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## How do you draw a diastereomer?

Ernest Z.
Featured 6 months ago

You change the configuration of one of the chiral centres.

#### Explanation:

Diastereomers are stereoisomers that are not superimposable and are not mirror images of each other.

A compound must have at least two chiral centres to have diastereomers.

Let's look at a Compound $\text{A}$ with two chiral centres.

$\text{C2}$ and $\text{C3}$ are chiral centres.

If we change the configuration at $\text{C2}$, we get Compound $\text{B}$.

This is not a mirror image of $\text{A}$, nor is it superimposable with $\text{A}$. It is a diastereomer of $\text{A}$.

Now, let's change the configuration of just $\text{C3}$ to get Compound $\text{C}$.

This is also a diastereomer of $\text{A}$.

BUT, if we change the configuration of both carbon atoms, we get Compound $\text{D}$.

This is nonsuperimposable on $\text{A}$.

However, it is a mirror image of $\text{A}$, because both $\text{OH}$ groups are on dashes instead of on wedges.

$\text{A}$ and $\text{D}$ are enantiomers.

## Why do dipoles form?

Monzur R.
Featured 4 months ago

They form due to the movement and attractions of electrons in atoms and molecules.

#### Explanation:

A dipole is the separation of two opposite charges, or, in this case, partial charges.

To answer your question, we have to distinguish between the different types of dipoles. There are three different types of dipole:

• Permanent
• Oscillating
• Induced

Permanent dipoles exist in molecules with covalent bonding where one atom is more electronegative than the other. The atom which is more electronegative attracts the bonded pair of electrons to it, increasing its electron density. It thus becomes slightly negative ($\delta$ negative). On the other end of the bond, the other atom loses electron density and becomes slightly positive ($\delta$ positive). The molecule now has a permanent dipole.

Oscillating dipoles occur by chance due to the random movement of electrons in atoms. At any point, the electrons in an atom can all be concentrated on one end, reducing the electron density of the other. This causes one end of the atom to become $\delta$ positive and the other to become $\delta$ negative - the atom now has dipoles. At another time, the electrons will be concentrated on the other end, so the dipoles will shift. The dipoles will constantly be shifting due to the random movement of electrons. This is called oscillating dipoles.

Induced dipoles form when a molecule with a permanent or oscillating dipole approaches a non-polar molecule (or the other way around). As the non-polar molecule approaches the polar one, its electrons will be attracted to the $\delta$ positive end of the molecule. Thus, a dipole has been induced into the non-polar molecule.

This sort of dipole can also form when a non-polar molecule approaches an ionic molecule.

## How can you know if there is a doublet of doublets by looking at a structure?

Ernest Z.
Featured 2 months ago

Here's how I would do it.

#### Explanation:

A doublet of doublets (dd) is a pattern of four lines of approximately equal intensity that results from coupling to two different protons.

I can think of two situations in which I would expect to see dd splitting patterns:

• vinyl groups
• 1,3,4-trisubstituted benzenes

A. Vinyl groups

Typical coupling constants in alkenes are ${J}_{\text{trans" ≈ "16 Hz}}$, ${J}_{\text{cis" ≈ "10 Hz}}$, and ${J}_{\text{gem" ≈ "2 Hz}}$.

Consider the $\text{^1"H}$-NMR spectrum of methyl acrylate:

(From organic spectroscopy international)

Each proton on the vinyl group is split into a doublet of doublets by its neighbours.

1,3,4-Trisubstituted benzenes

Typical coupling constants in substituted benzenes are ${J}_{\text{ortho" ≈ "8 Hz}}$, ${J}_{\text{meta" ≈ "2 Hz}}$, and ${J}_{\text{para" ≈ "0 Hz}}$.

Consider the $\text{^1"H}$-NMR spectrum of 3,4-dichlorobenzoyl chloride:

(From www.chem.wisc.edu)

The proton that has ortho and meta neighbours (on carbon 6) will be a doublet of doublets.

It appears at δ 7.95 (J = 8.5, 2.3 Hz).

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

Truong-Son N.
Featured 3 weeks 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.

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

Ernest Z.
Featured 3 weeks 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.

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

Ernest Z.
Featured 2 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.

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