An organic compound with the molecular formul | vedclass

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(D) $1$. The molecular formula $C_6H_6O$ corresponds to the degree of unsaturation $U = 6 - \frac{6}{2} + 1 = 4$, which suggests an aromatic ring.$2$.

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Phenol

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(D) $1$. The molecular formula $C_6H_6O$ corresponds to the degree of unsaturation $U = 6 - \frac{6}{2} + 1 = 4$, which suggests an aromatic ring.$2$. The compound dissolves in $NaOH$, indicating it is acidic in nature.$3$. It gives a characteristic violet colour with neutral $FeCl_3$, which is a standard test for phenolic groups.$4$. It reacts with bromine water to form a tri-bromo derivative $(2,4,6-	ext{tribromophenol})$, which is characteristic of phenol due to the strong activating effect of the $-OH$ group.$5$. Therefore, the compound is phenol.

List-$I$ (Reactants)	List-$II$ (Products)
$A$. Phenol,$Zn / \Delta$	$I$. Salicylaldehyde
$B$. Phenol,$CHCl_3, NaOH, HCl$	$II$. Salicylic acid
$C$. Phenol,$CO_2, NaOH, HCl$	$III$. Benzene
$D$. Phenol,Conc. $HNO_3$	$IV$. Picric acid

An organic compound with the molecular formula $C_6H_6O$ dissolves in $NaOH$, gives a characteristic colour with neutral $FeCl_3$ and on treatment with bromine water gives a tri-bromo derivative. Which is that compound among the following?

Similar Questions

What is the name of the following reaction?

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

List-$I$ (Reactants) List-$II$ (Products)

$A$. Phenol,$Zn / \Delta$ $I$. Salicylaldehyde

$B$. Phenol,$CHCl_3, NaOH, HCl$ $II$. Salicylic acid

$C$. Phenol,$CO_2, NaOH, HCl$ $III$. Benzene

$D$. Phenol,Conc. $HNO_3$ $IV$. Picric acid

An organic compound $A$ with empirical formula $C_6H_6O$ gives a sooty flame on burning. Its reaction with bromine solution in a low polarity solvent results in a high yield of $B$. $B$ is:

The reaction is given as: $X \xrightarrow{Y} \text{Benzoquinone}$. Identify $X$ and $Y$ in the above reaction.

Phenol on heating with $CHCl_3$ and $NaOH$ gives salicylaldehyde. The reaction is called

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