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


9 Questions MCQ Test Chemistry for JEE Advanced | Retro (Past 13 Year) IIT-JEE Advanced (Carboxylic Acids And Acid Derivatives)


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QUESTION: 1

Different possible thermal decomposition pathways for peroxyesters are shown below. Match each pathway from Column I with an appropriate structure from Column II and select the correct answer using the code given below the lists.


Solution:

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 :


*Answer can only contain numeric values
QUESTION: 2

The total number of carboxylic acid groups in the product P is 

(2013Adv., integer Type)


Solution:

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.

QUESTION: 3

The compound that does not liberate CO2, on treatment with aqueous sodium bicarbonate solution, is

(2013 Adv.,Only One Option Correct Type)

Solution:


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.

QUESTION: 4

Passage for Q. Nos. (4-5)

P and Q are isomers of dicarboxylic acid C4H4O4. Both decolourise Br2 / H2O. On heating, P forms the cyclic anhydride.
Upon treatment with dilute alkaline KMnO4. P as well as Q could produce one or more than one form S, Tand U.

Q. 

Compounds formed from P and Q respectively are,

Solution:

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.

QUESTION: 5

P and Q are isomers of dicarboxylic acid C4H4O4. Both decolourise Br2 / H2O. On heating, P forms the cyclic anhydride.
Upon treatment with dilute alkaline KMnO4. P as well as Q could produce one or more than one form S, Tand U.

Q. 

In the following reaction sequences V and W respectively are

Solution:

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

QUESTION: 6

The compound that undergoes decarboxylation most readily under mild condition is

(2012,Only One Option Correct Type)

Solution:

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]. 

QUESTION: 7

With reference to the scheme given, which of the given statement (s) about T U, V and W is (are) correct ?

(2012, One or More than One Options Correct Type)

Solution:

(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.

*Multiple options can be correct
QUESTION: 8

Identify the binary mixture(s) that can be separated into individual compounds, by differential extraction, as shown in the given scheme.

(2012, One or More than One Options Correct Type)

Solution:

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.

QUESTION: 9

In the following reaction sequence, the correct structures of E, F and G are

(2008,Only One Option Correct Type)

Solution: