10.6: Stability of the Allyl Radical- Resonance Revisited

Make certain that you can define, and use in context, the key terms below.

You will have encountered the concept of resonance if you have taken general first-year chemistry course. You should also briefly review Section 2.5.

When we can represent a species by two or more different Lewis or Kekulé structures, neither of which represents the true structure of the species, these structures are referred to as resonance forms. A common example used in general chemistry courses to illustrate the concept of resonance is ozone, O3. The two resonance forms of ozone may be represented as follows:

The concept of resonance is quite important, and will be used frequently throughout the remainder of this course. The guidelines below may assist you in drawing resonance contributors.

1. Resonance occurs whenever a molecule, radical or ion can be represented by two or more structures differing only in the arrangement of electrons (no atoms may be moved).
2. The true structure of a species is a hybrid of the resonance contributors and is more stable (i.e., lower in energy) than any of the contributors.
3. The most important contributors are those containing the most covalent bonds. Another way of saying the same thing is that the most important contributors have the least amount of charge separation.
4. Contributors in which all the atoms (except hydrogen) have a complete octet (i.e., are surrounded by eight electrons) are particularly important.

In the previous section we discussed the allylic bromination of a symmetrical alkene with NBS such as this cyclopentene, which affords one product.

However, with an unsymmetrical alkene and the delocalized unpaired electron forming various allylic resonances, several products are possible. For example, the NBS bromination of 4-methyl-cyclohexene leads to three products.

Cube Roots

The cube root The number that, when used as a factor with itself three times, yields the original number it is denoted with the symbol 3 . of a number is that number that when multiplied by itself three times yields the original number. Furthermore, we denote a cube root using the symbol 3 , where 3 is called the index The positive integer n in the notation n that is used to indicate an nth root. . For example,

The product of three equal factors will be positive if the factor is positive and negative if the factor is negative. For this reason, any real number will have only one real cube root. Hence the technicalities associated with the principal root do not apply. For example,

In general, given any real number a, we have the following property:

When simplifying cube roots, look for factors that are perfect cubes.

Example 3: Find the cube root.

Example 4: Find the cube root.

c. − 1 27 3 = ( − 1 3 ) 3 3 = − 1 3

It may be the case that the radicand is not a perfect cube. If an integer is not a perfect cube, then its cube root will be irrational. For example, 2 3 is an irrational number which can be approximated on most calculators using the root button. Depending on the calculator, we typically type in the index prior to pushing the button and then the radicand as follows:

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In relating a particle's energy to its wavelength, two equations are used. The first is the kinetic energy equation:

Equation Number One: KE = (1/2) mv 2

(The second equation is down the page a bit.)

There are three symbols in this equation:

I would like to emphasize that these are symbols, which are standing in place of the actual numerical values. Another example of an equation using symbols is PV = nRT. I would also like to emphasize that these symbols are not the units. For example, the symbol P stands for the pressure and the unit on the numerical value for pressure is atm.

Next, some comments on the units attached to the numerical values represented by the symbols in the kinetic energy equation.

The unit on mass is kilograms. That means, if you get a mass in grams given in the problem, you must convert it to kilograms.

The unit on velocity is meters per second, most usually written m s¯ 1 (it can also be written m/s). One possible point of confusion: The unit m means meters, the symbol m means mass. You have to keep the two separate.

The unit on KE is kg m 2 s¯ 2 . Remember that the velocity (symbol = v) is squared. That means the unit (m/s) gets both the length (unit = m, meaning meters) and the time (unit = s, meaning seconds) get squared.

Here's one last confusing thing to make you feel better: the name (as opposed to the symbol) of the KE unit is Joule.

You really do need to keep the difference between a symbol and a unit clear in your mind.

Now, let's manipulate the KE equation. I'm going to use E instead of KE:

E = (1/2) mv 2

2E = mv 2

2Em = m 2 v 2

2Em = (mv) 2

Here's a key point from physics: mv is a particle's momentum. The standard symbol for momentum is p:

Now comes the second equation. It is one of two related equations called the de Broglie equations. You can read more about de Broglie's work here. He received the 1929 Nobel Prize in Physics for this work. (I will discuss the second de Broglie equation below the following example problems.)

There are three symbols in this equation:

Just above, we developed that p = √(2Em). We can now substitute that into the de Broglie equation:

We will use this result in the example problems that follow.

You may skip the following discussion of the second de Broglie equation. I do not use it in any of the examples problems to follow. it is here in case you migt be interested in it.

The second de Broglie equation is this: &nu = E/h

There are three symbols in this equation:

Suppose an electron has momentum equal to p, then its wavelength is &lambda = h/p and its frequency is f = E/h. (ChemTeam comment: note the use of the two de Broglie equations.)

