Which of the following compounds undergoes ma | vedclass

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(A) The extent of hydration of carbonyl compounds depends on the electrophilicity of the carbonyl carbon and steric hindrance. $1$. Electron-withdrawi

Vedclass · Accepted Answer

$CF_3-CO-CF_3$

Vedclass · Answer

(A) The extent of hydration of carbonyl compounds depends on the electrophilicity of the carbonyl carbon and steric hindrance. $1$. Electron-withdrawing groups (like $-F$ or $-Cl$) increase the electrophilicity of the carbonyl carbon by the inductive effect ($-I$ effect),making it more susceptible to nucleophilic attack by water. $2$. The compound $CF_3-CO-CF_3$ (hexafluoroacetone) has two highly electronegative $CF_3$ groups,which exert a strong $-I$ effect,significantly increasing the positive charge on the carbonyl carbon. $3$. Consequently,$CF_3-CO-CF_3$ forms a stable hydrate (gem-diol) most readily compared to the other options. Therefore,the correct option is $A$.

List-$I$ Reaction	List-$II$ Reagents
$A$. Hoffmann Degradation	$I$. Conc. $KOH, \Delta$
$B$. Clemmensen reduction	$II$. $CHCl_3, NaOH / H_3O^{+}$
$C$. Cannizzaro reaction	$III$. $Br_2, NaOH$
$D$. Reimer-Tiemann reaction	$IV$. $Zn-Hg / HCl$

Which of the following compounds undergoes maximum hydration?

Similar Questions

Given the reactions:
$CaOCl_2 + H_2O \longrightarrow Ca(OH)_2 + X$
$X + CH_3CHO \longrightarrow Y$
$Y + Ca(OH)_2 \longrightarrow CHCl_3$
What is '$Y$'?

Match List-$I$ with List-$II$.

List-$I$ Reaction List-$II$ Reagents

$A$. Hoffmann Degradation $I$. Conc. $KOH, \Delta$

$B$. Clemmensen reduction $II$. $CHCl_3, NaOH / H_3O^{+}$

$C$. Cannizzaro reaction $III$. $Br_2, NaOH$

$D$. Reimer-Tiemann reaction $IV$. $Zn-Hg / HCl$

The product $(A)$ of the reaction is:

The reaction sequence is: $P$ $\xrightarrow{\Delta } X$ $\xrightarrow[{HCl}]{{Zn-Hg}} Y$. Here,$P$ is $2-$oxocyclohexanecarboxylic acid. Identify $Y$.

Consider the following sequence of reactions:
$2-\text{benzylbenzoic acid}$ $\xrightarrow[2. \text{anhyd. } AlCl_3]{1. SOCl_2} A$ $\xrightarrow{Zn-Hg, \text{Conc. } HCl} B$
The end product $(B)$ is:

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