CUET UG Chemistry Booster Test -2 Nomenclature and Bond Nature
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QUESTION 1 OF 20
Consider the properties of lower members of carbonyls:
1. Methanal is a gas at room temperature.
2. Ethanal is a volatile liquid.
3. Propanone is a solid at room temperature.
4. Lower aldehydes have a sharp pungent odour.
QUESTION 2 OF 20
Why are most higher members of aldehydes and ketones liquid or solid at room temperature?
QUESTION 3 OF 20
Arrange the following compounds in increasing order of their boiling points:
(A) n-Pentane
(B) Ethoxyethane
(C) Butanal
(D) Butan-1-ol
QUESTION 4 OF 20
Butan-1-ol has a significantly higher boiling point than butanal because:
QUESTION 5 OF 20
Read the statements about solubility:
1. Lower aldehydes are miscible in water due to hydrogen bond formation with water.
2. Increasing the alkyl chain length increases the solubility in water.
3. All aldehydes are fairly soluble in organic solvents.
4. Higher carboxylic acids are miscible in cold water.
QUESTION 6 OF 20
Match the following:
| List-I | List-II |
|---|---|
| 1. Propanone in water | a. Miscible in all proportions |
| 2. Increasing length of alkyl chain | b. Decreases solubility rapidly |
| 3. Aldehydes and ketones in ether | c. Fairly soluble |
| 4. Lower aldehydes | d. Hydrogen bonding with water |
QUESTION 7 OF 20
As the size of the aldehyde molecule increases, the characteristic sharp pungent odour:
QUESTION 8 OF 20
The unit of percentage for formalin, a well-known formaldehyde solution used to preserve biological specimens, is:
QUESTION 9 OF 20
Based on the passage, the nucleophile attacks the carbonyl carbon from which specific direction?
QUESTION 10 OF 20
Based on the passage, what specific geometric intermediate is produced during the hybridization change?
QUESTION 11 OF 20
Arrange the following compounds in decreasing order of their reactivity in nucleophilic addition reactions:
(A) p-Nitrobenzaldehyde
(B) Benzaldehyde
(C) p-Tolualdehyde
(D) Acetophenone
QUESTION 12 OF 20
Why is benzaldehyde less reactive than propanal in nucleophilic addition?
1. The carbon atom of the carbonyl group of benzaldehyde is more electrophilic.
2. The carbon atom of the carbonyl group of benzaldehyde is less electrophilic due to resonance.
3. The polarity of the carbonyl group is reduced in benzaldehyde.
4. Benzaldehyde has two large substituents attached to the carbonyl carbon.
QUESTION 13 OF 20
Identify the reaction step/type correctly describing the role of base in cyanohydrin formation:
QUESTION 14 OF 20
Why does the equilibrium for sodium hydrogensulphite addition lie largely to the left for most ketones?
QUESTION 15 OF 20
The cyclic products formed when ketones react with ethylene glycol under dry hydrogen chloride conditions are named:
QUESTION 16 OF 20
Match List-I (Carbonyl reactant + Alcohol) with List-II (Role of dry HCl catalyst):
| List-I | List-II |
|---|---|
| 1. Aldehyde + 1 equivalent of alcohol | a. Increases electrophilicity of carbonyl carbon by protonating oxygen |
| 2. Ketone + Ethylene glycol | b. Acts as a reducing agent |
| 3. Aldehyde + 2 equivalents of alcohol | c. Neutralizes the basic medium |
| 4. Carbonyl compound + HCN | d. Generates CN⁻ ion |
QUESTION 17 OF 20
Identify the reaction type for the intermediate step that converts the initial ammonia derivative adduct into the final >C=N-Z product:
QUESTION 18 OF 20
The name of the yellow, orange, or red solid derivative formed when an aldehyde reacts with 2,4-Dinitrophenylhydrazine:
QUESTION 19 OF 20
Match List-I (Reduction Process) with List-II (Reagents):
| List-I | List-II |
|---|---|
| 1. Clemmensen reduction | a. Hydrazine followed by heating with KOH in ethylene glycol |
| 2. Wolff-Kishner reduction | b. Zinc-amalgam and concentrated hydrochloric acid |
| 3. NaBH₄ reduction | c. Sodium borohydride |
| 4. LiAlH₄ reduction | d. Lithium aluminium hydride |
QUESTION 20 OF 20
Identify the overall reaction type taking place when an aldehyde is treated with Fehling's reagent (resulting in a reddish brown precipitate):
Test Complete!
