2
Active contributors today

## What is a delocalized pi bond?

Ernest Z.
Featured 2 months ago

A delocalized π bond is a π bond in which the electrons are free to move over more than two nuclei.

#### Explanation:

In a molecule like ethylene, the electrons in the π bond are constrained to the region between the two carbon atoms.

We say that the π electrons are localized.

Even in penta-1,4-diene, the π electrons are still localized.

The ${\text{CH}}_{2}$ group between the two π orbitals prevents them from overlapping.

(From iverson.cm.utexas.edu)

However, in buta-1,3-diene, the two orbitals can overlap, and the π electrons are free to spread over all four carbon atoms.

We say that these π electrons are delocalized.

In benzene, the π electrons are delocalized over all six atoms of the ring.

In β-carotene, the π electrons are delocalized over 22 carbon atoms!

## What determines carbocation stability?

Ernest Z.
Featured 9 months ago

The three factors that determine carbocation stability are adjacent (1) multiple bonds; (2) lone pairs; and (3) carbon atoms.

An adjacent π bond allows the positive charge to be delocalized by resonance.

Thus, $\text{H"_2"C=CHCH"_2^"+}$ is more stable than $\text{CH"_3"CH"_2"CH"_2^"+}$.

Resonance delocalization of the charge through a larger π cloud makes the cation more stable.

A lone pair on an adjacent atom stabilizes a carbocation.

Thus, $\text{CH"_3"OCH"_2^"+}$ is more stable than $\text{CH"_3"CH"_2"CH"_2^"+}$ because of resonance.

The cation is more stable because the charge is spread over two atoms.

${\text{CH"_3"O"stackrel(+)("C")"H"_2 ⟷ "CH"_3stackrel(+)("O")"=CH}}_{2}$

The stability of carbocations increases as we add more carbon atoms to the cationic carbon.

For example, $\text{CH"_3"CH"_2^"+}$ is more stable than $\text{CH"_3^"+}$

The stabilization is explained by a type of resonance called hyperconjugation.

${\text{H-CH"_2"-"stackrel("+")("C")"H"_2 ⟷ stackrel("+")("H") " CH"_2"=CH}}_{2}$

The ${\text{sp}}^{3}$ orbitals of the adjacent $\text{C-H}$ bonds overlap with the vacant $\text{p}$ orbital on the carbocation.

The delocalization of charge stabilizes the carbocation.

Thus, the more alkyl groups there are surrounding the cationic carbon, the more stable it becomes.

That gives us the order of stability: 3° > 2° >1° > 0°

## What are products of these Hoffman eliminations?

Ernest Z.
Featured 3 months ago

Here's what I think.

#### Explanation:

The Hofmann elimination is a process in which an amine is converted to a quaternary ammonium salt by treatment with excess methyl iodide and then reacted with silver oxide and water to form an alkene.

The silver oxide and water form hydroxide ions which eliminate a β-hydrogen.

For example,

$\text{CH"_3"CH"_2"CH"("CH"_3)"NH"_2 stackrelcolor(blue)("MeI"color(white)(mm))(→) "CH"_3"CH"_2"CH"("CH"_3) stackrelcolor(blue)("+")("N")"Me"_2 stackrelcolor(blue)("Ag"_2"O", "H"_2"O", Δ color(white)(mm))(→ )"CH"_3"CH"_2"CH=CH"_2 + "Me"_2"NH}$

The major product is the least substituted alkene.

Hofmann elimination of 1-azabicyclo[4.3.0]nonane

I would expect the quaternary ammonium salt to give a mixture of two alkenes.

One is N-methyl-2-(prop-2-en-1-yl)piperidine, formed by elimination from
the 5-membered ring.

The other is 2-(but-3-en-1-yl)-N-methylpyrollidine, formed by elimination from
the 6-membered ring.

I would not expect indole to undergo Hofmann elimination because it has no aliphatic β-hydrogens to be eliminated.

## What is the product of the reaction and what is the mechanism of the reaction?

Truong-Son N.
Featured 3 months ago

All of these involve bulky substrates, so these will all contain some mixture of ${S}_{N} 1$ or ${E}_{1}$ in some capacity.

