23: Naming Organic Compounds

Naming Organic Compounds/IUPAC Nomenclature:

Alkyl Group: a functional group attached to another hydrocarbon that contains only carbon and hydrogen atoms. Here, we'll only deal with the first two, methyl and ethyl.

Methyl: –CH₃



The blue shading is the main chain of carbons and the pink shading is methyl group. On a skeletal structure, it looks like one extra line protruding from the main chain, to show another carbon to carbon link with the three hydrogens implied on its end. Here is an example of a main chain with the structure of hexane with a methyl group attached to it:



Ethyl: –C₂H₅



Again, the blue is the main chain and the pink is the ethyl group. On a skeletal structure, it looks like an extra line with a kink in it protruding from the main chain. Here is an example of a main chain with the structure of heptane with an ethyl group attached to it:





(Rules for naming taken off of the handout from class)

Naming Alkanes:

1. Find parent chain: longest continuous chain of carbons. It will not always be straight across. In some of the examples, what appears to be an ethyl group is actually the end of the parent chain. The parent chain is named according to how many carbons it has: butane if it has four carbons, pentane if it has five carbons, etc. This will go at the very end of the name when we're done.

2. Find and identify the alkyl groups (methyl or ethyl as far as this class goes).

3. Number the carbons on the chain starting from the side closest to an alkyl group. Each alkyl group will get a prefix of the number associated with the carbon it is bonded to, even if there is more than one of that kind of alkyl group. But instead of writing out “methyl” two times if the alkane has two methyl groups, it is numbered with the appropriate Greek prefix (in this example, it would be di- for two). The numbers associated with the positions of the carbons are written in front, with commas separating them from each other and a dash separating them from words.

4. When both of the alkyl groups we're dealing with are present, number each group and list them in alphabetical order in the name.

Examples:








Naming Cycloalkanes:

Cycloalkanes are saturated hydrocarbons with rings. Their skeletal structures are geometric shapes where the corners represent carbon atoms assumed to be bonded with hydrogens.



1. As far as this class goes, there will either be an ethyl group or a methyl group (or multiples) attached to the ring. If there is only one, there is no need to give it a prefix.

2. When there are two alkyl groups, alphabetize them. The one that comes first alphabetically is given the first spot, and the carbons are numbered in the direction that will give the smallest number for the second substituent (so number towards the shortest path to the second alkyl group).

3. If there are more than two substituents, number in a way that yields the smallest numbers possible in the finished name. If there are identical sets of numbers, use alphabetical priority.

Examples:




Naming Haloalkanes:

Also called alkylhalides/halogenated alkanes. They are alkanes with one or more halogens attached (F, Cl, Br, I).

1.
F = fluoro
Cl = chloro
Br = bromo
I = iodo

2. Halogen and alkyl substituents are considered of equal rank, go by alphabetical priority.

Examples:




Naming Alkenes and Cycloalkenes:

Alkenes and cycloalkenes have one or more carbon double bonds.

1. Replace the suffix -ane with -ene.

2. The parent chain will be the longest continuous chain of carbon atoms that contains both carbons of the double bond.

3. Number the parent chain from the end closest to the double bond.

4. Two numbers will be associated with the double bond, assign it the smaller one. This number will appear as a prefix to the parent chain name, with a dash separating numbers and words.

5. With cycloalkenes, number in the direction that gives the lowest numbers for substituents.

Examples:




Naming Alkynes:

Alkynes have carbon triple bonds. The rules are exactly the same as they are for the alkenes, except the new suffix is -yne.

Example:




Naming Alcohols:

Alcohols have hydroxyl groups: –OH

1. Name the longest parent chain that includes the hydroxyl group. The suffix for alcohols is -anol.

2. Number starting at the end closest to the hydroxyl group. The number associated with the hydroxyl group will appear as a prefix to the parent chain name, with a dash separating numbers and words.

3. Name and number the other substituents the same as before.

Examples:

23: Introduction to Organic Chemistry

Organic Chemistry: The Chemistry of Hydrocarbons and Hydrocarbon Derivatives
Carbon can bond in four different ways. All of these bonds are very strong, especially if the bond is to another carbon.



Definitions:
Hydrocarbon: compound containing only carbon and hydrogen.
Hydrocarbon Derivatives: compounds containing carbon and hydrogen and one or more different elements.
Saturated Hydrocarbon: hydrocarbons that contain only single bonds between the carbon atoms.
Unsaturated Hydrocarbon: hydrocarbons that contain one or more multiple bonds (double or triple bonds) between the carbon atoms.
Cyclic Hydrocarbon: a hydrocarbon in which a chain of carbon atoms has formed a ring.
Acyclic Hydrocarbon: a hydrocarbon that does not contain a ring of carbon atoms.

Alkanes:


These are the molecular and structural formulas for the first six. Following the same pattern, after hexane comes heptane, octane, nonane, and decane.

It is important to remember that carbon atoms form four bonds. Do not give your carbon atom more than four things to carry, because he only has four arms.


Condensed Structural Formula: formulas where the bonds around each carbon atom in the compound are not explicitly written.

Example:
Butane has a chain of four carbons, two of which attach to three hydrogens each, and two of which attach to two hydrogens each.

