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3

## Why does a 1:1 mixture of benzene and hexafluorobenzene experience an elevated melting point?

Ernest Z.
Featured 1 month ago

The high melting point is caused by π-π stacking of the aromatic rings.

#### Explanation:

In organic chemistry, π–π stacking refers to attractive interactions between the π clouds of aromatic rings.

There are various types of stacking.

Neither benzene nor hexafluorobenzene has a dipole moment.

However, they have strong quadrupole moments, caused by the π clouds above and below the rings.

For example, in benzene, the π clouds are negatively charged and the plane of the ring is positively charged.

The situation is reversed in hexafluorobenzene, because the electronegative fluorine atoms withdraw electron density from the ring.

You can see the charge distribution better in this image:

Both benzene and hexafluorobenzene are destabilized by sandwich stacking, because areas with the same charge are placed next to each other.

However, benzene and hexafluorobenzene are strongly stabilized by sandwich stacking, because areas with opposite charge are placed next to each other.

Theoretical calculations put the stabilization energy at about 20 kJ/mol.

That makes the attractions as strong as many hydrogen bonds and dipole-dipole interactions.

Thus, the strong intermolecular quadrupole attractions cause a 1:1 mixture of benzene and hexafluorobenzene to have a high melting point.

2

## What are internal standards?

anor277
Featured 1 month ago

Do you mean with respect to NMR spectrometers.......?

#### Explanation:

""^1H and ""^13C{""^1H} $\text{NMR spectroscopies}$ remain a direct and very powerful tool of analysis for observation and identification of organic species in solution. As with any form of spectroscopy, a standard is usually used to calibrate the observed spectrum, so the measurement can be repeated and verified on different instruments.

This is typically the residual proton residue of the deuterated solvent that is used to acquire the spectrum (i.e. because we typically want to observe protons, we use a solvent such as $d - \text{chloroform}$ or ${d}_{6} - \text{benzene}$, i.e. C""^2HCl_3 or C_6""^2H_6; because these so-called deuterated solvents contain NEXT to NO protons, i.e. ""^1H $\text{nuclei}$, they are transparent in the ""^1H $\text{NMR}$ spectrum of the sample.

The residual protons in the sample (rarely can we afford to use 100% deuterated solvents) act as an internal reference with which we can calibrate the observed ""^1H $\text{NMR spectrum}$.

And thus in the ""^1H $\text{NMR spectrum}$, we have an internal standard; i.e. residual ""^1H resonances of the solvent; i.e. ${C}_{6} {D}_{5} H = 7.15 \cdot \text{ppm}$; $C {D}_{2} H C l = 7.24 \cdot \text{ppm}$; ${C}_{6} {D}_{5} C {D}_{2} H = 2.15 \cdot \text{ppm}$. And thus we can precisely and $\text{internally}$ calibrate the observed spectrum.

The use of deuterons rather than protons, is important in another respect. Modern NMR spectrometers usually have THREE channels; (i) for observing ""^1H nuclei; (ii) for observing heteroatoms, i.e. ""^13C, ""^31P, ""^15N, which occur at frequencies to which the spectrometer can conveniently be tuned; and (iii) the $\text{deuterium}$ or locking channel. The spectrometer can detect absorptions of ""^2H nuclei, and measurement can be calibrated by $\text{locking}$ onto this channel. The use of deuterium in labelling experiments given an active hydrogen is also widespread.

I am told these days, the modern operators sometimes do not even bother to use the locking channel. The magnetic fields generated are so stable, that they do not vary so much, and ""^1H spectra can often be run $\text{unlocked}$ (and even without spinning the sample!). Anyway, all of this should be known by a second year or third year organic student. At A-level you should know that ""^1H and ""^13C{""^1H} $\text{NMR spectroscopy}$ are a very powerful and direct means to characterize organic compounds in real time in solution.

2

## When you have a doublet of doublets how many neighboring Hydrogens will there be?

Ernest Z.
Featured 1 month ago

There will be two neighbouring hydrogens.

