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Answer:

Carbocations are generated when a carbon atom in a compound does not have a full octet, causing it to gain a positive charge.

Explanation:

Carbocations are generated when one of the bonds previously shared by a carbon atom is broken (such as to a hydrogen or halogen atom), leaving it with an incomplete octet. Carbon needs eight valence electrons to have a full octet, and adopts a #+1# positive charge when it shares only six.

Consider this #S_N1# mechanism for the reaction between ( R )-2-idodopentane and methanol.

enter image source here

The leaving group (iodine) leaves with its bonding electrons, resulting in a carbocation as carbon now shares only six valence electrons. The electronegative oxygen atom of methanol will then attack the electrophilic carbocation, attracted to its positive charge (methanol is the nucleophile).

enter image source here

I've chosen not to show stereochemistry in the product, as both stereoisomers (inversion and retention of original) will be present in the product mixture. Also note that another methanol molecule will deprotonate the methanol which initially attacks the carbocation, which is why we end up with a neutral #O# atom without any hydrogen.

The formation of a carbocation can allow for rearrangement to take place in reactions, resulting in more stable products. In this case, the most stable product by #S_N1# mechanism has already been formed (#2^o#).

Though this is an important concept discussed in organic chemistry and relevant to many types of reactions, I have used only the #S_N1# mechanism as an example. This is by no means the only reaction mechanism that the formation of a carbocation is relevant to (unimolecular elimination, addition, etc.).

Answer:

Here's one way to do it.

Explanation:

Draw the Fischer projection of your sugar, e.g. glucose.

4.bp.blogspot.com

Now, rotate the Fischer projection 90° clockwise.

Rotated
(Adapted from http://4.bp.blogspot.com)

Next, draw a template for a Haworth structure.

Template

Look at the #"OH"# groups on carbons, 2, 3, and 4 of the Fischer projection.

Those on the right will be "down" in the Haworth structure. Those on the left will be "up".

If the sugar is a D-sugar, the #"CH"_2"OH"# group will be "up".

The OH on carbon 1 can be "down" for the α-anomer or "up" for the β-anomer.

And you have the Haworth structure of glucose.

upload.wikimedia.org

Answer:

Here are some of the rules for naming alkynes.

Explanation:

Alkynes are unsaturated hydrocarbons containing a triple bond, #"R-C≡C-R"#.

You name them the same way as you name alkanes, but you replace the ending -ane with -yne.

When there is more than one triple bond, you use the multiplying prefixes to
get -diyne, -triyne, etc.

Step 1. Find the longest continuous chain of carbons containing the triple bond.

For example, #"CH"_3"CH"_2"CH"_2"C≡CCH"_3# has six carbon atoms, so its base name is hexyne.

Step 2. Number the carbon atoms in the main chain from the end closer to the triple bond.

#stackrelcolor(blue)(6)("C")"H"_3stackrelcolor(blue)(5)("C")"H"_2stackrelcolor(blue)(4)("C")"H"_2stackrelcolor(blue)(3)("C")≡stackrelcolor(blue)(2)("C")stackrelcolor(blue)(1)("C")"H"_3#

Insert the first number of the alkyne carbons as close as possible before the
ending -yne.

The name becomes hex-1-yne.

Step 3. If there are other substituents, list them alphabetically with their locating numbers.

#"H"stackrelcolor(blue)(1)("C")"≡"stackrelcolor(blue)(2)("C")stackrelcolor(blue)(3)("C")"H"("CH"_3)stackrelcolor(blue)(4)("C")"≡"stackrelcolor(blue)(5)("C")stackrelcolor(blue)(6)("C")"H"_3#

For example, the compound above is 3-methylhexa-1,4-diyne.

Answer:

The difference lies in the structure of the molecules and the manner in which the oxygen and hydrogen atoms are arranged. See below...

Explanation:

An oxyacid is one that will contain, in addition to hydrogen and another element (such as nitrogen, sulfur or phosphorus), a number of oxygen atoms.

Examples would include sulfuric acid #H_2SO_4#, and also sulfurous acid #H_2SO_3#, phosphoris acid #H_3PO_4#, nitric acid #HNO_3# to name just a few.

An organic acid is one which contains, in its structure, the particular arrangement of atoms called a carboxyl group #-COOH#

Examples include formic acid #HCOOH#, acetic acid (in vinegar) #CH_3COOH# and many more.