The condensed structural formula for butane is: CH₃CH₂CH₂CH₃ or CH₃(CH₂)₂CH₃

Think of the CH₃ on each end like the front and back of a train, and the CH₂ in the middle keeps getting added on to as you move to the next molecule in the series, just as you would add extra cars to a train.


Constitutional Isomerism: compounds with the same molecular formula but different structural formulas.



Methane, ethane, and propane all only have one isomer. They only have one structural formula each. Butane has two isomers. Pentane has three.


Remember a straight-chain can bend, but if the bonds on the carbon atoms haven't changed, it's not a new isomer.


Line-Angle Structural Formula (Skeletal Structure):


1. Each intersection, endpoint, and kink is a carbon.
2. The carbon to carbon bonds are shown, as well as the bonds between carbon and anything that's not a hydrogen.
3. Every remaining bond on carbon is assumed to be a hydrogen.
4. When counting the carbons in chain, the very beginning of the chain is where you count the first carbon.

22.7: Crystal Field Theory

Crystal Field Theory:
Remember what the 3d orbitals look like:


Simplify the ligands into negatively charged or slightly negatively charged point particles. A complex ion with six ligands will have them arranged octahedrally around the metal atom (because they will repel each other). The arrangement of these negative charges will lie on the x, y, and z axes, as represented by the red spheres below:


Notice that the arms of the d_z² and d_{x² – y²} orbitals also lie along the axes where the octahedrally arranged ligands (negative point particles) lie (the other orbitals lie between the axes). Those electrons will have a stronger repulsive effect on the ligands, creating a situation where the five 3d orbitals are no longer of the same energy.

These two orbitals that lie on the axes will have higher energy, and the other three will have lower energy.


Where Δ = crystal field splitting = the difference in energy between the two sets of d orbitals on a central metal ion that arises from the interaction of the orbitals with the electric field of the ligands.


High and Low Spin Complexes:

Pairing Energy: the energy required to put two electrons into the same orbital, denoted P. When an orbital already has an electron in it, there needs to be an input of energy to overcome the repulsion these two electrons will have for each other, this is the pairing energy.

Imagine the electron configuration of a metal ion in a complex, specifically the 3d orbitals. There are five to fill. Three have lower energy than the other two, which have higher energy. So the three low energy orbitals get filled first. But say there are four electrons in the configuration, not three, how do we know whether to start pairing the electrons in the low energy orbitals or to go ahead and start filling the empty, higher energy orbitals?


It depends on P and Δ. If P is greater than Δ, this means it requires more energy to pair an unpaired electron in a lower energy orbital than it requires to boost that electron up to the higher energy orbital. So, if P is greater than Δ, that fourth electron goes in the first empty high energy orbital, and it is called a high spin complex:



If Δ is greater than P, this means it requires more energy to boost that electron up to the higher energy orbital than it does to pair an unpaired electron in a lower energy orbital. So, if Δ is greater than P, that fourth electron pairs with the first unpaired electron in the low energy orbital, and it is called a low spin complex:


Here is a complete filling of the 3d orbitals, comparing high spin and low spin:


Paramagnetic substances have at least one unpaired electron in their orbitals. Most of the configurations in the diagram above show paramagnetism. Diamagnetic substances have no unpaired electrons in their orbitals. Diamagnetism is shown in the diagram above at six electrons at low spin, and at ten electrons at both high and low spin.

Spectrochemical Series: an arrangement of ligands according to the relative magnitudes of crystal field splittings they induce in the d orbitals of a metal ion.

As Δ increases, ligands bond more strongly (found on page 957).

Example:
Which one is high spin, [CrBr₆]^3– or [Cr(CN)₆]^3–?

From the spectrochemical series, we see that Br has a relatively small Δ, so it is high spin.
CN has a relatively large Δ, so it is low spin.


Visible Spectra of Transition-Metal Complexes:

Many transition-metal substances are colored, and the color is related to Δ through the wavelength, λ, of light absorbed:

Δ = hυ = hc / λ
or
λ = hc / Δ

where:
Δ = crystal field splitting
λ = wavelength
υ = frequency
h = Planck's Constant = 6.626068 x 10^-34 J · s
c = the speed of light = 3.00 x 10⁸ m / s
Note: With these problems, wavelength might be given in nanometers, but the rest of the units are in meters. To convert: 1 nm = 1 x 10^–9 m

Example:
For a certain substance, if λ = 555 nm, what is Δ?

Convert from nanometers to meters: 555 nm = 555 x 10^–9 m = λ
h = 6.626068 x 10^-34 J · s
c = 3.00 x 10⁸ m / s

Δ = hc / λ
Δ = [(6.626068 x 10^-34 J · s)(3.00 x 10⁸ m / s)] / (555 x 10^–9 m)
Δ = 3.58 x 10^-19J

Example:
For a certain substance, if Δ = 2.00 x 10^–19, what is λ?

λ = hc / Δ
λ = [(6.626068 x 10^-34 J · s)(3.00 x 10⁸ m / s)] / (2.00 x 10^–19 J)
λ = 9.95 x 10^–7 m = 995 nm

22.6: Valence Bond Theory of Complexes

Valence Bond Theory of Complexes:

Remember from Chapter 10 that a covalent bond is usually formed by the overlap of the a bonding orbital from each atom, each containing one electron. The subsequent bond then has two electrons.

