CUET Chemistry Set-5: Organic Chemistry - Functional Groups I
Question 1
Which of the following is the correct IUPAC name for $CH_3CH_2CH(OH)CH_2CH_3$?
(1) Pentan-2-ol
(2) Pentan-3-ol
(3) 2-Pentanol
(4) 3-Pentanol
Solution: The hydroxyl group is on the third carbon of a 5-carbon chain, hence pentan-3-ol.Show Answer
Answer: (2)
Question 2
The reagent used to distinguish between $1^\circ$, $2^\circ$ and $3^\circ$ alcohols is
(1) Lucas reagent
(2) Tollens’ reagent
(3) Fehling’s solution
(4) Benedict’s solution
Solution: Lucas reagent (conc. HCl + ZnCl₂) gives immediate turbidity with $3^\circ$, delayed with $2^\circ$, and no reaction with $1^\circ$ alcohols at room temperature.Show Answer
Answer: (1)
Question 3
Which of the following alcohols will give a positive iodoform test?
(1) Ethanol
(2) Propan-1-ol
(3) Butan-1-ol
(4) Pentan-1-ol
Solution: Ethanol has the $CH_3CH(OH)$- group, which gives a positive iodoform test.Show Answer
Answer: (1)
Question 4
The dehydration of alcohols to alkenes is catalysed by
(1) $H_2SO_4$
(2) $NaOH$
(3) $KMnO_4$
(4) $LiAlH_4$
Solution: Conc. $H_2SO_4$ acts as a dehydrating agent, removing water to form an alkene.Show Answer
Answer: (1)
Question 5
The order of reactivity of alcohols towards dehydration is
(1) $1^\circ > 2^\circ > 3^\circ$
(2) $3^\circ > 2^\circ > 1^\circ$
(3) $2^\circ > 3^\circ > 1^\circ$
(4) $1^\circ = 2^\circ = 3^\circ$
Solution: $3^\circ$ carbocations are most stable, so $3^\circ$ alcohols dehydrate fastest.Show Answer
Answer: (2)
Question 6
Phenol is more acidic than ethanol because
(1) Phenoxide ion is resonance-stabilised
(2) Ethoxide ion is resonance-stabilised
(3) Phenol has higher molar mass
(4) Ethanol is more soluble in water
Solution: The negative charge on the phenoxide ion is delocalised over the benzene ring via resonance, increasing acidity.Show Answer
Answer: (1)
Question 7
Which of the following will not give effervescence with NaHCO₃?
(1) Benzoic acid
(2) Phenol
(3) Acetic acid
(4) Formic acid
Solution: Phenol is weaker than carbonic acid and hence does not decompose NaHCO₃.Show Answer
Answer: (2)
Question 8
The product of Kolbe’s reaction (sodium phenoxide + CO₂) is
(1) Salicylaldehyde
(2) Salicylic acid
(3) p-Hydroxybenzoic acid
(4) Phenyl acetate
Solution: Ortho-carboxylation occurs yielding salicylic acid.Show Answer
Answer: (2)
Question 9
Ether formation from alcohols in presence of conc. $H_2SO_4$ at $140^\circ C$ is called
(1) Williamson synthesis
(2) Hydration
(3) Intermolecular dehydration
(4) Hydroboration
Solution: Two alcohol molecules lose water to form an ether.Show Answer
Answer: (3)
Question 10
The IUPAC name of $CH_3OCH_2CH_3$ is
(1) Methoxyethane
(2) Ethoxymethane
(3) 1-Methoxyethane
(4) Methyl ethyl ether
Solution: Alphabetical order of alkyl groups gives methoxyethane.Show Answer
Answer: (1)
Question 11
Which reagent converts an alkyl halide to an ether?
