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 \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{2} \mathrm{Br}\) undergoes hydrolysis faster than \(\mathrm{CH}_{3}-\mathrm{O}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{Br}\) and Statement 2 The carbocation formed by ionization of \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OCH}_{2} \mathrm{Br}\) is stabilized by resonance.

Short Answer

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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. Answer: a) Statement- 1 is True, Statement- 2 is True; Statement- 2 is a correct explanation for Statement- 1.

Step by step solution

01

Examine the Hydrolysis Reaction

The hydrolysis reaction typically involves breaking down a compound (in this case an organic compound) in the presence of water. We need to compare the reaction rates of CH3-CH2-O-CH2Br and CH3-O-CH2-CH2Br in the hydrolysis reaction.
02

Compare Reaction Rates

To compare the reaction rates, we must consider the following factors: - Stability of the intermediates formed in the reactions - Electronic effects that can stabilize the intermediates Comparing CH3-CH2-O-CH2Br and CH3-O-CH2-CH2Br, we can see that the difference between these two compounds is the position of the oxygen atom in the organic chain.
03

Analyze the Carbocation Formation of Statement 2

Let's focus on the carbocation formation of CH3-CH2-OCH2Br and study its stability. When CH3-CH2-OCH2Br undergoes hydrolysis and loses a bromide ion, it forms a carbocation: \((CH_3-CH_2-O-CH_2)^+\). This carbocation can be stabilized through resonance as the positive charge can be delocalized between the oxygen and carbon atoms, which increases the stability of the ions and leads to a faster hydrolysis reaction.
04

Determine the Correctness and Relation of the Statements

Based on our analysis, we can determine the following: - Statement 1 is true as CH3-CH2-O-CH2Br undergoes hydrolysis at a faster rate due to the factors we analyzed. - Statement 2 is also true as the carbocation formed by ionization of CH3-CH2-OCH2Br is stabilized by resonance. Now, we must determine how these statements are related. Since the stability of the carbocation through resonance explains why CH3-CH2-O-CH2Br undergoes hydrolysis faster than CH3-O-CH2-CH2Br, we can conclude that Statement 2 is a correct explanation for Statement 1. Therefore, the correct choice is (a) Statement- 1 is True, Statement- 2 is True; Statement- 2 is a correct explanation for Statement- 1.

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Most popular questions from this chapter

It is feasible to prepare tertiary butyl ethyl ether by treating ethyl bromide with sodium tertiary butoxide, rather than treating tertiary butyl bromide with sodium ethoxide because (a) sodium ethoxide will not react with t-butyl bromide. (b) the first reaction is faster and gives more yield. (c) t-butyl bromide undergoes dehydrohalogenation to give alkene in presence of sodium ethoxide. (d) t-butyl bromide undergoes rearrangement.

Which of the following alcohols will give a yellow coloured precipitate with iodine and sodium hydroxide?

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Write the steps involved in the conversion of the following (i) Ethanol to but-1-yne (ii) Ethyl bromide to butan- 2 - one (iii) 2 -chloropropane to propan-1-ol

Directions: This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. The correct decreasing order of reactivity of the following alcohols towards con. HCl is

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