The velocity of a de Broglie wave is:

Substituting for E, we obtain:

ChemTeam comment (including the indented equations just below): remember that p is momentum and that p = mv. Look at (1/2) p 2 /m and substitute mv for the p:

E = (1/2) (mv) 2 /m

E = (1/2) m 2 v 2 /m

E = (1/2) mv 2

Now comes the question: Should not the velocity of an electron be v = p/m? (ChemTeam comment: replace p with mv to get mv/m = v.) The posted answer was this:

At this point, we are beyond what the ChemTeam understands. To those of you reading this, best wishes with your continued studies!

Now, on to the example problems.

The deBroglie Equation: Example Problems

Problem #1: What is the wavelength of an electron (mass = 9.11 x 10¯ 31 kg) traveling at 5.31 x 10 6 m/s?

1) The first step in the solution is to calculate the kinetic energy of the electron:

KE = (1/2)mv 2

x = (1/2) (9.11 x 10¯ 31 kg) (5.31 x 10 6 m/s) 2

x = 1.28433 x 10¯ 17 kg m 2 s¯ 2 (I kept some guard digits)

When I use this value just below, I will use J (for Joules).

2) Next, we will use the de Broglie equation to calculate the wavelength:

Just to be sure about two things: (1) the unit on Planck's Constant is Joule-seconds, both are in the numerator and (2) there are three values following the radical in the denominator. All three of them are under the radical sign.

I'd like to compare this wavelength to ultraviolet light, if I may. Let's use 4000 Å = 4000 x 10¯ 8 cm = 4 x 10¯ 7 m.

Our electron's wavelength is almost 3000 times shorter than our ultraviolet example and its wavelength puts it in the X-ray region of the electromagnetic spectrum.

This turned out to be very important because one could then take a beam of electrons and perform experiments with detectable results. You can't do that with the short wavelengths of heavier particles (see examples below).

In 1926, de Broglie predicted that matter had wave-like properties. In 1927, experiments were done that showed electrons behaved as a wave (by showing the property of diffraction and interference patterns). In 1937, the Nobel Prize in Physics was awarded to Clint Davisson and George Thomson (son of J.J. Thomson) for this work.

I would like to pause briefly to analyze the units in the above problem. Specifically this equation:

The units on Planck's Constant are J ⋅ s, but I would like to use this instead:

I realize that seconds would cancel (leaving a s -1 ). Set that aside for the moment, but remember that a unit of seconds does cancel in the above set of units.

In the denominator, there are units on E and m, as follows:

I realize that kilograms would be a square, but I want to emphasize the energy times mass formulation.

Remembering that the units in the denominator are all under the radical sign, we apply the radical sign to the units in the denominator, arriving at this:

Everything cancels, except for a meter in the numerator, which is exactly what we want.

Problem #2: What is the wavelength in meters of a proton traveling at 255,000,000 m/s (which is 85% of the speed of light)? (Assume the mass of the proton to be 1.673 x 10¯ 27 kg.)

1) Calculate the kinetic energy of the proton:

KE = (1/2)mv 2

x = (1/2) (1.673 x 10¯ 27 kg) (2.55 x 10 8 m/s) 2

x = 5.43934 x 10¯ 11 J

2) Use the de Broglie equation:

This wavelength is comparable to the radius of the nuclei of atoms, which range from 1 x 10¯ 15 m to 10 x 10¯ 15 m (or 1 to 10 fm).

Problem #3: Calculate the wavelength (in nanometers) of a H atom (mass = 1.674 x 10 -27 kg) moving at 698 cm/s

2) Calculate the kinetic energy of the proton:

KE = (1/2)mv 2

x = (1/2) (1.674 x 10¯ 27 kg) (6.98 m/s) 2

x = 5.84226 x 10¯ 27 J

3) Use the de Broglie equation:

5) Comment: the absolute temperature of the H atom moving at 6.98 m/s can be calculated:

KE = (3/2) kT

T = 2KE / 3k

T = [(2) (5.84226 x 10¯ 27 J) / [(3) (1.38065 x 10¯ 23 J/K)]

T = 0.000304 K

Pretty cold, if you ask me!

Problem #4: What is the wavelength of a 5.00 ounce baseball traveling at 100.0 miles per hour? (5.00 oz = 0.14175 kg and 100 mph = 44.70 m/s)

1) Calculate the kinetic energy of the baseball:

KE = (1/2)mv 2

x = (1/2) (0.14175 kg) (44.70 m/s) 2

x = 141.6146 J (as always, some guard digits)

2) Use the de Broglie equation:

Yep, pretty short wavelength! Compare it to the Planck length.

Problem #5: An atom of helium has a de Broglie wavelength of 4.30 x 10¯ 12 meter. What is its velocity?

1) Use the de Broglie equation to determine the energy (not momentum) of the atom [note the appearence of the mass (in kg) of a He atom]:

&lambda = h/p

&lambda = h/√(2Em)

4.30 x 10¯ 12 m = 6.626 x 10¯ 34 J s / √[(2) (x) (6.646632348 x 10¯ 27 kg)]

I dropped the units.

