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Chapter 2: Problem 169

The maximum number of optical isomers produced on monochlorination of 2 -methylbutane will be (a) 2 (b) 3 (c) 4 (d) 1

Short Answer

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Step by step solution

01

Draw the structure of 2-methylbutane

Draw the structure of 2-methylbutane, which is also known as iso-pentane. The compound will have five carbon atoms, with four carbon atoms in a straight chain and one methyl group (-CH3) attached to the second carbon atom.
02

Write the Monochlorination Reaction

Monochlorination of 2-methylbutane involves replacing one hydrogen atom with a chlorine atom in the molecule. We need to consider all the possible ways chlorine could replace a hydrogen atom and form new compounds.
03

Identify Chiral Centers

A chiral center is an atom in a molecule that carries four different atoms or groups, making the molecule optically active. Check for chiral centers in the products obtained in step 2 and count the total number of chiral centers.
04

Determine the number of Optical Isomers

Now that you have identified the possible monochlorinated products and their chiral centers, you can find the number of optical isomers by determining the optically active products. Optical isomers are the isomers with non-superimposable mirror images. The total number of optically active isomers will correspond to the correct answer choice.
05

Match with the answer choice

After determining the number of optical isomers, compare your result with the answer choices provided (a, b, c, or d) and choose the correct option.

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Most popular questions from this chapter

Directions: Each question contains Statement-1 and Statement-2 and has the following choices (a), (b), (c) and (d), out of which ONLY ONE is correct. (a) Statement- 1 is True, Statement- 2 is True; Statement- 2 is a correct explanation for Statement-1 (b) Statement- 1 is True, Statement- 2 is True; Statement- 2 is NOT a correct explanation for Statement-1 (c) Statement-1 is True, Statement- 2 is False (d) Statement- 1 is False, Statement- 2 is True Statement 1 Erythro-2,3-dibromobutane on reaction with zinc and alcohol gives trans 2 -butene. and Statement 2 Elimination requires trans periplanar geometry.

Identify the product in the given reaction. \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2}+\mathrm{NOCl} \rightarrow(\mathrm{A})\) (a) \(\mathrm{CH}_{3}-\mathrm{CHCl}-\mathrm{CH}_{2} \mathrm{NO}\) (b) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHClNO}\) (d) \(\mathrm{CH}_{3} \mathrm{NOCH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\)

A reaction in which an alkene is formed is (a) 1 -bromo-1-methylcyclohexane is treated with \(\mathrm{NaOH}\) in the presence of acetone. (b) 1 -bromo- 1 -methylcyclohexane is treated with \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{~N}\). (c) Chlorocyclohexane is heated with \(\left(\mathrm{CH}_{3}\right)_{3}\) CONa in t-butanol (d) 1 -bromo-1-phenylbutane is treated with aqueous \(\mathrm{NaOH}\)

Which of the following reactions give an alkyne? (a) sodium fumarate \(\frac{\text { Kolbe's }}{\text { electrolysis }}\) (b) \(\mathrm{CH}_{3} \mathrm{CBr}_{2} \mathrm{CHBr}_{2} \longrightarrow \underset{\mathrm{Zn} \text { /alcohol } / \Delta}{\longrightarrow}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHBr}_{2} \stackrel{\mathrm{NaNH}_{2} / \Delta}{\longrightarrow}\) (d) \(\mathrm{CH}_{3} \mathrm{CHBr}-\mathrm{CH}_{2} \mathrm{Br} \stackrel{\mathrm{NaNH}_{2}}{\longrightarrow}\)

Question contains Statement-1 and Statement-2 and has the following choices (a), (b), (c) and (d), out of which ONLY ONE is correct. (a) Statement- 1 is True, Statement- 2 is True; Statement- 2 is a correct explanation for Statement-1 (b) Statement- 1 is True, Statement- 2 is True; Statement- 2 is NOT a correct explanation for Statement-1 (c) Statement- 1 is True, Statement- 2 is False (d) Statement-1 is False, Statement- 2 is True Statement 1 Alkynes are more reactive than alkenes in electrophilic addition reactions. and Statement 2 Electrophilic addition to both alkenes and alkynes involve a three membered cyclic intermediate and the bromonium ion type intermediate is less stable for an alkyne than for an alkene.
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