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5

## Ortho,para-bromoanisole + NaNH_2 + Liquid NH_3 =? How do you predict the product?

Ernest Z.
Featured 1 month ago

Here's how I would do it.

#### Explanation:

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 $\text{H}$ atom that is ortho to $\text{C3}$, generating a carbanion.

Step 2. Loss of $\text{Br"^"-}$ to form a benzyne intermediate.

The elimination is by an E2cb pathway.

Step 3. Addition of $\text{NH"_2^"-}$

The strain caused by a triple bond in a benzene ring can be relieved by a nucleophilic addition ($\text{Ad"_"N}$) of $\text{NH"_2^"-}$.

The methoxy group is electron-withdrawing by induction, so the nucleophile will attack $\text{C3}$ to place the carbanion as close as possible to the methoxy group.

Step 4. Protonation of the carbanion.

The product is meta-methoxyaniline.

para-Bromoanisole

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.

3

## What is the difference between an oxyacid and an organic acid? What are examples of each?

Dwight
Featured 1 month ago

The difference lies in the structure of the molecules and the manner in which the oxygen and hydrogen atoms are arranged. See below...

#### Explanation:

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 ${H}_{2} S {O}_{4}$, and also sulfurous acid ${H}_{2} S {O}_{3}$, phosphoris acid ${H}_{3} P {O}_{4}$, nitric acid $H N {O}_{3}$ to name just a few.

An organic acid is one which contains, in its structure, the particular arrangement of atoms called a carboxyl group $- C O O H$

Examples include formic acid $H C O O H$, acetic acid (in vinegar) $C {H}_{3} C O O H$ and many more.

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.

3

## How do you tell "S"_"N"1 and "S"_"N"2 reactions apart?

Morgan
Featured 1 month ago

See below.

#### Explanation:

There are several key differences between ${S}_{N} 1$ and ${S}_{N} 2$ reactions. I've outlined a comparison below, with the assumption that the reader has some basic knowledge of both reaction types.

${S}_{N} 2$

${S}_{N} 2$ 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.

${S}_{N} 2$ 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. $N a O C {H}_{3}$ is a strong nucleophile, as it breaks apart into $N {a}^{+}$ and $O C {H}_{3}^{-}$ 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.

${S}_{N} 2$ 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.

${S}_{N} 2$ 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.

${S}_{N} 2$ reactions prefer polar aprotic solvents, where polar protic solvents hinder ${S}_{N} 2$ reactions. Examples include $D M S O$ and acetone.

${S}_{N} 2$ 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 ${S}_{N} 2$ reaction. This is the big barrier for ${S}_{N} 2$ 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 ${S}_{N} 2$ reaction, with the reaction coordinate on the $x$-axis and energy on the $y$-axis. The reactants are represented by $\textcolor{b l u e}{N {u}^{-}} + R \textcolor{g r e e n}{L G}$. This symbolizes the nucleophile $\left(\textcolor{b l u e}{N {u}^{-}}\right)$ plus the leaving group $\left(\textcolor{g r e e n}{L G}\right)$ attached to some $R$. The transition state or rate-determining step is represented by ${\left[\textcolor{b l u e}{N u c} - - - R - - - \textcolor{g r e e n}{L G}\right]}^{-}$, 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.

${S}_{N} 1$

${S}_{N} 1$ 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.

${S}_{N} 1$ requires a weak nucleophile. This is because the ${S}_{N} 1$ 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 ${S}_{N} 1$ reactions, meaning that the nucleophile and the solvent are the same. A common example is $C {H}_{3} O H$. You might also see heat is used, given by $\Delta$.

${S}_{N} 1$ 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 ${S}_{N} 1$ 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 ${S}_{N} 2$ reactions, no rearrangement is possible.

${S}_{N} 1$ reactions prefer polar protic solvents. Examples include $C {H}_{3} O H$ and acetic acid.