4.30 x 10¯ 12 times √[(2) (x) (6.646632348 x 10¯ 27 ) = 6.626 x 10¯ 34 ]

√[(2) (x) (6.646632348 x 10¯ 27 )] = 6.626 x 10¯ 34 / 4.30 x 10¯ 12

I divided the right side and then squared both sides.

(2) (x) (6.646632348 x 10¯ 27 ) = 2.374466 x 10¯ 44

x = 1.786217333 x 10¯ 18 J

2) Use the kinetic energy equation to get the velocity:

KE = (1/2)mv 2

1.786217333 x 10¯ 18 = (1/2) (6.646632348 x 10¯ 27 ) v 2

v 2 = 5.3748 x 10 8

v = 2.32 x 10 4 m/s

Problem #6: Calculate the velocity of an electron (mass = 9.10939 x 10¯ 31 kg) having a de Broglie wavelength of 269.7 pm

2) Use the de Broglie equation to determine the energy (not momentum) of the atom:

&lambda = h/p

&lambda = h/√(2Em)

2.697 x 10¯ 10 m = 6.626 x 10¯ 34 J s / √[(2) (x) (9.10939 x 10¯ 31 kg)]

I dropped the units.

2.697 x 10¯ 10 times √[(2) (x) (9.10939 x 10¯ 31 ) = 6.626 x 10¯ 34 ]

√[(2) (x) (9.10939 x 10¯ 31 )] = 6.626 x 10¯ 34 / 2.697 x 10¯ 10

I divided the right side and then squared both sides.

(2) (x) (9.10939 x 10¯ 31 ) = 6.035885 x 10¯ 48

x = 3.313 x 10¯ 18 J

3) Use the kinetic energy equation to get the velocity:

KE = (1/2)mv 2

3.313 x 10¯ 18 = (1/2) (9.10939 x 10¯ 31 ) v 2

v 2 = 7.2738 x 10 12

v = 2.697 x 10 6 m/s

Problem #7: Calculate the velocity of a neutron with a wavelength of 65 pm:

2) Use the de Broglie equation to determine the energy (not momentum) of the atom (the mass of the neutron is in example #8):

&lambda = h/p

&lambda = h/√(2Em)

6.5 x 10 -11 m = 6.626 x 10¯ 34 J s / √[(2) (x) (1.67493 x 10¯ 27 kg)]

Algebra!

x = 3.102055 x 10¯ 20 J

3) Use the kinetic energy equation to get the velocity (I dropped the units since I know m/s results):

KE = (1/2)mv 2

3.102055 x 10¯ 20 = (1/2) (1.67493 x 10¯ 27 ) (v) 2

v = 6086 m/s

Problem #8: Calculate the de Broglie wavelength of a neutron (mass = 1.67493 x 10¯ 27 kg) moving at one five-hundredth of the speed of light (c/500).

1) Determine the speed of the neutron:

2) Calculate the kinetic energy of the neutron (I used Joule for the energy unit):

KE = (1/2)mv 2

KE = (1/2) (1.67493 x 10¯ 27 kg) (6.00 x 10 5 m/s) 2

KE = 5.02479 x 10 ¯ 22 J

3) Use the de Broglie equation:

Problem #9: Calculate the wavelength of an object weighing 100.0 kg and moving at 160 km/hour.

160 km/hr = 160,000 m/hr

160,000 m/hr = 160,000 m/3600 s = 44.44 m/s

2) Solve for the kinetic energy:

K.E. = (1/2) (100.0 kg) (44.44 m/s) 2

K.E. = 9.87654 x 10 4 J

Problem #10: What is the de Broglie wavelength (in nm) of a molecule of Return to Electrons in Atoms menu

Seasonal variation characteristics of hydroxyl radical pollution and its potential formation mechanism during the daytime in Lanzhou

3.1 Part 1 #4, 5ae, 6, 7, 8ace, 9a
3.1 Part 2 #3ace, 10ace, 11ac, 13, 15ace, 16ace, 18
3.2 #4ac, 5ac, 6ac, 7a, 8a, 10, 14
Assess Your Understanding p.149 #1a, 2a, 3a, 4a, 5, 8, 10
3.3 #7, 8, 10
3.5 #9, 10, 11, 14, 17, 20, 21
3.6 #8, 9, 12, 13, 15, 18, 19 (2 days – decomposition method (day 2) for 18 and 19)
Assess Your Understanding p. 180 #2aii,iv,vi,viii, 5ace, 6ace, 7ac, 8ac, 9ac
3.7 #4bd, 5bd, 8bd, 9b, 13b, 14, 17
3.8 #4ag, 8af, 10, 11ace, 12ace, 13bdf,18
Review#1b, 2b, 3b, 5b, 6b, 10, 11b, 12, 13, 14b, 18b, 19b, 25b, 27, 29b, 32, 33, 34