Remember also that the bonds in complexes are coordinate covalent bonds, where both electrons are donated from the ligand to the metal ion. With the coordinate covalent bond in a complex, a ligand orbital with two electrons overlaps an empty orbital on the metal.



This bonding requires hybrid orbitals.

Example:

What are the hybridized orbitals on Co in [Co(H₂O)₆]³⁺?

First we need to find the configuration of the metal ion. Start by writing the configuration of the cobalt atom. Here's a helpful table to remember how to do it:



Co: [Ar]4s²3d⁷

If our ion has a charge of +3, and H₂O is a neutral molecule, we know that the charge on the cobalt ion is +3. To reflect this in the configuration, remove 3 electrons. Remember the outer s electrons are removed first:

Co³⁺: [Ar] 3d⁶

Now draw the orbital diagram of the ion. Remember to follow Hund's Rule, where each electron is placed in a separate orbital with the same spin before pairing it with an electron of opposite spin:



In the complex ion [Co(H₂O)₆]³⁺ each water molecule wants to donate one pair of electrons to the metal. In order to do this, it needs an empty orbital. Since there are six H₂O, it needs six empty orbitals. Starting with the first empty orbital, work forward until you get six, and these are the hybridized orbitals:



It is called sp³d² because one orbital comes from s, three come from p, and two come from d.

Example:

What are the hybridized orbitals on Mn in [Mn(NH₃)₆]⁴⁺?

First we need to find the configuration of the metal ion. Start by writing the configuration of the manganese atom.

Mn: [Ar] 4s²3d⁵

If our ion has a charge of +4, and NH₃ is a neutral molecule, we know that the charge on the manganese ion is +4. To reflect this in the configuration, remove 4 electrons. Remember the outer s electrons are removed first:

Mn⁴⁺: [Ar] 3d³

Now draw the orbital diagram of the ion, following Hund's Rule:



In the complex ion [Mn(NH₃)₆]⁴⁺ each NH₃ molecule wants to donate one pair of electrons to the metal. In order to do this, it needs an empty orbital. Since there are six, it needs six empty orbitals:



It is called d²sp³ because two orbitals come from d, one comes from s, and three come from p.

22.5: Structure and Isomerism in Coordination Compounds

Structure and Isomerism in Coordination Compounds:

Constitutional Isomers: isomers that differ in how the atoms are joined together, specifically, in the order that they are bonded to each other.

Stereoisomers: isomers in which the atoms are bonded to each other in the same order but differ in the precise arrangement of the atoms in space.


Constitutional Isomers:

Example: Ion-Exchange Isomer

[Co(NH₃)₅(SO₄)]Br
[Co(NH₃)₅Br]SO₄

In the first isomer, SO₄ is attached to the Cobalt and is part of the complex ion (the cation), with Br as the anion. In the second isomer, Br is attached to the cobalt as part of the complex and SO₄ is acting as the anion.

Example: Coordination Isomer

[Cu(NH₃)₄][PtCl₄]
[Pt(NH₃)₄][CuCl₄]

Here, both the cation and anion are complex ions. In the first isomer, NH₃ is attached to the copper and the Cl^– are attached to the platinum. In the second isomer, they have swapped.

Example: Linkage Isomer

[FeCl₅(NO₂)]^3–
[FeCl₅(ONO)]^3–

Here, the difference is in how the ligand bonds to the metal. In the first isomer, the ligand bonds to the metal through an electron pair on the nitrogen. In the second isomer, the ligand bonds to the metal through an electron pair on one of the oxygen atoms. It's easier to see it:




Stereoisomerism:

Geometric Isomers: isomers in which the atoms are joined to one another in the same way but differ because some atoms occupy different relative positions in space.



In the cis arrangement, two like ligands are placed next to each other. In the trans arrangement, two like ligands are placed across from each other.




(helpful source)

(info on mer/fac isomers)


Chiral vs. Achiral:



Achiral Molecules: these isomers are superimposable mirror images of one another. A molecule and it's mirror image can completely overlap.

Chiral Molecules: If a molecule and a molecule that appears as its mirror image cannot completely overlap, they are chiral molecules. These isomers are non-superimposable mirror images of one another. Also called optical isomers, optically active, or enantiomers.

Enantiomers have identical properties, but are differentiated by their effect on plane-polarized light.



Dextrorotatory: a compound whose solution rotates the plane of polarized light to the right (when looking toward the source of light).

Levorotatory: a compound whose solution rotates the plane of polarized light to the left (when looking toward the source of light).

Racemic Mixture: a mixture of equal amounts of optical isomers. Because the two isomers rotate the plane of polarized light by the same angle in opposite directions, they cancel each other out and have no net effect.

