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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

This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. A compound with the molecular formula, \(\mathrm{C}_{4} \mathrm{H}_{6}\), does not exist in cis or trans form. It does not produce a precipitate with ammoniacal \(\mathrm{AgNO}_{3}\) solution. The compound as such is optically inactive but a monohalogen derivative can be optically active. The compound is (a) \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\) (b) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{C}=\mathrm{CH}_{2}\) (c) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}\) (d)

When propyne is treated with excess chlorine water, the final product obtained is (a) 1,1 -dichloro propanone (b) 2, 3 dichloro dihydroxy propane (c) 2,2 -dichloropropanal (d) 1 -hydroxy-3-chloro propanone

Four reagents and specific reactions in which they are involved are given. Identify the wrongly matched option among the following (a) Hot alcoholic KOH \(-\) dehydrohalogenation (b) \(\mathrm{H}^{+} / \mathrm{H}_{2} \mathrm{O}\) - hydration (c) \(\mathrm{H}^{+} / \Delta\) - dehydration (d) Unsymmetrical alkene - anti Markovnikov addition

The major product obtained when 2 -phenylpropene is treated with water containing acid is (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{OH}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CHOHCH}_{3}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHOH} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{OH}\)

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.
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