Answer Review
1 Consider the properties of lower members of carbonyls:
1. Methanal is a gas at room temperature.
2. Ethanal is a volatile liquid.
3. Propanone is a solid at room temperature.
4. Lower aldehydes have a sharp pungent odour.
�� Methanal is a gas. �� Ethanal is a volatile liquid. �� Lower aldehydes have pungent odour.
Methanal exists as a gas at room temperature, while ethanal is a volatile liquid. Lower aldehydes generally have a sharp pungent smell. Propanone is not a solid at room temperature; it is a liquid.
- �� Options B, C and D include incorrect statement 3.
Used
- Statement Analysis
- Methanal gas, ethanal volatile liquid.
2 Why are most higher members of aldehydes and ketones liquid or solid at room temperature?
�� Higher members have larger molecular size. �� Intermolecular forces increase. �� Hence they become liquids or solids.
As molecular size increases, van der Waals forces become stronger. Aldehydes and ketones also show dipole-dipole interactions due to the polar carbonyl group, making higher members liquids or solids.
- �� Option A: Molecular mass increases, not decreases.
- �� Option C: Aldehydes and ketones do not show intermolecular hydrogen bonding among themselves.
- �� Option D: Carbonyl polarity does not decrease as the main reason.
Used
- Concept MCQ
- Bigger molecule = stronger attraction = liquid/solid.
3 Arrange the following compounds in increasing order of their boiling points:
(A) n-Pentane
(B) Ethoxyethane
(C) Butanal
(D) Butan-1-ol
�� Alkane has lowest boiling point. �� Ether has weak polarity. �� Aldehyde has dipole-dipole forces. �� Alcohol has hydrogen bonding.
Increasing boiling point follows increasing intermolecular force strength: n-Pentane < Ethoxyethane < Butanal < Butan-1-ol
- �� Options B, C and D do not follow the correct intermolecular force trend.
Used
- Ordering
- Alkane < Ether < Aldehyde < Alcohol.
4 Butan-1-ol has a significantly higher boiling point than butanal because:
�� Alcohols form intermolecular hydrogen bonds. �� Aldehydes lack O–H bond. �� Stronger attraction increases boiling point.
Butan-1-ol contains an –OH group and forms strong intermolecular hydrogen bonds. Butanal has dipole-dipole interactions but cannot form hydrogen bonds with itself.
- �� Option A: Butanal is polar.
- �� Option C: Molecular masses are comparable.
- �� Option D: Butanal does not show this as the main reason.
Used
- Concept MCQ
- Alcohol = H-bond = high boiling point.
5 Read the statements about solubility:
1. Lower aldehydes are miscible in water due to hydrogen bond formation with water.
2. Increasing the alkyl chain length increases the solubility in water.
3. All aldehydes are fairly soluble in organic solvents.
4. Higher carboxylic acids are miscible in cold water.
�� Lower aldehydes form H-bonds with water. �� Longer alkyl chain reduces water solubility. �� Aldehydes dissolve in organic solvents.
Lower aldehydes are soluble in water because the carbonyl oxygen forms hydrogen bonds with water. However, solubility decreases as the alkyl chain length increases. Aldehydes are generally fairly soluble in organic solvents.
- �� Statement 2 is incorrect.
- �� Statement 4 is not correct in this context.
Used
- Statement Analysis
- Longer chain = less water solubility.
