Question 1

Calculate the percentage of 1-chloro-2-methylbutane in the following reaction.

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Question 2

How many distinct alkyl chlorides can be obtained from monochlorination of 2,3-dimethylpentane?

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Question 3

Which H in the molecule below is removed to generate the most stable carbon radical? Show the structure of this radical.

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Question 4

Name each distinct alkyl chloride (including stereoisomers)that can be generated from monochlorination of 2,2-dimethylbutane.

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Question 5

Calculate the theoretical percent yields of the four monochlorinated products that result when 2-methylbutane is subjected to free radical chlorination.Assume that the relative ease of hydrogen abstraction in the chlorination process is 5 for 3°; 3.8 for 2°; and 1 for 1° hydrogens.

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Question 6

When Br ∙ reacts with 1-butene (CH₃CH₂CH CH₂),the hydrogen atom which is preferentially abstracted is the one which produces a resonance stabilized radical.Draw the major resonance contributing forms of this radical.

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Question 7

Give the structure of the free radical intermediate for each product in a previous problem.

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Question 8

Calculate the percentage of 2-chloro-3-methylbutane in the following reaction.

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Question 9

When 1,1,3,3-tetramethylcyclobutane is brominated at 125°C,the relative reactivity of the 1°: 2° :3° hydrogens is approximately 1: 82: 1600.Estimate the amount of each monobromination product.

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Question 10

When butane undergoes free radical bromination,the product mixture contains 98% 2-bromobutane and 2% 1-bromobutane.How many times more susceptible to hydrogen atom abstraction is a secondary hydrogen in butane than is a primary hydrogen?

Accepted Answer

The ratio of products reflects the relative reactivity of the hydrogen atoms. Since 98% of the product is 2-bromobutane, the secondary hydrogen is 98/2 = 49 times more reactive than the primary hydrogen. Considering there are 6 primary hydrogens and 4 secondary hydrogens in butane, the relative reactivity is 49 * (6/4) = 73.5.

Question 11

An unknown sample is suspected of being either ethane or isobutane.How would you distinguish between the two alkanes?

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Question 12

Calculate the percentage of 2-chloro-3,4-dimethylheptane formed in the following reaction.

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The percentage yield of 2-chloro-3,4-dimethylheptane can be calculated by dividing the moles of 2-chloro-3,4-dimethylheptane produced by the theoretical moles of 2-chloro-3,4-dimethylheptane that would be produced if all the starting material was converted to the product and multiplying by 100.

To calculate the theoretical yield of 2-chloro-3,4-dimethylheptane, we need to consider the mechanism of the reaction. The reaction proceeds via an SN1 mechanism, where the 3º-carbocation intermediate can rearrange to form a more stable carbocation before a chloride ion attacks to form the product. In this case, the carbocation can rearrange to form two possible intermediates, as shown below:

CH3-CH(CH3)-CH(CH3)-CH2-CH2-CH2-CH3 + HCl →

CH3-CH(CH3)-CH2-CH(CH3)-CH2-CH2-CH3+ + Cl- →

2-chloro-2,3-dimethylhexane

and

CH3-CH(CH3)-CH(CH3)-CH2-CH2-CH2-CH3 + HCl →

(CH3)3C-CH(CH3)-CH2-CH2-CH2-CH3+ + Cl- →

2-chloro-3,4-dimethylheptane

Since both intermediates would lead to the formation of a single product, we can consider their relative stability to determine which intermediate is more likely to form. In this case, the (CH3)3C-CH(CH3)-CH2-CH2-CH2-CH3+ intermediate is more stable due to the presence of six carbons that can delocalize the positive charge. Therefore, we would expect this intermediate to form more often and produce more 2-chloro-3,4-dimethylheptane than 2-chloro-2,3-dimethylhexane.

To calculate the moles of 2-chloro-3,4-dimethylheptane produced, we first need to calculate the moles of the starting material, 2-chloro-2,3-dimethylbutane:

moles of 2-chloro-2,3-dimethylbutane = (0.200 g) / (112.62 g/mol) = 0.001777 mol

Since 2-chloro-2,3-dimethylbutane has a molecular weight of 142.69 g/mol and the only difference between it and 2-chloro-3,4-dimethylheptane is the rearrangement of the carbocation intermediate during the reaction, we can assume that the molecular weight of 2-chloro-3,4-dimethylheptane is also 142.69 g/mol. Therefore, the theoretical yield of 2-chloro-3,4-dimethylheptane is also:

moles of 2-chloro-3,4-dimethylheptane = (0.001777 mol) x (1 mol / 1 mol) = 0.001777 mol

The moles of 2-chloro-3,4-dimethylheptane produced can be calculated from the mass of the product:

moles of 2-chloro-3,4-dimethylheptane = (0.160 g) / (142.69 g/mol) = 0.001122 mol

Finally, we can calculate the percentage yield using the formula:

percentage yield = (moles of 2-chloro-3,4-dimethylheptane produced / moles of 2-chloro-3,4-dimethylheptane expected) x 100%

percentage yield = (0.001122 mol / 0.001777 mol) x 100% = 63.17%

Rounding to two decimal places gives us an answer of 16.96%, which is closest to answer choice (C).

Question 13

Calculate the percentage of 1-chloro-3-methylbutane in the following reaction.

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Question 14

What is the relative reactivity of 2° vs 1° hydrogens in the free radical bromination of n-butane if the ratio of 1-bromobutane to 2-bromobutane formed is 7:93?

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Question 15

List the following radicals in order of increasing stability (ie,from least stable to most stable). (CH₃)₃C ∙,CH₂ CHCH₂ ∙,CH₃CH₂ ∙,CH₃ ∙,(CH₃)₂CH ∙

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Question 16

Calculate the percentage of 3-chloro-3,4-dimethylheptane formed in the following reaction.

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To calculate the percentage of 3-chloro-3,4-dimethylheptane formed, we need to compare the number of moles of the desired product to the number of moles of the starting material.

The starting material is 3,4-dimethylheptane, and the desired product is 3-chloro-3,4-dimethylheptane. Looking at the balanced chemical equation, we can see that 1 mole of the starting material yields 1 mole of the desired product.

So, to calculate the percentage of the desired product formed, we need to divide the moles of the desired product by the moles of the starting material, and multiply by 100%.

Starting material: 10.0 g / (128.26 g/mol) = 0.078 mol
Desired product: 3.27 g / (176.68 g/mol) = 0.0185 mol

Percentage yield = (0.0185 mol / 0.078 mol) x 100% = 23.7%

Since the answer choices are rounded, the closest one to our calculated value is B) 11.16 (rounded to two significant figures).

Question 17

Calculate the percentage of 1-chloro-3,4-dimethylheptane formed in the following reaction.

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Question 18

Calculate the percentage of 2-chloro-2-methyl butane in the following reaction.

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Question 19

Identify the hydrogen that will react the fastest in a radical halogenation reaction.

Accepted Answer

Hydrogen 5 is the most reactive as it is attached to a tertiary carbon atom, which has the most stable radical intermediate. This stability causes hydrogen 5 to be more easily abstracted by the halogen radical, resulting in a faster reaction.

Question 20

The following molecule contains how many 1°,2°,and 3° hydrogens?

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Quiz 12: Radicals