Answer:
Carbon forms the large numbers of compound due to the following reasons
How many valence electrons does Carbonic Acid contain?
1 answer:
You might be interested in
Answer:The product formed on reaction with hydroxide ion as nucleophile is 2R-hexane-2-ol.
The product formed on reaction with water would be a 50:50 mixture of
2S-hexane-2-ol. and 2R-hexane-2-ol.
Explanation:
2S-iodohexane on reactiong with hydroxide ion would undergo SN² substitution reaction that is substitution bimolecular. Hydroxide ion has a negative charge and hence it is a quite good nucleophile .
The rate of a SN² reaction depends on both the substrate and nucleophile . Here the substrate is a secondary carbon center having Iodine as a leaving group.SN² reaction takes place here as hydroxide ion is a good nucleophile and it can attack the secondary carbon center from the back side leading to the formation of 2R-hexane-2-ol.
In a SN² reaction since the the nucleophile attacks from the back-side so the product formation takes place with the inversion of configuration.
When the same substrate S-2-iodohexane undergoes a substitution reaction with water as a nucleophile then the reaction occurs through (SN¹) substitution nucleophilic unimolecular mechanism .
The rate of a SN¹ reaction depends only on the nature of substrate and is independent of the nature of nucleophile.
The SN¹ reaction is a 2 step reaction , in the first step leaving group leaves leading to the formation of a carbocation and once the carbocation is formed then any weaker nucleophile or even solvent molecules can attack leading the formation of products.
In this case a secondary carbocation would be generated in the first step and then water will attack this carbocation to form the product in the second step.
The product formed on using water as a nucleophile would be a racemic mixture of R and S isomers of hexane -2-ol in 50:50 ratio. The two products formed would be 2R-hexane-2-ol and 2S-hexane-2-ol.
Kindly refer the attachment for reaction mechanism and structure of products.
This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:
Substituting in our values, we get
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.