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## What is mass spectrometry (MS)?

Truong-Son N.
Featured 2 months ago

Mass Spectroscopy (MS), in the most basic sense, is for tracing the fragmentation patterns of molecular ions in order to identify them. This tends to be more useful when coupled with other processes, such as Gas Chromatography and Liquid Chromatography.

MS has an interesting process by which we do the following (in a vacuum):

• Inject liquid sample (might be a few $\mu L$ if you are injecting into a GC-MS setup; depends on the injection method)
• Vaporize sample (must be a gas to minimize undesirable fragmentation)
• Ionize sample (must be an ion to interact with electric/magnetic fields) to facilitate fragmentation
• Accelerate fragments into field (electric and/or magnetic) to separate ions by $\frac{m}{z}$ ratio
• Detect the ions to get a count for the abundance of each ion
• Acquire mass spectrum

This is the essential process of Mass Spectroscopy.

In further depth:

Inject the sample via some method, such as the 30-ft long tube you use in Gas Chromatography (i.e. the GC is interfaced with MS and you have a GC-MS setup), for example.

Some sort of sample vaporization occurs so that you have a gaseous sample. This may be done with, perhaps, a coil with electric current flowing through it, or maybe a hot flame ($> 3000 K$ or so), for example.

Some sort of ionization occurs (which facilitates fragmentation), such as:

• Chemical Ionization (soft/indirect ionization via the presence of ions in the system)
• Matrix-Assisted Laser-Desorption Ionization (a sample matrix is hit with a laser, ionized, and the matrix itself soft-ionizes the sample embedded within and shielded by the matrix)
• Electron "Impact", where an electron beam allows electrons to interact with the sample to knock off an electron and thus ionize it (hard ionization).

The point of having both soft and hard ionization is that soft ionization better-retains the parent peak during fragmentation, so that you can find the peak that corresponds to the molecular mass of the original ion.

The fragments are then accelerated into an electric and/or magnetic field for the purpose of separating it by a mass-to-charge ratio, $m \text{/} z$. The ions then separate and then spiral towards some sort of collection surface that counts ions. This separation may be done with, for example, a quadrupole filter ("quadrupole" literally means "four [magnetic] poles").

The ions must reach a detector, such as a Faraday Plate or Faraday Cup. Behind it would be some sort of transducer to convert/encode the number of ions that are counted into a current that acts as a signal for a computer to read, so that it can generate some sort of display to give you your mass spectrum.

## Why is only one of the enantiomers of Ibuprofen effective?

Ernest Z.
Featured 4 months ago

Only one of the isomers is effective because it is the one that fits the receptor site on the enzyme involved in pain perception.

#### Explanation:

What ibuprofen does

Ibuprofen is works by inhibiting two enzymes called COX-1 and COX-2.

They convert arachidonic acid to prostaglandin H2 (PGH2) which, in turn, is converted by other enzymes to other prostaglandins that activate the body's response to inflammation.

How ibuprofen does it

Ibuprofen is 2-(4-isobutylphenyl)propanoic acid. Its structure is

(From scienceline.ucsb.edu)

Note the chiral centre at $\text{C2}$ of the propanoic acid.

Why only the $S$ isomer works

The ($S$)-ibuprofen has the same shape as the molecules that activate the COX enzymes to produce prostaglandins.

If we use the lock-and-key theory of enzyme action, we say that ($S$)-ibuprofen is the key that fits the lock (the receptor site) of the enzyme.

When it occupies the receptor site, it blocks access to the COX activators.

(From www.dailymail.co.uk)

The production of PGH2 ceases along with the pain and fever caused by the body's inflammatory response.

## What are the symmetry elements found on these isomers of dichloro cyclobutane?

Truong-Son N.
Featured 3 months ago

Since you have not specified which isomers, I will simply list four of them (assuming they are on different carbons):

From left to right, we have the cis-1,3, trans-1,3, trans-1,2, and cis-1,2 isomers. I will label them from left to right, isomers (1), (2), (3), and (4).

(1): $E$, ${C}_{2}$, ${\sigma}_{v} \left(x z\right)$, ${\sigma}_{v} \left(y z\right)$
(2): $E$, ${C}_{2}$, $i$, ${\sigma}_{h}$
(3): $E , {C}_{2}$
(4): $E , {\sigma}_{h}$

So basically:

(1): Identity, one rotation axis, two reflection planes
(2): Identity, one rotation axis, one point of inversion, one reflection plane
(3): Identity, one rotation axis
(4): Identity, one reflection plane

DISCLAIMER: Long answer! Possibly difficult to visualize...

The possible, primary symmetry elements overall (for any compound) are:

• $E$, the identity element, for completeness's sake.
• ${C}_{n}$, the principal rotation axis, where a rotation of ${360}^{\circ} / n$ about this axis returns the original molecule. This is defined to require the largest rotation angle to return the original molecule.
We define this to be along the $\boldsymbol{z}$ axis.
• ${C}_{n} '$, any other rotation axis, defined similarly; just something that isn't the previously-defined ${C}_{n}$. It could be perpendicular.
• ${\sigma}_{v}$, the vertical reflection plane, colinear with the principal rotation axis ${C}_{n}$. Generally, crosses through atoms.
• ${\sigma}_{h}$, the horizontal reflection plane, generally on the plane of a cyclic molecule that is perpendicular to the ${C}_{n}$ axis.
If no ${C}_{n}$ axis exists except for the trivial ${C}_{1}$ (${C}_{1} = E$), then we assign a reflection plane as ${\sigma}_{h}$.
• ${\sigma}_{d}$, the dihedral reflection plane, generally in between two atoms, bisecting a bond (this is not common unless ${\sigma}_{v}$ and ${\sigma}_{h}$ are also identified).
• $i$, the point of inversion. Basically you take the coordinates $\left(x , y , z\right)$ and swap them with the coordinates $\left(- x , - y , - z\right)$.

