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In diffuse sunlight, chlorine replaces the hydrogen atoms of an alkane one at a time by a free…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
In diffuse sunlight, chlorine replaces the hydrogen atoms of an alkane one at a time by a free radical chain. Hydrogen atoms that sit in equivalent places in a molecule give the same monochloro product. The figure shows the three alkanes of formula , labelled I to III. In II, the five carbon atoms carry the letters a to e.
In one experiment, a hydrogen atom on a tertiary carbon is replaced times as fast as one on a primary carbon, and a hydrogen atom on a secondary carbon times as fast. Take the amount of each product to be proportional to the number of hydrogen atoms that give it, times their relative rate. Count structural isomers only and ignore stereoisomers.
- How many monochloro products does I give?
- How many monochloro products does II give?
- How many monochloro products does III give?
- What percentage of the monochloro mixture from II has the chlorine on carbon b?
- What percentage of the same mixture has the chlorine on carbon c?
A good plan is to sort the hydrogen atoms of each alkane into sets first.
The worked solution
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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Plan. Sort the hydrogen atoms into sets that are alike by symmetry. One set gives one product. Then weight each set by its number of hydrogen atoms and its rate.
(a) Pentane, I. The two end carbons are alike, the second and fourth carbons are alike, and the middle carbon stands alone. That gives products.
(b) 2-Methylbutane, II. Carbons a and e are both methyl groups on carbon b, so they are alike. Carbon b has one tertiary hydrogen. Carbon c is a with two secondary hydrogens. Carbon d is a joined to c, not to b, so it is not like a. That gives products.
(c) 2,2-Dimethylpropane, III. All four methyl groups sit on the same central carbon, which has no hydrogen. All twelve hydrogen atoms are alike, so there is product.
(d) and (e) The shares in II. Primary hydrogens on a and e: . Primary hydrogens on d: . Secondary hydrogens on c: . The tertiary hydrogen on b: . The sum is . So the share on b is , and on c it is .
Answer: (a) . (b) . (c) . (d) . (e) .
Check. The hydrogen atoms in the sets of II add up to , all the hydrogens of . The four shares, , , and , add up to .
Watch out. The tertiary hydrogen is replaced fastest, yet its product is not the largest share: there is only one such hydrogen. If you use the rates alone and forget the number of hydrogen atoms, you get for (d).
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