6 Match the following:
| List-I | List-II |
|---|---|
| 1. Propanone in water | a. Miscible in all proportions |
| 2. Increasing length of alkyl chain | b. Decreases solubility rapidly |
| 3. Aldehydes and ketones in ether | c. Fairly soluble |
| 4. Lower aldehydes | d. Hydrogen bonding with water |
�� Propanone is miscible with water. �� Longer alkyl chain reduces solubility. �� Ether dissolves aldehydes and ketones.
Propanone is a lower ketone and is miscible with water in all proportions. Increasing alkyl chain length reduces water solubility. Aldehydes and ketones are fairly soluble in organic solvents like ether. Lower aldehydes dissolve in water because they can form hydrogen bonds with water.
- �� Options B, C and D contain incorrect matching.
Used
- Match the Following
- Lower carbonyls mix with water; longer chains reduce solubility.
7 As the size of the aldehyde molecule increases, the characteristic sharp pungent odour:
�� Lower aldehydes are pungent. �� Larger aldehydes are less pungent. �� Many have pleasant fragrance.
Lower aldehydes have sharp pungent odours. With increase in molecular size, the odour becomes less pungent and more pleasant or fragrant.
- �� Options A, B and D are not standard odour trends.
Used
- Concept MCQ
- Small aldehyde = pungent; large aldehyde = fragrant.
8 The unit of percentage for formalin, a well-known formaldehyde solution used to preserve biological specimens, is:
�� Formalin is aqueous formaldehyde. �� It contains about 40% formaldehyde. �� Used for preserving specimens.
Formalin is a commercial aqueous solution of formaldehyde containing approximately 40% formaldehyde. It is commonly used to preserve biological specimens.
- �� Options A, B and D are not the standard formalin concentration.
Used
- Data Recall
- Formalin = 40% formaldehyde.
9 Based on the passage, the nucleophile attacks the carbonyl carbon from which specific direction?
�� Carbonyl carbon is sp² hybridised. �� Nucleophile attacks from outside the plane. �� Attack is approximately perpendicular.
The passage states that the nucleophile attacks the electrophilic carbon atom of the polar carbonyl group from a direction approximately perpendicular to the plane of the sp² hybridised orbitals.
- �� Option A: Not the stated direction.
- �� Option B: Attack is not within the plane.
- �� Option D: Not the correct description.
Used
- Passage-Based MCQ
- Carbonyl attack happens from above or below the plane.
10 Based on the passage, what specific geometric intermediate is produced during the hybridization change?
�� Carbon changes from sp² to sp³. �� sp³ geometry is tetrahedral. �� Alkoxide intermediate is formed.
During nucleophilic addition, carbonyl carbon changes from sp² to sp³ hybridisation. This produces a tetrahedral alkoxide intermediate, which later accepts a proton to give the neutral product.
- �� Option A: Starting carbonyl is trigonal planar, not the intermediate.
- �� Option C: sp³ carbon is not linear.
- �� Option D: Octahedral geometry is not involved.
Used
- Passage-Based MCQ
- sp² carbonyl → sp³ tetrahedral alkoxide.
11 Arrange the following compounds in decreasing order of their reactivity in nucleophilic addition reactions:
(A) p-Nitrobenzaldehyde
(B) Benzaldehyde
(C) p-Tolualdehyde
(D) Acetophenone
�� Electron-withdrawing groups increase carbonyl reactivity. �� Electron-donating groups decrease reactivity. �� Ketones are less reactive than aldehydes.
p-Nitrobenzaldehyde is most reactive because the –NO₂ group withdraws electron density and increases the electrophilic nature of the carbonyl carbon. Benzaldehyde is less reactive than p-nitrobenzaldehyde. p-Tolualdehyde is still less reactive because the –CH₃ group donates electron density. Acetophenone is least reactive because ketones are more sterically hindered and electronically less reactive than aldehydes.
- �� Options B, C and D do not follow the correct electronic and steric reactivity trend.
Used
- Ordering
- –NO₂ increases reactivity; –CH₃ decreases reactivity; ketone least reactive.
