Category Archives: Chem 1

Nomenclature of organic compounds

Naming organic compounds can be a bit daunting, as there are a lot of rules and the names just get longer as the molecules get more complex, even if it is something familiar to you. Do you recognise the following names?

(10R,13R)-10,13-dimethyl-17-(6-methylheptan-2-yl)-2,3,4,7,8,9,11,12,14,15,16,17-dodecahydro-1H-cyclopenta[a]phenanthren-3-ol

Cholesterol

(5R)-[(1S)-1,2-dihydroxyethyl]-3,4-dihydroxyfuran-2(5H)-one

Ascorbic acid, or vitamin C

(5α,6α)-7,8-didehydro-4,5-epoxy-17-methylmorphinan-3,6-diol diacetate

Heroin

Once you get the hang of it, naming organic compounds is really very easy, and I will demonstrate with the following molecule:

This one

First thing to do is find the longest carbon chain.

Longest carbon chain has 7 carbons

Then identify what functional groups, (if any) are present:

We have one methyl group and one chloro group

Now we number the carbons on the chain, so that the largest number of substituents have the lowest number:

This way the chloro is number 2 and the methyl is number 5. If I’d numbered it the other way, the methyl would be 3 and the chloro 6.

Now we put it all together remembering that we order the substituents alphabetically, a hyphen goes between a number and a letter, a comma goes between two numbers and there shouldn’t be any spaces, (except when the functional group requires it).

So… that gives us: 2-chloro-5-methylheptane!

Easy!

Here are some links to some practice questions:

http://www.chembio.uoguelph.ca/educmat/chm19104/nomenclature/quizes.html

http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Questions/Nomencl/nomencl.htm

The order of precedence of functional groups can be found here

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Cycloalkanes

Naming cycloalkanes:

Nomenclature of cycloalkanes is simple, and follows the same rules as for any other alkane. The prefix cyclo- is added to the basic alkane name, and any substituents are added in the front.

For example:

Name the following cycloalkanes:

Questions taken from Solomans "Fundamentals of Organic Chemistry" 1994

 

Cast your minds back to types of isomerism, and you’ll remember that we discussed conformational isomers. These are particularly important in cycloalkanes, as the molecule is locked in a cyclic arrangement and cannot rotate out of an unfavourable conformation.

 

The most stable conformation for cyclohexane is the “chair conformation”, called so because it resembles a reclining chair. This conformation is the most stable as all carbon-carbon bond angles are 109.5 degrees, which minimises strain and, as the Newman projection shows, all of the substituents are in staggered conformations.

See? If you squint a bit, it looks like a cyclohexane ring

The most comfortable conformation

By simple rotation, the chair conformation can become the boat conformation, which is also free of strain, but is less stable as there are interactions between the substituents. The Newman projection shows that the boat conformation is an eclipsed conformation, and therefore less stable.

Avast! A more strained conformation

 

 

At room temperature, cyclohexanes rapidly flip, via the boat conformation, from one chair conformation to another. This ring flipping is important, because when the ring flips, all of the bonds that were axial become equatorial and vice versa.

If all of the substituents are hydrogens, this doesn’t matter too much as both chair conformations are equivalent and the ring will flip equally between the two. However, if there is a non-hydrogen substituent, the chair conformation that has the substituent in an equatorial position will be the most stable.

 

 

 

If there is more than one substituent on the ring, you want the bulkiest group to be in an equatorial position. The gold standard example of this is a tertiary butyl group:

This group is, comparatively, huge, (try making a cyclohexane ring with a tert-butyl group on it with a molecular modelling kit. You’ll see that the tert-butyl group is almost as large as the ring itself), and must always be in the equatorial position.

 

 

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Alkenes

Alkenes are hydrocarbons that contain a double bond between two carbons, and are sometimes referred to by their old name, “olefin”. You might see this in some older, American textbooks.

Double bonded carbons show sp2 hybridisation, rather than sp3, (alkanes). In sp2 atoms, the s orbital has hybridised with 2 of the 3 available p orbitals, forming a trigonal planar arrangement of 3 sp2 orbitals.