22.4: Naming Coordination Compounds

Naming Coordination Compounds:

Ligands That Are Part of a Complex: (never (rarely?) positively charged)

Ligands With No Charge:
H₂O = aqua
NH₃ = ammine
CO = carbonyl
en = ethylenediamine

Ligands With a Charge of –1:
CN^– = cyano
F^– = fluoro
Br^– = bromo
Cl^– = chloro
I^– = iodo
OH^– = hyrdoxo


Ligands With a Charge of –2:
O^2– = oxo
C₂O₄^2– = oxalato (only other bidentate ligand)
SO₄^2– = sulfato

With the exceptions of en and oxalato, all of these ligands are monodentate (one “bite”).

Rules For Naming:

1. The name of the cation comes first, the anion comes second. The name of the complex consists of two parts written together as one word. Ligands are named first and the metal atom is second.

2. Name and number the ligands in alphabetical order.
Prefixes for Numbering:
1 = mono
2 = di
3 = tri
4 = tetra
5 = penta
6 = hexa
7 = hepta
8 = octa
9 = nona/ennea
10 = deca

3. Metal name stays intact if it is a cation. Add -ate to the end otherwise.

Name/Anion Name:
Copper/Cuprate
Gold/Aurate
Iron/Ferrate
Lead/Plumbate
Silver/Argentate
Tin/Stannate

4. Add “ion” if it is one.

Example:

What is the name of the following complex ion?

[Co(NH₃)₆]³⁺

First, name and number the ligands in alphabetical order. There is only one ligand, the NH₃, so no need to worry about alphabetizing here. This ligand is called ammine.

There are six of them, so it is hexaammine.

We know that the NH₃ molecule has no charge, and we see that the ion has a +3 overall charge, which means this must come from the cobalt, so we know the name of the metal is Cobalt(III). And since it is the cation, we can use this name instead of changing it.

The name of our complex ion then is hexaamminecobalt(III) ion.

What is the Coordination Number of this ion?

NH₃ is monodentate, and there are six of them. So C.N. = 6


Example:

[Fe(H₂O)₂(CN)₄]^2–

Name and number the ligands in alphabetical order:

(H₂O)₂ = diaqua
(CN)₄ = tetracyano

Name the metal:

H2O has no charge. CN has a –1 charge, and there are 4 of them. And the total charge on the complex ion is –2. So we have:

Charge on Fe + 0 + (–4) = –2
Charge on Fe = +2

So we know we have Iron(II). And since the complex ion is an anion, it's called ferrate(II).

The name of the complex ion is diaquatetracyanoferrate(II) ion.

All six ligands are monodentate, so C.N. = 6.


Example:

Na₃[Ti(O)₂(OH)F₂(SO₄)]

The name of the cation, Sodium, comes first.

Name and number the ligands in alphabetical order:

F₂ = difluoro
OH = hydroxo (the mono is unnecessary)
O₂ = dioxo
SO₄ = sulfato

Name the metal:

O has a charge of –2 and there are two of them. OH has a charge of –1. F has a charge of –1 and there are two of them. SO4 has a charge of –2. And since Na has a charge of +1, and it takes 3 of them to neutralize the anion, we know the charge on the anion must be –3. So:

Charge on Ti + (–2 x 2) + (–1) + (–1 x 2) + (–2) = –3
Charge on Ti = +6

So it is Titanium(VI). Since it is part of the anion, it is called titinate(VI).

The name of the complex ion is Sodium difluorohydroxodioxosulfatotitinate(VI).

It has six monodentate ligands, so the C.N. = 6.


Example:

[Os(C₂O₄)₂Cl₂]Cl₂

Notice that Cl₂ appears twice. The first time it is part of the complex. The second time, it is the anion. And just like with other ionic compounds, the Cl₂ denotes two Cl^– and not Cl₂ gas. And as the anion, chloride will be named after the complex ion.

Name and number the ligands in alphabetical order:

Cl₂ = dichloro
(C₂O₄)₂ = dioxalato

Name the metal:

C2O4 has a charge of –2 and there are two of them. Cl has a charge of –1 and there are two of them. And we know that it takes two of the Cl^– anion to neutralize the charge on the complex ion, which must be +2. So:

Charge on Osmium + (–2 x 2) + (–1 x 2) = +2
Charge on Osmium = +8

So it is Osmium(VIII). As part of the cation, it gets to keep this name.

So the name of the complex is dichlorodioxalatoosmium(VIII) chloride.


Example:

[Cd(en)₃][PtCl₆]

Here we have a complex ion as both the anion and the cation.

Special case with en:

When the name of the ligand also has a number prefix, as in ethylenediamine, there is a new set of prefixes that needs to be used to name it:
2 en = bis-
3 en = tris-
4 en = tetrakis-

Remember that Cadmium has only one oxidation state, and can only have a charge of +2, and en has no charge. So the charge on the cation is +2, which means the charge on the anion is –2. Cl has a charge of –1 and there are six of them. So:

Charge on Platinum + (–1 x 6) = –2
Charge on Platinum = +4

So, following the rules, the name of our complex compound is trisethylenediaminecadmium(II) hexachloroplatinate(IV).

