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## How do Lewis acids and bases differ from Bronsted/Lowry acids and bases?

Kayla
Featured 7 months ago

Lewis acids and bases are defined in terms of being able to accept or donate electron pairs. While Bronsted Lowry acids and bases are defined in terms of being able to accept or donate hydrogen ions (${H}^{+}$).

#### Explanation:

•A Brønsted-Lowry acid is any substance (molecule or ion) that can donate a hydrogen ion (${H}^{+}$).

•A Brønsted-Lowry base is any species that can accept a hydrogen ion (${H}^{+}$).

$\textcolor{b l u e}{\text{Take the reaction below for example:}}$

$H N {O}_{2} + {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + N {O}_{2}^{-}$

Nitrous acid ($H N {O}_{2}$) is the Brønsted-Lowry acid because it donates a hydrogen ion to water. Water is the Brønsted-Lowry base because it accepts the hydrogen ion.

$\textcolor{red}{\text{On the other hand}}$, Lewis acids and bases are defined in terms of either being an electron-pair donor or an electron-pair acceptor.

• A Lewis acid is defined as an electron-pair acceptor. This means that acids can accept a lone pair of electrons from a Lewis base because the acid has vacant valence orbitals.

• A Lewis acid must have a vacant valence orbital and it can be a cation, such as $A {l}^{3 +}$, or a neutral molecule, such as $C {O}_{2}$.

• A Lewis base is an electron-pair donor. This means that a Lewis base has the ability to donate two of its electrons to a Lewis acid.

Here's a general depiction of a Lewis acid-base reaction:

In the diagram above, $A$ is a Lewis acid because it is accepting a pair of electrons from $B$ as denoted by the curved arrow.

$B$ is a Lewis base it is donating its electrons to $A$.

## Why are pi bonds formed?

Ernest Z.
Featured 6 months ago

π bonds are formed because the side-by-side overlap of $p$ orbitals leads to a lower-energy state than if there were no overlap.

#### Explanation:

Consider the formation of ethylene, ${\text{H"_2"C=CH}}_{2}$.

The $\text{C}$ atoms are $s {p}^{2}$ hybridized.

The $\text{C-C}$ σ bond is formed by the overlap of the $\text{C}$ $s {p}^{2}$ orbitals, and the $\text{C-H}$ σ bonds are formed by the overlap of the $\text{H}$ $1 s$ orbitals with the $\text{C}$ $s {p}^{2}$ orbitals.

This leaves an unhybridized p orbital containing one electron on each $\text{C}$ atom.

When you have two $p$ orbitals on adjacent $\text{C}$ atoms, they can overlap sideways to form a region of electron density that is not directly between the two nuclear centres but which still contributes to bonding.

We call this a π bond.

It forms because the π electrons are as close as they can get to the two nuclei and are therefore in their lowest energy state.

A π bond can only form only after a σ bond has already formed. It is always part of a double or triple bond.

## How is acetophenone phenylhydrazone catalyzed into 2-phenylindole?

Ernest Z.
Featured 5 months ago

Here's my proposal for the mechanism for the Fischer indole synthesis of 2-phenylindole.

#### Explanation:

Note: For convenience, I am using $\text{HCl}$ as the acid and $\text{Cl"^"-}$ as the conjugate base, but you should use the acid and base that were used in your synthesis.

Step 1. Protonation of the phenylhydrazone

Step 2. Tautomerization to an enamine

Step 3. Cyclic [3,3]-sigmatropic rearrangement

This forms a diimine and results in the loss of aromaticity.

Step 4. Re-aromatization

Step 5. Protonation of the imine

Step 6. Cyclization

Step 7. Deprotonation of the ammonium ion

Step 8. Protonation of the primary amine

Step 9. Elimination of ammonia

This forms a stable, aromatic indole.

And that is my proposed mechanism for the formation of 2-phenylindole.

## What is hybridization in organic chemistry?

Ernest Z.
Featured 4 months ago

Hybridization is the mixing of atomic orbitals to form new orbitals with different energies and shapes than the original orbitals.

#### Explanation:

Hybrid orbitals are mixtures of atomic orbitals in various proportions.

For example, the hybrid orbitals on the $\text{C}$ atom of methane consist of one-fourth $\text{s}$ character and three-fourths $\text{p}$ character.

We say they are ${\text{sp}}^{3}$ ("s-p-three") hybridized.

The four new hybridized orbitals have both $\text{s}$ and $\text{p}$ character.

The dumbbell shape reflects the $\text{p}$ character and (despite the picture) the big lobe is nearly spherical like an $\text{s}$ orbital.

The new orbitals all have the same energy but they point in different directions (towards the corners of a tetrahedron).

This leads to the most stable molecules when the $\text{C}$ atom forms bonds to four other atoms.

## How are carbocations generated?

Morgan
Featured 3 months ago

Carbocations are generated when a carbon atom in a compound does not have a full octet, causing it to gain a positive charge.

#### Explanation:

Carbocations are generated when one of the bonds previously shared by a carbon atom is broken (such as to a hydrogen or halogen atom), leaving it with an incomplete octet. Carbon needs eight valence electrons to have a full octet, and adopts a $+ 1$ positive charge when it shares only six.

Consider this ${S}_{N} 1$ mechanism for the reaction between ( R )-2-idodopentane and methanol.

The leaving group (iodine) leaves with its bonding electrons, resulting in a carbocation as carbon now shares only six valence electrons. The electronegative oxygen atom of methanol will then attack the electrophilic carbocation, attracted to its positive charge (methanol is the nucleophile).

I've chosen not to show stereochemistry in the product, as both stereoisomers (inversion and retention of original) will be present in the product mixture. Also note that another methanol molecule will deprotonate the methanol which initially attacks the carbocation, which is why we end up with a neutral $O$ atom without any hydrogen.

The formation of a carbocation can allow for rearrangement to take place in reactions, resulting in more stable products. In this case, the most stable product by ${S}_{N} 1$ mechanism has already been formed (${2}^{o}$).

Though this is an important concept discussed in organic chemistry and relevant to many types of reactions, I have used only the ${S}_{N} 1$ mechanism as an example. This is by no means the only reaction mechanism that the formation of a carbocation is relevant to (unimolecular elimination, addition, etc.).

## Explain the Michael reaction?

Truong-Son N.
Featured 2 months ago

The Michael reaction is an addition reaction, usually of an enolate to an $\alpha , \beta$-unsaturated ketone (a conjugated enone), to form a diketone.

The case you're familiar with is this one:

where I've highlighted the molecular fragment that gets attached. The mechanism is a bit visually challenging, but it's not too bad:

You can think of it as a variation on the aldol addition, where we've replaced the aldehyde with a conjugated enone.

1. A base deprotonates the $\boldsymbol{\alpha}$-carbon on the ketone, which forms the enolate. Sometimes, LDA, lithiumdiisopropylamide, is used instead to give a better yield (though it is too reactive for aldehydes).
2. The enolate then acts as a good nucleophile to attack the $\beta$ position on the enone. Draw the resonance structure, and you should find that the carbonyl oxygen, which is electron-withdrawing and ${\delta}^{-}$, causes the $\beta$-carbon to be ${\delta}^{+}$.
3. The reaction finishes by tautomerizing the enolate back into the ketone by deprotonating the water molecule that was made in step 1.

This is usually the first step in a Robinson Annulation. Here is an example of an annotated Robinson Annulation (notice the heat added in step 7):

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