Chapter Exam

Unit 17: Organic Compounds Containing Oxygen — Online MCQ Test

Q: The product of all solutions of the equation 5^(log_e x)^2 + 3^(log_e x)^2 = x^8, x > 0 is:

Answer: A. e — Taking logs or substituting t = (ln x)^2, the equation simplifies to find log values. Product of roots corresponds to sum of exponents equal to 2, leading to e^2.

Q: The compounds which give positive Fehling's test are: (A) HOCH2-CO-(CHOH)3-CH2-OH ... [Other structures omitted in text]. Choose the correct answer from the options given below:

Answer: B. (C), (D) and (E) Only — Fehling's test is positive for aliphatic aldehydes and alpha-hydroxy ketones (ketoses).

Q: Match List-I with List-II.

Answer: D. (4) A-II, B-IV, C-I, D-III — A(H2/Pd, BaSO4) = Rosenmund Reduction (II). B(SnCl2, HCl) = Stephen Reaction (IV). C(CrO2Cl2, CS2) = Etard Reaction (I). D(CO, HCl, Anhyd. AlCl3) = Gatterman-Koch Reaction (III).

Q: Match List-I (Reagents involving aldehydes) with List-II (Reaction Name): List-I: A. H2, Pd/BaSO4 B. SnCl2, HCl C. CrO2Cl2, CS2 D. CO, HCl / Anhyd. AlCl3 List-II: I. Etard Reaction II. Rosenmund Reduction III. Gatterman-Koch Reaction IV. Stephen Reaction Choose the correct answer:

Answer: D. (4) A-II, B-IV, C-I, D-III — A matches with II (Rosenmund Reduction), B matches with IV (Stephen Reaction), C matches with I (Etard Reaction), and D matches with III (Gatterman-Koch Reaction).

Q: Grignard reagent RMgBr (P) reacts with water and forms a gas (Q). One gram of Q occupies 1.4 dm3 at STP. (P) on reaction with dry ice in dry ether followed by H3O+ forms a compound (Z). 0.1 mole of (Z) will weigh ____ g. (Nearest integer)

Answer: C. 6 — Molar mass of gas Q = (mass/volume) * 22.4 = (1/1.4) * 22.4 = 16 g/mol. Gas Q is methane (CH4). Therefore, RMgBr is CH3MgBr. CH3MgBr + CO2 -> CH3COOH (Z). Molar mass of CH3COOH = 60 g/mol. 0.1 mole = 6 g.

Q: Given below are two statements: Statement I: Phenol on treatment with CHCl3/aq. KOH under refluxing condition, followed by acidification produces p-hydroxy benzaldehyde as the major product and o-hydroxy benzaldehyde as the minor product. Statement II: The mixture of p-hydroxybenzaldehyde and o-hydroxybenzaldehyde can be easily separated through steam distillation. In the light of the above statements, choose the correct answer from the options given below:

Answer: D. Statement I is false but Statement II is true — Statement I is false because Reimer-Tiemann reaction produces o-hydroxybenzaldehyde (salicylaldehyde) as the major product due to intramolecular hydrogen bonding. Statement II is true because ortho-isomer is volatile due to intramolecular H-bonding and can be separated from para-isomer by steam distillation.

Q: Incorrect method of preparation for alcohols from the following is:

Answer: C. Ozonolysis of alkene. — Hydroboration-oxidation of alkenes gives alcohols, and alkyl halides with aqueous $\mathrm{NaOH}$ undergo nucleophilic substitution to alcohols. Ketones react with $\mathrm{RMgBr}$ followed by hydrolysis to give tertiary alcohols. Ozonolysis of alkenes, however, cleaves the double bond to give carbonyl compounds, not alcohols. Therefore the incorrect method is ozonolysis of alkene.

Q: The water gas on reacting with cobalt as a catalyst forms

Answer: B. Methanol — This is the catalytic hydrogenation of water gas ($\mathrm{CO + H_2}$) over cobalt catalyst to give methanol: $\mathrm{CO + 2H_2 \rightarrow CH_3OH}$. Therefore the product is methanol.