Now try to write the notation for the complex ion from the name:

Example:

dicarbonyltrioxalatoaurate(IV) ion

When writing it this way, the metal comes first. Aurate(IV) = Au

dicarbonyl = (CO)₂
trioxalato = (C₂O₄)₃

charge on Au(IV) = +4
charge on (CO)₂ = 0 (neutral molecule)
charge on (C2O4)3 = –6
charge on ion = –2

[Au(CO)₂(C₂O₄)₃]^2–

22.3: Formation and Structure of Complexes

Formation and Structure of Complexes:

Transition-metal atoms often function as a Lewis acid in a chemical reaction. Remember that a Lewis acid accepts electron pairs from molecules or ions.

Example:

Complex ions are not held together by ionic bonds, but coordinate covalent bonds. Remember that coordinate covalent bonds are bonds between atoms where both electrons involved in the bond come from only one of the atoms.

In water, the Fe²⁺ ion ultimately bonds to six H₂O, forming the [Fe(H₂O)₆]²⁺ ion. The final charge on the ion can be found by adding up all the charges of its components.

Example:

Fe = +2
H₂O = 0

2 + (0 x 6) = 2

Charge on the ion = +2

Definitions:

Complex Ion: A metal ion with Lewis bases attached to it through coordinate covalent bonds.

Complex/Coordination Compound: a compound consisting either of complex ions and other ions of opposite charge or a neutral complex species (such as the anti-cancer drug in the chapter opening). Has a cation and an anion.

Example:

[FeOH₂]Cl₂ ← means 2 Cl^–, not Cl₂ gas. Two negatively charged chlorine ions neutralize the +2 charge on the complex ion. Everything in the brackets in front of the Cl₂ is the cation.

Ligands: the Lewis bases attached to the metal atom in a complex. Electron pair donors. Can be neutral (such as H₂O or NH₃) or anions (such as Cl^–).

Coordination Number: the total number of bonds the metal atom forms with ligands. The number of lone pairs that are attached to the metal via the coordinate covalent bond.

Example:

In [Fe(H₂O)₆]²⁺, the metal atom, Fe, bonds to each oxygen atom in six water molecules.

Coordination Number (C.N.) = 6

Example:

[Ag(NH₃)₂]⁺ → [H₃N꞉Ag꞉NH₃]⁺

One coordinate covalent bond between the Ag and each N of the two NH₃ molecules.
C.N. = 2

Polydentate Ligands: ligands that can bind with two or more atoms to a metal atom. Think of the word “bite” when you hear dentate. Monodentate means it forms one bond with its atoms (one bite), bidentate means it forms two bonds with its atoms (two bites), etc.

22.1: Periodic Trends in the Transition Elements

Transition Elements:



Transition Elements: those metallic elements that have an incompletely filled d subshell or easily give rise to common ions that have incompletely filled d subshells. Highlighted in the table above. They are the basis for Inorganic Chemistry (every other element except carbon).

Periodic Trends of the Transition Elements:

1. They are all metals.

2. Mostly have high melting points and high boiling points and are hard solids. This is due to metallic bonding.



The negatively charged electrons form an “electron sea” around the positively charged nuclei of the metal atoms and are shared as they move about the sea.

3. They have more than one oxidation state (exceptions: Cd²⁺, Ag⁺, Zn²⁺).

Example:

Fe²⁺, Fe³⁺; Mn²⁺, Mn³⁺, Mn⁴⁺, Mn⁷⁺.

4. Transition metal ion solutions have a color.




Atomic Radii:

For the main group elements, covalent radii decrease in size across a row due to effective nuclear charge: the positive pull of the nucleus on the negative electrons. This is the positive nuclear charge that acts on the covalent electrons to pull them in, diminished (shielded) by any electrons between them.



Going down from the fourth period to the fifth period, the atomic radii increase. Further down to the sixth period, atomic radii are approximately the same as they are in the fifth period.

This similarity is due to lanthanide contraction. The lanthanides are the first row of the inner transition elements (placed at the bottom of the table), from cerium to lutetium. Here, the 4f subshell is filled, and by the time we get back to this row of transition elements with hafnium, the covalent radius is practically the same as the elements in the row above it.

Interesting Articles Mentioned In Class:

Just How Small Is the Proton?

Galaxy-Scale Bubbles Extending From the Milky Way

CERN Physicists Trap Antimatter Atoms

20.4-20.7: Radioactive Rate of Decay-Nuclear Fission and Nuclear Fusion

Radioactive Rate of Decay:

half-life: the time it takes from one-half of the nuclei in a sample to decay. Independent of the amount of sample, denoted t_½.

First Order Half Life:
t_½ = 0.693 / k

Example:

k = 1.45 x 10^-12 s^-1 for Ta-183, find t_½ in years.

t_½ = 0.693 / 1.45 x 10^-12 s^-1
t_½ = 4.78 x 10¹¹ s

Now convert to years:

(4.78 x 10¹¹ s) x (1h / 3600s) x (1 day / 24h) x (1 year / 365 days) = 1.52 x 10⁴ years

Once we know the decay constant (k) for a radioactive isotope, we can calculate the fraction of the radioactive nuclei left after a given time by the following equation:

ln (N_t / N_o) = –kt

N_o = the number of nuclei in the original sample
N_t = the number of nuclei left after time t

Never really need to find N_t or N_o, it is the fraction that is important.

Example:

When t_½ = 10.76 years, what is the fraction remaining after 25 years?

