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2

It depends on your ionization method. After the ionization, the analyte ions are in general sent to an electromagnetic sector of the mass spectrometer for separation by #m"/"z# ratio.

It may help if you look at this overview of mass spectroscopy.


There are many methods, but some common ionization methods are:

  • Electrospray ionization
  • Matrix-Assisted Laser Desorption Ionization
  • Electron Ionization

Often the method is chosen in favor of so-called soft ionization (rather than hard ionization).

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.

ELECTROSPRAY IONIZATION (ESI)

http://www.chm.bris.ac.uk/

Basically...

  1. Analyte solution is passed through the spray needle, which is connected to a counterelectrode with a large potential difference in between them.
  2. It comes out as a charged droplet that evaporates on its way to the source-sampling counterelectrode.
  3. Solvent evaporation stresses the surface tension of the droplet until it breaks into smaller charged droplets (reaching the "Rayleigh" limit). This rapidly repeats until analyte ions are left.

These analyte ions have little residual energy, so this ionization method is soft.

MATRIX-ASSISTED LASER DESORPTION IONIZATION (MALDI)

http://www.chm.bris.ac.uk/

The general process was:

  1. An easily-sublimable solid sample matrix surrounds the analyte molecules, isolating them from each other.
  2. A laser of the appropriate wavelength excites the matrix surface, ejecting clusters in which the analyte molecules are surrounded by the matrix ions.
  3. The (presumably) easily-sublimed matrix ions then vaporize off, leaving gaseous analyte ions for sample analysis.

This is also a soft ionization method, since the laser does not directly ionize the sample itself (which would have overly fragmented some of the sample).

ELECTRON IONIZATION (EI)

http://www.chm.bris.ac.uk/

Sometimes labeled the "classical" method. The general idea is:

  1. Previously-vaporized analyte solution is sent into the electron ionization chamber.
  2. A hot (rhenium or tungsten) filament sends a beam of spiraling electrons towards an electron trap.
  3. What happens most often is that energy is transferred from the electron beam into the analyte gas, fragmenting it into ions.

This is a hard ionization method, since energy is directly transferred from the electron beam and generally little sample consists of intact molecular ions.


After the ionization, the analyte ions are in general sent to an electromagnetic sector of the mass spectrometer for separation by #m"/"z# ratio.

2

Answer:

You could have given us a bit more to go on.........

Explanation:

It is relatively straightforward to assign chirality to carbon compounds. Of course we have to have a depiction of the molecule.........and we ain't got one here. I can give you a few tips.

https://en.wikipedia.org/wiki/Chirality_(chemistry)

ANY carbon centre in a tetrahedral array that has four different substituents can generate a pair of optical isomers based on the geometry of its substituents. The one on the left (as we face it) is the #S# isomer, and the one on the right (as we face it) is the #R# isomer.

How? Well for each depicted stereoisomer, the substituent of least priority, the hydrogen, projects into the page. #Br# takes priority over #Cl# takes priority over #F#, and this orientation proceeds COUNTERCLOCKWISE around the chiral carbon, and hence a sinister geometry. The stereoisomer on the right hand side exhibits the opposite geometry, i.e. the #"R-stereoisomer"#.

Given such a stereoisomer, the interchange of ANY two substituents at one optical centre results in the enantiomer. Interchange again, (and it need not be the original two substituents) and you get the mirror-image of a mirror-image, i.e. the enantiomer of an enantiomer, that is the ORIGINAL isomer.

The big mistake that students tend to make with these problems IS NOT TO USE MOLECULAR MODELS, which are allowed examination materials in every test of organic and inorganic chemistry. It is hard to assess stereochemistry with a model; it is even harder to assess it without a model. And I assure if you go to the office of your organic prof., his or her desk will have a set of molecular models representing molecules in several stages of construction. The prof will have a fiddle with the models as an idea strikes them.

Of course you need to practise how to represent such models on the printed page. Good luck.

2

It can be considered as basically spatial crowding.


