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Taya2010 [7]
3 years ago
10

An alkene with the molecular formula C8H16 undergoes ozonolysis to yield a mixture of (CH3)2C=O and (CH3)3CCHO. The alkene is:

Chemistry
1 answer:
aalyn [17]3 years ago
8 0

Answer:

2,4,4-trimethyl-2-pentene yields mixture of (CH_{3})_{2}C=O and (CH_{3})_{3}CHO

Explanation:

In ozonolysis (hydrolysis step involve a reducing agent such as Zn, Me_{2}S etc.), a pi bond is broken to form ketone/aldehyde.

Ketone is formed from di-substituted side of double bond and aldehyde is formed from mono-substituted side of double bond.

Ozoznolysis involves two consecutive steps : (1) formation of ozonide, (2) hydrolysis of ozonide.

Hydrolysis can be done with/without using reducing agent. Carboxylic acid/carbon dioxide/ketone is produced when hydrolysis is done without using reducing agent.

Here, 2,4,4-trimethyl-2-pentene yields mixture of (CH_{3})_{2}C=O and (CH_{3})_{3}CHO

Reaction steps are shown below.

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A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
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Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

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