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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.

The #"CH"_2# group between the two π orbitals prevents them from overlapping.


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 #m"/"z# ratio.

It may help if you look at this overview of mass spectroscopy.

There are many methods, but some common ionization methods are:

  • Electrospray ionization
  • Matrix-Assisted Laser Desorption Ionization
  • Electron Ionization

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.



  1. Analyte solution is passed through the spray needle, which is connected to a counterelectrode with a large potential difference in between them.
  2. It comes out as a charged droplet that evaporates on its way to the source-sampling counterelectrode.
  3. Solvent evaporation stresses the surface tension of the droplet until it breaks into smaller charged droplets (reaching the "Rayleigh" limit). This rapidly repeats until analyte ions are left.

These analyte ions have little residual energy, so this ionization method is soft.


The general process was:

  1. An easily-sublimable solid sample matrix surrounds the analyte molecules, isolating them from each other.
  2. A laser of the appropriate wavelength excites the matrix surface, ejecting clusters in which the analyte molecules are surrounded by the matrix ions.
  3. The (presumably) easily-sublimed matrix ions then vaporize off, leaving gaseous analyte ions for sample analysis.

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).


Sometimes labeled the "classical" method. The general idea is:

  1. Previously-vaporized analyte solution is sent into the electron ionization chamber.
  2. A hot (rhenium or tungsten) filament sends a beam of spiraling electrons towards an electron trap.
  3. What happens most often is that energy is transferred from the electron beam into the analyte gas, fragmenting it into ions.

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 #m"/"z# ratio.



See below


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: #"cyclopentane"#. Pentane means it's a 5 carbon alkane and the prefix cyclo- means the parent chain is cyclic. It looks like this without any alkyl chains attached to it.

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.

  • #color(orange)"4-ethyl"#
  • #color(blue)"2-isopropyl"#
  • #color(magenta)"1-methyl"#

If we take our cyclopentane and first number all the carbons,


we can then, starting in numerical order, attach the substituents one-by-one.


#ul"Adding Substituents"#

Looking at #color(magenta)"1-methyl"#, we can see we have a methyl group attached at carbon number #1#. Placing it at the appropriate position, we get the following structure:


Next, we take a look at #color(blue)"2-isopropyl"#. We have an isopropyl group attached at carbon number #2#. Placing this alkyl group at its appropriate position we get the following structure thus far:


Lastly, looking at #color(orange)"4-ethyl"#, we place an ethyl group at carbon number #4#. Doing so, our finished structure looks like this:

enter image source here


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.



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 #"36 kcal/mol"#. If there were a perfect extensivity with regards to resonance stabilization, we would have expected the amount to be

#"Total Resonance Energy"#

#~~ "Number of Benzene Rings" xx "Resonance Energy"#

But you can see in the above diagram that it isn't:

  • Napthalene has #bb"11 kcal/mol"# less resonance energy than #2xx"benzene rings"#.
  • Anthracene has #bb"25 kcal/mol"# less resonance energy than #3xx"benzene rings"#.
  • Phenanthrene has #bb"17 kcal/mol"# less resonance energy than #3xx"benzene rings"#.

From this, we could postulate that in general, the more extended the #pi# system, the less resonance stabilization is afforded. The most likely reason for this is probably the volume of the system.


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)#,

where #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.


How is this relevant?

Well, the HOMO and LUMO are both required in electrophilic addition reactions. For example, with adding #"Br"_2#,

  • The HOMO donates #pi# electrons to polarize #"Br"_2# and break the #"Br"-"Br"# bond.
  • The LUMO accepts electrons from the #stackrel(delta^(+))("Br")#.

And this forms the so-called bromonium complex:

(Here, the HOMO contained the #pi# electrons in the double bond, and the LUMO accepted the electrons from the bottom #"Br"#.)