Steric hindrance is a kinetic factor that limits the ease to which a nucleophile (electron pair donor) can approach an electrophile (electron pair acceptor).

As an example, consider the reaction seen below, which one might hope is #"S"_N2#:

http://www.masterorganicchemistry.com/

Here, the nucleophile is cyanide (#""^(-):"CN"#) and the electrophile is the central carbon on the alkyl halide (tert-butyl bromide).

We should predict that reaction does not work via an #"S"_N2# mechanism, based on the idea of steric hindrance---the three #"CH"_3# groups are blocking the cyanide from performing its backside-attack.

https://qph.ec.quoracdn.net/

(LEFT: steric hindrance; RIGHT: reduced steric hindrance)

That reduces the ease to which a successful collision can occur between two reactants, and slows down the first step in a given substitution mechanism (making it the rate-limiting step).

Hence, this reaction proceeds more easily as a first-order mechanism (e.g. #"S"_N1# or #E1#), since the #"Br"^(-)# has the time (during the slow step) to come off on its own.

The rate law for this could then be approximated as first-order based on the slow step, since it dominates the extent of the reaction time:

#r(t) ~~ k_1["Br"^(-)]#

(Depending on the choice of solvent, either #"S"_N1# or #E1# could occur.)

2

Answer:

Three drugs that contain a pyrrole ring are Atorvastatin, Tolmetin, and Atorolac.

Explanation:

Pyrrole is a five-membered nitrogen-containing aromatic ring.

www.sigmaaldrich.com

The pyrrole ring is found in many drugs.

Atorvastatin (Lipitor) is a drug known as a statin.

It is often used to control the production of cholesterol in the body.

Here is its structure and its 3D shape.

upload.wikimedia.org

Atorvastatin

Tolmetin (Tolectin) is a nonsteroidal anti-inflammatory drug (NSAID).

It is used primarily to reduce the pain, swelling, and stiffness of arthritis.

upload.wikimedia.org

Tolmetin

Ketorolac (Toradol) is an NSAID that is often used to treat eye pain and the itchiness of seasonal allergies.

upload.wikimedia.org

Ketorolac

2

Answer:

The cis isomer will form 3- and 4-methylcyclopentene in roughly equal amounts. The trans isomer will form 4-methylcyclopentene as the major product.

Explanation:

What type of reaction?

I predict an #"E2"# elimination because:

  • The substrate is a secondary alkyl halide.
  • The substrate has at least one β-hydrogen: possible elimination.
  • The nucleophile (#"OH"^"-"#) is a strong base: possible elimination.
  • There is (presumably) a high base concentration: favouring #"E2"#.
  • Water is a polar protic solvent: favouring elimination.
  • The high temperature favours elimination.

cis-1-Chloro-3-methylcyclopentane

The structure of the substrate is

Cis

Elimination requires a trans arrangement of the β-hydrogen and the leaving group.

We see appropriate β-hydrogens at #"C2"# and #"C5"#.

Elimination of the hydrogen from #"C2"# forms 3-methylcyclopentene.

3-Methyl

Elimination of the hydrogen from #"C5"# forms 4-methylcyclopentene.

4-Methyl

These two isomers would probably be formed in roughly equal amounts.

trans-1-Chloro-3-methylcyclopentane

The structure of the substrate is

Trans

Again, we see trans β-hydrogens at #"C2"# and #"C5"#.

However, elimination will be slower in each case because of steric hindrance by the methyl group.

Elimination of the hydrogen from #"C5"# will be faster than from #"C2"# because of its greater distance from the methyl group.

Thus, the major product will be 4-methylcyclopentene.

2


I would start from the parent name, heptane. The prefix "hept" means #7#, so the main chain contains #7# carbons:

#color(white)(...........)|color(white)(......)|color(white)(......)|color(white)(......)|color(white)(......)|#
#"H"_3"C"-"C"-"C"-"C"-"C"-"C"-"CH"_3#
#color(white)(...........)|color(white)(......)|color(white)(......)|color(white)(......)|color(white)(......)|#

Then I would work from right to left in the name. The #2,2,3-"trimethyl"# suggests that:

  • #"tri" -># three duplicate branching groups
  • #"methyl" -> "H"_3"C"-# groups
  • The first two are on carbon-2 and the third one is on carbon-3.

