Here's how I would do it.
These are the reaction conditions for generating benzyne intermediates.
The reaction of ortho-bromoanisole with potassium amide in liquid ammonia (b. p. -33 °C) is extremely rapid.
Step 1. The amide ion attacks the
Step 2. Loss of
The elimination is by an E2cb pathway.
Step 3. Addition of
The strain caused by a triple bond in a benzene ring can be relieved by a nucleophilic addition (
The methoxy group is electron-withdrawing by induction, so the nucleophile will attack
Step 4. Protonation of the carbanion.
The product is meta-methoxyaniline.
The reaction with para-bromoanisole also follows a benzyne mechanism.
The nucleophile can attack either end of the triple bond, and the methoxy group is far enough away that its inductive effects are minimal.
The product is a mixture of para- and meta-methoxyaniline.
The difference lies in the structure of the molecules and the manner in which the oxygen and hydrogen atoms are arranged. See below...
An oxyacid is one that will contain, in addition to hydrogen and another element (such as nitrogen, sulfur or phosphorus), a number of oxygen atoms.
Examples would include sulfuric acid
An organic acid is one which contains, in its structure, the particular arrangement of atoms called a carboxyl group
Examples include formic acid
The difference comes in the actual structure of the molecule. In the oxyacid, the oxygen atoms are all bonded to the nitrogen or sulfur, or whatever it happens to be, with hydrogen atoms bonded to one or more of these oxygens.
In a carboxylic acid (the organic variety), a carbon is doubly bonded to one oxygen atom and singly bonded to a second oxygen. This second oxygen has the H atom bonded to it. So, a very particular structure.
"R" just represents the rest of the molecule.
There are several key differences between
#S_N2#stands for "substitution, nucleophilic, bimolecular," or bimolecular nucleophilic substitution. This implies that there are two molecules (bimolecular) involved in the transition state or rate-determining (slow) step. This also tells us that the rate of the reaction depends upon both reactants (nucleophile and electrophile); if you double the concentration of one of the reactants, you double the rate of the reaction.
#S_N2#reactions require a strong nucleophile. Strong nucleophiles are strong bases, so it may be easier to identify them this way at first. For example, strong nucleophiles bear a negative charge. #NaOCH_3#is a strong nucleophile, as it breaks apart into #Na^+#and #OCH_3^-#in solution.
This strong nucleophile forces what is called a backside attack. This is fairly literal. The nucleophile attacks the carbon opposite the leaving group as the two repel each other.
#S_N2#products show inversion of stereochemistry, a result of the backside attack. For example, if the leaving group was once represented as a wedge in the perspective drawing of the molecule, the nucleophile which replaces it will now be shown as a dash. The stereochemistry of any other substituents are left alone.
#S_N2#reactions are concerted, which means that the nucleophile attacks the electrophilic carbon at the same time that the leaving group leaves. There is no intermediate.
#S_N2#reactions prefer polar aprotic solvents, where polar protic solvents hinder #S_N2#reactions. Examples include #DMSO#and acetone.
#S_N2#reactions favor electrophilic carbon atoms which are least highly substituted, so #1^o>2^o>3^o#. You won't see a tertiary carbon undergo an #S_N2#reaction. This is the big barrier for #S_N2#reactions. It is due to the fact that the reaction is concerted, and a backside attack must take place. The steric hinderance on a tertiary carbon is too great to allow this.
This is a reaction diagram for a general
#S_N2#reaction, with the reaction coordinate on the #x#-axis and energy on the #y#-axis. The reactants are represented by #color(blue)(Nu^-)+Rcolor(green)(LG)#. This symbolizes the nucleophile #(color(blue)(Nu^-))#plus the leaving group #(color(green)(LG))#attached to some #R#. The transition state or rate-determining step is represented by #[color(blue)(Nuc)---R---color(green)(LG)]^-#, which symbolizes the concerted reaction, the backside attack of the nucleophile while the leaving group leaves. Note that the transition state cannot be isolated! This is just a visualization of what is happening at this point. We see two molecules involved here: the nucleophile and the compound it attacks (electrophile). Finally, we have the products all the way to the right of the diagram.
#S_N1#stands for "substitution, nucleophilic, unimolecular," or unimolecular nucleophilic substitution. This implies that there is only one molecule (unimolecular) involved in the transition state or rate-determining (slow) step. This also tells us that the rate of the reaction depends upon only one reactant at a time.
#S_N1#requires a weak nucleophile. This is because the #S_N1#reaction is step-wise, or occurs in two steps. First, the slow step: the leaving group leaves and the formation of the carbocation. If a strong nucleophile is present, this slow step does not occur because the nucleophile quickly attacks the electrophile. Additionally, because the carbocation formed is such a reactive electrophile, a weak nucleophile is all that is required. We also see solvolysis in #S_N1#reactions, meaning that the nucleophile and the solvent are the same. A common example is #CH_3OH#. You might also see heat is used, given by #Delta#.
#S_N1#products show both inversion and retention of stereochemistry. You will usually get a mixture of stereoisomers in your products.
Because a carbocation is formed in
#S_N1#reactions, rearrangement is possible. Rearrangement will only occur if a more substituted product is possible through a hydride or alkyl shift (must be adjacent). Because no carbocation is formed in #S_N2#reactions, no rearrangement is possible.
#S_N1#reactions prefer polar protic solvents. Examples include #CH_3OH#and acetic acid.
#S_N1#reactions favor electrophilic carbon atoms which are most highly substituted, so #3^o>2^o>1^o#. You will not see a primary carbon undergo an #S_N1#reaction.