1) tert-butyl bromide with ethanol

The major product forms from ${S}_{N} 1$; the alkyl bromide is tertiary, so it is bulky/sterically-hindered, and is more likely to lose ${\text{Br}}^{-}$ as a leaving group before managing to get attacked by a nucleophile (ethanol). The proton on the attached ethanol then dissociates on its own.

That is characteristic of a first-order process, where the slow step is the departure of the leaving group and a carbocation intermediate forms (here, a tertiary carbocation).

There is a minor product from $E 1$, but this is very minor unless the temperature is sufficiently high. This would be a $\beta$ elimination, i.e. the alcohol extracts a proton from the carbon adjacent to $\text{Br}$ and patches up the carbocation intermediate.

2) 1-chloro-1-ethylcyclopentane with methanol

Basically the same idea as in $\left(1\right)$, except...

The only difference is a mixture of elimination minor products. Either alkene is equally-substituted (both have one doubly-substituted and one singly-substituted $s {p}^{2}$ carbon), so by Zaitsev's rule, either minor product is likely to occur.

3) (1-bromo-1-methylethyl)-cyclohexane with acetic acid

Basically the same idea as in $\left(1\right)$ and $\left(2\right)$, except now the so-called nucleophile, acetic acid, does not do its job well at all; it's quite bulky, and is better at being a Lewis base (ugh, a base? That'll take gusto, as it's an acid! Gotta wait for it to dissociate as much as possible!).

So here, $E 1$ is likely, while ${S}_{N} 1$ pretty much won't happen (i.e. some ridiculously bulky ester won't form).

By Zaitsev's rule, we expect the more substituted alkene to form, which is the left one (because it has two disubstituted $s {p}^{2}$ carbons, compared to the right-hand one).

4) tert-butyl chloride with a) deuterated water and b) monodeuterated cyclohexanol

Reaction with deuterated compounds is not that different from the undeuterated forms. It is just easier to tell which $\text{H}$ is which.

a) is a mixture of ${S}_{N} 1$ and ${E}_{1}$ since the alkyl halide is bulky, as in $\left(1\right)$. $E 1$ at a sufficiently high temperature.

b) The difference with respect to $\left(a\right)$ is that now the reactant added is more bulky. In this case, the major and minor products switch, because substitution would require the approach of a big molecule towards a bulky molecule.

Thus, the kinetically-slow monodeuterated cyclohexanol prefers to be a base and perform $E 1$ to give the thermodynamically-favored product.

## Why does pyrene undergo electrophilic addition easily even though it is an aromatic compound?

Truong-Son N.
Featured 3 months ago

I would think that it's because pyrene has less resonance stabilization than benzene does (increasing its HOMO-LUMO gap by less), due to its sheer size causing its energy levels to be so close together.

A smaller HOMO-LUMO gap means a more reactive system, despite it having resonance throughout.

EXAMINING THE EXTENSIVITY OF RESONANCE STABILIZATION

Consider napthalene, anthracene, and phenanthrene (if you add one benzene ring to the upper-right of phenanthrene, you have pyrene):

The resonance stabilization that one benzene ring gets is $\text{36 kcal/mol}$. If there were a perfect extensivity with regards to resonance stabilization, we would have expected the amount to be

$\text{Total Resonance Energy}$

$\approx \text{Number of Benzene Rings" xx "Resonance Energy}$

But you can see in the above diagram that it isn't:

• Napthalene has $\boldsymbol{\text{11 kcal/mol}}$ less resonance energy than $2 \times \text{benzene rings}$.
• Anthracene has $\boldsymbol{\text{25 kcal/mol}}$ less resonance energy than $3 \times \text{benzene rings}$.
• Phenanthrene has $\boldsymbol{\text{17 kcal/mol}}$ less resonance energy than $3 \times \text{benzene rings}$.

From this, we could postulate that in general, the more extended the $\pi$ system, the less resonance stabilization is afforded. The most likely reason for this is probably the volume of the system.

