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Answer:

Because it is not symmetric so the bond dipoles do not cancel out.

Explanation:

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:

enter image source here

However, take a look at 1,4-dicholoranthracene:

enter image source here

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:

enter image source here

Now there is a zero dipole moment because all the forces on the electrons of the aromatic bond are the same!

Answer:

Epoxides are carcinogenic because they disrupt our DNA.

Explanation:

Epoxides are three-membered cyclic ethers.

Epoxide

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.

upload.wikimedia.org

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.

Benzopyrene metabolism
(From researcgate.net)

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.

www.researchgate.net

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 #"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)#

https://files.mtstatic.com/

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"#.

Answer:

You could have given us a bit more to go on.........

Explanation:

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.

https://en.wikipedia.org/wiki/Chirality_(chemistry)

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.

Answer:

Try the bromine decolourization test and the Baeyer unsaturation tests.

Explanation:

Test 1. Bromine decolourization test

Add a few drops of a solution of #"Br"_2# in #"CCl"_4# to a solution of each hydrocarbon in #"CCl"_4#.

The butane will not react (left-hand test tube). The but-2-ene will decolourize the bromine solution (right-hand test tube).

anodosedu.gr

#"CH"_3"CH=CHCH"_3 + underbrace("Br"_2)_color(red)("brown") → underbrace("CH"_3"CHBrCHBrCH"_3)_color(red)("colourless")#

Test 2. The Baeyer test

Bubble each gas through a cold, dilute, aqueous solution of #"KMnO"_4#.

www.susanhornbuckle.com

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).

#"3CH"_3"CH=CHCH"_3 + underbrace("2MnO"_4^"-")_color(red)("purple") + 4"H"_2"O"#

#→ "3CH"_3"CH(OH)CH(OH)CH"_3 + underbrace("2MnO"_2)_color(red)("dark brown precipitate") + 2"OH"^"-"#

Answer:

Warning! Long Answer.
1. 4 #"C"_3# axes, 6 #σ_text(d)# mirror planes, and 3 #"S"_4# improper axes
2. 1 #"C"_text(3v)# axis and 2 #σ_text(v )# planes

Explanation:

Point group for #"CF"_4#

#"CF"_4# has a tetrahedral geometry. It belongs to point group #"T"_text(d)#, like methane.

CF4

It has four #"C"_3# axes, three #"C"_2# axes, three #"S"_4# improper axes, and six #σ_text(d)# mirror planes.

(a) #"C"_3 # axes

If you place the model on a table and look "down" the vertical #"C-F"# bond, we can see that each bond is a #"C"_text(3)# axis of rotation.

You can rotate the molecule 120°# about the axis and get an indistinguishable configuration of the molecule.

C3

(b) #"C"_2# axes

The image below shows one of the four #"C"_text(2)# axes.

C2

The axis bisects a #"C-F"# bond angle.

You can rotate the molecule 180° about this axis and get an indistinguishable configuration of the molecule.

(c) #"σ"_text(d)# planes

The image below shows the two of the six #σ_text(v)# planes.

Sigma

One is the plane of the screen. The other is a plane perpendicular to the screen and bisecting an #"F-C-F"# bond angle.

(d) #S"_4# axes.

An #"S"_text(n)# improper rotation is a rotation of #(360°)/n# about a #C"_n# axis, followed by reflection in a plane perpendicular to that axis.

S4
(Adapted from slideshare.net)

There are four of these axes in the molecule.

#bb(1.)color(white)(m)"CF"_4# to #"CClF"_3#

The structure of #"CClF"_3# is

CClF3

We see that the molecule has a #"C"_text(3v)# axis and two #σ_text(v )# planes of symmetry.

#"C"_text(3v)# axes

The three-fold axis is more obvious if you look down the #"C-Cl"# with the #"Cl"# atom closest to your eye.

C3v

You can rotate the molecule 120° left or right and get a new configuration that is indistinguishable from the original.

#sigma_text(v)# planes

3σv

Each #σ_text(v)# includes a #"Cl-C-F"# bond. Here the mirror plane is the plane of the screen.

A reflection converts the molecule into an indistinguishable form of the original.

In going from #"CCl"_4# to #"CClF"_3#, the molecule has lost four #"C"_3# axes, six #σ_text(d)# mirror planes, and three #"S"_4# improper axes

**#bb(2.)color(white)(m)"CClF"_3# to #"CBrClF"_2#

The structure of the molecule is

CBrClF2

Thus, it is in point group #"C"_text(s)#. It has only a #σ_text(v )# plane of symmetry.

#σ_h# plane

σh

The plane contains the #"Br-C-F"# bonds and bisects the #"Cl-C-F"# bond angle#.

One half of the molecule is a mirror image of the other half.

Thus, in going from #"CClF"_3# to #"CBrClF"_2#, the molecule has lost a #"C"_text(3v)# axis and two #σ_text(v )# planes of symmetry.

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