JEE Mains · Chemistry · STD 12 - 7. Alcohol, phenol and ethers
Consider the following reaction sequence:
\(\begin{array}{lll}\underset{\text {Compound }( x )}{ } \frac{\begin{array}{c}\text { (i) } CO _2, NaOH , 120^{\circ} C , \\ \text {high pressure }\end{array}}{\text { (ii) } H _3 O ^{+}} & \text {Compound }( y ) \\ {[76.6 \% C , 6.38 \% H ,} & \text { (Major Product) } \\ \text {vapour density } 47] & & \end{array}\)
Compound (y) develops characteristic colour with neutral \(FeCl _3\) solution.
Identify the INCORRECT statement from the following for the above sequence.
- A Both compounds x and y will dissolve in NaOH.
- B Compound y will dissolve in \(NaHCO _3\) and evolve a gas.
- C Compound x is more acidic than compound y.
- D Both compounds x and y will burn with sooty flame.
Answer & Solution
Correct Answer
(C) Compound x is more acidic than compound y.
Step-by-step Solution
Detailed explanation
\(( X ) \xrightarrow[(2)_3 H _3 O ^{\oplus}]{\text { (1) } CO _2 / NaOH } Y _{\text {Major }}\) \(76.6 \% C\) \(6.38 \% H\) Vapour Density 47 Salicylic acid Gives colour with \(FeCl _3\) Soluble in \(NaHCO _3\) Soluble in NaOH
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