#### Explanation:

A doublet of doublets (dd) occurs when a hydrogen atom is coupled to two non-equivalent hydrogens.

An example is the NMR spectrum of methyl acrylate.

(From Chemistry LibreTexts)

Each of the vinyl protons ${\text{H"_text(a), "H}}_{\textrm{b}}$ and ${\text{H}}_{\textrm{c}}$ is a dd.

Let's examine the ${\text{H}}_{\textrm{c}}$ signal at 6.21 ppm.

There are four separate peaks because ${\text{H}}_{\textrm{c}}$ is coupled to both ${\text{H}}_{\textrm{a}}$ and ${\text{H}}_{\textrm{b}}$, but with different coupling constants for each.

The result is a doublet of doublets.

The splitting diagram shows that ${\text{H}}_{\textrm{c}}$ is split into a doublet by ${\text{H}}_{\textrm{a}}$ with a large coupling constant ${J}_{\textrm{t r a n s}}$.

These peaks are each split into doublets by coupling with ${\text{H}}_{\textrm{a}}$ with a much smaller coupling constant ${J}_{\textrm{\ge m}}$ = 1.5 Hz.

2

## Explain why lactose shows mutarotation but sucrose does not?

Maxwell
Featured 1 month ago

Because lactose contains a free anomeric carbon whereas sucrose does not, enabling it to equilibriate with the anomeric forms of the cyclic sugar.

#### Explanation:

Let me first quickly go over a little bit about $\textcolor{red}{\text{Carbohydrate chemistry}}$ and then I will get to the answer.

$\textcolor{red}{\text{Lactose}}$ and $\textcolor{red}{\text{sucrose}}$ are both carbohydrates or more commonly referred to as sugars. More specifically, they are $\textcolor{red}{\text{disacharrides}}$ as they are made up of 2 sugar units.

color(blue)["Figure 1. The disacharrides Sucrose and Lactose in their ring forms"

Let's look at sucrose first. Sucrose is made by the condensation reaction of a glucose molecule and a fructose molecule. What essentially happens is the hemiacetal of the glucose molecule reacts with the hydroxyl group of the fructose which forms the $\textcolor{red}{\text{O-glycosidic bond}}$ [color(blue)["Figure 2"] (a reaction you should be familiar with from organic chemistry)

color(blue)["Figure 2: Sucrose formation"------------

color(blue)[---------------------
$\textcolor{w h i t e}{a}$

color(blue)["Figure 3: Hemiacetal/hemiketal and acetal/ketal general reaction"

color(blue)[---------------------
$\textcolor{w h i t e}{a}$

During the formation of sucrose, both the glucose and fructose molecules' anomeric carbons get involved to form the O-glycosidic bond. An $\textcolor{red}{\text{anomeric carbon}}$ is the carbon that was once a part of the carbonyl group of the sugar before it reacted internally with the hydroxyl group which eventually lead to the cyclic structure of the sugar.

color(blue)["Figure 4: Anomeric carbon of Glucose"--------

Looking at the straight chain form, the carbonyl carbon reacts internally with the hydroxyl group to form a 5 membered ring. The carbon labeled 1 (C1) is the anomeric carbon in the ring form
color(blue)[---------------------

The cool thing about anomeric carbons is that depending on their stereochemistry two $\textcolor{red}{\text{anomers}}$ (isomers) can exist - the $\textcolor{red}{\alpha}$ or the $\textcolor{red}{\beta}$ form. At equilibrium, the 2 anomers and the straight chain form will exist with the $\beta$ form predominating in solution. These two forms can interconvert spontaneously in a process called $\textcolor{red}{\text{mutarotation}} .$

color(blue)["Figure 5: Mutarotation where equilibrium exists between cyclic"
color(blue)["and straight chain forms of the sugars"

color(blue)[---------------------
$\textcolor{w h i t e}{a}$
In order for mutarotation to occur, there needs to a $\textcolor{red}{\text{free anomeric carbon end}}$ or an anomeric carbon that is not involved in any bond and is open to equilibriate with its two isomer or straight chain forms. Well, sucrose can't do that as its 2 anomeric carbons are already involved in the glycosidic linkage. Lactose on the other hand, can mutarotate since it has one.

color(blue)["Figure 6: Lactose and its free anomeric end"-------

color(blue)[---------------------

3

## What are products of these Hoffman eliminations?

Ernest Z.
Featured 1 month 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.

