The chlorination of pentane in the presence of UV light gives a mixture of three monochlorinated products.
Predict the ratios in which these monochlorinated products will be formed. Given that a chlorine atom abstracts a hydrogen atom from a secondary carbon about 4.5 times as fast as it abstracts a hydrogen atom from a primary carbon.
MY INITIAL ANALYSIS:
the 3 monochrlorinated products are 1-chloronpentane, 2-chloropentane and 3-chloropentane.
The secondary carbon refers to the C-2 and C-3, which the chlorine atom attacks 4.5 times faster than that on C-1.
So, for every two 1-chloropentane molecules formed, you will get nine 2-chloropentane and 3-chloropentane respectively.
However, there are 6 positions to get 1-chloropentane, 4 positions to get 2-chloropentane and 2 positions to get 3-chloropentane.
Therefore I came to the conclusion of the ratios 2:6:3 respectively for 1-chloropentane:2-chloropentane:3-chloropentane
I have another school of thought which gives me 6:18:9 which I shall not elaborate because I think it is wrong.
And the more I think about this, the more confusing it gets........headache!!!
Answer :
Ratio of no. of hydrogens substitutable to get 1-chloropentane vs 2-chloropentane vs 3-chloropentane = 6 : 4 : 2
Corresponding Reactivity (ie. stability of alkyl radical intermediate) factor = 1 : 4.5 : 4.5
Hence final ratio = 6x1 : 4x4.5 : 2x4.5 = 6:18:9 = 2:6:3
So you're correct. Did you notice that your two alternative answers are actually the same? (when reduced to lowest terms?)
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I've posted on this matter before; see my thread here :
http://www.sgforums.com/forums/2297/topics/320107?page=8#posts-9484328
I'll reproduce it here :
‘A’ Level Qn.
lianamaster asked :
Q1) What relative amounts of 1-bromobutane and 2-bromobutane are actually formed when butane is brominated?
Q2) If all C-H bonds had the same homolytic dissociation energy, what would be the relative amounts of 1-bromobutane and 2-bromobutane produced in the bromination of butane?
My reply / the Solution :
For Q1.
The reactivity (radical intermediate stability) factor depends on temperature, but bromination is a lot more selective than chlorination.
At 125 deg C, a tertiary radical is formed 1600x faster, and a secondary radical 82x faster, compared to primary radical.
You needn’t memorize these values; the exam question will provide the relevant values if they require you to factor this in.
Relative amount of 1-bromobutane :
Number of H atoms substitutable x reactivity (radical stability)
= 6×1 = 6
Percentage yield = 6/(328+6) = 1.80%
Relative amount of 2-bromobutane :
Number of H atoms substitutable x reactivity (radical stability)
= 4×82 = 328
Percentage yield = 6/(328+6) = 98.2%
For Q2.
Assuming all C-H bonds had the same homolytic dissociation energy, in other words, ignoring the reactivity (radical stability) factor, simply count the number of hydrogens substitutable to obtain the two isomeric monobrominated products :
Relative amount of 1-bromobutane :
Number of H atoms substitutable
= 6
Percentage yield = 6/(6+4) = 60%
Relative amount of 2-bromobutane :
Number of H atoms substitutable
= 4
Percentage yield = 4/(6+4) = 40%
youphoria queried :
is q1 an a level qn? i have never seen it before. particularly regarding “reactivity (radical stability)” factor. only qn 2 seen before.
My reply :
It’s actually University level Organic Chemistry.
At ‘A’ levels, the exam question will usually allow you to explain and apply only the simpler factor of number of hydrogens substitutable.
If the exam question at ‘A’ levels requires you to factor in primary, secondary, tertiary radical stability (or equivalently, dissociation energies of primary, secondary, tertiary C-H bonds), the question will clue you in with relevant data (a brief explanation, together with relative rates of free radical substitution involving primary, secondary, tertiary alkyl radical intermediates, eg. 5.0 > 3.8 > 1.0 for chlorination at 298K).
For the 2007 ‘A’ level H2 Chemistry exam, for instance, the question required the candidate to “describe, explain and calculate based on any ONE factor that determines isomeric product distribution of monochlorination of butane” (where the two factors are of course, number of hydrogens substitutable, and stability of alkyl radical intermediates).