Two hydrocarbons, \(\mathrm{CH_3CHC(CH_3)CH_3}\) and \(\mathrm{CH_3CH(CH_3)CH_2CH_3}\), react separately with chlorine in the presence of ultraviolet light.
In each reaction, free-radical substitution occurs.
Which row is correct?
| Identity of the hydrocarbon that can also undergo electrophilic addition | A termination stage of the free-radical substitution of the saturated hydrocarbon | |
|---|---|---|
| A | \(\mathrm{CH_3CHC(CH_3)CH_3}\) | \(\mathrm{CH_3CH(CH_3)CH_2CH_2\bullet+Cl\bullet\rightarrow CH_3CH(CH_3)CH_2CH_2Cl}\) |
| B | \(\mathrm{CH_3CHC(CH_3)CH_3}\) | \(\mathrm{CH_3CHC(CH_3)CH_2\bullet+Cl\bullet\rightarrow CH_3CHC(CH_3)CH_2Cl}\) |
| C | \(\mathrm{CH_3CH(CH_3)CH_2CH_3}\) | \(\mathrm{CH_3CH(CH_3)CH_2CH_2\bullet+Cl\bullet\rightarrow CH_3CH(CH_3)CH_2CH_2Cl}\) |
| D | \(\mathrm{CH_3CH(CH_3)CH_2CH_3}\) | \(\mathrm{CH_3CHC(CH_3)CH_2\bullet+Cl\bullet\rightarrow CH_2CHC(CH_3)CH_2Cl}\) |
Correct Answer: \( \boxed{\mathrm{A}} \)
The hydrocarbon \(\mathrm{CH_3CHC(CH_3)CH_3}\) is an alkene, so it can undergo electrophilic addition reactions.
The other hydrocarbon, \(\mathrm{CH_3CH(CH_3)CH_2CH_3}\), is a saturated alkane and undergoes free-radical substitution.
A termination step occurs when two radicals combine to form a stable molecule, for example:
\(\mathrm{CH_3CH(CH_3)CH_2CH_2\bullet+Cl\bullet\rightarrow CH_3CH(CH_3)CH_2CH_2Cl}\)
Therefore, the correct answer is (A).
Two hydrocarbons, \(\mathrm{R{-}CH_3}\) and \(\mathrm{R'{-}CH_3}\), react separately with bromine in the presence of ultraviolet light. One of these hydrocarbons is unsaturated.
In each case, free radical substitution reactions occur.
\( \mathrm{R} \) is \(\mathrm{CH_3CHC(CH_3)_2}\).
\( \mathrm{R’} \) is \(\mathrm{CH_3CH(CH_3)CH_2}\).
Which row is correct?
| A propagation stage for the saturated hydrocarbon | A termination stage for the unsaturated hydrocarbon | |
|---|---|---|
| A | \(\mathrm{R{-}CH_2^{\bullet}+Br^{\bullet}\rightarrow R{-}CH_2Br}\) | \(\mathrm{R'{-}CH_2^{\bullet}+Br^{\bullet}\rightarrow R'{-}CH_2Br}\) |
| B | \(\mathrm{R{-}CH_2^{\bullet}+Br_2\rightarrow R{-}CH_2Br+Br^{\bullet}}\) | \(\mathrm{R'{-}CH_2^{\bullet}+Br_2\rightarrow R'{-}CH_2Br+Br^{\bullet}}\) |
| C | \(\mathrm{R{-}CH_3+Br^{\bullet}\rightarrow R{-}CH_2^{\bullet}+HBr}\) | \(\mathrm{2R'{-}CH_2^{\bullet}\rightarrow R'{-}CH_2CH_2{-}R’}\) |
| D | \(\mathrm{2R{-}CH_2^{\bullet}\rightarrow R{-}CH_2CH_2{-}R}\) | \(\mathrm{R{-}CH_3+Br^{\bullet}\rightarrow R{-}CH_2^{\bullet}+HBr}\) |
Correct Answer: \( \boxed{\mathrm{C}} \)
In free radical substitution:
The propagation step \(\mathrm{R{-}CH_3+Br^{\bullet}\rightarrow R{-}CH_2^{\bullet}+HBr}\) is correct.
The termination step \(\mathrm{2R'{-}CH_2^{\bullet}\rightarrow R'{-}CH_2CH_2{-}R’}\) correctly shows two radicals combining.
Therefore, the correct answer is (C).
Two steps in the free-radical substitution reaction between methane and chlorine are shown.
Step 1 \( \mathrm{CH_3^{\bullet} + Cl_2 \rightarrow CH_3Cl + Cl^{\bullet}} \)
Step 2 \( \mathrm{CH_3Cl + Cl^{\bullet} \rightarrow CH_2Cl^{\bullet} + HCl} \)
Which statement is correct?
Correct Answer: \( \boxed{\mathrm{D}} \)
In a free-radical substitution mechanism:
In Step 1, the methyl radical reacts with chlorine to produce chloromethane and a chlorine radical, so the radical is regenerated.
In Step 2, the chlorine radical reacts with chloromethane to form another radical, \( \mathrm{CH_2Cl^{\bullet}} \), so the chain continues.
Therefore, both steps are propagation, so the correct answer is (D).