2

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

Truong-Son N.
Featured 4 weeks 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.

4

## What is meant by +I & -I effects??

Ernest Z.
Featured 2 weeks ago

WARNING! Long answer! Inductive effects are the effects on rates or positions of equilibrium caused by the polarity of the bond to a substituent group.

#### Explanation:

A -I effect or negative inductive effect occurs when the substituent withdraws electrons.

A +I effect or positive inductive effect occurs when the substituent donates electrons.

Inductive effects

Consider a $\text{C-F}$ bond.

The highly electronegative $\text{F}$ atom will draw the electrons in the $\text{C-F}$ bond more closely toward itself.

The bond will be polarized, with the $\text{F}$ atom getting a partial negative (δ^"-") charge and the α-carbon atom getting a partial positive (δ^"+") charge.

underbrace(stackrelcolor(blue)(δδδδ^"+")("C"))_color(red)(δ)-underbrace(stackrelcolor(blue)(δδδ^"+")("C"))_color(red)(γ)-underbrace(stackrelcolor(blue)(δδ^"+")("C"))_color(red)(β)-underbrace(stackrelcolor(blue)(δ^"+")("C"))_color(red)(α)-stackrelcolor(blue)(δ^"-")("F")

The α-carbon will in turn withdraw some electron density from the β-carbon, giving it a smaller partial positive (δδ^"+") charge.

The inductive removal of electron density is passed with diminishing effect through the chain of $\text{C-C}$ σ-bonds until it is almost negligible at the δ-carbon.

$\text{C}$ is less electronegative than $\text{H}$, so alkyl groups are electron releasing.

underbrace(stackrelcolor(blue)(δδδδ^"-")("C"))_color(red)(δ)-underbrace(stackrelcolor(blue)(δδδ^"-")("C"))_color(red)(γ)-underbrace(stackrelcolor(blue)(δδ^"-")("C"))_color(red)(β)-underbrace(stackrelcolor(blue)(δ^"-")("C"))_color(red)(α)-stackrelcolor(blue)(δ^"+")("R")

Again, the effect is passed along the chain of carbon atoms but it dies out rapidly with distance.

-I effect

The strength of a carboxylic acid depends on the extent of its ionization: the more ionized it is, the stronger it is.

As an acid becomes stronger, the numerical value of its $\text{p} {K}_{\textrm{a}}$ drops.

Thus, for example. the order of acidity as shown by the $\text{p} {K}_{\textrm{a}}$ values is

${\underbrace{\text{Br-CH"_2"CH"_2"CH"_2"COO-H")_color(red)(4.59) < underbrace("Br-CH"_2"CH"_2"COO-H")_color(red)(4.01) < underbrace("Br-CH"_2"COO-H}}}_{\textcolor{red}{2.86}}$

The electronegative $\text{Br}$ atom removes electron density from the atoms next to it, eventually weakening the $\text{O-H}$ bond at the other end of the chain and making the compound more acidic.

Since this effect is caused by the inductive removal of electrons, it is called
a -I effect

+I effect

In the same way, an electron-donating alkyl group decreases the acidity of a carboxylic acid.

Thus, for example. the order of acidity as shown by the $\text{p} {K}_{\textrm{a}}$ values is

${\underbrace{\text{CH"_3"-CH"_2"COO-H")_color(red)(4.87) < underbrace("CH"_3"-COO-H")_color(red)(4.76) < underbrace("H-COO-H}}}_{\textcolor{red}{3.74}}$

Thus, acetic acid is weaker than formic acid, and propionic acid is weaker than acetic acid.