So far, we fill it in as:

#color(red)(color(white)(..........)"CH"_3color(white)(.)"CH"_3)#
#color(white)(...........)|color(white)(......)|color(white)(......)|color(white)(......)|color(white)(......)|#
#"H"_3"C"-"C"-"C"-"C"-"C"-"C"-"CH"_3#
#color(white)(...........)|color(white)(......)|color(white)(......)|color(white)(......)|color(white)(......)|#
#color(red)(color(white)(..........)"CH"_3)#

Lastly, we proceed to the leftmost part of the name. The #5,5-"diethyl"# suggests that:

  • #"di" -># two duplicate branching groups
  • #"ethyl" -> "H"_3"CCH"_2-# groups
  • These are both on carbon-5.

So we get:

#color(white)(..........)color(white)(..........)color(white)(..........)color(red)("CH"_3)#
#color(white)(..........)color(white)(..........)color(white)(...........)|#
#color(red)(color(white)(..........)"CH"_3color(white)(.)"CH"_3color(white)(.....)"CH"_2)#
#color(white)(...........)|color(white)(......)|color(white)(......)|color(white)(......)|color(white)(......)|#
#"H"_3"C"-"C"-"C"-"C"-"C"-"C"-"CH"_3#
#color(white)(...........)|color(white)(......)|color(white)(......)|color(white)(......)|color(white)(......)|#
#color(red)(color(white)(..........)"CH"_3color(white)(..............)color(red)("CH"_2))#
#color(white)(..........)color(white)(..........)color(white)(...........)|#
#color(white)(..........)color(white)(..........)color(white)(..........)color(red)("CH"_3)#

Lastly, since this is an alkane, the most saturated kind of hydrocarbon (relative to alkenes and alkynes), just fill in the rest of the branches with #"H"#.

#color(white)(..........)color(white)(..........)color(white)(..........)color(red)("CH"_3)#
#color(white)(..........)color(white)(..........)color(white)(...........)|#
#color(red)(color(white)(.........)"CH"_3color(white)(.)"CH"_3color(white)(.)"H"color(white)(...)"CH"_2color(white)(.)"H")#
#color(white)(...........)|color(white)(......)|color(white)(......)|color(white)(......)|color(white)(......)|#
#"H"_3"C"-"C"-"C"-"C"-"C"-"C"-"CH"_3#
#color(white)(...........)|color(white)(......)|color(white)(......)|color(white)(......)|color(white)(......)|#
#color(red)(color(white)(..........)"CH"_3color(white)(.)"H"color(white)(..)"H"color(white)(...)color(red)("CH"_2)color(white)(.)"H")#
#color(white)(..........)color(white)(..........)color(white)(...........)|#
#color(white)(..........)color(white)(..........)color(white)(..........)color(red)("CH"_3)#

And this in reality looks like:

3

It isn't! Phenol is more stable than phenoxide ion.

Consider the equilibrium between the two species:

#"C"_6"H"_5"OH(aq)" +"H"_2"O(l)" ⇌ "C"_6"H"_5"O"^"-""(aq)" + "H"_3"O"^"+""(aq)"#

You would expect two oppositely charged ions to attract and neutralize each other, if possible.

Thus, phenol is more stable than phenoxide ion.

The equilibrium constant, #K_text(a) = 1.6 × 10^"-10"#.

This shows that the position of equilibrium lies far to the left.

The #K_text(a)# value corresponds to a free energy difference of #55.9 color(white)(l) "kJ·mol"^"-1"#.

3

Answer:

Mass spectroscopy is a method to find a molecule weight.

Explanation:

I'm not so familiar, sorry...

