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Chapter 1: Problem 10

Optically active 2 -iodobutane on treatment with NaI, loses its optical activity. Explain why.

Short Answer

Expert verified
Based on the step-by-step solution, answer the following question: Explain the loss of optical activity in the reaction between optically active 2-iodobutane and NaI. In the reaction between 2-iodobutane and NaI, the loss of optical activity occurs due to the formation of a racemic mixture, which is a 1:1 mixture of two enantiomers. Despite containing optically active molecules, the racemic mixture does not exhibit optical activity because the rotations due to each enantiomer cancel each other out. This happens because the iodide ion can attack the chiral carbon atom from both the front and back sides, leading to the formation of both R and S configurations of 2-iodide butane. As a result, the overall optical activity of the reaction mixture is lost.

Step by step solution

01

Understand the concept of optical activity

Optical activity is a property of chiral molecules wherein they can rotate the plane of plane-polarized light. Chiral molecules do not possess a plane of symmetry and have at least one stereogenic center. In this case, 2-iodobutane consists of a stereogenic carbon atom bonded to four different groups, making it a chiral compound.
02

Write the given reaction

Write down the reaction between 2-iodobutane and NaI: 2-Iodobutane (R or S) + NaI -> 2-Iodide butane (R or S) + NaI
03

Understand the mechanism of the reaction

The reaction between 2-iodobutane and NaI involves the nucleophilic substitution (SN2) mechanism. In this mechanism, the iodide ion (I-) from NaI attacks the chiral carbon atom of 2-iodobutane and forms a bond. Simultaneously, the carbon-to-iodine bond (C-I) in 2-iodobutane breaks, resulting in the formation of a new compound, and the iodide ion is displaced.
04

Identify the product formation

In SN2 reactions, the nucleophile attacks the chiral carbon atom from the opposite side of the original leaving group. So, if the original compound is of R-configuration, then the iodide ion will attack from the opposite side, leading to a product with an S-configuration, and vice versa.
05

Explain the loss of optical activity

Since the iodide ion, in this case, can attack the chiral carbon atom from both the front and back sides, it results in the formation of both R and S configurations of 2-iodide butane. This process leads to the formation of a racemic mixture, which is a 1:1 mixture of two enantiomers. A racemic mixture, despite containing optically active molecules, does not exhibit optical activity because the rotations due to each enantiomer cancel each other out. Hence, the overall optical activity of the reaction mixture is lost.

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Most popular questions from this chapter

A chemical reaction is one in which old bonds are broken and new bonds are made. During the course of these changes a variety of intermediates are formed before a starting material is converted to final products. Formation of these intermediates depend on several factors like bond energies, kinetics of the reactions etc. In which of the following choices, the correct order is not indicated. (a) \(\mathrm{F}^{-}<\mathrm{Cl}^{-}<\mathrm{Br}^{-}<\mathrm{I}^{-}(\) the order of base strength \()\) (b) \(\mathrm{ClCH}_{2} \underset{\oplus}{\mathrm{C}}\left(\mathrm{CH}_{3}\right)_{2}<\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}<\left(\mathrm{CH}_{3}\right)_{2} \mathrm{COCH}_{3}\) (stability of carbocation) (c) \(\mathrm{O}_{2} \mathrm{~N}-\mathrm{CH}_{2}>\mathrm{CH}_{3}>\mathrm{CH}_{3} \mathrm{CH}_{2}\) (stability of carbanion) (d) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}<\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{3}<\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}\) (stability of alkenes)

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