This arrangement leaves a p orbital unhybridised, which sits perpendicular to the plane of the sp2 orbitals. This p orbital forms the pi bond of the double bond.

Naming alkenes:

The nomenclature of alkenes is much the same as alkanes:

  1. Find the longest carbon chain, which contains the double bond and name it as though it were an alkane, but change the end from –ane to –ene.
  2. Number the chain so as to include both carbon atoms of the double bond, and begin numbering at the end of the chain nearer the double bond. Use the number of the first carbon atom in the double bond to designate its position.
  3. Number the other substituents
  4. If you have a cycloalkene, number the carbons so the double bond has positions 1 and 2.
  5. Designate the geometry of the double bond with cis/trans or E/Z.

Name the following alkenes:

Questions taken from "Solomans Fundamentals of Organic Chemistry" 1994

E/Z isomerism:

Cis and trans can be used to designate the geometry of an alkene, provided it is disubstitued.

As soon as the alkene is tri or tetra substituted, cis and trans become meaningless, so E/Z is more correctly used for alkenes. (note: cis/trans is used for cycloalkanes).

E and Z designation follow the Cahn-Ingold-Prelog rules for determining priority of a substituent. The priority of each group on one carbon atom of the double bond is determined, then that is repeated for the second carbon atom. If the higher priority on both carbon atoms are on the same side of the double bond, (see cis), then it is Z (zusammen, German for together). If they are on opposite sides, (see: trans), it is E (entgegen, German for opposite).

Name the following alkenes, including E/Z and R/S geometries where required:

Questions taken from Solomans "Fundamentals of Organic Chemistry" 1994

http://www.chemguide.co.uk/basicorg/isomerism/geometric.html

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Unit conversion

I often get asked about unit conversion, and it is a common mistake made in exams and prac classes. Units are very important, your answer without a unit means absolutely nothing! So, it’s important to get it right.

In the metric system, you need to know some of the prefixes associated with orders of magnitude.

tera- (T-) 1012
giga- (G-) 109
mega- (M-) 106
kilo- (k-) 103
hecto- (h-) 102
deka or deca- (da-) 10
deci- (d-) 10-1
centi- (c-) 10-2
milli- (m-) 10-3
micro- (µ-) 10-6
nano- (n-) 10-9
pico- (p-) 10-12

You need to multiply by a factor of 1000 to go between each prefix shown above.

There are a series of tutorials below, many of which have worked examples and problems for you to try:

http://chemistry.about.com/od/convertcalculate/a/conversions.htm

http://www.tutor-homework.com/Chemistry_Help/Unit_Conversion.html

http://www.learnchem.net/tutorials/mathb.shtml

http://pages.towson.edu/ladon/unit.html

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Isomerism in organic molecules

Isomers are compounds, which have the same chemical formula, but a different structure. There are several classes of isomer.

Structural isomers (also called constitutional isomers):

Structural isomers are molecules that have the same formula, but are connected in different ways. This can include branching of a chain, different positions of functional groups or the splitting up of a functional group, (for example, a carboxylic acid may become a ketone and an alcohol).

Conformational isomers (also rotational isomers or “rotamers”):

Conformational isomers occur when there is rotation around a single bond. It is less important for small, chain molecules as they are constantly rotating, but it is very important for bulky or cyclic alkanes.

Stereoisomers, (also geometric or optical isomers):

This broad class of isomers includes any molecule with the same order of atoms, but different geometries, or different arrangement of the atoms in space. There are two different types of stereoisomers:

Enantiomers: non-superimposable mirror images.

Non-superimposable mirror images

Diastereomers: stereoisomers that are not mirror images. These can include cis-trans (or E-Z) isomers, and molecules with more than one stereocentre.

To get stereoisomerism, you need a chiral molecule – one that is not identical to its mirror image. The most common example of chirality is your hands – your left and right hands are mirror images of each other, but you can’t superimpose them.

Chiral molecules can be spotted by looking for a chiral (or stereo) centre. This will be a carbon with four different substituents attached to it.

