When 2-methyl-2,5-pentanediol is treated with sulfuric acid, dehydration occurs and 2,2-dimethyltetrahydrofuran is formed. Sugge
st a mechanism for this reaction. Which of the two oxygen atoms is most likely to be eliminated, and why?
1 answer:
Answer:The oxygen present at the tertiary carbon would be eliminated.The suggested mechanism of the reaction can be found in attachment
Explanation:
The Oxygen atom at the tertiary carbon atom would be eliminated because the removal of this oxygen in form of water after the protonation by sulphuric acid would lead to the formation of a stable tertiary carbocation which is vary stable.
The tertiary carbocation is stable on account of inductive effect of the methy groups.
The oxygen atom at the primary carbon would not be eliminated as its elimination would result in a primary carbocation which is unstable in nature,.
The mechanism of the overall reaction is following:
1. In the first step the OH group present at the tertiary carbocation is protonated by sulphuric acid and on account of this protonation the OH group turns into a good leaving group and leaves as (water) H₂O.
2. Once the H₂O molecule is eliminated it leads to the formation of a stable tertiary carbocation.
3. The tertiary carbocation so formed is electrophilic in nature and as there is one more OH group present at the primary carbon which is 3 carbons away . The OH group is weakly nucleophilic in nature and can appreciably attack the carbocation . The attack of OH at the carbocation leads to the formation of a 5-membered ring containing oxygen as heteroatom.
4.The 5-membered ring so formed has Oxygen as hetero atom which is protonated so the protonated oxygen atom is deprotonated using H₂O.
This further leads to the product formation.
Kindly refer the attachment for the complete reaction mechanism:
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Answer:
Explanation:
2NO₂ ⇌ N₂O₄
E/mol·L⁻¹: 0.058 0.012
K_{\text{eq}} = \dfrac{\text{[N$_{2}$O$_{4}$]}}{\text{[NO$_{2}$]$^{2}$}} = \dfrac{0.012}{0.058^{2}} = \mathbf{3.6} \\\\
\text{The $K_{\text{eq}}$ value would be $\boxed{\mathbf{3.6}}$}