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

Ernest Z.
Featured 8 months 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.

## What is the major product of the Stork enamine synthesis, that is, the reaction of cyclopentanone with pyrrolidine, heat, CH_3CH_2Br, H^+, and H_2O?

Ernest Z.
Featured 6 months ago

The major product is 2-ethylcyclopentanone.

#### Explanation:

The reaction is often called the Stork enamine synthesis.

Step 1. Formation of an enamine.

Step 2. ${\text{S}}_{\textrm{N}} 2$ alkylation

The double-bonded carbon in the enamine has a partial negative charge, so it can act as a nucleophile and displace the $\text{Br}$ from ethyl bromide and form an iminium salt.

Step 3. Hydrolysis of the iminium salt

The iminium salt is hydrolyzed to regenerate the ketone and the amine.

Note: The reaction is done this way because direct alkylation of the cyclopentanone enolate gives multiple alkylation, but the enamine stops at one alkylation.

## Why is piperidine a stronger base than morphine?

Ernest Z.
Featured 5 months ago

I suspect the reason involves steric hindrance.

#### Explanation:

The structure of piperidine is

Piperidine is a weak base with "pK_text(b) = 2.80.

It has a cyclohexane ring structure in which the lone pair on the nitrogen atom is quite accessible to an attacking acid.

We can see this in both the ball-and-stick model

and the space-filling model.

Morphine

The structure of morphine is

(From ResearchGate)

Ring D in its structure is a piperidine ring.

The methyl group on the nitrogen group should make morphine more basic than piperidine.

However, morphine is a weaker base, with "pK_text(b) = 6.13.

Morphine has a rigid pentacyclic ring structure, and the nitrogen atom is "buried" in the interior of the molecule, where it is less accessible to an attacking acid.

This becomes even clearer in a space-filling model of morphine.

Attack by an acid is strongly hindered on one side by Ring A.

We find basicities by measuring the position of an equilibrium:

$\text{B + HX ⇌ BH"^"+" + "X"^"-}$

If access to the base is hindered, the position of equilibrium lies further to the left, and we say that the base is weaker.

Thus, morphine is a weaker base than piperidine.

## What are two characteristics of a molecule that make it suitable for analysis by infrared spectroscopy?

Truong-Son N.
Featured 5 months ago

You are right that it has to have a dipole (not necessarily a net dipole). I'm not sure what other specific requirements you may be wondering about, but I can list several, and maybe one of them is what you are looking for.

REQUIREMENTS FOR VISIBILITY IN THE IR REGION

The main properties of a molecule that allow it to be analyzable via IR spectroscopy are:

1) It can experience a change in dipole moment, whether it is induced or permanent. We then must have that it is heteronuclear, if it is diatomic (therefore, ${\text{N}}_{2}$, ${\text{F}}_{2}$, etc. are invisible in the IR).

2) It should have resonant frequencies that are in the infrared frequency range of $100 - 4000$ ${\text{cm}}^{- 1}$.

3) Ideally, it should be not overly soluble in a nonpolar solvent, which is ideal since many analyses are carried out using a solvent such as ${\text{CS}}_{2}$ or ${\text{CCl}}_{4}$. If something is too soluble, one may saturate or overload the spectrometer.

CHANGE IN DIPOLE MOMENT

Requirement $\left(1\right)$ is usually easy to check; if you can imagine the molecule stretching or bending in a way that enforces asymmetry, you've more than likely changed its dipole moment, and thus that kind of vibrational motion is IR-active.

A somewhat tricky example is ${\text{CO}}_{2}$.

Even though it is symmetrical, it can stretch both oxygens in the same direction (as in $B$) to generate a nonzero dipole moment along the molecular axis. It can also bend its $\text{O"-"C"-"O}$ angle (as in $C$ and $D$) so that it generates a nonzero dipole moment through the oxygen.

RESONANT FREQUENCIES IN THE IR RANGE

This is usually satisfied automatically, just due to the general strengths of chemical bonds, but it couldn't hurt to check. The fundamental frequency can be estimated using the force constant $k$ of the bond.

$\boldsymbol{t i l \mathrm{de} \omega = \frac{1}{2 \pi c} \sqrt{\frac{k}{\mu}}}$,

where:

• $t i l \mathrm{de} \omega$ is the fundamental frequency in ${\text{cm}}^{- 1}$.
• $c$ is the speed of light, $2.998 \times {10}^{10} \text{cm/s}$.
• $k$ is the force constant in ${\text{kg/s}}^{2}$.
• $\mu = \frac{{m}_{1} {m}_{2}}{{m}_{1} + {m}_{2}}$ is the reduced mass of the two atoms in the bond, where each mass is in $\text{kg}$.

As an example, I randomly found on a quick Google search that $k \approx \text{2385 N/m}$ for ${\text{CO}}_{2}$. So:

tildeomega = 1/(2pi*2.998 xx 10^(10) "cm/s") sqrt(("2385 kg/s"^2)/((12.011*15.999)/(12.011 + 15.999) xx 10^(-3) "kg"/"mol" xx "mol"/(6.0221413 xx 10^(23) "molecules"))

$\approx {\text{2429 cm}}^{- 1}$

and the literature value for the fundamental frequency is ${\text{2349 cm}}^{- 1}$ from NIST.

This is found to be the strongest peak in its IR spectrum:

which you can estimate to be near ${\text{2350 cm}}^{- 1}$, in agreement with the literature value.

SOLUBILITY IN A NONPOLAR SOLVENT

Nice IR solvents that are commonly used are ${\text{CCl}}_{4}$ and ${\text{CS}}_{2}$. Since both are nonpolar, you would have the most "control" over the IR intensity if you had a reasonably soluble analyte. This is why IR tends to be suitable for organic molecules.

This kind of consideration is not necessary, but being too soluble may make it frustrating to get a spectrum without overloading the spectrometer. Being "somewhat" soluble in nonpolar solvents is probably fine.

## What are internal standards?

anor277
Featured 5 months 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.

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

Maxwell
Featured 4 months 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)[---------------------

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