(1) Sodium alkoxide
(2) $PCl_5$
(3) $SOCl_2$
(4) $KCN$
Solution: Williamson synthesis uses sodium alkoxide + alkyl halide.Show Answer
Answer: (1)
Question 12
Cleavage of ethers with HI gives
(1) Alcohol + alkyl iodide
(2) Only alcohol
(3) Only alkyl iodide
(4) Aldehyde + ketone
Solution: Ethers undergo nucleophilic cleavage to give an alcohol and an alkyl iodide.Show Answer
Answer: (1)
Question 13
The functional group present in carboxylic acids is
(1) $-CHO$
(2) $-CO-$
(3) $-COOH$
(4) $-OH$
Solution: Carboxylic acids contain the carboxyl $-COOH$ group.Show Answer
Answer: (3)
Question 14
The acid strength of $CH_3COOH, ClCH_2COOH, Cl_2CHCOOH, Cl_3CCOOH$ decreases in the order
(1) $Cl_3CCOOH > Cl_2CHCOOH > ClCH_2COOH > CH_3COOH$
(2) $CH_3COOH > ClCH_2COOH > Cl_2CHCOOH > Cl_3CCOOH$
(3) $ClCH_2COOH > Cl_2CHCOOH > Cl_3CCOOH > CH_3COOH$
(4) $Cl_2CHCOOH > Cl_3CCOOH > CH_3COOH > ClCH_2COOH$
Solution: Increasing –I effect of Cl stabilises the conjugate base, increasing acid strength.Show Answer
Answer: (1)
Question 15
Hell–Volhard–Zelinsky reaction converts carboxylic acid into
(1) $\alpha$-haloacid
(2) $\beta$-haloacid
(3) $\alpha$-hydroxyacid
(4) Ester
Solution: $P/X_2$ halogenates the $\alpha$-carbon.Show Answer
Answer: (1)
Question 16
Esterification is
(1) Acid + Alcohol $\rightarrow$ Ester + Water
(2) Acid + Base $\rightarrow$ Salt + Water
(3) Alcohol + HX $\rightarrow$ Alkyl halide + Water
(4) Acid + Amine $\rightarrow$ Amide + Water
Solution: Fischer esterification condenses acid and alcohol.Show Answer
Answer: (1)
Question 17
The reagent that reduces carboxylic acid to primary alcohol is
(1) $LiAlH_4$
(2) $NaBH_4$
(3) $KMnO_4$
((4) $Br_2/aq$
Solution: $LiAlH_4$ is a strong reducing agent for $-COOH$ to $-CH_2OH$.Show Answer
Answer: (1)
Question 18
Decarboxylation of sodium acetate gives
(1) Methane + $CO_2$
(2) Ethane + $CO_2$
(3) Ethene + $CO_2$
(4) Acetone + $CO_2$
Solution: Electrolytic decarboxylation yields methane and carbon dioxide.Show Answer
Answer: (1)
Question 19
The IUPAC name of $(CH_3)_2CHCOOH$ is
(1) 2-Methylpropanoic acid
(2) Butanoic acid
(3) 2,2-Dimethylacetic acid
(4) Isobutyric acid
Solution: Longest chain is 3 carbons with a methyl substituent at C-2.Show Answer
Answer: (1)
Question 20
Which of the following will give a silver mirror with Tollens’ reagent?
(1) Acetaldehyde
(2) Acetone
(3) Acetic acid
(4) Ethanol
Solution: Aldehydes reduce Tollens’ reagent to metallic silver.Show Answer
Answer: (1)
Question 21
Aldol condensation requires
(1) $\alpha$-hydrogens
(2) No $\alpha$-hydrogens
(3) Only ketones
(4) Only aromatic aldehydes
Solution: $\alpha$-hydrogens are necessary for enolate formation.Show Answer
Answer: (1)
Question 22
Cannizzaro reaction is shown by
(1) Aldehydes without $\alpha$-hydrogens
(2) Ketones with $\alpha$-hydrogens
(3) All aldehydes
(4) All ketones
Solution: Non-enolisable aldehydes undergo disproportionation.Show Answer
Answer: (1)
Question 23
The product of Rosenmund reduction is
(1) Primary alcohol
(2) Aldehyde
(3) Ketone
(4) Carboxylic acid
Solution: Acyl chloride + $H_2/Pd–BaSO_4$ gives aldehyde.Show Answer
Answer: (2)
Question 24
Which of the following has the highest boiling point?
(1) $CH_3CHO$
(2) $CH_3CH_2OH$
(3) $CH_3COCH_3$
(4) $CH_3COOH$
Solution: Carboxylic acids form strong hydrogen-bonded dimers.Show Answer
Answer: (4)
Question 25
The number of $\alpha$-hydrogens in acetophenone is
(1) 2
(2) 3
(3) 4
(4) 5
Solution: Methyl group attached to carbonyl has 3 $\alpha$-hydrogens.Show Answer
Answer: (2)
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