trans• Tri- and tetra-substituted alkenes are named

according to the E,Z system.

2. Double Bond – Consists of a bond and a bond bond: overlap of two sp2 hybrid lobes on carbon bond: overlap of two remaining p orbitals bond E (~65 kcal/mol); bond E (~108 kcal/mol)

Page 55: Vollhardt 6e Lecture PowerPoints - Chapter 11

Important Concepts11

3. Alkene Properties – • Flat, sp2 hybridization• Dipoles possible• Alkenyl hydrogen is relatively acidic.

4. NMR –• Alkenyl hydrogens and carbons appear at low field:• 1H (δ = 4.6 - 5.7 ppm); 13C (δ = 100 -140 ppm)• Jtrans > Jcis; jgeminal very small; Jallylic variable, small.

5. IR – Measures vibration excitation• 1-10 kcal/mol (2.5-16.7 μm; 600-4000 cm-1)• Characteristic peaks for stretching, bending and

other vibrational modes• Fingerprint region (<1500 cm-1)

Page 56: Vollhardt 6e Lecture PowerPoints - Chapter 11

Important Concepts11

6. Alkane IR –• C-H Stretching: 2840 to 3000 cm-1

• C=C Stretching: 1620 to 1680 cm-1

• Alkenyl C-H Stretching: ~3100 cm-1

• Bending Modes: below 1500 cm-1

• Alcohols: Broad O-H stretch: between 3200 and 3650 cm-1

7. Degree of Unsaturation – Number of rings + number of bonds:

• Degree of unsaturation = (Hsat – Hactual)/2• Hsat = 2nC + 2 – NX + NN (disregard oxygen and

sulfur)

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Important Concepts11

8. Heats of Hydrogenation – indicate relative stability of isomeric alkenes.

• Stability decreases with decreasing substitution.• Trans isomers are more stable than cis.

9. Eliminations of Haloalkanes (and other alkyl derivatives) –

• Follow the Sayzex rule (non-bulky base, internal alkene formation) or the Hofmann rule (bulky base, terminal alkene formation).

• Trans alkene products predominate over cis.• Elimination is stereospecific (dictated by the anti

transition state).

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Important Concepts11

10. Dehydration of Alcohols – Dehydration in the presence of strong acid results in a mixture of products (major constituent is the most stable alkene).