3-Ferrocenyl-4-(4-hydroxyphenyl)-hex-3-ene

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Molecules 2014, 19(7), 10350-10369; doi:10.3390/molecules190710350
http://www.mdpi.com/1420-3049/19/7/10350/htm


3-Ferrocenyl-4-(4-hydroxyphenyl)-hex-3-ene (4) 
Zinc powder (1.82 g, 27.9 mmol), titanium tetrachloride (2.04 mL, 18.6 mmol), propionylferrocene (1.13 g, 4.65 mmol), p-hydroxypropiophenone (0.70 g, 4.65 mmol). After column chromatography followed by preparative HPLC (acetonitrile/water 90/10), and recrystallization from a pentane/diethylether, 4 was obtained as orange crystals (0.50 g, 30% yield) consisting of a mixture of Z and E isomers (63/37). 

1H-NMR (300 MHz, CDCl3): 
δ 0.92 (t, J = 7.5 Hz, 3H, CH3), 
1.24 (t, J = 7.5 Hz, 3H, CH3), 
2.28 and 2.41 (q, J = 7.5 Hz, 2H, CH2), 
2.53 and 2.61 (q, J =7.5 Hz, 2H, CH2), 
3.70 and 4.24 (s, 2H, C5H4), 
3.97 and 4.35 (s, 2H, C5H4), 
4.03 and 4.15 (s, 5H, Cp), 
4.70 and 4.72 (s, 1H, OH), 
6.77 and 6.82 (d, J = 8.5 Hz, 2H, C6H4), 6.90 and 7.00 (d, J = 8.5 Hz, 2H, C6H4). 


13C-NMR (75.4 MHz, CDCl3): 
δ 13.0 and 13.2 (CH3), 
15.3 and 16.0 (CH3), 
26.7 and 28.4 (CH2), 
28.8 and 29.5 (CH2), 
67.8 and 67.9 (2CH, C5H4), 
69.0 and 69.1 (2CH, C5H4), 
69.1 and 69.2 (5CH, Cp), 
77.4 (C, C5H4), 
115.1 and 115.2 (2CH, C6H4), 
130.1 and 130.5 (2CH, C6H4), 
133.3 (C), 137.1 and 137.2 (C), 138.0 (C), 153.8 and 153.9 (C). 

IR (KBr, ν cm−1): 3418 (OH), 3094, 3062, 2929, 2871 (CH2, CH3). 

MS (EI, 70 eV) m/z: 360 [M]+., 331 [M-Et]+, 295 [M-Cp]+, 121 [CpFe]+. MS (CI, CH4) m/z: 361 [M+H]+, 360 [M]+., 295. 

HRMS (ESI, C22H24FeO: [M]+.) calcd: 360.11765, found: 360.11711.






http://www.mdpi.com/1420-3049/19/7/10350/htm

Atypical McMurry Cross-Coupling Reactions Leading to a New Series of Potent Antiproliferative Compounds Bearing the Key [Ferrocenyl-Ene-Phenol] Motif

Pascal Pigeon 1,2, Meral Görmen 1, Konrad Kowalski 3, Helge Müller-Bunz 4, Michael J. McGlinchey 4, Siden Top 2,* and Gérard Jaouen 1,2,*

1

Chimie ParisTech, 11 rue Pierre et Marie Curie, Paris F75231 Paris cedex 05, France; E-Mails:pascal.pigeon@chimie-paristech.fr (P.P.); meralgormen@gmail.com (M.G.)

2

Sorbonne Universités, UPMC univ Paris 6, Institut Parisien de Chimie Moléculaire (IPCM) - UMR 8232, 4 place Jussieu, 75252 Paris Cedex 05, France

3

Faculty of Chemistry, Department of Organic Chemistry, University of Łódź, Tamka 12, 91-403 Łódź, Poland; E-Mail: konkow@chemia.uni.lodz.pl

4

UCD School of Chemistry and Chemical Biology, University College Dublin, Belfield, Dublin 4 Ireland; E-Mails: helge.muellerbunz@ucd.ie (H.M.-B.); michael.mcglinchey@ucd.ie (M.J.M.)

