Reaction Mechanisms

To solve the problem, we need to determine the structure of product $D$ by following the sequence of reactions step by step.

Step 1: Analyze the given reaction
The starting compound is:

$ $%5Ctext%7BCH%7D_3-%5Cunderset%7B%5Ctext%7BCH%7D_3%7D%7B%5Cunderbrace%7B%5Ctext%7BC%7D%7D%7D-%5Ctext%7BH%7D%20%2B%20%5Ctext%7BO%7D_2%20%5Crightarrow%20%5Ctext%7BProduct%20A%7D$ $This%20is%20a%20benzene%20ring%20with%20a%20methyl%20group%20and%20an%20ethyl%20group.%20The%20reaction%20involves%20oxidation%20of%20the%20alkyl%20side%20chain.%20Reaction%201%3A%20Oxidation%20of%20the%20Alkyl%20Side%20Chain**%20The%20alkyl%20side%20chain%20($ \text{CH}_3-\text{CH}_2- $)%20undergoes%20oxidation%20in%20the%20presence%20of%20oxygen%20($ \text{O}_2 $).%20The%20primary%20alcohol%20formed%20is%3A$ $%5Ctext%7BCH%7D_3-%5Cunderset%7B%5Ctext%7BCH%7D_3%7D%7B%5Cunderbrace%7B%5Ctext%7BC%7D%7D%7D(%5Ctext%7BOH%7D)-%5Ctext%7BH%7D$ $This%20is%20the%20product$ A $.%20Reaction%202%3A%20Acid-Catalyzed%20Dehydration%20The%20product$ A $(a%20primary%20alcohol)%20is%20treated%20with%20acid%20($ \text{H}^+ $)%20and%20water%20($ \text{H}_2\text{O} $)%20to%20form%20two%20products%2C$ A' $and$ B' $.%20This%20is%20an%20acid-catalyzed%20dehydration%20reaction%2C%20which%20eliminates%20a%20molecule%20of%20water%20to%20form%20an%20alkene%20and%20a%20carboxylic%20acid.%20The%20primary%20alcohol%20undergoes%20dehydration%20to%20form%3A%20-$ A' $%3A%20An%20alkene.%20-$ B' $%3A%20A%20carboxylic%20acid.%20The%20structure%20of$ A' $is%3A$ $%5Ctext%7BCH%7D_3-%5Cunderset%7B%5Ctext%7BCH%7D_3%7D%7B%5Cunderbrace%7B%5Ctext%7BC%7D%7D%7D%3D%5Ctext%7BCH%7D_2$ $This%20is%20a%20conjugated%20diene%20(4-methylpent-1%2C3-diene).%20The%20structure%20of$ B' $is%3A$ $%5Ctext%7BCH%7D_3-%5Ctext%7BCOOH%7D$ $This%20is%20acetic%20acid.%20Step%202%3A%20Identify%20Product$ A' $The%20problem%20states%20that%20product$ A' $gives%20a%20white%20precipitate%20when%20treated%20with%20bromine%20water.%20This%20indicates%20that$ A' $is%20an%20alkene%2C%20as%20alkenes%20react%20with%20bromine%20water%20to%20form%20addition%20products%2C%20which%20can%20sometimes%20appear%20as%20white%20precipitates%20due%20to%20the%20formation%20of%20insoluble%20bromides.%20Thus%2C$ A' $is%3A$ $%5Ctext%7BCH%7D_3-%5Cunderset%7B%5Ctext%7BCH%7D_3%7D%7B%5Cunderbrace%7B%5Ctext%7BC%7D%7D%7D%3D%5Ctext%7BCH%7D_2$ $Step%203%3A%20Identify%20Product$ B' $The%20problem%20states%20that%20product$ B' $is%20treated%20with%20barium%20hydroxide%20($ \text{Ba(OH)}_2 $)%20to%20form%20product$ C $.%20Acetic%20acid%20($ \text{CH}_3\text{COOH} $)%20reacts%20with%20barium%20hydroxide%20to%20form%20barium%20acetate%20($ \text{Ba(CH}_3\text{COO)}_2 $)%20and%20water%20($ \text{H}_2\text{O} $)%3A$ $%5Ctext%7BCH%7D_3%5Ctext%7BCOOH%7D%20%2B%20%5Ctext%7BBa(OH)%7D_2%20%5Crightarrow%20%5Ctext%7BBa(CH%7D_3%5Ctext%7BCOO)%7D_2%20%2B%202%5Ctext%7BH%7D_2%5Ctext%7BO%7D$ $Thus%2C%20product$ C $is%20barium%20acetate.%20Step%204%3A%20Identify%20Product$ D $The%20problem%20states%20that%20product$ C $(barium%20acetate)%20is%20heated%20strongly%20to%20form%20product$ D $.%20When%20barium%20acetate%20is%20heated%20strongly%2C%20it%20decomposes%20to%20form%20acetic%20acid%20($ \text{CH}_3\text{COOH} $)%20and%20barium%20oxide%20($ \text{BaO} $)%3A$ $%5Ctext%7BBa(CH%7D_3%5Ctext%7BCOO)%7D_2%20%5Cxrightarrow%7B%5Ctext%7BHeating%7D%7D%20%5Ctext%7BCH%7D_3%5Ctext%7BCOOH%7D%20%2B%20%5Ctext%7BBaO%7D$ $However%2C%20the%20problem%20asks%20for%20the%20final%20product$ D $%2C%20which%20is%20the%20result%20of%20the%20initial%20dehydration%20reaction.%20From%20the%20earlier%20steps%2C%20we%20know%20that%20the%20dehydration%20of%20the%20primary%20alcohol%20forms%20the%20alkene%3A$ $%5Ctext%7BCH%7D_3-%5Cunderset%7B%5Ctext%7BCH%7D_3%7D%7B%5Cunderbrace%7B%5Ctext%7BC%7D%7D%7D%3D%5Ctext%7BCH%7D_2$ $This%20is%204-methylpent-1%2C3-diene.%20Final%20Answer%3A%20The%20product$ D $is%3A$ $%5Cboxed%7B%5Ctext%7B(A)%204-Methylpent-3-en-2-one%7D%7D$ $