${S}_{N} 1$ 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 ${S}_{N} 1$ reaction.

In this diagram for an ${S}_{N} 1$ 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 $S M$ stands for starting material and $P$ for products.

Here is a comparison chart:

2

## Are alkyl halides sterically hindered?

Morgan
Featured 1 month ago

Sometimes.

#### Explanation:

This depends on the specific alkyl halide. For example, take tert-butylbromide and bromopentane.

tert-butylbromide:

We see that this alkyl halide is tertiary $\left({3}^{o}\right)$, making it very sterically hindered. This makes a backside attack, as seen in the ${S}_{N} 2$ mechanism, virtually impossible. Inversely, the ${S}_{N} 1$ mechanism, for example, would favor this alkyl halide well under the appropriate reaction conditions.

Bromopentane:

This is a primary $\left({1}^{o}\right)$ alkyl halide, which has minimal sterical hinderance. This makes a backside attack very possible, and an ${S}_{N} 2$ mechanism would be favored.

Backside attack:

For an $S {N}_{2}$ mechanism to occur, a backside attack must be able to take place. We can see below that this would much more difficult on the tert-butylbromide than the bromopentane. I've used a strong base, $N a O C {H}_{3}$ (sodium methoxide) for the ${S}_{N} 2$ mechanism. I've shown the mechanism for bromopentane first.

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, $N {a}^{+}$ 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 ${S}_{N} 1$ and ${S}_{N} 2$ mechanisms in my examples. The same rules for elimination mechanisms still apply, and they are in competition with the corresponding substitution mechanisms.

Note:

It is possible to have a primary alkyl halide which is too sterically hindered for an ${S}_{N} 2$ mechanism to occur. Neopentyl bromide is an example:

The sterical hinderance of the adjacent carbon is enough to render an ${S}_{N} 2$ mechanism highly improbable.

1

## What is the mechanism for the hydrolysis of an anhydride into a carboxylic acid using HCl as a catalyst?

Truong-Son N.
Featured 1 month ago

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. $\text{O}$ in a carbonyl or $\text{N}$ in a nitrile) takes on a proton and becomes susceptible to nucleophilic attack.

It is perfectly acceptable to assume one starts with hydronium when a strong acid like $\text{HCl}$ is the acid catalyst.

1. The electron-dense carbonyl oxygen acquires a proton from the strong acid. Either carbonyl oxygen is fine, just not the ether oxygen.
2. Water then nucleophilically attacks the partially positive carbon, as oxygen withdraws electron density to break the carbonyl bond.
3. Proton transfer pt1.
4. Proton transfer pt2.
5. Tetrahedral collapse. You may choose which hydroxyl group does so, but it is usually the one that originated from the nucleophile that is typically used.
6. Regenerate the catalyst.

NOTE: You must replace ${R}_{1}$ and ${R}_{2}$ with $R$ groups that correspond to your specific substrate.

2

## Discuss the conformational analysis of propane with Newman projection, and also draw the energy diagram?

Ernest Z.
Featured 1 month ago

WARNING! Long answer! Here's what I get.

#### Explanation:

Newman projections

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), $\text{C-1}$ is the blue methyl group, $\text{C-2}$ is at the centre of the circle, and $\text{C-3}$ is hidden directly behind it.

Conformations

Conformations are the different arrangements that the atoms can take by rotating about the $\text{C-C}$ single bond.

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.

The $\text{H-C-C-H}$ and $\text{CH"_3"-C-C-H}$ dihedral angles are 0°.

In the staggered conformation, the groups on the two carbon atoms are as far from each other as they can get, and the $\text{H-C-C-H}$ and $\text{CH"_3"-C-C-H}$ dihedral angles are 60°.

Conformational analysis

The eclipsed conformation is a high-energy conformation because the negatively charged electrons in the $\text{C-H}$ and ${\text{C-CH}}_{3}$ bonds repel each other most when the bonds line up.

