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The reaction of (+) 2-iodobutane and Nal* in acetone was studied by measuring the rate of incorporation of I* (ki) and the rate of racemisation(kr)
(+) CH3CH(I)CH2CH3 + Nal* → CH3CH(I*)CH2CH3 + Nal
For this reaction, the relationship between kr and ki is
  • a)
    ki = 2kr
  • b)
    ki = (1/2)kr
  • c)
    ki = kr
  • d)
    ki = (1/3)kr
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
The reaction of (+) 2-iodobutane and Nal* in acetone was studied by me...
The given reaction is : (+)CH3CH(I)CH2CH3 + Nal* → CH3CH(I*)CH2CH3 + Nal

This is an example of an SN2 type reaction.
  • SN2 reaction involves the product formation with inversion of configuration. When one molecule of the original substances undergoes inversion, one molecule of product is formed with inversion of configuration.
  • This trier that when one original optically active molecule undergoes inversion, the inverted product and another origins, optically active molecules lose their optical activity due to the occurrence of racemisation i.e. when one molecule is inverted, actually two molecules are racemised.
  • Hence, it can be concluded that the rate of racemisation is double to the rate of inversion. Hence, mathematically, 2ki = kr
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Community Answer
The reaction of (+) 2-iodobutane and Nal* in acetone was studied by me...
The complete reaction equation is:

CH3CH(I)CH2CH3 + Nal* → CH3CH(I)CH(Nal*)CH3 + NaI

The rate of incorporation of I* (ki) is the rate at which the iodine atom from 2-iodobutane is incorporated into the CH3CH(I)CH(Nal*)CH3 product. This can be determined by monitoring the concentration of NaI formed during the reaction.

The rate of racemisation (kr) is the rate at which the product CH3CH(I)CH(Nal*)CH3 undergoes racemisation, i.e. the rate at which the chiral center at the alpha carbon (C*) changes from R to S or vice versa. This can be determined by monitoring the optical rotation of the product using a polarimeter.

Note: The * symbol indicates that the iodine and sodium atoms are part of the same molecule or ion.
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The reaction of (+) 2-iodobutane and Nal* in acetone was studied by measuring the rate of incorporation of I* (ki) and the rate of racemisation(kr)(+) CH3CH(I)CH2CH3 + Nal* → CH3CH(I*)CH2CH3 + NalFor this reaction, the relationship between kr and ki isa)ki = 2krb)ki = (1/2)krc)ki = krd)ki = (1/3)krCorrect answer is option 'B'. Can you explain this answer? for Chemistry 2025 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about The reaction of (+) 2-iodobutane and Nal* in acetone was studied by measuring the rate of incorporation of I* (ki) and the rate of racemisation(kr)(+) CH3CH(I)CH2CH3 + Nal* → CH3CH(I*)CH2CH3 + NalFor this reaction, the relationship between kr and ki isa)ki = 2krb)ki = (1/2)krc)ki = krd)ki = (1/3)krCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Chemistry 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The reaction of (+) 2-iodobutane and Nal* in acetone was studied by measuring the rate of incorporation of I* (ki) and the rate of racemisation(kr)(+) CH3CH(I)CH2CH3 + Nal* → CH3CH(I*)CH2CH3 + NalFor this reaction, the relationship between kr and ki isa)ki = 2krb)ki = (1/2)krc)ki = krd)ki = (1/3)krCorrect answer is option 'B'. Can you explain this answer?.
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