4.1 #1, 2ae, 3bf, 5d
4.2 #3, 4, 5, 10b, 13, 14, 17, 19
4.3 #4, 5, 7, 9, 10bdfh, 11bdfh, 12bdfh, 16, 20, 24
Mid Unit Review p. 221 #1ac, 2ac, 3, 4ace, 5ac, 7, 9ac, 11ac.
4.5 #3, 6, 7, 8, 9, 10, 13
4.6 #3bd, 4bd, 5bd, 6bd, 7bd, 9bdfh, 10, 11, 13, 14, 16, 21
Review Asn’t p. 246 #1-4, 6bdf, 8, 9, 11, 12, 17bd, 18bd, 19bd, 26, 28b, 29b, 30b, 32b

Contents

The unpaired electron of the hydroxyl radical is officially represented by a middle dot, ·, beside the O (or "cdot" in LaTeX). [5]

Hydroxyl radicals can occasionally be produced as a byproduct of immune action. Macrophages and microglia most frequently generate this compound when exposed to very specific pathogens, such as certain bacteria. The destructive action of hydroxyl radicals has been implicated in several neurological autoimmune diseases such as HAND when immune cells become over-activated and toxic to neighboring healthy cells. [6]

The hydroxyl radical can damage virtually all types of macromolecules: carbohydrates, nucleic acids (mutations), lipids (lipid peroxidation), and amino acids (e.g. conversion of phenylalanine to m-tyrosine and o-tyrosine). [7] The hydroxyl radical has a very short in vivo half-life of approximately 10 −9 seconds and a high reactivity. [8] This makes it a very dangerous compound to the organism. [9] [10] However, humans, animals and plants have evolved to coexist with hydroxyl radicals, and hydroxyl radicals cannot enter the blood stream or tissues within the body.

Unlike superoxide, which can be detoxified by superoxide dismutase, the hydroxyl radical cannot be eliminated by an enzymatic reaction. [9]

Effects on pathogens Edit

Hydroxyl radicals attack essential cell components and are therefore lethal to pathogenic viruses and bacteria (both gram -ve and +ve) – both in the air and on surfaces. Pathogenic viruses suffer from oxidation of their surface structures. Hydroxyl radicals disrupt the lipid envelope and/or capsid around the virus, causing lysing. They also penetrate the virus’s interior and disrupt the genome. These actions inactivate the virus. Hydroxyl radicals also pass through the outer cell wall structures of bacteria and oxidise the membrane responsible for electron transport, making the organism non-viable. [11]

Effects on allergens Edit

Hydroxyl radicals have been shown to modify the IgE-binding capacity in pollens, spores and pet dander through the degradation and modification of the tertiary structure and/or the induction of protein denaturation and/or aggregation, resulting in a modified allergen structure. Hydroxyl radicals instantly denature Der p1 and Der f1 (house dust mites). Hydroxyl radicals oxidise their protein structures, for example causing protein backbone damage due primarily to a hydrogen abstraction or oxygen addition. Both hydroxyl radical initiated oxidation mechanisms result in a modified allergen structure. Modified allergen structures are no longer recognised by the immune system and therefore histamine and other chemical mediators are not released. [12] [13] [14] [15]

Hydroxyl radicals play a key role in the oxidative destruction of organic pollutant using a series of methodologies collectively known as advanced oxidation processes (AOPs). The destruction of pollutants in AOPs is based on the non-selective reaction of hydroxyl radicals on organic compounds. It is highly effective against a series of pollutants including pesticides, pharmaceutical compounds, dyes, etc. [16] [17]

The atmospheric chemistry leading to hydroxyl radical creation is generally absent indoors. However, new technologies, pioneered by NASA (see Next Generation Hybrid Photo-Catalytic Oxidation (PCO) for Trace Contaminant Control (H-PCO)), have now made it possible to reproduce the outdoor effects of hydroxyl radicals indoors, enabling the continuous deactivation of viruses and bacteria, removal of toxic gases (such as ammonia, carbon monoxide and formaldehyde) and odours, and neutralisation of allergens throughout an inside space. In a similar development, Engineered Water Nanostructures (EWNS) are synthesized using two processes in parallel, namely, electrospraying and ionization of water. Pressurized water exits a hypodermic needle into an electric field (3KV-5KV) to produce a large number of reactive oxygen species (ROS), primarily hydroxyl (OH•) and superoxide (O2−) radicals. Good results were reported inactivating pathogens.