ENERGY GAPS AS A FUNCTION OF VOLUME (AND ENTROPY)

The energy gaps (and thus the HOMO-LUMO gap) in any molecule are a function of the system volume and entropy.

At constant entropy though (which means at a constant distribution of states amongst the energy levels), the trend of volume vs. energy gap can be examined.

To illustrate this, the following graph was generated and derived from Huckel MO Theory, for which we have:

${E}_{k} = \alpha + 2 \beta \cos \left(\frac{2 k \pi}{n}\right)$,

where $k$ is the energy level index and $n$ is the number of fused rings. $\alpha$ is the nonbonding energy and $\beta$ is the negative difference in energy from the nonbonding level.

We can see then that the HOMO-LUMO gap converges as the number of rings increases, i.e. as the system volume increases.

PARTICIPATION OF HOMO & LUMO IN ELECTROPHILIC ADDITION

How is this relevant?

Well, the HOMO and LUMO are both required in electrophilic addition reactions. For example, with adding ${\text{Br}}_{2}$,

• The HOMO donates $\pi$ electrons to polarize ${\text{Br}}_{2}$ and break the $\text{Br"-"Br}$ bond.
• The LUMO accepts electrons from the $\stackrel{{\delta}^{+}}{\text{Br}}$.

And this forms the so-called bromonium complex:

(Here, the HOMO contained the $\pi$ electrons in the double bond, and the LUMO accepted the electrons from the bottom $\text{Br}$.)

Since the HOMO-LUMO gap gets smaller when the system gets larger, it's very likely that the gap is so small for pyrene that the resonance stabilization (which increases this gap) isn't enough to make it unreactive towards electrophilic addition.

## How can we write IUPAC name of ether, ester and cyanide?

Ernest Z.
Featured 1 month ago

Here's what I get.

#### Explanation:

When there is more than one permissible IUPAC name, the first name is preferred.

Ethers

There are two systems of IUPAC names for ethers.

1. Substitutive names.

The ethers are named as alkoxyalkanes, with the senior component selected as the parent compound.

Thus, ${\text{CH"_3"OCH"_2"CH"_2"CH}}_{3}$ is 1-methoxypropane.

2. Functional group names.

The ethers are named as alkyl alkyl ethers, with the alkyl groups in alphabetical order followed by the class name ether, each as a separate word.

Thus, ${\text{CH"_3"OCH"_2"CH"_2"CH}}_{3}$ is methyl propyl ether.

Esters

Esters are named as alkyl alkanoates.

The name of the alkyl group is written first, followed by the name of the acid with the ending -ic acid replaced by the ending -ate.

Thus, ${\text{CH"_3"CH"_2"COO-CH}}_{3}$ is methyl propanoate (two words).

Cyanides (nitriles)

There are three systems of IUPAC names for nitriles.

1. As alkanenitriles.

The ending -nitrile is added to the name of the alkane with the same number of carbon atoms.

Thus, $\text{CH"_3"CH"_2"C≡N}$ is propanenitrile.

2. If the compound is considered to be formed from a carboxylic acid with a "trivial name" ($\text{RCOOH → RC≡N}$), the ending -ic acid is changed to -onitrile.

Thus, $\text{CH"_3"CH"_2"C≡N}$ is also called propionitrile (from propionic acid).

3. As alkyl cyanides.

The name of the alkyl group precedes the class name cyanide.

Thus, $\text{CH"_3"CH"_2"-C≡N}$ is ethyl cyanide (two words).

##### Questions
• · 2 hours ago
• · 2 hours ago
• · 6 hours ago
• · 6 hours ago
• · 10 hours ago
• · Yesterday
• · Yesterday
• Yesterday
• · Yesterday
• · Yesterday
• · 2 days ago
• 2 days ago · in Dipoles
• · 2 days ago
· 3 days ago
• · 3 days ago
• · 3 days ago
• · 3 days ago
• · 3 days ago
• · 4 days ago
• · 4 days ago
• 4 days ago · in Leaving Groups
• · 5 days ago
• · 5 days ago
• · 5 days ago
• · 6 days ago
• · 1 week ago
• · 1 week ago