Remember k = 0.693 / t_½:

k = 0.693 / (10.76 years)
ln (N_t / N_o) = –(0.693t / t_½)
ln (N_t / N_o) = –(0.693x 25years / 10.76 years)
ln (N_t / N_o) = –1.61
(N_t / N_o) = e^–1.61
(N_t / N_o) = 0.200

This means 20% is left.

Example:

How long until there is only 5% left?

(N_t / N_o) = 5% = 0.05
ln(0.05) = –(0.693t / 10.76 years)
t = 46.6 years


Radioactive Dating:

Carbon–14, t_½ = 5730 years

The upper atmosphere is constantly bombarded by cosmic ray radiations from space, which creates high energy neutrons when it reacts with Nitrogen–14 (the most abundant nitrogen nuclide).

¹⁴₇N + ¹₀n → ¹⁴₆C + ¹₁H

Then goes into carbon dioxide:

¹⁴₆CO₂ → plants → animals

There is a set amount of Carbon–14 per kg of living matter. When something dies, it no longer takes in Carbon–14. The Carbon–14 then decays:

¹⁴₆C → ¹⁴₇N + ⁰_-1e

In living matter, we measure 15.3 disintegrations per kg of C per minute. Anything lower → must be dead. In this way, this ratio of carbon isotopes becomes a clock measuring the time since death of an organism.

N_t / N_o = (less than 15.3 dis/kg • min) / (15.3 dis/kg • min)

Example:

You want to carbon date a caveman's foot, where you observe (4.50 dis/kg • min). When did the caveman lose his foot?

ln(N_t / N_o) = –(0.693t / t_½)
ln(4.50 / 15.3) = –(0.693t / 5730 years)
t = 10118 years


Applications of Radioactive Isotopes:

Chemical Analysis:

Radioactive Tracer: a very small amount of radioactive isotope added to a chemical, biological, or physical system to study the system.

Evidence of dynamic equilibrium can be seen using radioactive tracers. In a beaker, imagine a solution of PbI₂ in contact with the solid. This beaker contains only iodine atoms with nonradioactive isotopes. Now add a radioactive iodide ion to the solution. The solution is still saturated, and the amount of solid remains constant. But, after time, the solid lead iodide, which was initially nonradioactive, is radioactive. Nonradioactive iodide ions in the solid have swapped with radioactive ions in solution.

Tracers are also used to determine chemical mechanisms, like photosynthesis.

Isotope Dilution: a technique to determine the quantity of a substance in a mixture or of the total volume of solution by adding a known amount of an isotope to it. After time, removal of a sample of the mixture and measurement of the fraction by which the isotope has diluted provides a way to determine the quantity of substance or volume of solution.

Neutron Activation Analysis: an analysis of elements in a sample based on the conversion of stable isotopes to radioactive isotopes by bombarding a sample with neutrons. Example: measuring trace amounts of arsenic in human hair. The arsenic is converted into a metastable nucleus via bombardment with a neutron. As the arsenic returns to its ground state, it emits gamma rays. These measurements can be interpreted to find who the hair belongs to, or how much arsenic is in it.

⁷⁵₃₃As + ¹₀n → ^76m₃₃As

^76m₃₃As → ⁷⁵₃₃As + ⁰₀γ


Mass-Energy Equivalence:

When nuclei decay, they form products of lower energy. The change of energy is related to changes in mass, according to the mass-energy equivalence derived by Albert Einstein.

E = mc²

Where c = the speed of light = 3.00 x 10⁸ m/s
E = energy
m = mass

For any mass, there is an associated energy. For any energy, there is an associated mass.
When the energy changes by an amount ΔE, the mass changes by an amount Δm.

ΔE = Δmc²

When energy is given off, mass decreases.

Example:

Chemical reaction:

When carbon burns in oxygen, it releases heat energy:
C (gr) + O₂ (g) → CO₂ (g); ΔH = –393.5 kJ
ΔE = –3.935 x 10⁵ J (converted from kJ to J)
Recall that J = (kg•m² / s²)

Δm = ΔE / c²
Δm = –(3.935 x 10⁵ kg•m² / s²) / (3.00 x 10⁸ m/s)²
Δm = –4.37 x 10^-12 kg


Mass changes in nuclear reactions are approximately a milion times larger per mole of reactant thatn those in chemical reaction.

Example:

Nuclear Reaction:

Consider the alpha decay of Uranium–238 to Thorium–234

²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He

Use the nuclear mass in amu to find the change in mass for this reaction. You would think the mass of the reactants is the same as the mass of the products, but not so. There is a mass loss.

²³⁸₉₂U = 238.0003 amu
²³⁴₉₀Th = 233.9942 amu
⁴₂He = 4.00150 amu

Δm = (233.9942 amu + 4.00150 amu) – (238.00003 amu)
Δm = – 0.0046 amu

(Assume 1 mole of isotope.)

Δm = – 0.0046 amu = – 0.0046g = – 4.6 x 10^-6 kg
ΔE = Δmc² = (– 4.6 x 10^-6 kg)(3.00 x 10⁸ m/s)²
ΔE = – 4.14 x 10¹¹ J = – 4.14 x 10⁸ kJ


Nuclear Fission: spontaneous break up of isotopes, naturally occurring, does not require energy input. Powers nuclear reactors and bombs.

