4-hydroxypentanal reacts with one equivalent of methanol to form a cyclic acetal.
1 answer:
Answer:
Explanation:
The equation is given as:
CH3CHOHC2H4CHO + CH3OH --> CYCLIC ACETAL + H2O
This above equation is carried out in the presence of a strong acid. There are five mechanisms employed and they are:
Step 1:
Initial formation of the hemiacetal which takes several steps
Step 2:
Addition of a proton. The hemicetal is protonated on the hydroxyl group (-OH group)
Step 3:
As seen a bond is broken to give the H2O molecule and a resonance stabilized cation.
The carbonyl group on the cation is enriched with the oxygen-18 got from the H2O molecule as seen in the mechanism.
Step 4:
An attraction occurs between electrophile and nucleophile i.e the stabilised cation and the lone paids of the methanol.
Step 5:
Finally, a proton (+) is removed from the molecule by a lone pair of electron on the methanol.
Attached are the Steps 1 - 5 mechanism below
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Answer:
The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.
Explanation:
..[1]
..[2]
..[3]
..[4]
Using Hess's law:
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
2 × [4] = [2]- (3 ) × [1] - (2) × [3]
The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.
The acid dissociation constant (Ka) defines the difference between a weak and a strong acid. The % ionization of hypochlorous acid is 0.14%.
<h3>What is the acid dissociation constant?</h3>The acid dissociation constant is used to define the ionization constant of an acidic substance. It gives the quantitative measurement of the strength.
The ICE table is attached to the image below.
The acid dissociation constant (Ka) for the reaction is,
Ka = [H⁺][ClO⁻] ÷ [HClO]
= a² ÷ (0.015 - a)
= 3.0 x 10⁻⁸
Now, a² + 3.0 x 10⁻⁸ a - 4.5 × 10⁻¹⁰ = 0
So, a = 2.210 × 10⁻⁵
Solving further,
[H+] = a = 2.210 × 10⁻⁵ M
The percent ionization is calculated as,
[H+] ÷ [HClO] × 100
= 2.210 × 10⁻⁵ M ÷ 0.015 × 100
= 0.14 %
Therefore, 0.14 % is the percentage of hypochlorous ionization.
Learn more about acid dissociation constant here:
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Your question is incomplete, but most probably your full question was, The ka of hypochlorous acid (HClO) is 3.0 x 10⁻⁸ at 25.0°C. What is the % of ionization of hypochlorous acid in a 0.015 aqueous solution of HClO at 25.0C?