It is important to be able to distinguish between different enantiomers, so we use the R/S nomenclature. To designate R or S, you use the Cahn-Ingold-Prelog rules:

1. Prioritise the atoms directly attached to the chiral centre by atomic number.

2. If two or more atoms are the same, look at the next atom out, continuing until you find a point of difference.

3. Multiple bonds are counted as the same number of bonds to the same atom, (for example, a double bond to an oxygen counts as two single bonds to oxygen atoms)

Now, with the 4th priority substituent pointing into the page, draw a curved line from priority 1 to 2 to 3. If it is clockwise, the molecule is R, anticlockwise is S.

This molecule is S

Here are some links that might be useful:

http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/sterism3.htm

http://facultystaff.vwc.edu/~jeaster/courseinfo/Tutorials/stereochemistryl.html

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Acid-base equilibria

Acids and bases are easy to deal with when they are strong, as you can assume that any reaction goes to completion. However, weak acids and bases don’t always ionise completely, so you need to look at the equilibria to fully understand them.

Acids dissociate in water to give a proton, (or hydronium ion) and the conjugate base ion.

 HA ⇄ H+ + A

 We can put this into an equilibrium expression, which gives us Ka, the acid ionisation constant:

Ka is an indication of the strength of the acid, the stronger the acid, the larger the Ka.

The same thing applies to bases;

B + H2O ⇄ HB+ + OH

Leading to the Kb expression:

Ka and Kb are the inverse of one another for a given equilibrium.

 Buffers:

Most of the stuff for acid-base equilibria is the same as for any other chemical equilibrium problem, however buffers do require a bit more discussion.

A buffer is a solution that can resist changes in pH when small amounts of acid or base are added. They are very important in a broader sense as they regulate the pH of your body and the oceans. Buffers contain a conjugate pair of a weak acid or base, generally in equal parts, for example a carbonate buffer would look like this:

H2CO3 ⇄ H+ + HCO3

where you have added H2CO3 and NaHCO3 to water. The buffer can shift its equilibrium between the acid and conjugate base when small amounts of acid or base are added. The pH will be regulated, because you have such large amounts of H2CO3 and HCO3 to start with, any small change will not affect the overall equilibrium to a large degree.

The Henderson-Hasselbalch equation is used to calculate the pH of a given buffer, and is just a rearrangement of the Ka expression:

 

 Links:

http://www.shodor.org/unchem/basic/ab/

http://www.nauticus.org/chemistry/chemacidbase.html

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Chemical equilibrium

Not all chemical reactions go to completion, most go backwards and forwards around a point somewhere in the middle, eventually coming to a stop. Of course, the reaction doesn’t cease at this point, the reaction continues to hover around the equilibrium point, which is why it is known as dynamic equilibrium.

Technically, equilibrium is defined as the point at which the rates of the forward and reverse reactions are equal.

We express the point of equilibrium using the equilibrium expression:

for the reaction aA + bB ⇄ cC + dD. K is the equilibrium constant, the larger it is, the more the reaction favours the products. Obtaining the equilibrium constant is simply a matter of substituting the equilibrium concentrations into the equilibrium constant.

Alternatively, you may be given K and the initial concentrations and asked to find the equilibrium concentration. There is an example here,

because I can’t get the example to format here. The procedure to answer these problems is as follow:

  1. Set up a table of concentrations showing the initial, change and final concentrations for all products and reactants.
  2. Substitute the “equilibrium concentrations” (including the x) into your equilibrium expression.
  3. Solve!

And for those of you who have forgotten year 10 maths – the quadratic equation and a reminder of how to use it is here. For those of you who do remember, you can see a cat on a vacuum cleaner.

Links:

http://library.thinkquest.org/10429/low/equil/equil.htm

http://www.shodor.org/unchem/advanced/equ/index.html

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Thermodynamics and spontaneity

The first law of thermodynamics is basically the law of conservation of energy as relevant to thermodynamic systems. The change in internal energy of a system, (ΔU) is equal to the sum of the heat and work (q + w) of the system.
The second law of thermodynamics describes whether or not a change is spontaneous, expressing it in terms of entropy.

