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Because it is not symmetric so the bond dipoles do not cancel out.
You only get a zero dipole moment for an organic molecule when the bond dipoles cancel each other out. For example, in 1,4-dicholorbenzene the two chlorines will pull electrons from the aromatic ring, but those two pulls exactly cancel each other:
However, take a look at 1,4-dicholoranthracene:
Yes--across the molecule the two chlorines will cancel each other out, but the electrons in the rest of the aromatic rings will still be pulled toward the ring with the two chlorines on it.
You could fix this by substituting the three ring:
Now there is a zero dipole moment because all the forces on the electrons of the aromatic bond are the same!
Epoxides are carcinogenic because they disrupt our DNA.
Epoxides are three-membered cyclic ethers.
They are highly strained because the nominal bond angles are 60° instead of 109.5°.
They tend to react with other molecules to open the ring and reduce the strain.
A common pro-carcinogen is benzo[a]pyrene.
It is a constituent of automobile exhaust, smog, cigarette smoke and charcoal-cooked foods.
When the body absorbs benzo[a]pyrene, it tries to make it water-soluble so it can be excreted in the urine or feces.
It converts the benzo[a]pyrene into benzo[a]pyrene-7,8-epoxide.
This is hydrolyzed to the 7,8-dihydrodiol, which is further converted into benzo[a]pyrene-7,8-dihydrodiol-9,10-epoxide.
Once formed, the epoxide can π stack with the bases in DNA where the epoxide group reacts with guanine residues.
This distorts the structure of DNA and causes errors in DNA replication.
If these mutations occur in a gene that encodes a molecule that regulates the production of cells, the result may be cancer.
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.
Try the bromine decolourization test and the Baeyer unsaturation tests.
Test 1. Bromine decolourization test
Add a few drops of a solution of
The butane will not react (left-hand test tube). The but-2-ene will decolourize the bromine solution (right-hand test tube).
Test 2. The Baeyer test
Bubble each gas through a cold, dilute, aqueous solution of
With butane, there will be no colour change (top test tube).
With but-2-ene, the purple colour will disappear and a dark brown precipitate will form (bottom test tube).
Warning! Long Answer.
Point group for
It has four
If you place the model on a table and look "down" the vertical
You can rotate the molecule 120°# about the axis and get an indistinguishable configuration of the molecule.
The image below shows one of the four
The axis bisects a
You can rotate the molecule 180° about this axis and get an indistinguishable configuration of the molecule.
The image below shows the two of the six
One is the plane of the screen. The other is a plane perpendicular to the screen and bisecting an
(Adapted from slideshare.net)
There are four of these axes in the molecule.
The structure of
We see that the molecule has a
The three-fold axis is more obvious if you look down the
You can rotate the molecule 120° left or right and get a new configuration that is indistinguishable from the original.
A reflection converts the molecule into an indistinguishable form of the original.
In going from
The structure of the molecule is
Thus, it is in point group
The plane contains the
One half of the molecule is a mirror image of the other half.
Thus, in going from
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