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hodyreva [135]
3 years ago
14

Why is it easier to deprotonate alkynes compared to alkanes and alkenes?​

Chemistry
1 answer:
8090 [49]3 years ago
4 0

Answer:

The acetylinic hydrogen atom is more acidic than the olefinic hydrogen atoms or hydrogen atoms attached to alkanes.

Explanation:

The acidity of a terminal alkyne is as a result of the high level of s character in the sp hybridized orbital, which bonds with the s orbital of the hydrogen atom to form a single bond. This very high level of s character in an sp‐hybridized carbon causes the region of overlap the σ bond to shift much closer to the carbon atom. This effect leads to a polarization the bond, causing the hydrogen atom to become slightly positive. This slight positive charge makes the acetylinic hydrogen atom a weak proton, which can be removed by a strong base.

In the case of alkanes and alkenes, the s character in the hybridized carbon bonds is much less than that of the acetylinic bond, resulting in fewer electronegative carbon atoms and a corresponding lesser shift of electron density toward those atoms in the overlap region of the σ bond. The location of the overlap region makes the corresponding hydrogen atoms less electron deficient and thus less acidic than the acetylinic hydrogen. In reality, the hydrogen atoms bonded to alkanes and alkenes can be removed as protons, but much stronger nonaqueous bases are necessary.

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Read 2 more answers
How much heat is required to vaporize 31.5 gg of acetone (C3H6O)(C3H6O) at 25 ∘C∘C? The heat of vaporization for acetone at this
KiRa [710]

Answer:

≅ 16.81 kJ

Explanation:

Given that;

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heat of vaporization for acetone = 31.0 kJ/molkJ/mol.

Number of moles = \frac{mass}{molar mass}

Number of moles of acetone = \frac{31.5}{58.08}

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The heat required to vaporize 31.5 g of acetone can be determined by multiplying the number of moles of acetone with the heat of vaporization of acetone;

Hence;

The heat required to vaporize 31.5 g of acetone = 0.5424 mole × 31.0 kJ/mol

The heat required to vaporize 31.5 g of acetone = 16.8144 kJ

≅ 16.81 kJ

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3 years ago
a 250.0 ml buffer solution is 0.250 M in acetic acid and. 250M in sodium acetate. what is the ph after addition of. 0050 mol of
maw [93]
Hello there.

<span>A 250.0 ml buffer solution is 0.250 M in acetic acid and. 250M in sodium acetate. what is the ph after addition of. 0050 mol of HCL? what is the ph after the addition of. 0050 mol of NaOH?

Part 2 answer: </span><span>pH = 4.67 </span>
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