Cyclopentadiene is unusually acidic for a hydrocarbon. Why? Cyclopentadiene is aromatic. The conjugate base of cyclopentadiene i
s aromatic. Cyclopentadiene is an unstable diradical. The conjugate base of cyclopentadiene is an unstable diradical.
1 answer:
Answer: The unusual acidity of cyclopentadiene is because of the fact that on losing a proton it can attain aromaticity and be quite stable.
The conjugate base of cyclopentadiene is aromatic.
Explanation:
The acidity of a molecule can be explained in terms of stability of its conjugate base anion.
So greater the stabilization of conjugate base anion greater the acidity of the molecule.
In case of cyclopentadiene as soon as we abstract a proton from the sp³ hybridised carbon atom ,it leads to a formation of a carbanion and hybridisation of the carbon atom changes from sp³ to sp²and hence now the molecular becomes planar. The negative charge on the carbanion is in the p-orbital and hence now this negative charge can be in conjugation with the 4π electrons available in the system and be delcoalized over the whole ring and now there are 6π electrons available.
Intially the number of π eletcrons=4
after proton abstraction number of π eletcrons=6
The rule for aromaticity comes from the Huckels rule that the system must have 4n+2π electrons.
After the deprotonation molecule has 4n+2π electrons with n=1.
This delocalization of the negative charge over the whole ring leads to increment in the π electrons of the molecule and the molecule from being anti-aromatic becomes aromatic.
So the attainment of aromaticity after losing a proton leads to a high acidity of cyclopentadiene.
Kinly find in attachment the structure and electron delocalisation of π electrons.
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Answer:
To prepare 0,2 L of a 0,1M solution of a phosphate buffer to a pH of 7,2 you need to add 1,05 mL of 6M HCl, 2,722 g of K₂HPO₄ and complete 0,2 L with water.
Explanation:
The acid equilibrium of phosphate buffer for a pH of 7,2 is:
H₂PO₄⁻ ⇄ HPO₄²⁻+ H⁺ pka = 6,86
Using Henderson-Hasselbalch formula:
pH = pka + log₁₀
7,2 = 6,86 + log₁₀
2,18776 = <em> (1)</em>
You need to add 0,2L× 0,1M = 0,02 moles of phosphate buffer, that means:
0,02 moles = H₂PO₄⁻ + HPO₄²⁻ <em>(2)</em>
Replacing (2) in (1):
H₂PO₄⁻: <em>6,2739x10⁻³ moles</em>
Thus,
HPO₄²⁻: 0,013726 moles
K₂HPO₄ reacts with HCl thus:
K₂HPO₄ + HCl → KH₂PO₄ + KCl
Thus, you need to add 0,02 moles of K₂HPO₄ that will react with 6,2739x10⁻³ moles of HCl To produce the necessary moles for the buffer:
0,02 moles of K₂HPO₄× = <em>2,722 g</em>
6,2739x10⁻³ moles of HCl÷ 6 M = 1,05x10⁻³ L ≡<em> 1,05 mL of 6M HCl</em>
Thus, to prepare 0,2 L of a 0,1M solution of a phosphate buffer to a pH of 7,2 you need to add 1,05 mL of 6M HCl, 2,722 g of K₂HPO₄ and complete 0,2 L with water.