A delocalized π bond is a π bond in which the electrons are free to move over more than two nuclei.
In a molecule like ethylene, the electrons in the π bond are constrained to the region between the two carbon atoms.
We say that the π electrons are localized.
Even in penta-1,4-diene, the π electrons are still localized.
However, in buta-1,3-diene, the two orbitals can overlap, and the π electrons are free to spread over all four carbon atoms.
We say that these π electrons are delocalized.
In benzene, the π electrons are delocalized over all six atoms of the ring.
In β-carotene, the π electrons are delocalized over 22 carbon atoms!
It depends on your ionization method. After the ionization, the analyte ions are in general sent to an electromagnetic sector of the mass spectrometer for separation by
It may help if you look at this overview of mass spectroscopy.
There are many methods, but some common ionization methods are:
Often the method is chosen in favor of so-called soft ionization (rather than hard ionization).
Soft ionization better-retains the parent peak during fragmentation, so that you can find the peak that corresponds to the molecular mass of the original ion.
ELECTROSPRAY IONIZATION (ESI)
These analyte ions have little residual energy, so this ionization method is soft.
MATRIX-ASSISTED LASER DESORPTION IONIZATION (MALDI)
The general process was:
This is also a soft ionization method, since the laser does not directly ionize the sample itself (which would have overly fragmented some of the sample).
ELECTRON IONIZATION (EI)
Sometimes labeled the "classical" method. The general idea is:
This is a hard ionization method, since energy is directly transferred from the electron beam and generally little sample consists of intact molecular ions.
After the ionization, the analyte ions are in general sent to an electromagnetic sector of the mass spectrometer for separation by
Let's break this down a bit.
First, your job is to identify the parent chain, which is the longest carbon chain.
That's easy as the last part of the structure's name gives it away:
Next, you have to identify the substituents. Well, knowing that cyclopentane is the parent chain, whatever else that we have left are treated as the substituents. Let's write them down.
If we take our cyclopentane and first number all the carbons,
we can then, starting in numerical order, attach the substituents one-by-one.
Next, we take a look at
Lastly, looking at
I would think that it's because pyrene has less resonance stabilization than benzene does (increasing its HOMO-LUMO gap by less), due to its sheer size causing its energy levels to be so close together.
A smaller HOMO-LUMO gap means a more reactive system, despite it having resonance throughout.
DISCLAIMER: THOROUGH/LONG ANSWER!
EXAMINING THE EXTENSIVITY OF RESONANCE STABILIZATION
Consider napthalene, anthracene, and phenanthrene (if you add one benzene ring to the upper-right of phenanthrene, you have pyrene):
The resonance stabilization that one benzene ring gets is
#"Total Resonance Energy"#
#~~ "Number of Benzene Rings" xx "Resonance Energy"#
But you can see in the above diagram that it isn't:
From this, we could postulate that in general, the more extended the
ENERGY GAPS AS A FUNCTION OF VOLUME (AND ENTROPY)
The energy gaps (and thus the HOMO-LUMO gap) in any molecule are a function of the system volume and entropy.
At constant entropy though (which means at a constant distribution of states amongst the energy levels), the trend of volume vs. energy gap can be examined.
To illustrate this, the following graph was generated and derived from Huckel MO Theory, for which we have:
#E_k = alpha + 2betacos((2kpi)/n)#,
#k#is the energy level index and #n#is the number of fused rings. #alpha#is the nonbonding energy and #beta#is the negative difference in energy from the nonbonding level.
We can see then that the HOMO-LUMO gap converges as the number of rings increases, i.e. as the system volume increases.
PARTICIPATION OF HOMO & LUMO IN ELECTROPHILIC ADDITION
How is this relevant?
Well, the HOMO and LUMO are both required in electrophilic addition reactions. For example, with adding
And this forms the so-called bromonium complex:
(Here, the HOMO contained the
Since the HOMO-LUMO gap gets smaller when the system gets larger, it's very likely that the gap is so small for pyrene that the resonance stabilization (which increases this gap) isn't enough to make it unreactive towards electrophilic addition.
Well, these distribution graphs should correlate with the titration curve.
If we know the first
#"pH"_("1st half equiv. pt.") = "pKa"_1 + cancel(log\frac(["HA"^(-)])(["H"_2"A"]))^("Equal conc.'s, "log(1) = 0)#
#"pH"_("2nd half equiv. pt.") = "pKa"_2 + cancel(log\frac(["A"^(2-)])(["HA"^(-)]))^("Equal conc.'s, "log(1) = 0)#
We represent each stage of a diprotic acid as:
#"H"_2"A"(aq) rightleftharpoons overbrace("HA"^(-)(aq))^"singly deprotonated" + "H"^(+)(aq)#
#rightleftharpoons overbrace("A"^(2-)(aq))^"doubly deprotonated" + "H"^(+)(aq)#
The two midpoints shown are the first and second half-equivalence points, respectively.
