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## What is mass spectrometry (MS)?

Truong-Son N.
Featured 5 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 the #C_6H_8# molecule different from benzene with respect to resonance?

Ernest Z.
Featured 5 months ago

Benzene has two important equivalent resonance contributors, while any other ${\text{C"_6"H}}_{8}$ isomers have only one.

#### Explanation:

There are many molecules with the formula ${\text{C"_6"H}}_{8}$, but I suspect that you are thinking of hexa-1,3,5-triene or cyclohexa-1,3-diene

Benzene

Benzene has two equivalent resonance contributors, neither of which involves the separation of charge.

Hexa-1,3,5-triene

Hexa-1,3,5-triene has only one important resonance contributor.

(from www.chemsynthesis.com)

Any other contributor we might draw has a separation of charge. For example,

${\text{CH"_2"=CH-CH=CH-CH=CH"_2 ⟷ stackrel(-)("C")"H"_2"-"stackrel(+)("C")"H-CH=CH-CH=CH}}_{2}$

Cyclohexa-1,3-diene

Cyclohexa-1,3-diene is an "almost"-benzene.

But it is still a "linear" diene.

Like hexa-1,3,5-triene, it has only one important resonance contributor, with any other contributors having separation of charge.

## Why is it important to describe the resonance structures in molecules such as ozone, #O_3#, and the organic molecule benzene, #C_6H_6#?

anor277
Featured 5 months ago

Because it helps to explain and rationalize experiment; if you like it explains reality.

#### Explanation:

Ozone is a bent molecule. That's the experimental fact. A Lewis structure of $O = \stackrel{+}{O} - {O}^{-}$ in which there are 3 regions of electron-density around the central atom explains this geometry. It also explains the trigonal planar geometry of $C {O}_{3}^{2 -}$. The benzene molecule, ${C}_{6} {H}_{6}$, crops up persistently in organic chemistry. It is reactive under certain circusmtances, but not as reactive and not reactive in the same way as an olefin or an alkyne.

Our ideas of resonance can help rationalize the observed reactivity. The idea of $6$ $\pi$ electrons delocalized around a 6-membered ring can be extended to inorganic chemistry. Borazine, ${B}_{3} {N}_{3} {H}_{3}$, a benzene analogue, is isostructural and isoelectronic with benzene, and has similar aromatic chemistry.

## What is the product of the aldol condensation between cyclopentanone and 4-Hydroxybenzaldehyde?

Ernest Z.
Featured 5 months ago

The product depends on which reactant is in excess.

#### Explanation:

The aldol condensation

An aldol condensation is usually a base-catalyzed reaction in which an aldehyde or ketone with α-hydrogens reacts with a carbonyl compound to form
a β-hydroxyaldehyde or β-hydroxyketone, followed by dehydration to give a conjugated enone.

For example

$\text{RCH=O" + "H-CH"_2"COR" stackrelcolor(blue)("OH"^"-"color(white)(m))(→) "RCH(OH)CH"_2"COR}$

$\text{RCH(OH)CH"_2"COR" → "RCH=CHCOR"+ "H"_2"O}$

You can find the mechanism in this Socratic answer.

In this problem, 4-hydroxybenzaldehyde is the aldehyde and cyclopentanone is the ketone with the α-hydrogen.

If there is excess cyclopentanone

There will be only enough 4-hydroxybenzaldehyde to react at $\text{C-2}$ of the cyclopentanone.

The major product will be 2-(4-hydroxybenzylidene)cyclopentanone.

The product is probably a mixture of ($E$) and ($Z$) isomers.

If there is excess aldehyde

Carbon 5 of the cyclopentanone still has α-hydrogens at $\text{C-5}$, so it can undergo further condensation with the excess aldehyde to form 2,5-bis(4-hydroxybenzylidene)cyclopentanone.

The product is probably a mixture of ($E , Z$), ($E , E$), and ($Z , Z$) isomers.

## When is it better to draw Sawhorse projections in comparison with other projections?

Truong-Son N.
Featured 1 month ago

Sawhorse projections generally show when something is antiperiplanar or synperiplanar, more easily than something like a Newman projection or a basic line structure can.

Take ethane as an example.

An antiperiplanar conformation has a ${180}^{\circ}$ dihedral angle, i.e. the atoms of interest across one bond are on opposite sides along the vertical molecular plane.

A synperiplanar conformation has a ${0}^{\circ}$ dihedral angle, i.e. the atoms of interest across one bond are on the same side along the vertical molecular plane.

A sawhorse projection approximates this 3D structure extremely well, and allows one to judge whether an $E 2$ reaction is likely to occur or not (it requires an antiperiplanar conformation).

A Newman projection would not depict the dihedral angle correctly, because one would be viewing the important atoms from the front instead of an aerial view.

Thus a ${180}^{\circ}$ dihedral angle may very well look like a ${120}^{\circ}$ dihedral angle, unless one knows the precise bond lengths of interest to recognize when the bonds are not entirely opposite to each other.

## Why is piperidine a stronger base than morphine?

Ernest Z.
Featured yesterday

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.

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