Obviously, all four isomers have $\boldsymbol{E}$, because they are all themselves. That aside...

ISOMER (1)

• There is one ${C}_{n}$ axis through the ring, through the plane of the screen. That is a $\boldsymbol{{C}_{2}}$ principal rotation axis, along the $z$-axis. Therefore, the plane of the ring is the $x y$-plane.
• There is one $\boldsymbol{{\sigma}_{v} \left(x z\right)}$ vertical reflection plane bisecting the molecule through carbon-1 and carbon-3 (perpendicular to the ring).
• There is one more $\boldsymbol{{\sigma}_{v} \left(y z\right)}$ vertical reflection plane bisecting the molecule through carbon-2 and carbon-4 (perpendicular to the ring).

That, by the way, assigns this molecule to a $\boldsymbol{{C}_{2 v}}$ point group.

ISOMER (2)

Compared to (1), there is no longer a ${\sigma}_{v} \left(y z\right)$ reflection plane through carbon-2 and carbon-4, and there is no longer a ${C}_{2}$ axis through the plane of the screen.

• There is one $\boldsymbol{{C}_{2}}$ axis bisecting the molecule through carbon-2 and carbon-4. We now define that as its $z$-axis. Let the plane of the molecule be the $y z$-plane, then (so that the $x$ axis is through the plane of the screen).
• There is one point of inversion, $\boldsymbol{i}$. Try taking carbon-1 and carbon-3, and switching their coordinates. Your chlorines would exactly swap places.
• There is one $\boldsymbol{{\sigma}_{h} \left(x z\right)}$ vertical reflection plane bisecting the molecule through carbon-1 and carbon-3 (perpendicular to the ring). Since it is perpendicular to ${C}_{2}$, it is still considered "horizontal". We've just reoriented our axes.

That, by the way, assigns this molecule to a $\boldsymbol{{C}_{2 h}}$ point group.

ISOMER (3)

Compared to (2), there is no point of inversion $i$.

Using the same reasoning as in (1) and (2):

• There is a $\boldsymbol{{C}_{2}}$ principal rotation axis bisecting the ${C}_{1} - {C}_{2}$ bond, and we define that as the $z$-axis since $n$ is as low as possible (${C}_{1} = E$).

And I think that's it... I cannot find any reflection planes. There is no point of inversion as I mentioned before, since the chlorines are both on the same side.

So this is, by the way, assigned to a $\boldsymbol{{C}_{2}}$ point group, one of low symmetry.

ISOMER (4)

Compared to (3), there is actually a reflection plane, but no ${C}_{n}$ axis...

• All I see is a $\boldsymbol{{\sigma}_{h}}$ reflection plane, since there is no ${C}_{n}$ axis I can find. This is bisecting the molecule through the ${C}_{1} - {C}_{2}$ bond (perpendicular to the plane of the screen).

So this is, by the way, assigned to the $\boldsymbol{{C}_{s}}$ point group, which is of low symmetry.

Does this make sense? It may require you to pull out a model kit to visualize this.

## What is hybridization in organic chemistry?

Ernest Z.
Featured 3 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.

## Why are aromatic rings stable?

Ernest Z.
Featured 1 month ago

Aromatic rings are stable because they are cyclic, conjugated molecules.

#### Explanation:

Conjugation of π orbitals lowers the energy of a molecule.

Thus, the order of stability is ethene < buta-1,3-diene < hexa-1,3,5-triene ≪ benzene.

(From Kshitij IIT JEE Online Coaching)

Benzene is different because it is a cyclic conjugated molecule.

In a cyclic conjugated molecule, each energy level above the first occurs in pairs.

Thus, the total energy of the six π electrons in benzene is much less than the energy of three separate ethene molecules or even of a hexa-1,3,5-triene molecule.

A Frost circle reflects the pattern of orbital energies.

You place the ring inside a circle with one of its vertices pointing downwards.

Then you draw a horizontal line though the vertices. This represents the orbital energy levels.

If there isn't one already, you draw a horizontal dashed line through the circle's (and molecule's) centre. This represents the nonbonding energy level.

Qualitatively, a Frost circle explains the $\left(4 n + 2\right)$ rule.

You need 2 electrons to fill the first level and 4 electrons to fill each level above the first.

## What is the difference between an oxyacid and an organic acid? What are examples of each?

Dwight
Featured 2 weeks ago

The difference lies in the structure of the molecules and the manner in which the oxygen and hydrogen atoms are arranged. See below...

#### Explanation:

An oxyacid is one that will contain, in addition to hydrogen and another element (such as nitrogen, sulfur or phosphorus), a number of oxygen atoms.

Examples would include sulfuric acid ${H}_{2} S {O}_{4}$, and also sulfurous acid ${H}_{2} S {O}_{3}$, phosphoris acid ${H}_{3} P {O}_{4}$, nitric acid $H N {O}_{3}$ to name just a few.

An organic acid is one which contains, in its structure, the particular arrangement of atoms called a carboxyl group $- C O O H$

Examples include formic acid $H C O O H$, acetic acid (in vinegar) $C {H}_{3} C O O H$ and many more.

The difference comes in the actual structure of the molecule. In the oxyacid, the oxygen atoms are all bonded to the nitrogen or sulfur, or whatever it happens to be, with hydrogen atoms bonded to one or more of these oxygens.

In a carboxylic acid (the organic variety), a carbon is doubly bonded to one oxygen atom and singly bonded to a second oxygen. This second oxygen has the H atom bonded to it. So, a very particular structure.

"R" just represents the rest of the molecule.

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