*

Authors to whom correspondence should be addressed; E-Mails: siden.top@upmc.fr (S.T.);gerard.jaouen@upmc.fr (G.J.); Tel.: +33-1-44-27-66-99 (S.T.); +33-1-43-26-95-55 (G.J.); Fax: +33-1-43-26-00-61 (S.T.); +33-1-43-26-00-61 (G.J.).

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How the Mass Spectrometer Works

This page describes how a mass spectrum is produced using a mass spectrometer.

How a mass spectrometer works

The basic principle

If something is moving and you subject it to a sideways force, instead of moving in a straight line, it will move in a curve - deflected out of its original path by the sideways force.

Suppose you had a cannonball traveling past you and you wanted to deflect it as it went by you. All you've got is a jet of water from a hose-pipe that you can squirt at it. Frankly, its not going to make a lot of difference! Because the cannonball is so heavy, it will hardly be deflected at all from its original course.

But suppose instead, you tried to deflect a table tennis ball traveling at the same speed as the cannonball using the same jet of water. Because this ball is so light, you will get a huge deflection.

The amount of deflection you will get for a given sideways force depends on the mass of the ball. If you knew the speed of the ball and the size of the force, you could calculate the mass of the ball if you knew what sort of curved path it was deflected through. The less the deflection, the heavier the ball.

You can apply exactly the same principle to atomic sized particles.

An outline of what happens in a mass spectrometer

Atoms can be deflected by magnetic fields - provided the atom is first turned into an ion. Electrically charged particles are affected by a magnetic field although electrically neutral ones aren't.

The sequence is :

• Stage 1: Ionization: The atom is ionised by knocking one or more electrons off to give a positive ion. This is true even for things which you would normally expect to form negative ions (chlorine, for example) or never form ions at all (argon, for example). Mass spectrometers always work with positive ions.
• Stage 2: Acceleration: The ions are accelerated so that they all have the same kinetic energy.
• Stage 3: Deflection: The ions are then deflected by a magnetic field according to their masses. The lighter they are, the more they are deflected. The amount of deflection also depends on the number of positive charges on the ion - in other words, on how many electrons were knocked off in the first stage. The more the ion is charged, the more it gets deflected.
• Stage 4: Detection: The beam of ions passing through the machine is detected electrically.

A full diagram of a mass spectrometer

Understanding what's going on

The need for a vacuum

It's important that the ions produced in the ionization chamber have a free run through the machine without hitting air molecules.

Ionization

The vaporized sample passes into the ionization chamber. The electrically heated metal coil gives off electrons which are attracted to the electron trap which is a positively charged plate.

The particles in the sample (atoms or molecules) are therefore bombarded with a stream of electrons, and some of the collisions are energetic enough to knock one or more electrons out of the sample particles to make positive ions.

Most of the positive ions formed will carry a charge of +1 because it is much more difficult to remove further electrons from an already positive ion. These positive ions are persuaded out into the rest of the machine by the ion repeller which is another metal plate carrying a slight positive charge.

Acceleration

The positive ions are repelled away from the very positive ionization chamber and pass through three slits, the final one of which is at 0 volts. The middle slit carries some intermediate voltage. All the ions are accelerated into a finely focused beam.

Deflection

Different ions are deflected by the magnetic field by different amounts. The amount of deflection depends on:

• the mass of the ion. Lighter ions are deflected more than heavier ones.
• the charge on the ion. Ions with 2 (or more) positive charges are deflected more than ones with only 1 positive charge.

These two factors are combined into the mass/charge ratio. Mass/charge ratio is given the symbol m/z (or sometimes m/e).

For example, if an ion had a mass of 28 and a charge of 1+, its mass/charge ratio would be 28. An ion with a mass of 56 and a charge of 2+ would also have a mass/charge ratio of 28.

