Question

What are the major products X and Y respectively in the following reactions? \[ {(CH}_3{)}_3{COONa} + {CH}_2 {CH} {Br} \rightarrow X \] \[ {(CH}_3{)}_3{COCH}_2 {CH}_3 + {CH}_3{COO}{Na} \rightarrow Y \]

• A. \( {CH}_2 = {CH}_2, \, ({CH}_3)_3 {COCH}_2 {CH}_3 \)
• B. \( ({CH}_3)_3 {COCH}_2 {CH}_3, \, ({CH}_3)_3 {COCH}_2 {CH}_3 \)
• C. \( ({CH}_3)_3 {COCH}_2 {CH}_3, \, {CH}_2 = {CH}_2 \)
• D. \( ({CH}_3)_3 {COCH}_2 {CH}_3, \, {CH}_2 = {CH}_2 \)

Hint:

In esterification reactions involving alkylation with bromides, the halide is substituted, and in hydrolysis reactions, the ester is split to yield an alkene or alcohol.

Answer 1

The Correct Option is D

We are given two reactions, and we are to determine the major products X and Y. The reaction types can be identified as follows: 
Step 1: The reaction between \( {CH}_3 {COONa} \) (sodium acetate) and \( {CH}_2 {CH} {Br} \) (bromine-substituted ethene) leads to the substitution of the bromine atom, resulting in an alkylated product: \( ({CH}_3)_3 {COCH}_2 {CH}_3 \). Thus, product X is \( ({CH}_3)_3 {COCH}_2 {CH}_3 \). 
Step 2: The second reaction involves the addition of a strong base or another suitable agent to break the ester bond and form the final product Y, which is \( {CH}_2 = {CH}_2 \). Thus, the major products X and Y are \( ({CH}_3)_3 {COCH}_2 {CH}_3 \) and \( {CH}_2 = {CH}_2 \), corresponding to option (4).

Answer 2

The reactions given are examples of organic chemistry transformations involving either substitution or elimination reactions. Let's analyze each reaction individually:

1. First reaction: \((\text{CH}_3)_3\text{COONa} + \text{CH}_2=\text{CHBr} \rightarrow X\)

The reagent \((\text{CH}_3)_3\text{COONa}\) is a strong base, commonly used to deprotonate compounds or facilitate elimination reactions. In this case, \((\text{CH}_3)_3\text{COONa}\) will cause a dehydrohalogenation, an elimination reaction that removes HBr from the alkyl halide \(\text{CH}_2=\text{CHBr}\), forming an alkene. The expected product \(X\) is a simple alkene without halogen: \(\text{CH}_2=\text{CH}_2\).

2. Second reaction: \((\text{CH}_3)_3\text{COCH}_2\text{CH}_3 + \text{CH}_3\text{COONa} \rightarrow Y\)

Here, \((\text{CH}_3)_3\text{COCH}_2\text{CH}_3\) has no leaving group for elimination or substitution by the \(\text{CH}_3\text{COONa}\) reagent. Instead, the presence of sodium acetate (\(\text{NaOOCCH}_3\)) does not participate further, suggesting the product \(Y\) remains unchanged as \((\text{CH}_3)_3\text{COCH}_2\text{CH}_3\).

Based on these analyses, the major products for the reactions are:

\(X = (\text{CH}_3)_3\text{COCH}_2\text{CH}_3\)

\(Y = \text{CH}_2=\text{CH}_2\)

Thus, the correct answer is:

\((\text{CH}_3)_3\text{COCH}_2\text{CH}_3, \, \text{CH}_2=\text{CH}_2\)