Since this effect is caused by the inductive donation of electrons, it is called
a +I effect

3

## What is the structure for 4-ethyl-2-iso-propyl-1-methylcyclopentane?

Maxwell
Featured 2 weeks ago

See below

#### Explanation:

Let's break this down a bit.

First, your job is to identify the parent chain, which is the longest carbon chain.

That's easy as the last part of the structure's name gives it away: $\text{cyclopentane}$. Pentane means it's a 5 carbon alkane and the prefix cyclo- means the parent chain is cyclic. It looks like this without any alkyl chains attached to it.
$\textcolor{w h i t e}{a a a a a a a a a a a}$

Next, you have to identify the substituents. Well, knowing that cyclopentane is the parent chain, whatever else that we have left are treated as the substituents. Let's write them down.

• $\textcolor{\mathmr{and} a n \ge}{\text{4-ethyl}}$
• $\textcolor{b l u e}{\text{2-isopropyl}}$
• $\textcolor{m a \ge n t a}{\text{1-methyl}}$

If we take our cyclopentane and first number all the carbons,

we can then, starting in numerical order, attach the substituents one-by-one.

$\textcolor{w h i t e}{a a a a}$

$\underline{\text{Adding Substituents}}$

Looking at $\textcolor{m a \ge n t a}{\text{1-methyl}}$, we can see we have a methyl group attached at carbon number $1$. Placing it at the appropriate position, we get the following structure:

Next, we take a look at $\textcolor{b l u e}{\text{2-isopropyl}}$. We have an isopropyl group attached at carbon number $2$. Placing this alkyl group at its appropriate position we get the following structure thus far:

Lastly, looking at $\textcolor{\mathmr{and} a n \ge}{\text{4-ethyl}}$, we place an ethyl group at carbon number $4$. Doing so, our finished structure looks like this:

$\textcolor{w h i t e}{a a a a a} \text{4-ethyl-2-isopropyl-1-methylcyclopentane}$

2

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

Truong-Son N.
Featured 2 weeks 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.

3

## How to draw distribution graph if pKa of acid is 4,4 and pKa of base is 6,7?

Truong-Son N.
Featured 2 weeks ago

Well, these distribution graphs should correlate with the titration curve.

If we know the first $\text{pKa}$ is $4.4$ and the second $\text{pKa}$ is $6.7$, then we have an idea of where the half-equivalence points are (i.e. where the concentrations of acid and conjugate base are equal), because the $\text{pH}$ $=$ $\text{pKa}$ at those points:

"pH"_("1st half equiv. pt.") = "pKa"_1 + cancel(log\frac(["HA"^(-)])(["H"_2"A"]))^("Equal conc.'s, "log(1) = 0)

"pH"_("2nd half equiv. pt.") = "pKa"_2 + cancel(log\frac(["A"^(2-)])(["HA"^(-)]))^("Equal conc.'s, "log(1) = 0)

We represent each stage of a diprotic acid as:

${\text{H"_2"A"(aq) rightleftharpoons overbrace("HA"^(-)(aq))^"singly deprotonated" + "H}}^{+} \left(a q\right)$

rightleftharpoons overbrace("A"^(2-)(aq))^"doubly deprotonated" + "H"^(+)(aq)

The two midpoints shown are the first and second half-equivalence points, respectively.

• At midpoint 1, we have that $\left[{\text{H"_2"A"] = ["HA}}^{-}\right]$, and that $\text{pH} \approx 4.4$.

• At midpoint 2, we have that $\left[{\text{HA"^(-)] = ["A}}^{2 -}\right]$, and that $\text{pH} \approx 6.7$.

A distribution graph shows the change in concentration of each species in solution as the $\text{pH}$ increases. It correlates well with a base-into-diprotic-acid titration curve.

See below for an overlay of both:

Each species in solution is tracked in the bottom graph.

• The cross-over points on the distribution graph are the half-equivalence points on the titration curve.

• The maximum concentration for each species after the starting $\text{pH}$ correlate with the first few equivalence points, and the last species to show up dominates at high $\text{pH}$.