In the last diagram, ion stream A is most deflected - it will contain ions with the smallest mass/charge ratio. Ion stream C is the least deflected - it contains ions with the greatest mass/charge ratio.

It makes it simpler to talk about this if we assume that the charge on all the ions is 1+. Most of the ions passing through the mass spectrometer will have a charge of 1+, so that the mass/charge ratio will be the same as the mass of the ion.

Assuming 1+ ions, stream A has the lightest ions, stream B the next lightest and stream C the heaviest. Lighter ions are going to be more deflected than heavy ones.

Detection

Only ion stream B makes it right through the machine to the ion detector. The other ions collide with the walls where they will pick up electrons and be neutralised. Eventually, they get removed from the mass spectrometer by the vacuum pump.

When an ion hits the metal box, its charge is neutralised by an electron jumping from the metal on to the ion (right hand diagram). That leaves a space amongst the electrons in the metal, and the electrons in the wire shuffle along to fill it.

A flow of electrons in the wire is detected as an electric current which can be amplified and recorded. The more ions arriving, the greater the current.

Detecting the other ions

How might the other ions be detected - those in streams A and C which have been lost in the machine?

Remember that stream A was most deflected - it has the smallest value of m/z (the lightest ions if the charge is 1+). To bring them on to the detector, you would need to deflect them less - by using a smaller magnetic field (a smaller sideways force).

To bring those with a larger m/z value (the heavier ions if the charge is +1) on to the detector you would have to deflect them more by using a larger magnetic field.

If you vary the magnetic field, you can bring each ion stream in turn on to the detector to produce a current which is proportional to the number of ions arriving. The mass of each ion being detected is related to the size of the magnetic field used to bring it on to the detector. The machine can be calibrated to record current (which is a measure of the number of ions) against m/z directly. The mass is measured on the ¹²C scale.

What the mass spectrometer output looks like

The output from the chart recorder is usually simplified into a "stick diagram". This shows the relative current produced by ions of varying mass/charge ratio. The stick diagram for molybdenum looks like this:

You may find diagrams in which the vertical axis is labeled as either "relative abundance" or "relative intensity". Whichever is used, it means the same thing. The vertical scale is related to the current received by the chart recorder - and so to the number of ions arriving at the detector: the greater the current, the more abundant the ion.

As you will see from the diagram, the commonest ion has a mass/charge ratio of 98. Other ions have mass/charge ratios of 92, 94, 95, 96, 97 and 100. That means that molybdenum consists of 7 different isotopes. Assuming that the ions all have a charge of 1+, that means that the masses of the 7 isotopes on the carbon-12 scale are 92, 94, 95, 96, 97, 98 and 100.

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Molecular Ion and Nitrogen

Molecular Weight: Even or Odd?

There is one other element that can be detected easily just by looking at the molecular ion: nitrogen. Usually if there is a nitrogen present in the molecule, the molecular weight is odd, as you can see in the mass spectrum of triethylamine. That isn't true with other compounds.

• Molecular weights of organic compounds are almost always even.
• Odd molecular weights result when there is nitrogen in the compound.

Figure MS6. Mass spectrum of triethylamine. Source: SDBSWeb : http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology of Japan, 22 August 2008)

This phenomenon is a result of the fact that the most common elements in organic compounds, carbon and oxygen, have even atomic weights (12 and 16, respectively), so any number of carbons and oxygens will have even weights. The most common isotope of hydrogen has an odd molecular weight, but because carbon and oxygen both have even valences (carbon forms four bonds and oxygen forms two), there is always an even number of hydrogen atoms in an organic compound containing those elements, so they also add up to an even numbered weight.

Nitrogen has an even atomic weight (14), so any number of nitrogen atoms will add up to an even molecular weight. Nitrogen, however, has an odd valence (it forms three bonds), and as a result there will be an odd number of hydrogens in a nitrogenous compound, and the molecular weight will be odd because of the presence of an extra hydrogen.

