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netineya [11]
3 years ago
11

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?

Chemistry
1 answer:
Romashka [77]3 years ago
5 0

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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A titration of 25.0 mL of a solution of the weak base aniline, C6H5NH2, requires 25.67 mL of 0.175 M HCl to reach the equivalenc
Lorico [155]

Answer:

a. 0.180M of C₆H₅NH₂

b. 0.0887M C₆H₅NH₃⁺

c. pH = 2.83

Explanation:

a. Based in the chemical equation:

C₆H₅NH₂(aq) + HCl(aq) → C₆H₅NH₃⁺(aq) + Cl⁻(aq)

<em>1 mole of aniline reacts per mole of HCl</em>

Moles required to reach equivalence point are:

Moles HCl = 0.02567L ₓ (0.175mol / L) = 4.492x10⁻³ moles HCl = moles C₆H₅NH₂

As the original solution had a volume of 25.0mL = 0.0250L:

4.492x10⁻³ moles C₆H₅NH₂ / 0.0250L = 0.180M of C₆H₅NH₂

b. At equivalence point, moles of C₆H₅NH₃⁺ are equal to initial moles of C₆H₅NH₂, that is 4.492x10⁻³ moles

But now, volume is 25.0mL + 25.67mL = 50.67mL = 0.05067L. Thus, molar concentration of C₆H₅NH₃⁺ is:

[C₆H₅NH₃⁺] = 4.492x10⁻³ moles / 0.05067L = 0.0887M C₆H₅NH₃⁺

c. At equivalence point you have just 0.0887M C₆H₅NH₃⁺ in solution. C₆H₅NH₃⁺ has as equilibrium in water:

C₆H₅NH₃⁺(aq) + H₂O(l) → C₆H₅NH₂ + H₃O⁺

Where Ka = Kw / Kb = 1x10⁻¹⁴ / 4.0x10⁻¹⁰ =

<em>2.5x10⁻⁵ = [C₆H₅NH₂] [H₃O⁺] / [C₆H₅NH₃⁺]</em>

When the system reaches equilibrium, molar concentrations are:

[C₆H₅NH₃⁺] = 0.0887M - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in Ka formula:

2.5x10⁻⁵ = [X] [X] / [0.0887M - X]

2.2175x10⁻⁶ - 2.5x10⁻⁵X = X²

0 = X² + 2.5x10⁻⁵X - 2.2175x10⁻⁶

Solving for X:

X = -0.0015 → False solution. There is no negative concentrations.

X = 0.001477 → Right solution.

As [H₃O⁺] = X, [H₃O⁺] = 0.001477

Knowing pH = -log [H₃O⁺]

pH = -log 0.001477

<h3>pH = 2.83</h3>

7 0
2 years ago
When 50 ml of 1.000x10^-1m pb(no3)2 solution was added to 50 ml of 1.000x10^-1m nai solution?
podryga [215]

Balanced chemical reaction: Pb(NO₃)₂ (aq) + 2NaI(aq) → 2PbI₂(s) + 2NaNO₃(aq).

V(Pb(NO₃)₂) = 50 mL ÷ 1000 mL = 0.05 L, volume of solution.

c(Pb(NO₃)₂) = 0.1 mol/L; concentration of solution.

n(Pb(NO₃)₂) = c(Pb(NO₃)₂) · V(Pb(NO₃)₂).

n(Pb(NO₃)₂) = 0.1 mol/L · 0.05 L.

n(Pb(NO₃)₂) = 0.005 mol.

n(NaI) = c(NaI) · V(NaI).

n(NaI) = 0.1 mol/L · 0.05 L.

n(NaI) = 0.005 mol; amount of substance.

From chemical reaction: n(Pb(NO₃)₂) : n(NaI) = 1 : 2.

n(Pb(NO₃)₂) = 0.005 mol ÷ 2.

n(Pb(NO₃)₂) = 0.0025 mol; number of moles Pb(NO₃)₂ used.

n(NaI) = 0.005 mol; number of moles NaI used.

The limiting reagent is Pb(NO₃)₂.

n(PbI₂) = 0.005 mol.

m(PbI₂) = n(PbI₂) · M(PbI₂).

m(PbI₂) = 0.005 mol · 461 g/mol.

m(PbI₂) = 2.305 g; the theoretical yield of PbI₂.

3 0
3 years ago
Describe the relationship between temperature and kinetic energy.
RSB [31]
Temperature is a measure of the average kinetic energy of the particles in an object.
7 0
3 years ago
Due is 10 min please help
Nonamiya [84]

Answer:

its c hope this helped;)

Explanation:

5 0
2 years ago
The following equation represents the type of reaction called
Eva8 [605]

Answer: C REDUCTION

Explanation:

Guessed after knowing oxidation isn't the answer. Got right

7 0
3 years ago
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