The hydroxyl • OH radical is one of the main chemical species controlling the oxidizing capacity of the global Earth atmosphere. This oxidizing reactive species has a major impact on the concentrations and distribution of greenhouse gases and pollutants in the Earth atmosphere. It is the most widespread oxidizer in the troposphere, the lowest part of the atmosphere. Understanding • OH variability is important to evaluating human impacts on the atmosphere and climate. The • OH species has a lifetime in the Earth atmosphere of less than one second. [18] Understanding the role of • OH in the oxidation process of methane (CH4) present in the atmosphere to first carbon monoxide (CO) and then carbon dioxide (CO2) is important for assessing the residence time of this greenhouse gas, the overall carbon budget of the troposphere, and its influence on the process of global warming. The lifetime of • OH radicals in the Earth atmosphere is very short, therefore • OH concentrations in the air are very low and very sensitive techniques are required for its direct detection. [19] Global average hydroxyl radical concentrations have been measured indirectly by analyzing methyl chloroform (CH3CCl3) present in the air. The results obtained by Montzka et al. (2011) [20] shows that the interannual variability in • OH estimated from CH3CCl3 measurements is small, indicating that global • OH is generally well buffered against perturbations. This small variability is consistent with measurements of methane and other trace gases primarily oxidized by • OH, as well as global photochemical model calculations.

In 2014, researchers reported their discovery of a "hole" or absence of hydroxyl throughout the entire depth of the troposphere across a large region of the tropical West Pacific. They suggested that this hole is permitting large quantities of ozone-degrading chemicals to reach the stratosphere, and that this may be significantly reinforcing ozone depletion in the polar regions with potential consequences for the climate of the Earth. [21]

First interstellar detection Edit

The first experimental evidence for the presence of 18 cm absorption lines of the hydroxyl ( • OH) radical in the radio absorption spectrum of Cassiopeia A was obtained by Weinreb et al. (Nature, Vol. 200, pp. 829, 1963) based on observations made during the period October 15–29, 1963. [22]

Important subsequent detections Edit

Year Description
1967 • HO Molecules in the Interstellar Medium. Robinson and McGee. One of the first observational reviews of • OH observations. • OH had been observed in absorption and emission, but at this time the processes which populate the energy levels are not yet known with certainty, so the article does not give good estimates of • OH densities. [23]
1967 Normal • HO Emission and Interstellar Dust Clouds. Heiles. First detection of normal emission from • OH in interstellar dust clouds. [24]
1971 Interstellar molecules and dense clouds. D. M. Rank, C. H. Townes, and W. J. Welch. Review of the epoch about molecular line emission of molecules through dense clouds. [25]
1980 • HO observations of molecular complexes in Orion and Taurus. Baud and Wouterloot. Map of • OH emission in molecular complexes Orion and Taurus. Derived column densities are in good agreement with previous CO results. [26]
1981 Emission-absorption observations of HO in diffuse interstellar clouds. Dickey, Crovisier and Kazès. Observations of fifty eight regions which show HI absorption were studied. Typical densities and excitation temperature for diffuse clouds are determined in this article. [27]
1981 Magnetic fields in molecular clouds — • HO Zeeman observations. Crutcher, Troland, and Heiles. • OH Zeeman observations of the absorption lines produced in interstellar dust clouds toward 3C 133, 3C 123, and W51. [28]
1981 Detection of interstellar HO in the Far-Infrared. J. Storey, D. Watson, C. Townes. Strong absorption lines of • OH were detected at wavelengths of 119.23 and 119.44 microns in the direction of Sgr B2. [29]
1989 Molecular outflows in powerful HO megamasers. Baan, Haschick, and Henkel. Observations of • H and • OH molecular emission through • OH megamasers galaxies, in order to get a FIR luminosity and maser activity relation. [30]

Chemistry Edit

In order to study gas phase interstellar chemistry, it is convenient to distinguish two types of interstellar clouds: diffuse clouds, with T = 30–100 K and n = 10–1000 cm −3 , and dense clouds, with T = 10–30 K and density n = 10 4 – 10 3 cm −3 . Ion chemical routes in both dense and diffuse clouds have been established for some works (Hartquist, Molecular Astrophysics, 1990).

Production pathways Edit

The • OH radical is linked with the production of H2O in molecular clouds. Studies of • OH distribution in Taurus Molecular Cloud-1 (TMC-1) [31] suggest that in dense gas, • OH is mainly formed by dissociative recombination of H3O + . Dissociative recombination is the reaction in which a molecular ion recombines with an electron and dissociates into neutral fragments. Important formation mechanisms for • OH are:

(Dissociative recombination: 1a )

(Dissociative recombination: 1b )

(Dissociative recombination: 2a )

(Ion–molecular ion neutralization: 4a )

Destruction pathways Edit

Experimental data on association reactions of • H and • OH suggest that radiative association involving atomic and diatomic neutral radicals may be considered as an effective mechanism for the production of small neutral molecules in the interstellar clouds. [32] The formation of O2 occurs in the gas phase via the neutral exchange reaction between O and • OH, which is also the main sink for • OH in dense regions. [31]

Atomic oxygen takes part both in the production and destruction of • OH, so the abundance of • OH depends mainly on the H3 + abundance. Then, important chemical pathways leading from • OH radicals are:

Rate constants and relative rates for important formation and destruction mechanisms Edit

Rate constants can be derived from the dataset published in a website. [33] Rate constants have the form:

The following table has the rate constants calculated for a typical temperature in a dense cloud T = 10 K .

