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## What are the intermolecular forces of CHF3, OF2, HF, and CF4?

Dr. Cawas K.
Featured 8 months ago

The relative magnitude of the inter molecular forces are:
$C {F}_{4} < O {F}_{2} < C H {F}_{3} < H F .$

#### Explanation:

All molecules will have London dispersion forces which get stronger as the molecule gets heavier (more electrons causes a shift in electron cloud distribution resulting in a temporary dipole).

$C H {F}_{3}$ is a polar molecule. So it will have dipole - dipole interaction along with the weaker dispersion forces.

$O {F}_{2}$ is a polar molecule with a bent shape just like ${H}_{2} O$. There are two pairs of bonded electrons and two pairs of unbonded lone pair electrons. These lone pairs repel the bonded electrons resulting in a bent shape with an angle less than ${105}^{0}$. The dominant inter molecular forces would be dipole-dipole.

HF is a polar molecule. Hence the primary inter molecular forces would be dipole - dipole and hydrogen bond which is a special type of dipole - dipole interaction between the hydrogen atom and electronegative F atom.

$C {F}_{4}$ has a tetrahedral structure. It is non-polar molecule. The dominant inter molecular force would be London dispersion force.

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

Ernest Z.
Featured 6 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 would the product be if I add Br2 as step one and CH3OH in step 2?

Martin M.
Featured 3 months ago

For determining the product in these type of questions, we have to look at the functional groups of the reactants. In this case, we have a $B {r}_{2}$ molecule and a cyclohexane ring with a double bond side group. As we know, double bonds are pretty reactive, so we expect something to happen there.

Now the $B {r}_{2}$ molecule can line up with this double bond, creating partial charges. This is indicated in the figure below. Because of the partial positive charge (created on the side of the double bond), the electrons will attack there and create a cyclic structure with one of the $B r$ atoms.

The other $B r$ will be released.

Now we add $C {H}_{3} O H$. The free electrons on the $O$ can react with the structure. This is showed in the mechanism below.

The carbon atom of the cyclic bromine group is partial positively charged, as indicated in the image above. This is created by hyperconjugation. This carbon atom is attached to 3 other carbon atoms (tertiary carbon atom), which stabilises a charge on the carbon atom more easily. Therefore the electrons from methanol will attack at that carbon atom, 'pushing away' the electrons from the bromine bond.

In the last step, the $H$ atom will be released. This causes the oxygen to be neutral instead of positively charged.

And there you have your product!

## What makes so many epoxides so carcinogenic?

Ernest Z.
Featured 3 months ago

Epoxides are carcinogenic because they disrupt our DNA.

#### Explanation:

Epoxides are three-membered cyclic ethers.

They are highly strained because the nominal bond angles are 60° instead of 109.5°.

They tend to react with other molecules to open the ring and reduce the strain.

A common pro-carcinogen is benzo[a]pyrene.

It is a constituent of automobile exhaust, smog, cigarette smoke and charcoal-cooked foods.

When the body absorbs benzo[a]pyrene, it tries to make it water-soluble so it can be excreted in the urine or feces.

(From researcgate.net)

It converts the benzo[a]pyrene into benzo[a]pyrene-7,8-epoxide.

This is hydrolyzed to the 7,8-dihydrodiol, which is further converted into benzo[a]pyrene-7,8-dihydrodiol-9,10-epoxide.

Once formed, the epoxide can π stack with the bases in DNA where the epoxide group reacts with guanine residues.

This distorts the structure of DNA and causes errors in DNA replication.

If these mutations occur in a gene that encodes a molecule that regulates the production of cells, the result may be cancer.

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

Truong-Son N.
Featured 3 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.

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

Ernest Z.
Featured 3 months 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.

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