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 $\text{C1-C2}$ bond, and the ${\text{CH}}_{3}$ is coming off the back carbon.

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 $\text{C-H, C-H}$ repulsions and a ${\text{C-H, C-CH}}_{3}$ repulsion.

We know from the conformational analysis of ethane that one $\text{C-H, C-H}$ repulsion contributes 4 kJ/mol, so a ${\text{C-H, C-CH}}_{3}$ repulsion contributes about 6 kJ/mol to the torsional strain.

2

## What are the different types of optical isomers seen in coordination compounds? How do you draw optical isomers of coordination compounds?

Ernest Z.
Featured 1 month ago

Here's what I get.

#### Explanation:

Monodentate ligands

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.

4

## Why is the dipole moment of 1,4-dichloro-anthracene not 0?

Andy Wolff
Featured 3 weeks ago

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:

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!

4

## What is stereo structural IUPAC name of the following compound?

Ernest Z.
Featured 3 weeks ago

The name of the compound is ($2 S , 3 R$)-2-bromo-3-chloropropane.

#### Explanation:

Configuration at $\text{C-2}$

The Cahn-Ingold-Prelog priorities of the groups on $\text{C-2}$ are

$\text{Br = 1; C-3 = 2; C-1 = 3; H = 4}$

The $\text{H}$ atom is furthest from our eye, and the 1 → 2 → 3 direction is counterclockwise.

Thus, the configuration at $\text{C-2}$ is $S$.

Configuration at $\text{C-3}$

The Cahn-Ingold-Prelog priorities of the groups on $\text{C-3}$ are

$\text{Cl = 1; C-2 = 2; C-4 = 3; H = 4}$

The $\text{H}$ atom is closest to our eye, and the 1 → 2 → 3 direction is counterclockwise.

The configuration at $\text{C-3}$ appears to be $S$.

However, because the $\text{H}$ atom is closest to our eye, we invert the assignment.

The configuration at $\text{C-3}$ is $R$.

The name of the compound is, therefore, ($2 S , 3 R$)-2-bromo-3-chlorobutane.

NOTE:

Your image was the PubChem structure for ($2 S , 3 S$)-2-bromo-3-chloro-butane.

They give three structures.

Wedge-Dash

(From PubChem)

This is the ($2 S , 3 S$) isomer.

Wire Frame

This is the structure you give.

Amazingly, PubChem is incorrect. This is the ($2 S , 3 R$) isomer.

3-Dimensional

(From PubChem)

This is the correct ($2 S , 3 S$) structure.

3

## What are examples of dipoles?

Andrew B.
Featured 4 days ago

Any molecule that has a nonzero vector sum of dipole moments is said to be polar and have a dipole moment

#### Explanation:

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 ($\Delta E N$) determines how polar a bond is.

If:
$0 \le \Delta E N \le 0.4$, the bond is nonpolar.

$0.4 <$$\Delta E N$$\le 1.8$, the bond is polar

$\Delta E N > 1.8$, the bond is ionic

Consider the bonds in ${H}_{2} O$,

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
$E {N}_{O} = 3.44$
$E {N}_{H} = 2.20$
$\Delta E N = 3.44 - 2.20 = 1.24$

In the figure above, the $\delta$ symbol indicates an area of partial charge on the atom. Note that they are not full charges as in ions, but partial charges due to a difference in electron density at each "pole". The arrow in the figure indicates the direction of electron density and the slight negative charge ${\delta}^{-}$and the cross indicates an area of electron deficiency and the slight positive charge ${\delta}^{+}$. This difference in charges is called a dipole moment and it is a vector quantity; it has magnitude and direction.

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.

Consider $C {O}_{2}$,

As you can see, the $\Delta E N$ for the $C - O$ bond is within the polar range. However, since $C {O}_{2}$ is a linear molecule, the dipoles point in opposite directions and the vector sum of the two is equal to zero. $C {O}_{2}$ is nonpolar.