free radical substitution of propane - an unanswerable cape chemistry question related to unit 2...

ONE CAPE MULTIPLE CHOICE QUESTION WITH NO POSSIBLE ANSWERWITH NO POSSIBLE ANSWER

FINDING THE RATIO OFFINDING THE RATIO OF

1‐CHLOROPROPANE

TO

2‐CHLOROPROPANE2 CHLOROPROPANEWHEN PROPANE REACTS WITH CHLORINE GAS IN THE PRESENCE OF SUNLIGHT

FORFOR

CAPE UNIT 2 CHEMISTRY STUDENTS

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Overview Of Content

1. Using butane to show that probability l d t di t th ti falone does not predict the ratio of monochlorinated products.

2. Discussion of factors which affect ratio of d tproducts.

3 Application of these factors to explain3. Application of these factors to explain experimental results of propane’s hl i ti

2

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Overview Of Content

4. The CAPE question which has no answer.

3Compiled by Denison at Global in Cunupia. 739-2656.http://www.facebook.com/CAPEChemistryLessons

Concept Of Primary And Secondary

A primary carbon is one which is bonded to l th b tonly one other carbon atom.




A secondary carbon is one which is bonded to only two other carbon atoms.y




A secondary carbon is one which is bonded to only two other carbon atoms.y

(Tertiary carbons will not be encountered here and are thus not discussed.)


Primary Carbons In Butane

Primary carbon atoms are circled above in red.


Secondary Carbons In Butane

Secondary carbon atoms are circled above in green.


Primary And Secondary Hydrogens

A i h d i hi h i b d dA primary hydrogen is one which is bonded to a primary carbon atom.

A secondary hydrogen is one which is bonded y y gto a secondary carbon atom.


Primary Hydrogens In Butane

Primary hydrogen atoms are circled above in red.


Secondary Hydrogens In Butane

Secondary hydrogen atoms are circled above in green.


P i C H b d th b tPrimary C‐H bonds are those between a primary H and a primary C.

Secondary C‐H bonds are those between a secondary H and a secondary C.secondary H and a secondary C.


Monochlorination Of Butane

Two different alkyl halides are obtained from th hl i ti f b tthe monochlorination of butane.

Substitution of a hydrogen bonded to one of the terminal carbons produces p1‐chlorobutane, whereas substitution of a hydrogen bonded to one of the internalhydrogen bonded to one of the internal carbons forms 2‐chlorobutane.


The expected (statistical) distribution ofThe expected (statistical) distribution of products is:

60% 1‐chlorobutane

and

40% 2‐chlorobutane14

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Such a ratio is based on the id th tidea that:

i f b ’ h d bSix of butane’s 10 hydrogens can be substituted to form 1‐chlorobutane, whereas

only four can be substituted to form2‐chlorobutane.


This assumes, however, that all of the C‐HThis assumes, however, that all of the C H bonds in butane are equally easy to break.


This is an incorrectThis is an incorrect assumption!assumption!


If this were the case then the relativeIf this were the case then, the relative amounts of the two products would depend

l h b bili f hl i di lonly on the probability of a chlorine radical colliding with a primary hydrogen, compared with its colliding with a secondary hydrogen.


When we carry out the reaction in the laboratory and analyze the product howeverlaboratory and analyze the product, however, we find that it is:

29% 1‐chlorobutane

and

71% 2 chlorobutane71% 2‐chlorobutane.

Therefore, probability alone does not explain the ratio of products.


Because more 2‐chlorobutane is obtained than expected:than expected:



we conclude that it must be easier to substitute a hydrogen atom of a secondary carbon than of a primary carbon.carbon than of a primary carbon.


After experimentally determining the amount of each chlorination product obtainedof each chlorination product obtainedfrom various hydrocarbons, chemists were bl t l d th table to conclude that:


1. At room temperature it is 5.0 times easier for a chlorine radical to abstract afor a chlorine radical to abstract a hydrogen atom from a tertiary

b th f i bcarbon than from a primary carbon,


1. At room temperature it is 5.0 times easier for a chlorine radical to abstract afor a chlorine radical to abstract a hydrogen atom from a tertiary

b th f i bcarbon than from a primary carbon,

and

2. It is 3.8 times easier to abstract a hydrogen atom from a secondary carbonhydrogen atom from a secondary carbon than from a primary carbon.


To determine the relative amounts of productsobtained from radical chlorination ofobtained from radical chlorination ofan alkane,


To determine the relative amounts of productsobtained from radical chlorination ofobtained from radical chlorination ofan alkane,

both probability (the number of hydrogens that can be abstracted that will lead to the formation of the particular product) and p p )reactivity (the relative rate at which a particular hydrogen is abstracted)a particular hydrogen is abstracted)

must be taken into account.


When both factors are considered,



the calculated amounts

of 1‐chlorobutane and 2‐chlorobutane





agree with g





agree with g

the amounts obtained experimentally.the amounts obtained experimentally.


