0
Active contributors today

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.

How do atoms achieve stability in single covalent bonds?

David Drayer
Featured 6 months ago

Atoms achieve stability in a single covalent bond by sharing valence electrons to create filled electron shells which are stable.

Explanation:

"Co-" means to share. Valence electrons are the outermost electrons in an atom. By sharing the valence electrons, atoms can achieve a filled valence shell which is stable.

Examples:

• Neon has eight valence electrons in its outer shell. Neon is a very stable atom. Neon is part of family called the Noble gases because of their stability these atoms rarely interact with other atoms.
• Oxygen has six valence electrons in its outer shell. Eight electrons are needed to fill the second level to achieve stability. So oxygen needs two more electrons to be stable.
• Hydrogen has one electron in its outer shell, so two electrons are all that are needed in the first period (see helium, number two, a noble gas). Hydrogen needs one more electron to be stable.
• Two hydrogen atoms each share their electron with oxygen, giving oxygen a total of eight electrons, creating a filled shell for oxygen.

The oxygen in turn shares one of its valence electrons with each of the two hydrogens in $\setminus m a t h b f \left(\text{H"_2"O}\right)$ (water), providing each hydrogen with a filled shell. By sharing all the atoms are filled and "happy", like neon and helium.

Why are pi bonds formed?

Ernest Z.
Featured 5 months ago

π bonds are formed because the side-by-side overlap of $p$ orbitals leads to a lower-energy state than if there were no overlap.

Explanation:

Consider the formation of ethylene, ${\text{H"_2"C=CH}}_{2}$.

The $\text{C}$ atoms are $s {p}^{2}$ hybridized.

The $\text{C-C}$ σ bond is formed by the overlap of the $\text{C}$ $s {p}^{2}$ orbitals, and the $\text{C-H}$ σ bonds are formed by the overlap of the $\text{H}$ $1 s$ orbitals with the $\text{C}$ $s {p}^{2}$ orbitals.

This leaves an unhybridized p orbital containing one electron on each $\text{C}$ atom.

When you have two $p$ orbitals on adjacent $\text{C}$ atoms, they can overlap sideways to form a region of electron density that is not directly between the two nuclear centres but which still contributes to bonding.

We call this a π bond.

It forms because the π electrons are as close as they can get to the two nuclei and are therefore in their lowest energy state.

A π bond can only form only after a σ bond has already formed. It is always part of a double or triple bond.

Why do dipoles form?

Abdur R.
Featured 3 months ago

They form due to the movement and attractions of electrons in atoms and molecules.

Explanation:

A dipole is the separation of two opposite charges, or, in this case, partial charges.

To answer your question, we have to distinguish between the different types of dipoles. There are three different types of dipole:

• Permanent
• Oscillating
• Induced

Permanent dipoles exist in molecules with covalent bonding where one atom is more electronegative than the other. The atom which is more electronegative attracts the bonded pair of electrons to it, increasing its electron density. It thus becomes slightly negative ($\delta$ negative). On the other end of the bond, the other atom loses electron density and becomes slightly positive ($\delta$ positive). The molecule now has a permanent dipole.

Oscillating dipoles occur by chance due to the random movement of electrons in atoms. At any point, the electrons in an atom can all be concentrated on one end, reducing the electron density of the other. This causes one end of the atom to become $\delta$ positive and the other to become $\delta$ negative - the atom now has dipoles. At another time, the electrons will be concentrated on the other end, so the dipoles will shift. The dipoles will constantly be shifting due to the random movement of electrons. This is called oscillating dipoles.

Induced dipoles form when a molecule with a permanent or oscillating dipole approaches a non-polar molecule (or the other way around). As the non-polar molecule approaches the polar one, its electrons will be attracted to the $\delta$ positive end of the molecule. Thus, a dipole has been induced into the non-polar molecule.

This sort of dipole can also form when a non-polar molecule approaches an ionic molecule.

What determines carbocation stability?

Ernest Z.
Featured 1 month 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°#

Ortho,para-bromoanisole + #NaNH_2# + Liquid #NH_3# =? How do you predict the product?

Ernest Z.
Featured 2 weeks ago

Here's how I would do it.

Explanation:

These are the reaction conditions for generating benzyne intermediates.

The reaction of ortho-bromoanisole with potassium amide in liquid ammonia (b. p. -33 °C) is extremely rapid.

Step 1. The amide ion attacks the $\text{H}$ atom that is ortho to $\text{C3}$, generating a carbanion.

Step 2. Loss of $\text{Br"^"-}$ to form a benzyne intermediate.

The elimination is by an E2cb pathway.

Step 3. Addition of $\text{NH"_2^"-}$

The strain caused by a triple bond in a benzene ring can be relieved by a nucleophilic addition ($\text{Ad"_"N}$) of $\text{NH"_2^"-}$.

The methoxy group is electron-withdrawing by induction, so the nucleophile will attack $\text{C3}$ to place the carbanion as close as possible to the methoxy group.

Step 4. Protonation of the carbanion.

The product is meta-methoxyaniline.

para-Bromoanisole

The reaction with para-bromoanisole also follows a benzyne mechanism.

The nucleophile can attack either end of the triple bond, and the methoxy group is far enough away that its inductive effects are minimal.

The product is a mixture of para- and meta-methoxyaniline.

Questions
• · 18 minutes ago
• 20 minutes ago · in Dipoles
• · An hour ago · in Enantiomers
• · 3 hours ago · in Mass Spectometry
• · 5 hours ago · in Naming Functional Groups
• 6 hours ago · in Molecules to NMR
• 6 hours ago · in Cis and Trans
• 7 hours ago · in Lewis Dot Diagram
• 8 hours ago · in Dipoles
• · 8 hours ago
• · 9 hours ago · in Naming Aromatics
• 9 hours ago · in Dipoles
• · 10 hours ago · in Dipoles
• 10 hours ago · in Diastereomers