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Practice Unit 17: Organic Compounds Containing Oxygen with a free chapter-wise online MCQ test for JEE JEE MAINS CHEMISTRY. This chapter covers: General methods of preparation properties reactions and uses. ALCOHOLS PHENOLS AND ETHERS Alcohols: Identification of primary secondary and tertiary alcohols mechanism of dehydrati.... AI-generated questions from basic to board-exam level, with instant results and explanations.

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Unit 17: Organic Compounds Containing Oxygen — Important Questions & Answers (FAQ)

Frequently asked questions from JEE JEE MAINS CHEMISTRY — Unit 17: Organic Compounds Containing Oxygen, with answers and explanations. These are sample questions; the exam has its own separate question set.

The product of all solutions of the equation 5^(log_e x)^2 + 3^(log_e x)^2 = x^8, x > 0 is:

A. e ✓
B. e^(8/5)
C. e^(6/5)
D. e^2

Answer: A. e
Taking logs or substituting t = (ln x)^2, the equation simplifies to find log values. Product of roots corresponds to sum of exponents equal to 2, leading to e^2.

The compounds which give positive Fehling's test are: (A) HOCH2-CO-(CHOH)3-CH2-OH ... [Other structures omitted in text]. Choose the correct answer from the options given below:

A. (A), (D) and (E) Only
B. (C), (D) and (E) Only ✓
C. (A), (C) and (D) Only
D. (A), (B) and (C) Only

Answer: B. (C), (D) and (E) Only
Fehling's test is positive for aliphatic aldehydes and alpha-hydroxy ketones (ketoses).

Match List-I with List-II.

A. (1) A-II, B-III, C-IV, D-I
B. (2) A-IV, B-III, C-I, D-II
C. (3) A-IV, B-I, C-II, D-III
D. (4) A-II, B-IV, C-I, D-III ✓

Answer: D. (4) A-II, B-IV, C-I, D-III
A(H2/Pd, BaSO4) = Rosenmund Reduction (II). B(SnCl2, HCl) = Stephen Reaction (IV). C(CrO2Cl2, CS2) = Etard Reaction (I). D(CO, HCl, Anhyd. AlCl3) = Gatterman-Koch Reaction (III).

Match List-I (Reagents involving aldehydes) with List-II (Reaction Name): List-I: A. H2, Pd/BaSO4 B. SnCl2, HCl C. CrO2Cl2, CS2 D. CO, HCl / Anhyd. AlCl3 List-II: I. Etard Reaction II. Rosenmund Reduction III. Gatterman-Koch Reaction IV. Stephen Reaction Choose the correct answer:

A. (1) A-II, B-III, C-IV, D-I
B. (2) A-IV, B-III, C-I, D-II
C. (3) A-IV, B-I, C-II, D-III
D. (4) A-II, B-IV, C-I, D-III ✓

Answer: D. (4) A-II, B-IV, C-I, D-III
A matches with II (Rosenmund Reduction), B matches with IV (Stephen Reaction), C matches with I (Etard Reaction), and D matches with III (Gatterman-Koch Reaction).

Grignard reagent RMgBr (P) reacts with water and forms a gas (Q). One gram of Q occupies 1.4 dm3 at STP. (P) on reaction with dry ice in dry ether followed by H3O+ forms a compound (Z). 0.1 mole of (Z) will weigh ____ g. (Nearest integer)

A. 4
B. 5
C. 6 ✓
D. 7

Answer: C. 6
Molar mass of gas Q = (mass/volume) * 22.4 = (1/1.4) * 22.4 = 16 g/mol. Gas Q is methane (CH4). Therefore, RMgBr is CH3MgBr. CH3MgBr + CO2 -> CH3COOH (Z). Molar mass of CH3COOH = 60 g/mol. 0.1 mole = 6 g.

