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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


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:

6

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"^"-"#

3

The tropylium anion really has two valence electrons on the negative carbon:

The criterion for aromaticity are:

  • #4n+2# #pi# electrons IN the ring, where #n = 1, 2, 3, . . . #
  • Planar structure
  • Closed ring
  • Electron conjugation all around the ring.

If, however, there are #4n# #pi# electrons instead, but the other three criteria are satisfied, the compound is antiaromatic.

In the case of tropylium anion, the lone pair of electrons is in the ring (i.e. they are electrons in the #pi# system) because there is a hydrogen atom parallel to the ring in an #sp^2# hybridization:

This structure may or may not be planar, but assuming that it is, the lone pair would have to be in the ring, being within a pure #p# orbital perpendicular to the ring, delocalizing #bbpi# electron density throughout the ring:

Under this assumption that it is planar, and the ring is obviously closed, we satisfy three out of four conditions.

However, with #8 xx pi# electrons in the ring,

#4n = 8# #pi# electrons in the #pi# system, with #n = 2#,

and you cannot write #4n + 2 = 8# unless #n# is not an integer.

Therefore, tropylium anion is antiaromatic. However, if it is not planar in reality (that is, if the lone pair is sufficient to push the hydrogen out of the plane), then it is non-aromatic.


On the other hand, tropylium cation is aromatic:

1

Answer:

I predict that the product will be 1-chloro-2-nitrocyclohexane.

Explanation:

We would expect #"HCl"# to add to 1-nitrocyclohexene according to Markovnikov's rule and form 1-chloro-1-nitrocyclohexane:

Mark

However, this mechanism puts a positive charge on the carbon atom next to another positive atom:

Mech

In this case, the addition is more likely to be anti-Markovnikov.

anti-Mark

This avoids two adjacent + charges in the intermediate and forms a secondary carbocation.

I predict the product will be 1-chloro-2-nitrocyclohexane.

3

Answer:

The compound is 3-methyl-2-butanone.

Explanation:

To begin with the IR-abosorption, #1720# #cm^-1# peak shows that this compound has a carbonyl group.
The molecule of #C_5H_10O# has a double bonding in the carbonyl group, and has no #C=C# double bonding.

Then, let's proceed to the NMR spectrum.
[1] A peak near #1.10# ppm corresponds to #-CH_3# group. This is a doublet, so there will be a single proton next to it.
[2] A peak at #2.10# ppm is for the #-(C=O)-CH_3#. There is no proton in its neighbor.
[3] The fact that the peak at #2.50# ppm is a septet tells us there are six protons in the adjacent point. This might be for the #-(C=O)-Ccolor(red)H-(CH_3)_2#.

Therefore, the structure will be
#CH_3-(CO)-CH-(CH_3)_2#. The IUPAC name is 3-methyl-2-butanone.
https://pubchem.ncbi.nlm.nih.gov/compound/3-methyl-2-butanone#section=Top

Let's check the answer:
https://www.chemicalbook.com/SpectrumEN_563-80-4_1HNMR.htm

Here is a chemical shift table.(PDF)
https://staff.aub.edu.lb/~tg02/nmrchart.pdf#search=%27proton+NMR+table%27

5

Answer:

Warning! Long Answer. The compound is propionic anhydride.

Explanation:

Preliminary analysis

You know the formula is #"C"_6"H"_10"O"_3"#.

An alkane with six carbon atoms has the formula #"C"_6"H"_14"#.

The degree of unsaturation #U# is

#U = (14-10)/2 = 4/2 = 2#

Therefore, the compound contains two rings and/or double bonds.

#""^1"H NMR"#

The spectrum has 10 protons and only two peaks. The molecule must have a symmetrical structure.

A peak with 2 neighbours and aone with 3 neighbors corresponds to an ethyl group (#"C"_2"H"_5#, triplet-quartet pattern).

The #"6H:4H"# pattern tells us there are two ethyl groups.

However, I think you have the assignments reversed. The methyl group should have the smaller chemical shift.

andromeda.rutgers.edu

A methyl group is normally at 0.9 ppm. Something is pulling it downfield
to 1.2 ppm.

A methylene group is normally at 1.3 ppm. Something is pulling it downfield
to 2.5 ppm.

We see from the table that a #"CH"_3# next to a #"C=O"# is shifted downfield to 2.2 ppm (a shift of 1.3 ppm).

We could expect a similar 1.3 ppm shift for a #"CH"_2# group; from 1.2 ppm to 2.5 ppm.

This is just where the #"CH"_2# group appears.

We now know that the partial structure is

#"CH"_3"CH"_2"(C=O)"# + #"(O=C)CH"_2"CH"_3#

These fragments add up to #"C"_6"H"_10"O"_2"#.

There is only one #"O"# atom left to insert. It must go in the middle:

#"CH"_3"CH"_2"(C=O)-O-(O=C)CH"_2"CH"_3#

The compound is propionic anhydride.

Confirmation:

1. The compound has two double bonds.

2. Three #""^13"C"# NMR signals tell us there are three different carbon environments,

image.slidesharecdn.com

We should expect to see

  • #"CH"_3color(white)(mm)# at 10 ppm
  • #"CH"_2 color(white)(mm)# at 30 ppm
  • #"(C=O)O"# at 170 ppm

We see peaks at 8 ppm, 28 ppm, and 170 ppm. This is consistent with propionic anhydride.

3. The #""^13"C"# NMR spectrum of propionaldehyde is

www.chemicalbook.com