Practical Organic Chemistry

JEE Chemistry · 106 questions · Page 11 of 11 · Click an option or "Show Solution" to reveal answer

Q101

Match with .tg .tg (Technique) (Application

List - IList - II
(A) Distillation (I) Separation of glycerol from spent-lye
(B) Fractional distillation (II) Aniline - Water mixture
(C) Steam distillation (III) Separation of crude oil fractions
(D) Distillation under reduced pressure (IV) Chloroform - Aniline
A A-I, B-II, C-IV, D-III
B A-II, B-III, C-I, D-IV
C A-IV, B-I, C-II, D-III
D A-IV, B-III, C-II, D-I
Correct Answer
Option D
Solution

To match List I (Technique) with List II (Application), we need to understand the relevance of each distillation technique: 1.

Distillation is typically used for separating components of mixtures who have a significant difference in boiling points.

Application: Separation of glycerol from spent-lye.

2.

Fractional distillation is used for separating mixtures of liquids with closer boiling points by using a fractionating column.

Application: Separation of crude oil fractions.

3.

Steam distillation is generally used for compounds that decompose under high temperatures and are insoluble in water.

Application: Aniline - Water mixture.

4.

Distillation under reduced pressure (vacuum distillation) is used for compounds with very high boiling points to avoid decomposition.

Application: Chloroform - Aniline.

Therefore, the correct match is: Option D A-IV (Distillation - Chloroform - Aniline) B-III (Fractional distillation - Separation of crude oil fractions) C-II (Steam distillation - Aniline - Water mixture) D-I (Distillation under reduced pressure - Separation of glycerol from spent-lye)

Q102
Match List I with List II .tg .tg LIST I (Test) LIST II (Observation) A. Br2\mathrm{Br_2} water test I. Yellow orange or orange red precipitate formed B. Ceric ammonium nitrate test II. Reddish orange colour disappears C. Ferric chloride test III. Red colour appears D. 2, 4 - DNP test IV. Blue, Green, Violet or Red colour appear Choose the correct answer from the options given below:
A A-II, B-III, C-IV, D-I
B A-III, B-IV, C-I, D-II
C A-I, B-II, C-III, D-IV
D A-IV, B-I, C-II, D-III
Correct Answer
Option A
Solution

Let's match the given tests in List I with the correct observations in List II based on known chemical reactions and their typical outcomes: List I with List II A. Br2\mathrm{Br_2} water test This test is used to detect unsaturation in organic compounds.

If the compound has a double bond (alkenes) or a triple bond (alkynes), the Br2\mathrm{Br_2} in water (bromine water) will react and decolorize.

Correct Observation: Reddish orange color of bromine water disappears (II).

B.

Ceric ammonium nitrate test This test is used to detect alcohols.

Ceric ammonium nitrate forms a red-colored complex with alcohols.

Correct Observation: Red color appears (III).

C.

Ferric chloride test Commonly used to detect phenols, the ferric chloride test leads to the formation of a colored complex with phenols, typically varying shades like purple, green, blue, or even red, depending on the specific phenol.

Correct Observation: Blue, Green, Violet, or Red color appear (IV).

D.

2,4-DNP test This test is used for the detection of carbonyl compounds (aldehydes and ketones).

2,4-Dinitrophenylhydrazine (2,4-DNP) reacts with carbonyl groups to form a yellow, orange, or red precipitate.

Correct Observation: Yellow orange or orange red precipitate formed (I).

Conclusion: Based on these observations, the correct matching of tests with outcomes should be: A with II B with III C with IV D with I This makes the correct choice: Option A: A-II, B-III, C-IV, D-I

Q103
Match List I with List II .tg .tg LIST I (Test) LIST II (Identification) A. Bayer's test I. Phenol B. Ceric ammonium nitrate test II. Aldehyde C. Phthalein dye test III. Alcoholic-OH group D. Schiff's test IV. Unsaturation Choose the correct answer from the options given below :
A (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
B (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
C (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
D (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
Correct Answer
Option D
Solution

To match the tests in List I with the corresponding identifications in List II, let's analyze each test: Bayer's test: This test is used to test for unsaturation in compounds, such as alkenes and alkynes.