Entropy (S), is the thermodynamic quantity that describes the disorder, (randomness) in a system. The entropy is related to the number of states a molecule has available to it. A molecule at high temperature has more vibrational states available than one at a lower temperature, and therefore has a higher entropy. A crystal locks molecules into a certain configuration, whereas molecules in a gas are free to move about and therefore have higher entropy.

The second law of thermodynamics states that the total entropy of a system and its surroundings always increases for a spontaneous process. Generally, we refer to this is the entropy of the universe, as the sum of the entropies of the system and surroundings must increase, however one may decrease and the process may still be spontaneous.

ΔSuniverse = ΔSsystem +ΔSsurroundings

We can restate this law so it refers only to the system, and as heat flows into or out of the system, the entropy goes with it. So, at a certain temperature, the entropy is associated with heat q;

ΔS > q/T for a spontaneous process

Therefore, for a spontaneous process at a certain temperature, the change in the entropy must be greater than the heat divided by the absolute temperature. For systems that are at equilibrium, the entropy is equal to the heat over temperature.

Phase changes:

The entropy of a phase change is derived from the equation above.

ΔS=ΔH/T

Where the ΔH is the heat of the phase change, and the temperature at which the phase change occurs.

 Spontaneity:

Looking at entropy and enthalpy, we can determine whether or not a process is spontaneous, and we introduce the concept of free energy (sometimes called Gibbs’ free energy, ΔG), which is equal to:

ΔG = ΔH – TΔS

For a spontaneous process, ΔG ≤ ΔH – TΔS, (ie, negative). We want the TΔS term to be larger than the ΔH term, indicating that even if a reaction is endothermic, if ΔS is larger, the reaction will still proceed.

 

Some links:

http://www.learnchem.net/tutorials/spont.shtml

http://secondlaw.oxy.edu/

http://www.chem1.com/acad/webtut/thermo/entropy.html

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Introduction to thermodynamics

Thermochemistry, the study of heat and energy and its effects on chemical reactions and physical transformations is one of the first branches of the chemical sciences and took off in the 1780s.

In chemistry, we look at how systems change when we change the conditions they are in. In terms of thermodynamics, we want to know what happens to the energy of a system under certain conditions.

Two types of energy we concern ourselves with in this area are work and heat. They are related to one another, and can be transformed from one to another. For example, heat energy released by the combustion of fuel can be used to turn a turbine, or rubbing sticks together will heat them up.

Work (w):

In physics, work is defined as

w= F x d

 where F is the force exerted and d is the distance moved.

In chemistry we change the equation slightly. We are no longer talking about something moving in one dimension, but a three dimensional system, we change Δd to ΔV and the force being exerted is the pressure of the gas. Therefore, work becomes:

w= -PΔV

It is negative because in chemistry we define work as that done by the system, not on it.

Heat (q):

Heat energy is related to the motion of the molecules – the hotter it is the faster they move, and jiggle about.

We measure changes in heat energy by looking at changes in temperature, and the two are related via the heat capacity of the substance. The heat capacity is the amount of energy required to raise one unit of the substance by one K (or °C). Molar heat capacity is measured in J/mol.K and specific heat capacity is J/g.K.

 

 q = CΔT

Internal energy (ΔU):

The total internal energy of the system is the sum of the potential energy and the kinetic energy. In this case, the potential energy is the work energy, and kinetic is related to the heat energy:

ΔU = w + q

Enthalpy (ΔH):

Enthalpy is a bit weird. It is the internal energy of a reaction plus any work it needs to do against the constant pressure of the atmosphere. My analogy is to imagine there is a magician who is going to make a rabbit appear:

In order to create the rabbit, (assuming he is actually magical and not just concealing one in his hat), he has to use a certain amount of energy, (internal energy of a rabbit). Next, he needs to create enough space for the rabbit to appear into, so he needs to exert some work to push back the atmosphere in the rabbit-shaped space (work). The enthalpy is the total energy expended by our magician:

 ΔH = ΔU +PΔV

Or

Energy expended by magician = internal energy of a rabbit + work to move the atmosphere

Meanwhile, the rabbit is left with existential dilemmas for which we have no equations

The real trick to understanding how these things fit together, is to look at what happens under certain circumstances:

Constant temperature:

Under constant temperature, ΔT = 0. That is fairly self-explanatory. If there is no change in temperature, then the internal energy of the system is constant, ΔU= 0, so q = w. In other words, any change in heat energy is exactly countered by work being done by/on the system so there is no net temperature change.