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 #"pKa"# is #4.4# and the second #"pKa"# is #6.7#, then we have an idea of where the half-equivalence points are (i.e. where the concentrations of acid and conjugate base are equal), because the #"pH"# #=# #"pKa"# at those points:

#"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 #["H"_2"A"] = ["HA"^(-)]#, and that #"pH" ~~ 4.4#.

  • At midpoint 2, we have that #["HA"^(-)] = ["A"^(2-)]#, and that #"pH" ~~ 6.7#.

A distribution graph shows the change in concentration of each species in solution as the #"pH"# increases. It correlates well with a base-into-diprotic-acid titration curve.

See below for an overlay of both:

Titration Curve (Truong-Son N.) + Distribution Graph (Ernest Z.)

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 #"pH"# correlate with the first few equivalence points, and the last species to show up dominates at high #"pH"#.



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 #S# isomer, and the one on the right (as we face it) is the #R# isomer.

How? Well for each depicted stereoisomer, the substituent of least priority, the hydrogen, projects into the page. #Br# takes priority over #Cl# takes priority over #F#, and this orientation proceeds COUNTERCLOCKWISE around the chiral carbon, and hence a sinister geometry. The stereoisomer on the right hand side exhibits the opposite geometry, i.e. the #"R-stereoisomer"#.

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 #"Br"^"+"# ion generated by the reaction of the bromine molecule with ferric bromide, a Lewis acid.

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 #"FeBr"_4^−# to regenerate the #"FeBr"_3# catalyst and form the product HBr.

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.

Thus, #"CH"_3"OCH"_2"CH"_2"CH"_3# is 1-methoxypropane.

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.

Thus, #"CH"_3"OCH"_2"CH"_2"CH"_3# is methyl propyl ether.


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.

Thus, #"CH"_3"CH"_2"COO-CH"_3# is methyl propanoate (two words).

Cyanides (nitriles)

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.

Thus, #"CH"_3"CH"_2"C≡N"# is propanenitrile.

2. If the compound is considered to be formed from a carboxylic acid with a "trivial name" (#"RCOOH → RC≡N"#), the ending -ic acid is changed to -onitrile.

Thus, #"CH"_3"CH"_2"C≡N"# is also called propionitrile (from propionic acid).

3. As alkyl cyanides.

The name of the alkyl group precedes the class name cyanide.

Thus, #"CH"_3"CH"_2"-C≡N"# is ethyl cyanide (two words).


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 #"S"_N2#:

Here, the nucleophile is cyanide (#""^(-):"CN"#) and the electrophile is the central carbon on the alkyl halide (tert-butyl bromide).

We should predict that reaction does not work via an #"S"_N2# mechanism, based on the idea of steric hindrance---the three #"CH"_3# groups are blocking the cyanide from performing its backside-attack.

(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. #"S"_N1# or #E1#), since the #"Br"^(-)# has the time (during the slow step) to come off on its own.

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 #"S"_N1# or #E1# could occur.)



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 #"E2"# elimination because:

  • The substrate is a secondary alkyl halide.
  • The substrate has at least one β-hydrogen: possible elimination.
  • The nucleophile (#"OH"^"-"#) is a strong base: possible elimination.
  • There is (presumably) a high base concentration: favouring #"E2"#.
  • Water is a polar protic solvent: favouring elimination.
  • The high temperature favours elimination.


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 #"C2"# and #"C5"#.

Elimination of the hydrogen from #"C2"# forms 3-methylcyclopentene.


Elimination of the hydrogen from #"C5"# forms 4-methylcyclopentene.


These two isomers would probably be formed in roughly equal amounts.


The structure of the substrate is


Again, we see trans β-hydrogens at #"C2"# and #"C5"#.

However, elimination will be slower in each case because of steric hindrance by the methyl group.

Elimination of the hydrogen from #"C5"# will be faster than from #"C2"# because of its greater distance from the methyl group.

Thus, the major product will be 4-methylcyclopentene.