I assume you mean for the
For water, it's a combination band of the symmetric stretch (
This is a Raman spectrum, but this particular vibrational motion is active in both Raman and IR, so this is a suitable proof of the frequency range.
You can see that its range is somewhere around
The major product is 2-ethylcyclopentanone.
The reaction is often called the Stork enamine synthesis.
Step 1. Formation of an enamine.
The double-bonded carbon in the enamine has a partial negative charge, so it can act as a nucleophile and displace the
Step 3. Hydrolysis of the iminium salt
The iminium salt is hydrolyzed to regenerate the ketone and the amine.
Note: The reaction is done this way because direct alkylation of the cyclopentanone enolate gives multiple alkylation, but the enamine stops at one alkylation.
The naming of C is correct.
To answer this kind of question, you should draw the molecular structure. To do this we start with the end of the names. So for A, we start with pentane, which indicates a chain of 5 carbon atoms.
Then we add all the other stuff to this chain. We got a methyl group at carbon atom 3 and an isopropyl group at position 4 at the carbon chain. The same method applies to the other structures
The drawings are shown below.
We can now name this structures using the IUPAC rules. These rules are showed below this answer.
Now have a look at the structures above. We see that with A and B the longest carbon chain wasn't correctly chosen. I have provided the correct names of the structures in the figure. Also, the correct carbon chain is shown in the blue circle.
For the C the naming of the structure was correct.
Simplified IUPAC Rules
For determining the product in these type of questions, we have to look at the functional groups of the reactants. In this case, we have a
Now we add
The carbon atom of the cyclic bromine group is partial positively charged, as indicated in the image above. This is created by hyperconjugation. This carbon atom is attached to 3 other carbon atoms (tertiary carbon atom), which stabilises a charge on the carbon atom more easily. Therefore the electrons from methanol will attack at that carbon atom, 'pushing away' the electrons from the bromine bond.
In the last step, the
And there you have your product!
D-carbohydrates and L-carbohydrates are visually best seen in their straight chain forms.
When numbering, the
Both of these molecules are glucose. The only difference is that they both differ in their absolute configurations at their
Another property of enantiomers is that they are
Note: The D and L in the carbohydrates just refers to the direction the
I suspect the reason involves steric hindrance.
The structure of piperidine is
Piperidine is a weak base with
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.
The structure of morphine is
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
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:
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.
Epoxides are carcinogenic because they disrupt our DNA.
Epoxides are three-membered cyclic ethers.
They are highly strained because the nominal bond angles are 60° instead of 109.5°.
They tend to react with other molecules to open the ring and reduce the strain.
A common pro-carcinogen is benzo[a]pyrene.
It is a constituent of automobile exhaust, smog, cigarette smoke and charcoal-cooked foods.
When the body absorbs benzo[a]pyrene, it tries to make it water-soluble so it can be excreted in the urine or feces.
It converts the benzo[a]pyrene into benzo[a]pyrene-7,8-epoxide.
This is hydrolyzed to the 7,8-dihydrodiol, which is further converted into benzo[a]pyrene-7,8-dihydrodiol-9,10-epoxide.
Once formed, the epoxide can π stack with the bases in DNA where the epoxide group reacts with guanine residues.
This distorts the structure of DNA and causes errors in DNA replication.
If these mutations occur in a gene that encodes a molecule that regulates the production of cells, the result may be cancer.
Here's what I get.
Acetate is the carboxylate ion of acetic acid.
In acetoacetate, an α-hydrogen has been replaced by an aceto or acetyl group,
Acetyl-CoA is Coenzyme A in which the
This is Coenzyme A:
And this is acetyl-CoA:
Acetoacetyl-CoA is Coenzyme A in which the
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
A somewhat tricky example is
Even though it is symmetrical, it can stretch both oxygens in the same direction (as in
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
#bb(tildeomega = 1/(2pic)sqrt(k/mu))#,
#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
#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
This is found to be the strongest peak in its IR spectrum:
which you can estimate to be near
SOLUBILITY IN A NONPOLAR SOLVENT
Nice IR solvents that are commonly used are
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.
Tetrachloromethane or Carbon tetra chloride (
So testing only the inflammability they can easily be distinguished
Lucas' reagent is a solution of anhydrous zinc chloride in concentrated hydrochloric acid. This is used to distinguish between primary,secondary and tertiary alcohol lower molar.
They all produce alkyl halides when treated with this reagent through
In case of 2-methylpropan-2-ol the turbidity appears almost immediately but in case of 1-propanonl turbidity appears after long time.
It is soluble in dilute
It is insoluble in dilute
In carbileamine test the comound under test is treated with chloform and alcoholic
Both the compounds are unsaturated but the second one being aromatic one it is more stable and does not respond to test of un-saturation like dicolorization of red bromine water as done by the first one.