At midpoint 1, we have that
At midpoint 2, we have that
A distribution graph shows the change in concentration of each species in solution as the
See below for an overlay of both:
Each species in solution is tracked in the bottom graph.
The cross-over points on the distribution graph are the half-equivalence points on the titration curve.
The maximum concentration for each species after the starting
You could have given us a bit more to go on.........
It is relatively straightforward to assign chirality to carbon compounds. Of course we have to have a depiction of the molecule.........and we ain't got one here. I can give you a few tips.
ANY carbon centre in a tetrahedral array that has four different substituents can generate a pair of optical isomers based on the geometry of its substituents. The one on the left (as we face it) is the
How? Well for each depicted stereoisomer, the substituent of least priority, the hydrogen, projects into the page.
Given such a stereoisomer, the interchange of ANY two substituents at one optical centre results in the enantiomer. Interchange again, (and it need not be the original two substituents) and you get the mirror-image of a mirror-image, i.e. the enantiomer of an enantiomer, that is the ORIGINAL isomer.
The big mistake that students tend to make with these problems IS NOT TO USE MOLECULAR MODELS, which are allowed examination materials in every test of organic and inorganic chemistry. It is hard to assess stereochemistry with a model; it is even harder to assess it without a model. And I assure if you go to the office of your organic prof., his or her desk will have a set of molecular models representing molecules in several stages of construction. The prof will have a fiddle with the models as an idea strikes them.
Of course you need to practise how to represent such models on the printed page. Good luck.
Electrophilic aromatic substitution requires of an aromatic reactant, a pre-electrophile and an electrophile maker.
The reaction begins with a 'pre-step' which involves the electrophile maker turning the pre-electrophile into a very reactive electrophile.
Only reactive electrophiles can cause a reaction because aromatic rings are very stable.
The aromatic rings acts as a nucleophile.
I will take the bromination of benzene as an example:
1. The reaction starts with the production of a good electrophile by a catalyst.
For the bromination of benzene reaction, the electrophile is the
2. The electrophile attacks the π electron system of the benzene ring to form a non-aromatic carbocation intermediate.
3. The positive charge on the carbocation is delocalized throughout the molecule.
4. The aromaticity is restored by the loss of a proton from the atom to which the bromine atom (the electrophile) has bonded.
5. Finally, the proton reacts with the
Here's an energy diagram that corresponds for the bromination of benzene.
Here's what I get.
When there is more than one permissible IUPAC name, the first name is preferred.
There are two systems of IUPAC names for ethers.
1. Substitutive names.
The ethers are named as alkoxyalkanes, with the senior component selected as the parent compound.
2. Functional group names.
The ethers are named as alkyl alkyl ethers, with the alkyl groups in alphabetical order followed by the class name ether, each as a separate word.
Esters are named as alkyl alkanoates.
The name of the alkyl group is written first, followed by the name of the acid with the ending -ic acid replaced by the ending -ate.
There are three systems of IUPAC names for nitriles.
1. As alkanenitriles.
The ending -nitrile is added to the name of the alkane with the same number of carbon atoms.
2. If the compound is considered to be formed from a carboxylic acid with a "trivial name" (
3. As alkyl cyanides.
The name of the alkyl group precedes the class name cyanide.
It can be considered as basically spatial crowding.
Steric hindrance is a kinetic factor that limits the ease to which a nucleophile (electron pair donor) can approach an electrophile (electron pair acceptor).
As an example, consider the reaction seen below, which one might hope is
Here, the nucleophile is cyanide (
We should predict that reaction does not work via an
(LEFT: steric hindrance; RIGHT: reduced steric hindrance)
That reduces the ease to which a successful collision can occur between two reactants, and slows down the first step in a given substitution mechanism (making it the rate-limiting step).
Hence, this reaction proceeds more easily as a first-order mechanism (e.g.
The rate law for this could then be approximated as first-order based on the slow step, since it dominates the extent of the reaction time:
#r(t) ~~ k_1["Br"^(-)]#
(Depending on the choice of solvent, either
The cis isomer will form 3- and 4-methylcyclopentene in roughly equal amounts. The trans isomer will form 4-methylcyclopentene as the major product.
What type of reaction?
I predict an
The structure of the substrate is
Elimination requires a trans arrangement of the β-hydrogen and the leaving group.
We see appropriate β-hydrogens at
Elimination of the hydrogen from
Elimination of the hydrogen from
These two isomers would probably be formed in roughly equal amounts.
The structure of the substrate is
Again, we see trans β-hydrogens at
However, elimination will be slower in each case because of steric hindrance by the methyl group.
Elimination of the hydrogen from
Thus, the major product will be 4-methylcyclopentene.