New answers tagged aromatic-compounds
1
Let me bring clarity to the matter.
All hydrocarbons are divided into saturated and unsaturated. In turn, each of them is divided into aliphatic and cyclic hydrocarbons.
So, alkanes and cycloalkanes refer to saturated hydrocarbons, BUT: alkanes are aliphatic and cycloalkanes are cyclic. Alkenes refer to unsaturated aliphatic hydrocarbons. They are ...
0
What is wrong with the "definitions" that I have?
Yes, you are right. Employing common sense,
All alkenes are
unsaturated
hydrocarbons
the general formula, $\ce{C_nH_{2n}}$
Cycloalkane
is a hydrocarbon
is saturated
has the general formula, $\ce{C_{n}H_{2n}}$
All alkanes are
saturated
hydrocarbons
general formula, $\ce{C_nH_{2n+2}}$
There are ...
-1
The formula $\ce{C_{n}H_{2n+2}}$ is useful only for saturated, aliphatic alkanes. Not for cycloalkanes.
Cycloalkanes belong to the homologous series with general formula $\ce{C_{n}H_{2n}}$.
For example, the following is a straight chain alkane:
Here, the two terminal alkanes have 3 bonded hydrogens and 1 bond with carbon, while the other intermediary ...
0
Elimination does occur, but only to form an intermediate.
As TRC has mentioned above, this involves a benzyne mechanism.
$\ce{NH2+}$ is a very strong base; it is the conjugate base of ammonia, which itself is already a base.
Under such strong conditions, the aromaticity of the benzene ring can be broken with the $\ce{NH2+}$ eliminating a hydrogen from the ...
6
Naphthalene is more reactive towards electrophilic substitution reactions than benzene.
On a quick glance you might think that as 10 pi electrons are delocalized on 10 carbon atoms in case of naphthalene, it should have resonance energy per bond similar to that of benzene and thus making both equally active towards electrophiles. But in practise it is ...
0
I read that Friedel Craft's alkylation of naphthalene by alkyl halides having more than 2 carbons occurs at 2-position. On the other hand, with smaller halides, it occurs at the 1-position.
The above statement from OP clearly indicate that the substitution selection is due to steric reasons. I'd say the above belief is derived by the most of literature ...
2
The 1-isomer is the generally favoured product as it is formed via a less energetic intermediate, whereas the 2-isomer is formed via a more energetic one.
The stabilization of the 1-isomer intermediate is due to the higher extent of resonance. Substitution Reactions of Polynuclear Aromatic Hydrocarbons[1] nicely sums up the point.
Substitution usually ...
18
Does folic acid contain a benzyl or a phenyl group?
This is the question asked in the title. At the first glance to the structure, one would say folic acid consists of phenyl function but not benzyl function because the question did not define what is the phenyl group. In reality, phenyl group $(\ce{Ph})$ is $\ce{C6H5}$. As Poutnik pointed out in the ...
3
Determination of residual solvents is a routine exercise in pharmaceutical industries and benzene is one of them. The technique is called gas chromatography. When you have a very complex mixture, such as sunscreen, one might use a technique called headspace gas chromatography. You would heat a sample in a very small tightly closed vial, at a desired ...
7
This can be explained by comparing the stability of the carbocations formed in Friedel-Crafts acylation and Friedel-Crafts alkylation using chloroacetylchloride. This is because the rate determining step in both reactions is the formation of the carbocation.
As you can see, the acyl carbocation is resonance stabilized, whereas the carbocation formed in the ...
5
No the products shouldn’t be reversed as, to put simply, acylation is preferred over alkylation.
This is a conclusion based on various experimental results and there are many reasons that can be given for the same.
As said in the concluding lines of this article from Chemistry Libretexts,
Friedel-Crafts Acylations offer several synthetic advantages over ...
7
There are three common classes of methods:
Separate the components of the mixture, then detect the amounts of the substances.
Use a fancy method that can identify multiple substances in a mixture (such as training a dog to smell different substances, or nuclear magnetic resonance)
Use a method that detects only your substance of interest, even when in a ...
6
Most likely chromatography. Gas chromatography is often used for detecting volatile compounds. You vaporize the sample at the inlet of a very long, thin tube (the column). The sample then gets pushed along by a carrier gas (the mobile phase). The different components of the mixture interact differently with the stationary phase inside the column and so take ...
9
It is complicated, as this paper from Olah et al. here shows, but this passage offers a possible explanation:
With the reactive nitronium salts, the isomer distribution of the nitration of anisole shows the highest ortho/para ratio (2.7-2.4), reflecting the "early" (i.e., starting aromatic-like) nature of the transition state of highest energy. ...
9
The correct order is in fact $(\bf{d}) \gt (\bf{b}) \gt (\bf{c}) \gt (\bf{a})$, the reason for this is as follows.
$(\bf{a})$ is definitely last in this order of basicity since its lone pair is delocalised by the phenyl ring. Now we need to compare the other three.
We can split this into two parts, a comparison between
$(\bf{d})$ and $(\bf{b})$
$(\bf{b})$ ...
3
We know that effect of neither hyperconjugation nor resonance is observed if the groups are attached in meta position. Hence, stability of structures I and IV will be decided on the basis of inductive effect only. Alkyl groups exert +I effect, and ethyl group exerts greater +I than methyl, hence stabilizing the electron deficient benzylic carbocation more. ...
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