Reaction k at T = 10 K (cm 3 ·s −1 )
1a 3.29 × 10 −6
1b 1.41 × 10 −7
2a 4.71 × 10 −7
3a 5.0 × 10 −11
4a 1.26 × 10 −6
5a 2.82 × 10 −6
1A 7.7 × 10 −10
2A 3.5 × 10 −11
3A 1.38 × 10 −10
4A 1.0 × 10 −10
5A 3.33 × 10 −14

Formation rates rix can be obtained using the rate constants k(T) and the abundances of the reactants species C and D:

where [Y] represents the abundance of the species Y. In this approach, abundances were taken from The UMIST database for astrochemistry 2006, and the values are relatives to the H2 density. Following table shows the ratio rix/r1a in order to get a view of the most important reactions.

r1a r1b r2a r3a r4a r5a
r1a 1.0 0.043 0.013 0.035 3.6 × 10 −5 0.679

The results suggest that (1a) reaction is the most prominent reaction in dense clouds. It is in concordance with Harju et al. 2000.

The next table shows the results by doing the same procedure for destruction reaction:

r1A r2A r3A r4A r5A
r1A 1.0 6.14 × 10 −3 0.152 3.6 × 10 −5 4.29 × 10 −3

Results shows that reaction 1A is the main sink for • OH in dense clouds.

Interstellar observations Edit

Discoveries of the microwave spectra of a considerable number of molecules prove the existence of rather complex molecules in the interstellar clouds, and provides the possibility to study dense clouds, which are obscured by the dust they contain. [34] The • OH molecule has been observed in the interstellar medium since 1963 through its 18-cm transitions. [35] In the subsequent years • OH was observed by its rotational transitions at far infrared wavelengths, mainly in the Orion region. Because each rotational level of • OH is split in by lambda doubling, astronomers can observe a wide variety of energy states from the ground state.

Tracer of shock conditions Edit

Very high densities are required to thermalize the rotational transitions of • OH, [36] so it is difficult to detect far-infrared emission lines from a quiescent molecular cloud. Even at H2 densities of 10 6 cm −3 , dust must be optically thick at infrared wavelengths. But the passage of a shock wave through a molecular cloud is precisely the process which can bring the molecular gas out of equilibrium with the dust, making observations of far-infrared emission lines possible. A moderately fast shock may produce a transient raise in the • OH abundance relative to hydrogen. So, it is possible that far-infrared emission lines of • OH can be a good diagnostic of shock conditions.

In diffuse clouds Edit

Diffuse clouds are of astronomical interest because they play a primary role in the evolution and thermodynamics of ISM. Observation of the abundant atomic hydrogen in 21 cm has shown good signal-to-noise ratio in both emission and absorption. Nevertheless, HI observations have a fundamental difficulty when they are directed at low mass regions of the hydrogen nucleus, as the center part of a diffuse cloud: the thermal width of the hydrogen lines are of the same order as the internal velocities of structures of interest, so cloud components of various temperatures and central velocities are indistinguishable in the spectrum. Molecular line observations in principle do not suffer from this problem. Unlike HI, molecules generally have excitation temperature TexTkin, so that emission is very weak even from abundant species. CO and • OH are the most easily studied candidate molecules. CO has transitions in a region of the spectrum (wavelength < 3 mm) where there are not strong background continuum sources, but • OH has the 18 cm emission, line convenient for absorption observations. [27] Observation studies provide the most sensitive means of detections of molecules with subthermal excitation, and can give the opacity of the spectral line, which is a central issue to model the molecular region.

Studies based in the kinematic comparison of • OH and HI absorption lines from diffuse clouds are useful in determining their physical conditions, especially because heavier elements provide higher velocity resolution.

Masers Edit

• OH masers, a type of astrophysical maser, were the first masers to be discovered in space and have been observed in more environments than any other type of maser.

In the Milky Way, • OH masers are found in stellar masers (evolved stars), interstellar masers (regions of massive star formation), or in the interface between supernova remnants and molecular material. Interstellar • OH masers are often observed from molecular material surrounding ultracompact H II regions (UC H II). But there are masers associated with very young stars that have yet to create UC H II regions. [37] This class of • OH masers appears to form near the edges of very dense material, place where H2O masers form, and where total densities drop rapidly and UV radiation form young stars can dissociate the H2O molecules. So, observations of • OH masers in these regions, can be an important way to probe the distribution of the important H2O molecule in interstellar shocks at high spatial resolutions.