Nuclear Fusion: Fusing of two nuclei into a larger nucleus. Takes GIGANTIC amounts of energy.

²₁H ³₁Th→ ⁴₂He + ¹₀n

20.2-20.3: Nuclear Bombardment Reactions-Radiations and Matter

Transuranium Elements:

DOES NOT INCLUDE URANIUM!!!

Includes elements with an atomic number greater than that of uranium (Z = 92), the naturally occurring element of greatest Z.


Radiation Counters:
Because nuclear radiations can ionize molecules and break chemical bonds, they adversely affect biological organisms. So it's good to know when they're there.

Geiger Counter: a kind of ionization counter used to count particles emitted by radioactive nuclei, consists of a metal tube filled with gas, such as argon. Can detect both alpha and beta particles, and under special circumstances, neutrons.

¹₀n + ¹⁰₅B → ⁷₃Li + ⁴₂He



Scintillation Counter: a device that detects nuclear radiation from flashes of light generated in a material by the radiation. A phosphor is a substance that emits flashes of light when struck by radiation. Can detect both beta and gamma particles using a photomultiplier tube.




One e^– can produce 10⁶ e^–.

This vintage reel on Scintillation Counters is way better than a diagram:


Radiation counters measure the number of nuclear disintegrations in a radioactive sample.

Activity: (activity of a radioactive source) is the number of nuclear disintegrations per unit time occurring in a radioactive material.

Curie: Denoted Ci, is a unit of activity equal to 3.700 x 10¹⁰ disintegrations per second.

Nuclear disintegrations / second = nuc / s = Ci


Biological Effects of Radiation Dosage:

Radiation dosage effects can be quite damaging. DNA is especially affected, which interferes with cell division.

rad → (radiation absorbed dose) the dosage of radiation that deposits 1 x 10^-2 J of energy per kg of tissue.

Depends not only on the amount of energy deposited in the tissue, but also on the particle. Neutrons are more destructive than gamma rays of the same radiation dosage measured in rads.

rem → a unit of radiation dosage used to relate various kinds of radiation in terms of biological destruction. It equals the rad times a factor for the type of radiation, called the relative biological effectiveness, or RBE (basically, how efficient it is at killing you).

rems = rads x RBE → total damage

RBE = 1 for γ (gamma) and β (beta).
RBE = 5 for n^o
RBE = 10 for α (alpha)

Here's that movie mentioned in class:


It really is an awesome movie, go rent it!


Rate of Radioactive Decay:

The rate equation for radioactive decay has the same form as the rate law for a first order chemical reaction.

R = kN_t

R = Rate = Activity
N_t = the number of radioactive nuclei at time t
k = radioactive decay constant, the rate constant for radioactive decay

Example:

A 2.50 mg sample of Tc-99 has an activity of 2.70 x 10^-5 Ci and is decaying by beta emission. What is the decay constant?

R = kN_t

First convert the activity to nuc/s:

R = 2.70 x 10^-5 Ci x (3.700 x 10¹⁰ nuc/s / 1s) = 9.99 x 10⁵ nuc/s

Now convert the given sample to N_t:

2.50 mg Tc-99 = 2.50 x 10^-3g Tc-99
(2.50 x 10^-3g Tc-99) x (1 mol Tc-99 / 99g Tc-99) x (6.023 x 10²³ nuclei / 1 mol Tc-99) = 1.52 x 10¹⁹ nuclei = N_t

k = Rate / N_t
k = (9.99 x 10⁵ nuc/s) / ( 1.52 x 10¹⁹ nuclei)
k = 6.57 x 10^-14 s^-1

Units for k will always be inverted seconds because it is first order.

20.1: Radioactivity-Nuclear Bombardment Reactions

Nuclear Stability:

Why don't all the protons in the nucleus repel each other and cause the nucleus to blow apart?

Nuclear Force: a strong force of attraction between nucleons that acts only at very short distances, overwhelms the electrostatic force (the positive charges repelling each other.)

Shell Model of the Nucleus: a nuclear model in which protons and neutrons exist in levels, or shells, analogous to the shell structure that exists for electrons in an atom.



First Test For Stability:

Magic Numbers: the number of nuclear particles in a completed shell of protons or neutrons. Think of the way full electron shells make noble gases stable, this is similar. If a nucleus has two magic numbers, it is very stable.

Magic Numbers for p⁺ → 2, 8, 20, 28, 50, 82, (114)
Magic Numbers for n^o → 2, 8, 20, 28, 50, 82, 126

Example:

⁴₂He
p⁺ → 2
n^o → 2
Both are magic numbers, very stable.

Example:

Which is more stable?
¹⁰²₅₁Sn or ¹⁰¹₅₁Sn?

¹⁰²₅₁Sn → p⁺ = 51, n^o = 51
¹⁰¹₅₁Sn → p⁺ = 51, n^o = 50

¹⁰¹₅₁Sn has a magic number and ¹⁰²₅₁Sn doesn't.

Second Test for Stability:

Even/Odd Test:



Nuclei with an even number of protons and an even number of neutrons are very likely to be stable.
Nuclei with an odd number of protons and an odd number of neutrons are not at all likely to be stable.
Nuclei with variably even and odd numbers of protons or neutrons may or may not be stable.