In this diagram for an
#S_N1#reaction, we see that there are two separate transition states. One represents the formation of the carbocation, which is the slow, rate-determining step (notice it requires more energy), while the second represents the nucleophilic attack on the newly-formed carbocation. This is the fast step. Note that #SM#stands for starting material and #P#for products.
Here is a comparison chart:
This depends on the specific alkyl halide. For example, take tert-butylbromide and bromopentane.
We see that this alkyl halide is tertiary
#(3^o)#, making it very sterically hindered. This makes a backside attack, as seen in the #S_N2#mechanism, virtually impossible. Inversely, the #S_N1#mechanism, for example, would favor this alkyl halide well under the appropriate reaction conditions.
This is a primary
#(1^o)#alkyl halide, which has minimal sterical hinderance. This makes a backside attack very possible, and an #S_N2#mechanism would be favored.
Note this stereochemistry could be flipped. I simply chose one possible orientation for the example.
Note the inversion of stereochemistry, a product of the backside attack. Of course,
#Na^+#is present in solution. It is a counter ion and has not been shown.
For tert-butylbromide, the backside attack isn't plausible, as there are too many other substrates bonded to the same carbon as the halide. It is literally too crowded for this mechanism to take place.
Note that I only considered
It is possible to have a primary alkyl halide which is too sterically hindered for an
The sterical hinderance of the adjacent carbon is enough to render an
This is a fairly standard mechanism (or mechanistic pattern) that you should get to know.
It is practically identical for anhydrides as it is for acyl halides (particularly acyl chlorides) and esters, and similar variations are seen in the acid-catalyzed hydrolysis of nitriles, amides, etc, wherein the electron-dense atom (e.g.
It is perfectly acceptable to assume one starts with hydronium when a strong acid like
NOTE: You must replace
WARNING! Long answer! Here's what I get.
In a Newman projection, we are looking down a carbon-carbon bond axis so that the two atoms are one-behind-the-other.
In the Newman projection of propane (below),
Conformations are the different arrangements that the atoms can take by rotating about the
There are an infinite number of conformations, but there are two important ones.
In the eclipsed conformation, the substituents on the two carbon atoms are as close to each other as they can get.
In the staggered conformation, the groups on the two carbon atoms are as far from each other as they can get, and the
The eclipsed conformation is a high-energy conformation because the negatively charged electrons in the
The staggered conformation is the most stable because the bonds are furthest away from each other and the electron repulsions are minimal.
The energy difference between the two conformations is called torsional strain.
Conformational analysis is the study of the energy changes that occur during the rotations about σ bonds.
In the conformational energy diagram below, we are looking down the
We start with the molecule in the eclipsed conformation.
As we rotate the back carbon clockwise, the molecule reaches an energy minimum with a staggered conformation at 60°.
Rotating another 60°, the molecule reaches an energy maximum with an eclipsed conformation.
The pattern repeats twice more as the bond rotates a full 360°.
The energy difference between the maxima and minima is 3.4 kcal/mol (14 kJ/mol).
This represents the total repulsion of three bond pairs, two
We know from the conformational analysis of ethane that one
Here's what I get.
Complexes with monodentate ligands are classified as C (clockwise) or A (anticlockwise).
An example is the hypothetical complex ion
(From Department of Chemistry, UWI, Mona)
You assign priorities to the ligands per the usual Cahn-Ingold-Prelog rules.
Arrange the complex so the highest-priority ligand is at the top.
Viewing from the top, you look at the ligands in the horizontal plane.
You designate the isomers as C or A according to whether the direction from the highest to the next-highest priority ligand in the plane is clockwise or anticlockwise.
(From Department of Chemistry, UWI, Mona)
Our hypothetical complex was the C isomer.
Δ and Λ isomers
Optically active bis- and tris-bidentate complexes are said to have a screw chirality.
(From CSB | SJU Employees Personal Web Sites - College of Saint Benedict)
You arrange them so they look like left- or right-handed screws.
If you must rotate them clockwise to screw them into the paper, they are classified as Delta Δ (right-handed).
If you must rotate them counterclockwise, they are Lambda Λ (left-handed).
An example is the trisoxalatoferrate(III) ion.
You must twist the left-hand image to the left to screw it into the paper, so it is the Λ isomer.
The other image is like a right-hand screw, so it is the Δ isomer.
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!
The name of the compound is (
The Cahn-Ingold-Prelog priorities of the groups on
Thus, the configuration at
The Cahn-Ingold-Prelog priorities of the groups on
The configuration at
However, because the
The configuration at
The name of the compound is, therefore, (
Your image was the PubChem structure for (
They give three structures.
This is the (
This is the structure you give.
Amazingly, PubChem is incorrect. This is the (
This is the correct (
Any molecule that has a nonzero vector sum of dipole moments is said to be polar and have a dipole moment
A dipole moment refers to slight opposite charges on opposite sides of a bond. The resulting bond is said to be polar; it has a positive pole and a negative pole, much like a bar magnet.
In order to determine if a particular bond is polar or not, one must look for the electronegativity of each atom. Pauling's electronegativity is a measure of how strong a particular atom pulls electrons towards it in a bond. The value of the difference between their electronegativities (
Consider the bonds in
There is one oxygen bonded to two hydrogens in one water molecule. Based on the difference in electronegativites for the bonds, it is clearly a polar molecule
In the figure above, the
Notice that the water molecule has an overall dipole moment that points straight up towards the oxygen. This is because a dipole moment of a molecule depends on the vector sum of the bond dipoles.
As you can see, the