Chapter 9: Reaction Mechanisms, Pathways, Bioreactions and Bioreactors

An active intermediate is a molecule that is in a highly energetic and reactive state It is short lived as it disappears virtually as fast as it is formed. They are short lived CA 10 -14 s and present in very low concentrations. That is, the net rate of reaction of an active intermediate, A*, is zero.

The assumption that the net rate of reaction is zero is called the Pseudo Steady State Hypothesis (PSSH)

The active intermediates reside in the trough of the reaction coordinate as shown below for in the reaction studied by Zewoil.

has an elementary rate law

However. Look what happens to the rate as the temperature is increased.

Why does the rate law decrease with increasing temperature?

 (1) (2) (3)

The PSSH assumes that the net rate of species A * (in this case, NO3 * ) is zero.

This result shows why the rate decreases as temperature increases.

Michaelis-Menten Kinetics

Enzymes are protein like substances with catalytic properties.

Enzyme unease. [From Biochemistry, 3/E by Stryer, copywrited 1988 by Lubert Stryer. Used with permission of W.H. Freeman and Company.]

It provides a pathway for the substrate to proceed at a faster rate. The substrate, S, reacts to form a product P.

A given enzyme can only catalyze only one reaction. Urea is decomposed by the enzyme urease, as shown below.

It has been proposed that an artificial kidney to remove urea from the blood could contain encapsulated enzymes and be worn externally.

The corresponding mechanism is:

Types of Enzyme Inhibition

Uncompetitive Substrate Inhibition

The Uncompetitive Substrate Inhibition rate law is

Data from Laboratory of H.S. Fogler taken by P.h.D Candidate Barry Wolf.

Polymath Setup

d(Cs)/d(t) = D*(Cso - Cs) - Ysc*rg - m*Cc

rg = (((1 - (Cp/Cpstar))**0.52) * mumax*(Cs/(Ks + Cs))*Cc

1.) Neglect Death Rate and Cell Maintenance

How does this figure relate to drinking a lot of fluids when you have an infection or cold?

Alcohol Metabolism

Water tissue volumes, flow rates, and perfusion rates for this model

Rate-law parameters for this model

Drug Delivery

See the Professional Reference Shelf's Pharmacokinetic Section

Polymers are macromolecules built up by the linking together of large numbers of much smaller molecules. The smaller molecules are called monomers and they repeat many times.

A polymer is a molecule made up of repeating structural (monomer) units.

Structural Repeating Unit (mer)

High density:
Plastic cups

Low density:
Sandwich bags

Poly (2-hydroxyethyl methacrylate)

Poly (tetra fluoro ethylene)

Poly (ethylene teraphthalate)

Polymers that are synthesized from a single monomer are named by adding the prefix poly such as polyethylene. However, a parenthesis is placed after the prefix poly when the monomer has a substituted parent name or multiword name such as poly (acrylic acid) or poly (vinyl alcohol).

Homopolymers consist of a single repeating unit. All of the above are examples of homopolymers.

Can Crystallize.

 a. Botactic = isotatic = same side b. Syndiotatic = alternating c. Atactic = random

3. Branched Type A: Long Branches Off Backbone

Branched Type B: Short Branches Off the Backbone

Branched Type C: Branches on Branches Off the Backbone

For example, copolymers used to make records.

PVC - hard - irrigation pipes, hard to engrave
PVAc - easy to engrave
PVC + PVAc copolymer phonograph records (these are a thing of the past)

 Alternating QSQSQS Poly (vinyl acetate-alt-vinylchloride) Block QQQSSS Poly (vinyl acetate-b-vinyl chloride) Graph QQQQQ . . . . . | . . . . . SSSS Poly (vinyl acetate-g-vinyl chloride Random QSSQQQSQSSS Poly (VA c -co-VC) Statistical QSSQSQQSS

D. What affects polymer properties

• Chemistry

• Molecular Weight () and Molecular Weight Distribution

Weight Average Molecular Weight

Molecular Weight Distribution

• Crystalinity

Amorphous Phase (Non-crystalline Phase) no order or orientation

T g - characteristic of amorphous state

Below glass transition temperature, T g , there is a cessation of virtually all molecular motion (vibration , rotation).

Crystalline Phase gives an order to the structure.

Order means crystallinity

Above the crystalline melting temperature, T m , thre is no order. Fraction of total polymer that is in the crystalline state is the degree of crystallinity

• Branching

• Tacticity

E. Molecular Weight (MW)

Number average molecular weight

Weight average molecular weight

Hence gives a truer picture of the average molecular weight.

TWO TYPES OF HOMOGENEOUS POLYMERIZATION: STEP AND CHAIN

Step Polymerization. Monomer must be bifunctional. Polymerization proceeds by the reaction of two different functional groups. Monomer disappears rapidly, but molecular weight builds up slowly.