Propane – Primary And Secondary Carbons

There are two primary carbon atoms inThere are two primary carbon atoms in propane, circled in red above.


Propane – Primary And Secondary Carbons

There is one secondary carbon atom inThere is one secondary carbon atom in propane, circled in green above.


Propane – Primary And Secondary Hydrogens

There are SIX primary hydrogen atoms inThere are SIX primary hydrogen atoms in propane, circled in red above.


Propane – Primary And Secondary Hydrogens

There are TWO secondary hydrogen atoms inThere are TWO secondary hydrogen atoms in propane, circled in green above.


Keep the numbers 6 and 2 closeKeep the numbers 6 and 2 close to the front of your mind because ythey will come into play very soon in a calculation similar to the one

done for butane previouslydone for butane previously.


Chlorination Of Propane

When propane undergoes chlorination, ti it diff b t d dreactivity differences between secondary and

primary hydrogens come into play.


Monosubstituted Propanes

Looking only at monochlorinated products, we th t th t 1 hl dsee that there are two: 1‐chloropropane and

2‐chloropropane.


If h i i f i d d C HIf the reactivity of primary and secondary C‐Hbonds were identical, then the statistical ratio of 1‐chloropropane to 2‐chloropropane should be 3:1 because there are 6 primary hydrogen p y y gatoms and 2 secondary hydrogen atoms.


This is not soThis is not so.


The actual ratio of 1‐chloropropane

toto 2‐chloropropanep p

(as determined by experiment)

is 1 : 1.2243Compiled by Denison at Global in Cunupia. 739-2656.

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It can be calculated as follows:It can be calculated as follows:

Relative amount of 1-chloropropane

Number of hydrogens × reactivityNumber of hydrogens × reactivity6 × 1 = 6




Number of hydrogens × reactivity



Number of hydrogens × reactivity2 × 3.8 = 7.6








This gives a ratio of 1 : 1.27








This gives a ratio of 1 : 1.27

Which agrees closely with the experimentally determined ratio of 1 : 1 22

47

determined ratio of 1 : 1.22Compiled by Denison at Global in Cunupia. 739-2656.http://www.facebook.com/CAPEChemistryLessons

h di h liThe preceding theory applies to free radical chlorination at RTP infree radical chlorination at RTP in the presence of sunlight/uv light.p g g


O l h dOnly when temperatures exceed 300 °C do both primary and300 C do both primary and

secondary reactivities both take a yvalue of 1.


The Un‐Answerable CAPE Question

2010 Q4

When propane reacts with Cl2(g) in the presence of sunlight, two possible monosubstituted products, 1‐chloropropane and 2‐chloropropane are formed. What is the ratio of 1‐chloropropane to 2‐chloropropane?

(A) 1 : 2(B) 1 3(B) 1 : 3(C) 2 : 1(D) 3 : 1

*I saw this question in a few places online during the recently concluded examination period and was told about it by students who wrote exams in 2010.


examination period, and was told about it by students who wrote exams in 2010.

As we saw previously, the ratio is 1 : 1.27, so the first number in the ratio must be 1 and the 2‐chloropropane must be present in a larger

amount Therefore C and D are outamount Therefore C and D are out.

2010 Q42010 Q4

Wh t ith Cl i th f li ht t iblWhen propane reacts with Cl2(g) in the presence of sunlight, two possible monosubstituted products, 1‐chloropropane and 2‐chloropropane are formed. What is the ratio of 1‐chloropropane to 2‐chloropropane?

(A) 1 : 2(B) 1 : 3(C) 2 : 1(C) 2 : 1(D) 3 : 1


2010 Q42010 Q4


(A) 1 : 2(B) 1 : 3


What now?

2010 Q42010 Q4


(A) 1 : 2(B) 1 : 3


1 : 3 could never be obtained by rounding 1 : 1 22 or 1 : 1 27 so B is out1 : 1.22 or 1 : 1.27 so B is out.

2010 Q42010 Q4


(A) 1 : 2(B) 1 : 3


2010 Q42010 Q4


(A) 1 : 2


But this can’t be the answer either since 1 : 2 also cannot be obtained byalso cannot be obtained by rounding 1 : 1.22 or 1 : 1.27.

2010 Q42010 Q4


(A) 1 : 2


What do we do?

2010 Q42010 Q4


(A) 1 : 2(B) 1 : 3(C) 2 : 1(C) 2 : 1(D) 3 : 1


Choose D and assume that they wanted you to wrongly use only probability in calculatingto wrongly use only probability in calculating

the answer.2010 Q42010 Q4


(A) 1 : 2(B) 1 : 3(C) 2 : 1(C) 2 : 1(D) 3 : 1


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free radical substitution of propane - an unanswerable cape chemistry question related to unit 2...

Education

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