[Step 1] Molecules are ionized and crashes into fragments.
[Step2] Then, the ions are accelerated and induced into magnetic field.
[Step3] In the magnetic field, the ions are affected by Lorentz force #F(N)# and their obrit is bended.

#F=q(vxxB)# (#q(C)# the charge of ion: #v(m"/"s)#: velocity of ion, #xx#: cross product, #B(T)#: magnetic flux density).

And of course...
#F=ma# (m(#kg#): mass of ion, #a(m"/"s^2#): acceleration) and you can conclude:
#a=(q(vxxB))/m#

The point is that #color(red)"orbit of lighter ions or fractions are bended"#
#color(red)"more easily than that of hevier ions."#
Acceleration rate #a# is proportional to #q/m#.

[Step4] Sort ions or fractions by #m/q#(i.e. the reciprocal of #q/m#) and record the relative abundance of ions. This is how you can find a mass of the molecule or mass of fragments.

enter image source here
This is a mass spectrum for 1-phenyl-1-butanone(#C_6H_5COC_3H_7#). Cited from http://www2.odn.ne.jp/had26900/topics_&_items2/about_MS1.htm

What you can find from the result…
148: The mass of the whole molecule, and this is exactly the same as #C_6H_5COC_3H_7#.
77: The mass is same as the phenyl group, #C_6H_5-#. This implies that the molecule might have benzene ring.
105: #105-77=28# is the mass of carbonyl group(#-CO-#). Thus we can expect the molecule to have #C_6H_5CO-# group.

4

You'll have to be able to draw these out to name them. It is impossible to name by inspection of their chemical formulas.

  1. I've highlighted the longest hydrocarbon chain, which you would have had to identify. #(a)# is therefore a kind of hexane, while #(b)# is a kind of heptane (hex = six, hept = seven carbons in the main chain).

  2. Now you simply count from each end to determine the lowest possible set of carbon indices, and account for duplicate functional groups using prefixes. Alphabetize them afterwards.

NOTE: the smallest index should be the first point of difference between two sets of numberings.

There are four methyl groups in #(a)#, while there is one methyl and one ethyl group in #(b)#.

Counting from the left:

#(a)# is then denoted #ul"2,2,3,3-tetramethylhexane"#.

#(b)# is then denoted #"3-ethyl-6-methylheptane"#.

Counting from the right:

#(a)# is then denoted #"4,4,5,5-tetramethylhexane"#.

#(b)# is then denoted #ul"5-ethyl-2-methylheptane"#.

The first name for #(a)# is correct since its indices are minimized.

The second name for #(b)# is correct since the first encountered functional group with the lowest index is methyl, but since ethyl comes before it alphabetically, we rearrange these while keeping the indices as they are.

#(b)# is more difficult, so remember that example.

3

Answer:

Try the bromine decolourization test and the Baeyer unsaturation tests.

Explanation:

Test 1. Bromine decolourization test

Add a few drops of a solution of #"Br"_2# in #"CCl"_4# to a solution of each hydrocarbon in #"CCl"_4#.

The butane will not react (left-hand test tube). The but-2-ene will decolourize the bromine solution (right-hand test tube).

anodosedu.gr

#"CH"_3"CH=CHCH"_3 + underbrace("Br"_2)_color(red)("brown") → underbrace("CH"_3"CHBrCHBrCH"_3)_color(red)("colourless")#

Test 2. The Baeyer test

Bubble each gas through a cold, dilute, aqueous solution of #"KMnO"_4#.

www.susanhornbuckle.com

With butane, there will be no colour change (top test tube).

With but-2-ene, the purple colour will disappear and a dark brown precipitate will form (bottom test tube).

#"3CH"_3"CH=CHCH"_3 + underbrace("2MnO"_4^"-")_color(red)("purple") + 4"H"_2"O"#

#→ "3CH"_3"CH(OH)CH(OH)CH"_3 + underbrace("2MnO"_2)_color(red)("dark brown precipitate") + 2"OH"^"-"#