Constant volume:

If ΔV = 0, then there can be no work done, (w=0), so q = ΔU.

Constant pressure:

This is the most common condition for chemical reactions, as the external pressure (ie, the atmosphere) remains constant for anything done in an open container, (like, say, a beaker). At constant pressure, ΔH = q.

Some links:

http://www.ulb.ac.be/sma/testcenter/Test/problems/problems.html

http://www.shodor.org/unchem/advanced/thermo/

http://www.chem1.com/acad/webtext/energetics/

 

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Kinetic theory of gases

We can derive the ideal gas law by looking at the empirical laws of Charles, Boyle and Avogadro, but we can also find it theoretically using the kinetic theory of gases.

Hooke, an assistant of Boyle was the first to propose the idea that gas pressure is a result of collisions of randomly moving gas molecules/atoms with the sides of the container, (the explanation we use today). He didn’t advertise this idea though, as Newton had proposed an alternative, (and wrong) theory and Newton was a man that you simply didn’t disagree with. Hooke’s reputation isn’t all that great though – he may have been a brilliant man, (he developed the microscope, coined the term “cell” to describe… cells, developed Hooke’s law to describe the forces in a spring as well as contributions to astronomy, mechanics and the theory of gravitation. It was the latter that got him into trouble with Newton) but was apparently not a very nice one.

The kinetic theory is based on the idea that gas molecules/atoms are constantly, randomly moving and relies on the following postulates:

1. Gases are composed of molecules whose size is negligible, (compared to the distance between them).
2. Molecules move about randomly, but in straight lines and all at different speeds.
3. They don’t interact (attract or repulse) with each other.
4. When they do collide, the interaction is elastic (the total kinetic energy is maintained)
5. The average kinetic energy (KEav= 3/2RT) is proportional to the temperature.

To get the ideal gas equation, we remember that pressure is the result of a collision with the sides of the container and is therefore proportional to the frequency of the collisions and the average force exerted by the molecule during a collision.

P α frequency of collisions x force

The force depends on its momentum (its mass (m) x its velocity (u) – the bigger and faster it is, the more force it exerts. In other words, it hurts more to be hit with a fast bowling ball than a slow table tennis ball). The frequency of the collisions is dependant on the size of the container – a larger container will have fewer collisions, and on the speed of the molecules – the faster they are, the more often they will collide with the walls. Also, obviously, the more molecules you put in the container, the more often they will collide.

So… putting all of those things together:

P α (u x 1/V x N) x mu

If you put volume on the left:

PV α Nmu2

Kinetic energy depends on the mass and velocity of the molecules too (KE = 1/2mu2, which is also proportional to temperature).

PV α nT

(We can change N (number of molecules) to n (moles of molecules)) and when we put a proportionality constant in there we get the ideal gas equation!

PV = nRT.

Ta da!

The postulates for kinetic theory and the ideal gas equation are all well and good for ideal gases but they fall down for real gases. We know from experience that Boyle’s law is not valid had high pressure, (the gas will liquefy and the relationship between pressure and volume will no longer hold) and Charles’ law requires temperatures above the point where the gas will become liquid. These are true because in real life, gas particles will interact with one another, and are generally attracted to each other. They also have a real, measurable volume and cannot be compressed indefinitely. When considering real gases, we need to take these things into account and the ideal gas equation becomes the Van der Waals equation:

This version of the ideal gas equation has a correction factor (a) for the attractive forces between the molecules/atoms and (b) the real volume of the molecules/atoms.

 

Links:

http://www.science.uwaterloo.ca/~cchieh/cact/c120/gaskinetics.html

http://chemed.chem.wisc.edu/chempaths/GenChem-Textbook/Kinetic-Theory-of-Gases-Postulates-of-the-Kinetic-Theory-938.html

http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Gases/Kinetic_Theory_of_Gases

 

 

 

 

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