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Please can someone explain Q5's part 4 in the from the following paper: http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w11_qp_43.pdf
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Chemistry help
- Thread starter bravegirl123
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Here we go!
For J we can see that the reaction is R3CH --> R3CCl
This reaction is given a relative rate of 21.
For K we can see that the reaction is RCH3 -> RCH2Cl
This reaction is given a relative rate of 1.
For J, that reaction can only be done at one place, since we only have one group which can undergo that reaction, so the relative rate stays at 21.
For K, we can see that this reaction can be done on 9 different positions, 3 for each of the hydrogens in a CH3. Since we have 3 CH3´s, you get a total relative rate of 9 (1*3*3). Hence the ratio J/K is 21:9, which you may simplify to 7:3 or to 2.3:1
Hope it helps.
For J we can see that the reaction is R3CH --> R3CCl
This reaction is given a relative rate of 21.
For K we can see that the reaction is RCH3 -> RCH2Cl
This reaction is given a relative rate of 1.
For J, that reaction can only be done at one place, since we only have one group which can undergo that reaction, so the relative rate stays at 21.
For K, we can see that this reaction can be done on 9 different positions, 3 for each of the hydrogens in a CH3. Since we have 3 CH3´s, you get a total relative rate of 9 (1*3*3). Hence the ratio J/K is 21:9, which you may simplify to 7:3 or to 2.3:1
Hope it helps.