Of course, if there are two nitrogens in a molecule, there will be two extra hydrogens, so the molecular weight will actually be even. That means the rule about molecular weight and nitrogen should really be expressed as:

• odd numbers of nitrogen atoms in a molecule in an odd molecular weight.

What about those other atoms that sometimes show up in organic chemistry, such as the halogens? Halogens all have odd molecular weights (19 amu for fluorine, 35 or 37 for chlorine, 79 or 81 for bromine, and X for iodine). However, halogens all have a valence of 1, just like hydrogen. As a result, to add a halogen to methane, we would need to erase one of the hydrogen atoms and replace it with the halogen. Since we are just substituting one odd numbered atomic weight for another, the total weight remains even.

Problem MS6. 

Calculate molecular weights for the following compounds.

The M+1 Peak

This page explains how the M+1 peak in a mass spectrum can be used to estimate the number of carbon atoms in an organic compound.

What causes the M+1 peak?

If you had a complete (rather than a simplified) mass spectrum, you will find a small line 1 m/z unit to the right of the main molecular ion peak. This small peak is called the M+1 peak.

The carbon-13 isotope

The M+1 peak is caused by the presence of the ¹³C isotope in the molecule. ¹³C is a stable isotope of carbon - don't confuse it with the ¹⁴C isotope which is radioactive. Carbon-13 makes up 1.11% of all carbon atoms.

If you had a simple compound like methane, CH₄, approximately 1 in every 100 of these molecules will contain carbon-13 rather than the more common carbon-12. That means that 1 in every 100 of the molecules will have a mass of 17 (13 + 4) rather than 16 (12 + 4).

The mass spectrum will therefore have a line corresponding to the molecular ion [¹³CH₄]⁺ as well as [¹²CH₄]⁺.

The line at m/z = 17 will be much smaller than the line at m/z = 16 because the carbon-13 isotope is much less common. Statistically you will have a ratio of approximately 1 of the heavier ions to every 99 of the lighter ones. That's why the M+1 peak is much smaller than the M+ peak.

Using the M+1 peak

What happens when there is more than 1 carbon atom in the compound?

Imagine a compound containing 2 carbon atoms. Either of them has an approximately 1 in 100 chance of being ¹³C.

There's therefore a 2 in 100 chance of the molecule as a whole containing one ¹³C atom rather than a ¹²C atom - which leaves a 98 in 100 chance of both atoms being ¹²C. That means that the ratio of the height of the M+1 peak to the M+ peak will be approximately 2 : 98. That's pretty close to having an M+1 peak approximately 2% of the height of the M+ peak.

Using the relative peak heights to predict the number of carbon atoms

If there are small numbers of carbon atoms

If you measure the peak height of the M+1 peak as a percentage of the peak height of the M+ peak, that gives you the number of carbon atoms in the compound. We've just seen that a compound with 2 carbons will have an M+1 peak approximately 2% of the height of the M+ peak. Similarly, you could show that a compound with 3 carbons will have the M+1 peak at about 3% of the height of the M+ peak.

With larger numbers of carbon atoms

The approximations we are making won't hold with more than 2 or 3 carbons. The proportion of carbon atoms which are ¹³C isn't 1% - it's 1.11%. And the appoximation that a ratio of 2 : 98 is about 2% doesn't hold as the small number increases.

Consider a molecule with 5 carbons in it. You could work out that 5.55 (5 x 1.11) molecules will contain 1 ¹³C to every 94.45 (100 - 5.55) which contain only ¹²C atoms. If you convert that to how tall the M+1 peak is as a percentage of the M+ peak, you get an answer of 5.9% (5.55/94.45 x 100). That's close enough to 6% that you might assume wrongly that there are 6 carbon atoms.

Above 3 carbon atoms, then, you shouldn't really be making the approximation that the height of the M+1 peak as a percentage of the height of the M+ peak tells you the number of carbons - you will need to do some fiddly sums!