An organic compound $A$ $(C_{6}H_{6}O)$ gives a dark green coloration with ferric chloride. On treatment with $CHCl_{3}$ and $KOH$,followed by acidification,it gives compound $B$. Compound $B$ can also be obtained from compound $C$ on reaction with pyridinium chlorochromate $(PCC)$. Identify $A, B$ and $C$.
- A$A$ = Phenol,$B$ = Salicylaldehyde,$C$ = Salicyl alcohol
- B$A$ = Salicylaldehyde,$B$ = Salicyl alcohol,$C$ = Phenol
- C$A$ = Salicyl alcohol,$B$ = Salicylaldehyde,$C$ = Phenol
- D$A$ = Phenol,$B$ = Salicylaldehyde,$C$ = Salicyl alcohol
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Similar Questions
Identify $A$ in the following reaction.
Difficult
View SolutionWhich of the following reactions gives alcohol or phenol as a product?
$(i)$ $C_2H_5COOH \xrightarrow{LiAlH_4} C_2H_5CH_2OH$
$(ii)$ $C_2H_5Br$ $\xrightarrow[\text{then } H_2O]{\text{Mg, dry ether}} C_2H_5MgBr$ $\xrightarrow{H_2O} C_2H_6$
$(iii)$ $\text{Benzene diazonium chloride} \xrightarrow{H_3PO_2, H_2O} \text{Benzene}$
$(iv)$ $4-\text{Chlorotoluene} \xrightarrow[300 \text{ atm}]{\text{NaOH, } 623 \text{ K}} 4-\text{Methylphenol}$
$(i)$ $C_2H_5COOH \xrightarrow{LiAlH_4} C_2H_5CH_2OH$
$(ii)$ $C_2H_5Br$ $\xrightarrow[\text{then } H_2O]{\text{Mg, dry ether}} C_2H_5MgBr$ $\xrightarrow{H_2O} C_2H_6$
$(iii)$ $\text{Benzene diazonium chloride} \xrightarrow{H_3PO_2, H_2O} \text{Benzene}$
$(iv)$ $4-\text{Chlorotoluene} \xrightarrow[300 \text{ atm}]{\text{NaOH, } 623 \text{ K}} 4-\text{Methylphenol}$
DifficultTS EAMCET 2020
View SolutionMatch List-$I$ with List-$II$.
Choose the correct answer from the options given below:
Choose the correct answer from the options given below:
Which of the following reactions will yield $2-propanol$? Select the correct answer from $a, b, c$ and $d$.
$I. \, CH_2 = CH - CH_3 + H_2O \xrightarrow{H^+} $
$II. \, CH_3 - CHO \xrightarrow[(ii) \, H_2O]{(i) \, CH_3MgI} $
$III. \, CH_2O \xrightarrow[(ii) \, H_2O]{(i) \, C_2H_5MgI} $
$IV. \, CH_2 = CH - CH_3 \xrightarrow{KMnO_4} $
$I. \, CH_2 = CH - CH_3 + H_2O \xrightarrow{H^+} $
$II. \, CH_3 - CHO \xrightarrow[(ii) \, H_2O]{(i) \, CH_3MgI} $
$III. \, CH_2O \xrightarrow[(ii) \, H_2O]{(i) \, C_2H_5MgI} $
$IV. \, CH_2 = CH - CH_3 \xrightarrow{KMnO_4} $
Medium
View Solution$1.12 \, mL$ of a gas is produced at $STP$ by the action of $4.12 \, mg$ of alcohol with methyl magnesium iodide. The molecular mass of the alcohol is:
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