For the monochlorination of hydrocarbons, the following ratio of reactivities has been found: \(3^{\circ}>2^{\circ}>\) \(1^{\circ}, \quad 4.3: 3: 1 . \quad\) How many different monochloro derivatives of 2 -methylbutane are possible, and what percentage of each would you expect to find?

Short Answer

Expert verified
There are 5 different possible monochloro derivatives of 2 -methylbutane. The expected percentages are approximately 46.63% for monochlorination on \(\,^1H\), 31.09% on \(\,^2H\), and 22.28% on \(\,^3H\).

Step by step solution

01

Identify possible positions for Chlorine

There are five potential sites for chlorination of 2-methylbutane. Three of these are \(\,^1H\,\) positions, one is a \(\,^2H\,\) position, and one is a \(\,^3H\,\) position.
02

Consider reactivity ratios and calculate numbers of hydrogen

We're told the reactivity ratios are \(4.3: 3: 1\) for \(\,^3H\), \(\,^2H\), and \(\,^1H\) respectively. Now we can calculate the number of each type of hydrogen: there are 9 \(\,^1H\)'s, 2 \(\,^2H\)'s, and 1 \(\,^3H\).
03

Calculate the total reactivity

The sum of the total reactivities can be calculated as follows: \( (9 × 1) + (2 × 3) + (1 × 4.3) = 9 + 6 + 4.3 = 19.3 .\) This is the sum of all individual reactivities, which we will use to determine the percentages.
04

Determine percentages for each type

Divide each individual reactivity by the total reactivity, then multiply the result by 100% to get the percentage. For \(\,^1H\): \( (9 × 1 / 19.3) × 100% = 46.63%\), for \(\,^2H\): \( (2 × 3 /19.3 ) × 100% = 31.09%\), and for \(\,^3H\): \( (1 × 4.3 / 19.3) × 100% = 22.28% \). Remember to perform each step of the mathematical calculation carefully.
05

Determine the number of different monochloro derivatives and summarize results

The number of different monochloro derivatives of 2-methylbutane is equal to the number of unique sites for chlorination, which is 5. Review the calculations to confirm the percentages for each derivative.

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Most popular questions from this chapter

Predict the products of the following reactions. Assume that \([\mathrm{O}]\) represents \(\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4}:\) (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{OH} \stackrel{[\mathrm{O}]}{\longrightarrow}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCOOH} \frac{\mathrm{H}^{+}}{\Delta}\) (c) 3 -methyl- 2 -butanol \(+\mathrm{NaCl} \stackrel{\mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} \frac{\mathrm{H}_{2} \mathrm{SO}_{4}}{\Delta}\)

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