Example:

Which is more likely to be stable, ⁶²₃₁Ga or ⁶⁴₃₂Ge?

⁶⁴₃₂Ge because it has both an even number of protons and an even number of neutrons.

Third Test for Stability:

Band of Stability: the region in which stable nuclides lie in a plot of number of protons against number of neutrons.




When it turns out to be unstable, what does the nucleus do about it?

The Six Types of Radioactive Decay:

1. Alpha Emission: emission of a ⁴₂He nucleus, or alpha particle from an unstable nucleus. Happens for very large nuclei.

Example:

²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂He

2. Beta Emission: emission of a high-speed electron (β^–) from an unstable nucleus.

Example:

¹⁴₆C → ¹⁴₇N + ⁰_-1β

Equivalent to the conversion of a neutron to a proton.

¹₀n → ¹₁p + ⁰_-1e

3. Positron Emission: emission of a positron (β⁺, or ⁰₁e) from an unstable nucleus.

Example:

⁹⁵₄₃Tc → ⁹⁵₄₂Mo + ⁰₁e

Equivalent to the conversion of a proton to a neutron.

¹₁p → ¹₀n + ⁰₁e

Professor said a positron is an electron traveling backward in time, but I have no idea what that means.

4. Electron Capture: the decay of an unstable nucleus by capturing, or picking up, an electron from an inner orbital of an atom.

Example:

⁴⁰₁₉K + ⁰_-1e → ⁴⁰₁₈Ar

In effect, a proton is changed to a neutron, as in positron emission:

¹₁p → ⁰₁e^- + ¹₀n

5. Gamma Emission: emission from an excited nucleus of a gamma photon (denoted γ). Often, gamma emission occurs very quickly after radioactive decay. In some cases, however, an excited state has a significant lifetime before it emits a gamma photon.

Metastable Nucleus: a nucleus in an excited state with a lifetime of at least one nanosecond (10^-9s). In time, the metastable nucleus decays by gamma emission.

Example:

^99m₄₃Tc → ⁹⁹₄₃Tc + ⁰₀γ

^99m₄₃Tc → the m denotes metastable, excited state for nucleus
⁹⁹₄₃Tc → ground state

6. Spontaneous Fission: the spontaneous decay of an unstable nucleus in which a heavy nucleus of mass number greater than 89 splits into lighter nuclei and energy is released.

Example:

²³⁶₉₂U → ⁹⁶₃₉Y + ¹³⁶₃₃I + 4¹₀n

When A is greater than 89 → spontaneous fission

Recall the Band of Stability:



*To the left of the band, nuclides have a neutron to proton ratio larger than that needed for stability, so they tend to decay by beta emission.
*To the right of the band, nuclides have a smaller neutron to proton ratio that that needed for stability, so they tend to decay by either positron emission or electron capture.
*As the curve follows Z as it becomes larger than 83, decay is often by alpha emission.


Radioactive Decay Series: a sequence in which one radioactive nucleus decays to a second, which then decays to a third, and so forth. Eventually, a stable nucleus, which is an isotope of lead, is reached.

Each step will involve an alpha or a beta decay.



Example:

What is the nucleus formed from Uranium–238 after six alpha (⁴₂He) and two beta (⁰_-1β) emissions?

²³⁸₉₂U → ^□_□□ + 6⁴₂He + 2⁰_-1e

^□_□□ = ²¹⁴₈₂Pb


Nuclear Bombardment Reactions:

Alchemists → wanted to turn lead to gold and live forever

Transmutation: the change of one element to another by bombarding the nucleus of the element with nuclear particles or nuclei.

Notation:

Target Nucleus + Subatomic Particle → Product Nucleus + Subatomic Particle

Example:

¹⁴₇N +⁴₂He → ¹⁷₈O + ¹₁H

¹⁴₇N(⁴₂He, ¹₁H)¹⁷₈O

Example:

⁹₄Be +⁴₂He → ¹²₆C + ¹₀n

⁹₄Be(⁴₂He, ¹₀n)¹²₆C


Elements with large atomic numbers deflect alpha particles (large positive nucleus deflect positive alpha particles.) To shoot large particles into heavy nuclei, charged particles must be accelerated.

Particle Accelerator: a device used to accelerate electrons, protons, and alpha particles and other ions to very high speeds. The kinetic energies of these particles is units of electron volts.

Electron Volt: Denoted eV, the quantity of energy that would have to be imparted to an electron to accelerate it by one volt potential difference.

1 eV = 1.602 x 10^-19 J

Cyclotron: a type of particle accelerator consisting of two hollow, semicircular metal electrodes called dees (because the shape resembles the letter D), in which charged particles are accelerated by stages to higher and higher kinetic energies.



Ions introduced at the center of the cyclotron are accelerated in the space between the two dees. A magnetic field keeps the ions moving in a spiral path.



The dees are connected to a high-frequency electric current that changes their polarity so that each time the ion moves in the space between the dees, it is accelerated. Then it leaves the cyclotron at high speed and hits its target.

This guy really loves cyclotrons: