reaction
ralative rate
RCH3 to
RCH2Cl
1
R2CH2 to R2CHCl
7
R3CH to
R3CCl
21
Using this information, and considering the number of hydrogen
atoms of each type
(primary, secondary or tertiary) within the molecule, predict the
relative ratio of the
two possible products J and K from the chlorination of
2-methylpropane. Explain your answer.
CH3
CH3
CH3
CH3 CH CH3
yielding CH3
C Cl CH3 CH
CH3
CH3
CH2Cl
2-methylpropane
J
K
SOLUTION:
ratio J / K = 21 / 1
EXPALNATION:
According to the information given tertiary hydrogen atom
will be repalced in 2-methylpropane to produce J in higher
quantity where as there is only ONE primary replaceable hydrogen in
the minor product K. Hence the ratio would be
J:K = 21 : 1
Is this the right expalnation or am I missing something
UltimaOnline?
This question requires the student to consider both factors in
his/her calculations : number of H atoms substitutable and
stability of alkyl radical intermediates.
To get J :
(number of H atoms substitutable) x (stability of alkyl
radical intermediates)
= 1 x 21 = 21
To get K :
(number of H atoms substitutable) x (stability of alkyl
radical intermediates)
= 9 x 1 = 9
Hence ratio of products J to K = 21 / 9 or 7J :
3K
If Cambridge asks "what factors affect product distribution in
free radical substitution?" the student should qualitatively
explain both factors.
But if Cambridge asks "what is the relative percentages of the
expected product distribution?", then :
If the Cambridge exam question does not provide data
for the relative stabilities of alkyl radical intermediates
(which can be phrased as "relative rates of reaction to generate
primary, secondary and tertiary alkyl halides" as seen in the
original question above), then the student need only give a
quantitative answer based on calculations that only consider
"number of H atoms substitutable".
If the Cambridge exam question does provide data
for the relative stabilities of alkyl radical intermediates
(such as in the original question above), then the student needs to
consider both factors "number of H atoms substitutable" and
"stability of alkyl radical intermediates" by multiplying the
values associated with both factors, to obtain the true product
distribution.
Most schools only teach the simpler factor of "number of H atoms
substitutable", but this is misleading because in practice, both
factors (which you would notice tend to contradict each other :
tertiary alkyl halides have the least no. of H atoms to substitute
away, but yet tertiary alkyl radicals are the most stabilized by
hyperconjugation from the alkyl groups) are equally important and
teaches the student that in Chemistry, just as in Real
Life, it is myopic and does oneself injustice, to
simply consider one side of the story or one particular viewpoint
only; one must consider all sides of the story or all relevant
viewpoints to understand and appreciate the truer, bigger,
more beautiful picture of the Universe and of Life.