Step 1: Understanding the Question:
The question asks for the rate of a first-order reaction at a specific concentration (

0.01%20%5Ctext%7B%20M%7D

). To find this, we first need to determine the rate constant (

k

) using the time taken for a given change in concentration.
Step 2: Key Formula or Approach:
For a first-order reaction:
1. Integrated rate law:

k%20%3D%20%5Cfrac%7B2.303%7D%7Bt%7D%20%5Clog%5Cleft(%5Cfrac%7B%5BR%5D_0%7D%7B%5BR%5D_t%7D%5Cright)

2. Rate law:

%5Ctext%7BRate%7D%20%3D%20k%5BR%5D

3. Relation between

k

and half-life

t_%7B1%2F2%7D

k%20%3D%20%5Cfrac%7B0.693%7D%7Bt_%7B1%2F2%7D%7D

Step 3: Detailed Explanation:
1. Finding the Rate Constant ( $k$ ):
Initial concentration,

%5BR%5D_0%20%3D%200.1%20%5Ctext%7B%20M%7D

.
Final concentration after 40 min,

%5BR%5D_t%20%3D%200.025%20%5Ctext%7B%20M%7D

.
Notice that

%5Cfrac%7B%5BR%5D_t%7D%7B%5BR%5D_0%7D%20%3D%20%5Cfrac%7B0.025%7D%7B0.1%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D

.
Since the concentration becomes one-fourth, two half-lives have passed in 40 minutes (

0.1%20%5Crightarrow%200.05%20%5Crightarrow%200.025

).
Therefore,

2%20%5Ctimes%20t_%7B1%2F2%7D%20%3D%2040%20%5Ctext%7B%20min%7D%20%5Cimplies%20t_%7B1%2F2%7D%20%3D%2020%20%5Ctext%7B%20min%7D

.
Now, calculate the rate constant

k

k%20%3D%20%5Cfrac%7B0.693%7D%7Bt_%7B1%2F2%7D%7D%20%3D%20%5Cfrac%7B0.693%7D%7B20%20%5Ctext%7B%20min%7D%7D%20%3D%200.03465%20%5Ctext%7B%20min%7D%5E%7B-1%7D

2. Calculating the Rate of Reaction:
We need the rate when

%5BR%5D%20%3D%200.01%20%5Ctext%7B%20M%7D

%5Ctext%7BRate%7D%20%3D%20k%20%5Ctimes%20%5BR%5D

%5Ctext%7BRate%7D%20%3D%200.03465%20%5Ctext%7B%20min%7D%5E%7B-1%7D%20%5Ctimes%200.01%20%5Ctext%7B%20M%7D

%5Ctext%7BRate%7D%20%3D%200.0003465%20%5Ctext%7B%20M%20min%7D%5E%7B-1%7D%20%3D%203.465%20%5Ctimes%2010%5E%7B-4%7D%20%5Ctext%7B%20M%20min%7D%5E%7B-1%7D

Rounding to three significant figures, we get

3.47%20%5Ctimes%2010%5E%7B-4%7D%20%5Ctext%7B%20M%20min%7D%5E%7B-1%7D

.
Step 4: Final Answer:
The rate is

3.47%20%5Ctimes%2010%5E%7B-4%7D%20%5Ctext%7B%20M%20min%7D%5E%7B-1%7D

, which corresponds to option (B).

About Reaction Mechanisms - KCET

Reaction Mechanisms is a vital chapter for KCET aspirants. Mastering the concepts covered in this chapter is essential for securing a top rank.

By rigorously practicing the previous year questions associated with this chapter, you can identify high-yield topics, understand the examiner's perspective, and boost your confidence during the actual exam.

Frequently Asked Questions

Why focus on Reaction Mechanisms PYQs?

Analyzing PYQs for this specific chapter reveals the most frequently tested concepts and the typical complexity of questions, allowing you to tailor your study plan efficiently.

How to best use this analysis?

Review the topic breakdown to see which sub-topics within Reaction Mechanisms carry the most weight. Then, tackle the questions iteratively to solidify your understanding.

High-Yield Trend

Chapter Questions
2 MCQs

Official Solution

Official Solution

About Reaction Mechanisms - KCET

Frequently Asked Questions

Why focus on Reaction Mechanisms PYQs?

How to best use this analysis?

Reaction Mechanisms

High-Yield Trend

Chapter Questions 2 MCQs

Official Solution

Official Solution

About Reaction Mechanisms - KCET

Frequently Asked Questions

Why focus on Reaction Mechanisms PYQs?

How to best use this analysis?

Chapter Questions
2 MCQs