All species are treated as polymers. Mostly used to produce polyesters and polyamides.

Chain Polymerization. Requires an initiator. Molecular weight builds up rapidly. Growing chains require 0.0001 to 1 to 10 seconds to terminate. Have high molecular weight polymers right at the start.

I. Step Polymerization

1. Different functional groups on each end of monomer.

Structural Unit

Here the structural unit is the repeating unit.

2. Same functional groups on each end. Example: diamines and diols

B. Polymerization Mechanism

Monomer dimer ----> trimer ----> tetrameter ----> Pentamer ---->

The number of structural units equals the number of bifunctional monomers present.

1. Monomers with different functional groups - one structural unit.

Here the repeating unit is the structural unit.

 Let p = fraction of functional groups of either A or B that have reacted. Let M = concentration of either A or B functional groups at time t. Let M 0 be the concentration of either A or B functional groups initially Let N = total number (concentration) of polymer molecules present at time t. Let N 0 = total number of polymer molecules initially Let M A = number of functional groups of A at time t. Let M A0 = number of functional groups of A initially. = number average degree of polymerization. It is the average number of structural units per chain.

the number average molecular weight.

Where is the mean molecular weight of the structural units and is the molecular weight of the end group.

D. Monomers with Same End Group

For a stoichiometric feed the number of A and B functional groups the same.

E. Stoichiometry Imbalance in the Feed

1. Stoichiometry Imbalance Type 1: Monomers with thesame end group and r not equal to 1

The maximum number average chain length is greatly reduced if the initial feed is not exactly stoichiometric

2. Stoichiometry Imbalance Type 2: Monomers with different end groups. Monofunctional Monomer Present

3. Stoichiometry Imbalance Type 3: Monomers with different end groups. Monofunctional Monomer Present

REACTION BETWEEN A DIOL (HOROH) AND A DIBASIC ACID (HOOCR1COOH)

Assume Reaction (1) is essentially in equilibrium

Case 1: The acid itself acts as a strong acid catalyst:

[HA] º [COOH] and Stoichiometric Feed.

As the reaction proceeds and more ester is produced, the solution becomes less polar. As a result the uncatalyzed carboxylic acid becomes the major catalyst for the reaction, and the overall reaction order at high conversion is well described by a third order reaction (Case 1). The high conversion region is of primary importance because this region is where the high molecular weight polymers are formed.

At low conversions the solution is more polar and the proton, H + is the more effective catalyst (Case 2) than the unionized carboxylic acid. Under these conditions, the reaction is self catalyzed and the reaction is 5/2 order.

Case 2: Self catalyzed but acid acts as a weak acid catalyst, not completely dissociated

Case 3: External Acid Catalyzed H + is constant

F. Kinetics of Step Polymerization

 (1)

k is defined wrt the reactants

Why 2k? Because there are two ways A and B can react (thus, 2k)

 (2)

 (3)

For all reactions of P1

In general for j ? 2

Mole balance on polymer of length j, in terms of the concentration P j in a batch system

If we proceed further it can be shown that

Total number of polymer molecules (i.e. functional groups of either A or B) =

This is the Flory Distribution for the mole fraction of molecules with chain length j.

The weight fraction is just

G. Flory Distribution-Probability Approach

Rule: The probability of several events occurring successively in a particular way equals the product of the probabilities that each event happens that way.

P = probability that an A group will has reacted

(1-P) = probability group has not reacted.

A - R - B

HO - R C OO - H

Mo = number of functional groups initially (no. of molecules)

M = number of functional groups remaining

 Number distribution function.

 Weight distribution function

On a number average basis there will always be more monomer than polymer.

II. Chain Polymerization

Example: Polyethylene

Back biting

Branched Polyethylene

resulting low density (0.92)

B. Cationic Polymerization

D. Ziegler-Natta Polymerization

Ziegler-Natta Catalyst

Steps in Polymer Chain Growth

(4) Desorption from active site

+

to produce linear polymer: Eq. High Density Polyethylene (0.98) (HDPE)

Chain polymerizations require an initiator.

1. The Reaction

INITIATION

This reaction produces the formation of the Primary Radical

The distance and midpoint formulas

We want to calculate the distance between the two points (-2, 1) and (4, 3). We could see the line drawn between these two points is the hypotenuse of a right triangle. The legs of this triangle would be parallel to the axes which mean that we can measure the length of the legs easily.

We'll get the length of the distance d by using the Pythagorean Theorem

This method can be used to determine the distance between any two points in a coordinate plane and is summarized in the distance formula

The point that is at the same distance from two points A (x1, y1) and B (x2, y2) on a line is called the midpoint. You calculate the midpoint using the midpoint formula

We can use the example above to illustrate this

$m =left ( frac<4+(-2)> <2> ight ),: : left ( frac<3+1> <2> ight )=$