The difference comes in the actual structure of the molecule. In the oxyacid, the oxygen atoms are all bonded to the nitrogen or sulfur, or whatever it happens to be, with hydrogen atoms bonded to one or more of these oxygens.

In a carboxylic acid (the organic variety), a carbon is doubly bonded to one oxygen atom and singly bonded to a second oxygen. This second oxygen has the H atom bonded to it. So, a very particular structure.

enter image source here

"R" just represents the rest of the molecule.

Answer:

I suspect the reason involves steric hindrance.

Explanation:

The structure of piperidine is

upload.wikimedia.org

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

ball and stick

and the space-filling model.

Space filling

Morphine

The structure of morphine is

Morphine
(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.

ball and stick

This becomes even clearer in a space-filling model of morphine.

Space-filling

Attack by an acid is strongly hindered on one side by Ring A.

We find basicities by measuring the position of an equilibrium:

#"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.

You are right that it has to have a dipole (not necessarily a net dipole). I'm not sure what other specific requirements you may be wondering about, but I can list several, and maybe one of them is what you are looking for.


REQUIREMENTS FOR VISIBILITY IN THE IR REGION

The main properties of a molecule that allow it to be analyzable via IR spectroscopy are:

#1)# It can experience a change in dipole moment, whether it is induced or permanent. We then must have that it is heteronuclear, if it is diatomic (therefore, #"N"_2#, #"F"_2#, etc. are invisible in the IR).

#2)# It should have resonant frequencies that are in the infrared frequency range of #100 - 4000# #"cm"^(-1)#.

#3)# Ideally, it should be not overly soluble in a nonpolar solvent, which is ideal since many analyses are carried out using a solvent such as #"CS"_2# or #"CCl"_4#. If something is too soluble, one may saturate or overload the spectrometer.

CHANGE IN DIPOLE MOMENT

Requirement #(1)# is usually easy to check; if you can imagine the molecule stretching or bending in a way that enforces asymmetry, you've more than likely changed its dipole moment, and thus that kind of vibrational motion is IR-active.

A somewhat tricky example is #"CO"_2#.

http://wag.caltech.edu/

Even though it is symmetrical, it can stretch both oxygens in the same direction (as in #B#) to generate a nonzero dipole moment along the molecular axis. It can also bend its #"O"-"C"-"O"# angle (as in #C# and #D#) so that it generates a nonzero dipole moment through the oxygen.

RESONANT FREQUENCIES IN THE IR RANGE

This is usually satisfied automatically, just due to the general strengths of chemical bonds, but it couldn't hurt to check. The fundamental frequency can be estimated using the force constant #k# of the bond.

#bb(tildeomega = 1/(2pic)sqrt(k/mu))#,

where:

  • #tildeomega# is the fundamental frequency in #"cm"^(-1)#.
  • #c# is the speed of light, #2.998 xx 10^(10) "cm/s"#.
  • #k# is the force constant in #"kg/s"^2#.
  • #mu = (m_1m_2)/(m_1 + m_2)# is the reduced mass of the two atoms in the bond, where each mass is in #"kg"#.

As an example, I randomly found on a quick Google search that #k ~~ "2385 N/m"# for #"CO"_2#. So:

#tildeomega = 1/(2pi*2.998 xx 10^(10) "cm/s") sqrt(("2385 kg/s"^2)/((12.011*15.999)/(12.011 + 15.999) xx 10^(-3) "kg"/"mol" xx "mol"/(6.0221413 xx 10^(23) "molecules"))#

#~~ "2429 cm"^(-1)#

and the literature value for the fundamental frequency is #"2349 cm"^(-1)# from NIST.

This is found to be the strongest peak in its IR spectrum:

http://webbook.nist.gov/

which you can estimate to be near #"2350 cm"^(-1)#, in agreement with the literature value.

SOLUBILITY IN A NONPOLAR SOLVENT

Nice IR solvents that are commonly used are #"CCl"_4# and #"CS"_2#. Since both are nonpolar, you would have the most "control" over the IR intensity if you had a reasonably soluble analyte. This is why IR tends to be suitable for organic molecules.

This kind of consideration is not necessary, but being too soluble may make it frustrating to get a spectrum without overloading the spectrometer. Being "somewhat" soluble in nonpolar solvents is probably fine.

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