Retro (Past 13 Year) IIT-JEE Advanced (Carboxylic Acids And Acid Derivatives) - JEE MCQ

This problem can be solved by using the stability of radical obtained after fragmentation of peroxyester.
Allylic radical are more stable than alkyl radical, so when there is a possibility of formation of allyl radical, it will undergo fragmentation through formation of allyl radical, i.e. fragmentation produces stable radical.
On the basis of stability of radical, fragm entation can be done as :

Plan Reactant is cyclic anhydride and changes to dicarboxylic acid on hydrolysis. Also there is decarboxylation on heating if there is keto group w.r.t —COOH group. Ozonolysis cleaves   bo nd and H₂O₂ oxidises —CHO to —COOH group.

Thus, number of —COOH groups in P = 2.

is decomposed by acid releasing CO²

If acid is stronger than then CO₂ is released.
Phenol is less acidic and thus, does not liberate CO₂ with NaHCO₃.

Plan Alkenes decolourise Br2 water c/s-isomer Meso isomers by syn addition trans-isomer (+) and / (-) isomers by syn addition thus, racemic mixture.
Formation of anhydride from dicarboxylic acid indicates c/s-isomer. P and Q are isomers of dicarboxylic acids.

T and U (in 1 :1 molar ratio) form optically inactive (racemic mixture) due to external compensation.

Plan Ni/H₂ reduces(C = C )bond.
Benzene undergoes Friedel-Crafts reaction Zn-Hg/HCI reduces carbonyl group (Clemmensen reduction)

It is αβ-keto acid which undergo decarboxylation in very mild condition i.e. on simple heating. This occur through a six m embered cyclic transition state as :

Note (i) Ordinary carboxylic acid require soda lime catalyst for decarboxylation.
(ii) Final step of decarboxylation in the above shown mechanism involve tautomerism, therefore, for decarboxylation of β-keto acid by above mechanism, the acid must contain an α-H]. 

(a) Undergoes an ester hydrolysis in hot aqueous alkali as 

(b) LiAIH4 reduces ester to alcohol as

"U" No chiral carbon optically inactive.
(c) U on treatment with excess of acetic anhydride forms a diester as

(d) U on treatment with CrO₃H⁺ undergo oxidation to diacid which gives effervescence with NaHCO₃.

For separation be differential extraction one of the component must form salt with the given base so that the salt will be extracted in aqueous layer leaving other component in organic layer.
(a) Both phenol and benzoic acid form salt with NaOH, hence this mixture cannot be separated.
(b) Benzoic acid forms salt with NaOH while benzyl alcohol does not, hence the mixture can be separated using NaOH. Also benzoic acid forms salt with NaHCO₃ but benzyl alcohol does not, hence NaHCO₃ can be used for separation.
(c) Neither benzyl alcohol nor phenol forms salt with NaHCO₃, mixture cannot be separated using NaHCO₃.
(d) C₆H₅CH₂COOH forms salt with NaOH, C₆H₅CH₂OH does not, hence mixture can be separated using NaOH-C₆H₅CH₂COOH forms salt with NaHCO₃- C₆H₅CH₂OH does not, hence mixture can be separated using NaHCO₃.