The presence of unsaturation is indicated by the decolorization of potassium permanganate solution.

Ceric ammonium nitrate test: This test is used to identify alcohols.

A positive result is indicated by a color change.

Phthalein dye test: This test is used to identify phenols.

Phenols react with phthalic anhydride to form a colored compound.

Schiff's test: This test is used for the detection of aldehydes.

A positive result indicates the formation of a magenta or pink colored compound.

Based on the information above, the correct matching is: A.

Bayer's test - IV.

Unsaturation B.

Ceric ammonium nitrate test - III.

Alcoholic-OH group C.

Phthalein dye test - I.

Phenol D.

Schiff's test - II.

Aldehyde Thus, the correct answer is: Option D: (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

Q104

 Match with .  \text{ Match with . } $$

List - IList - II
(B)  Fractional distillation  \text{ Fractional distillation } (I) Diesel + Petrol
(C)  Distillation under reduced pressure  \text{ Distillation under reduced pressure } (II) Aniline + Water
(D)  Steam distillation  \text{ Steam distillation } (III) Chloroform + Aniline
() (IV) Glycerol + Spent-lye $$ \text {
A (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
B (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
C (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
D (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
Correct Answer
Option A
Solution

The correct matching is: (A) Simple distillation → (III) Chloroform + Aniline (B) Fractional distillation → (I) Diesel + Petrol (C) Distillation under reduced pressure → (IV) Glycerol + Spent-lye (D) Steam distillation → (II) Aniline + Water So the answer is Option A.

Brief rationale: Simple distillation (A): •

bp(CHCl3)=61C,  bp(C6H5NH2)=184Cbp(\text{CHCl}_3)=61^\circ\text{C},\;bp(\text{C}_6\text{H}_5\text{NH}_2)=184^\circ\text{C}

• Δbp is large, so the lower-boiling chloroform distils off cleanly.

Fractional distillation (B): • Petrol (40–200°C) and diesel (180–360°C) have overlapping but distinct boiling-range cuts.

• A fractionating column resolves them efficiently.

Vacuum distillation (C): • Glycerol boils ≈290°C at 1 atm—too high for normal distillation.

• Lowering the pressure brings its boiling point down, letting you separate it from the spent-lye.

Steam distillation (D): • Aniline (bp≈184°C) can be co-distilled with water around 98°C, avoiding thermal decomposition and separating it from the aqueous phase.

Q105
The correct match between Item-I and Item-II is : .tg .tg Item-I (drug) Item-II (test) A Chloroxylenol P Carbylamine test B Norethindrone Q Sodium hydrogen carbonet test C Sulphapyridine R Ferric chloride test D Penicillin S Bayer's test
A A \to R; B \to P; C \to S; D \to Q
B A \to Q; B \to S; C \to P; D \to R
C A \to R; B \to S; C \to P; D \to Q
D A \to Q; B \to P; C \to S; D \to R
Correct Answer
Option C
Solution

(1) To react with FeCl3, with benzene ring there should be a -OH group attached.

(2) For carbylamine test there should be -NH2 group in a compound.

(3) NaHCO3 reacts when there is -COOH group in a compound.

(4) For Bayer's test there should be double bond or tripple bond between two carbon.

Q106
The correct match between item 'I' and item 'II' is : .tg .tg Item 'I' (compound) Item 'II' (reagent) (A) Lysine (P) 1-naphthol (B) Furfural (Q) ninhydrin (C) Benzylalcohol (R) KMnO4 (D) Styrene (S) Ceric ammonium nitrate
A (A) \to (Q); (B) \to (R); (C) \to (S); (D) \to (P)
B (A) \to (Q); (B) \to (P); (C) \to (S); (D) \to (R)
C (A) \to (R); (B) \to (P); (C) \to (Q); (D) \to (S)
D (A) \to (Q); (B) \to (P); (C) \to (R); (D) \to (S)
Correct Answer
Option B
Solution

Lysine is an amino acid which gives nindydrin test.

Furfural has aldehyde group that is why it reacts with 1-napthol to give purple colouration.

Benzyl alcohol undergoes reaction with ceric ammonium nitrate to give pink colouration.

Styrene reacts with KMnO4 and gives benzoic acid and brown MnO2.

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