Given below are two statements: Statement I: Phenol on treatment with CHCl3/aq. KOH under refluxing condition, followed by acidification produces p-hydroxy benzaldehyde as the major product and o-hydroxy benzaldehyde as the minor product. Statement II: The mixture of p-hydroxybenzaldehyde and o-hydroxybenzaldehyde can be easily separated through steam distillation. In the light of the above statements, choose the correct answer from the options given below:

A. Both Statement I and Statement II are false
B. Statement I is true but Statement II is false
C. Both Statement I and Statement II are true
D. Statement I is false but Statement II is true ✓

Answer: D. Statement I is false but Statement II is true
Statement I is false because Reimer-Tiemann reaction produces o-hydroxybenzaldehyde (salicylaldehyde) as the major product due to intramolecular hydrogen bonding. Statement II is true because ortho-isomer is volatile due to intramolecular H-bonding and can be separated from para-isomer by steam distillation.

The major product 'P' formed in the given reaction is

A. Compound with two $COOH$ groups and one side-chain $CH_2COOH$, plus reduced aromatic ring substituents
B. A fused-ring dicarboxylic acid with two $COOH$ groups and one $CH_2COOH$ group
C. Fused-ring compound containing $CH_2COOH$ and $COCH_3$ side chains
D. A fused-ring tricarboxylic acid with three $COOH$ groups ✓

Answer: D. A fused-ring tricarboxylic acid with three $COOH$ groups
Hot alkaline $KMnO_4$ oxidizes all oxidizable side chains on aromatic rings to $COOH$ groups. Here, the benzylic $CH_2CH_3$ group is oxidized to $COOH$, the vinyl side chain $CH=CH_2$ is cleaved and oxidized to $COOH$, and the ester side chain $CH_2COOCH_3$ is hydrolyzed under the strongly oxidative/alkaline conditions and the side chain is ultimately converted to $COOH$ after acidification. Thus the product is the tricarboxylic acid corresponding to the fused aromatic framework, which matches option D.

Incorrect method of preparation for alcohols from the following is:

A. Hydroboration-oxidation of alkene.
B. Reaction of ketone with $\mathrm{RMgBr}$ followed by hydrolysis.
C. Ozonolysis of alkene. ✓
D. Reaction of alkyl halide with aqueous $\mathrm{NaOH}$.

Answer: C. Ozonolysis of alkene.
Hydroboration-oxidation of alkenes gives alcohols, and alkyl halides with aqueous $\mathrm{NaOH}$ undergo nucleophilic substitution to alcohols. Ketones react with $\mathrm{RMgBr}$ followed by hydrolysis to give tertiary alcohols. Ozonolysis of alkenes, however, cleaves the double bond to give carbonyl compounds, not alcohols. Therefore the incorrect method is ozonolysis of alkene.

$A$ and $B$ in the above reactions are:

A. $A = \mathrm{CO_2H}$-terminated keto acid; $B = \mathrm{CO_2H}$-terminated alkane
B. A dicarboxylic acid derivative and B a hydrazone-containing product
C. $A$ is a keto acid and $B$ is the corresponding open-chain acid after Wolff-Kishner reduction
D. $A$ is a keto acid and $B$ is the corresponding hydrocarbon with one carbonyl reduced to $\mathrm{CH_2}$ ✓

Answer: D. $A$ is a keto acid and $B$ is the corresponding hydrocarbon with one carbonyl reduced to $\mathrm{CH_2}$
Hot alkaline $\mathrm{KMnO_4}$ oxidatively cleaves the side chain of the cyclohexanone derivative, giving ring opening to a keto acid $A$ with a terminal $\mathrm{CO_2H}$ group. Then treatment with $\mathrm{NH_2NH_2}$, $\mathrm{KOH}$ followed by $\mathrm{H_3O^+}$ performs Wolff-Kishner reduction, converting the ketone carbonyl in $A$ to $\mathrm{CH_2}$ while leaving the carboxylic acid unchanged. Therefore $B$ is the corresponding alkyl carboxylic acid, i.e. the carbonyl has been reduced to methylene.

The water gas on reacting with cobalt as a catalyst forms

A. Methanal
B. Methanol ✓
C. Methanoic acid
D. Ethanol

Answer: B. Methanol
This is the catalytic hydrogenation of water gas ($\mathrm{CO + H_2}$) over cobalt catalyst to give methanol: $\mathrm{CO + 2H_2 \rightarrow CH_3OH}$. Therefore the product is methanol.

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