12 Why is benzaldehyde less reactive than propanal in nucleophilic addition?
1. The carbon atom of the carbonyl group of benzaldehyde is more electrophilic.
2. The carbon atom of the carbonyl group of benzaldehyde is less electrophilic due to resonance.
3. The polarity of the carbonyl group is reduced in benzaldehyde.
4. Benzaldehyde has two large substituents attached to the carbonyl carbon.
�� Benzaldehyde shows resonance with benzene ring. �� This reduces carbonyl electrophilicity. �� Reduced polarity lowers nucleophilic addition reactivity.
In benzaldehyde, the carbonyl group is conjugated with the benzene ring. Due to resonance, electron density is delocalised and the carbonyl carbon becomes less electrophilic. The polarity of the carbonyl group is also reduced compared to aliphatic aldehydes like propanal, making benzaldehyde less reactive toward nucleophilic addition.
- �� Statement 1 is incorrect because benzaldehyde carbonyl carbon is less electrophilic.
- �� Statement 4 is incorrect because benzaldehyde has one phenyl group and one hydrogen, not two large substituents.
Used
- Statement Analysis
- Benzene resonance reduces carbonyl reactivity.
13 Identify the reaction step/type correctly describing the role of base in cyanohydrin formation:
�� HCN is weakly ionised. �� Base helps generate CN⁻. �� CN⁻ attacks carbonyl carbon.
In cyanohydrin formation, hydrogen cyanide reacts with aldehydes or ketones in the presence of a base. The base helps produce cyanide ion (CN⁻), which is a stronger nucleophile and attacks the electrophilic carbonyl carbon.
- �� Option A: HCN is not oxidised.
- �� Option C: Nucleophile attacks carbonyl carbon, not electrophilic addition to oxygen.
- �� Option D: Water elimination is not the key step.
Used
- Reaction Role Identification
- Base makes CN⁻; CN⁻ attacks C=O.
14 Why does the equilibrium for sodium hydrogensulphite addition lie largely to the left for most ketones?
�� Ketones have two alkyl groups. �� These groups cause steric hindrance. �� Addition product formation becomes less favourable.
Most ketones do not readily form stable sodium hydrogensulphite addition products because the two alkyl groups around the carbonyl carbon create steric hindrance. This makes nucleophilic addition less favourable, so the equilibrium lies largely to the left.
- �� Option A: Methyl group absence is not the reason.
- �� Option C: Ketones are generally less electrophilic than aldehydes.
- �� Option D: Precipitation would drive equilibrium to the right, not left.
Used
- Concept MCQ
- Ketones are crowded, so bisulphite addition is difficult.
15 The cyclic products formed when ketones react with ethylene glycol under dry hydrogen chloride conditions are named:
�� Ketones react with ethylene glycol. �� Dry HCl acts as acid catalyst. �� Cyclic ketals are formed.
Ketones react with ethylene glycol in the presence of dry HCl to form cyclic ketals known as ethylene glycol ketals. These are useful for protecting carbonyl groups.
- �� Option A: Formed by HCN addition.
- �� Option B: Formed from aldehyde and one equivalent of alcohol.
- �� Option D: Formed with hydroxylamine.
Used
- Naming
- Ketone + Ethylene glycol = Ketal.
16 Match List-I (Carbonyl reactant + Alcohol) with List-II (Role of dry HCl catalyst):
| List-I | List-II |
|---|---|
| 1. Aldehyde + 1 equivalent of alcohol | a. Increases electrophilicity of carbonyl carbon by protonating oxygen |
| 2. Ketone + Ethylene glycol | b. Acts as a reducing agent |
| 3. Aldehyde + 2 equivalents of alcohol | c. Neutralizes the basic medium |
| 4. Carbonyl compound + HCN | d. Generates CN⁻ ion |
�� Dry HCl protonates the carbonyl oxygen. �� Protonation increases electrophilicity of carbonyl carbon. �� CN⁻ is the active nucleophile in cyanohydrin formation.
In reactions of aldehydes and ketones with alcohols, dry HCl acts as an acid catalyst by protonating the carbonyl oxygen. This increases the electrophilic character of the carbonyl carbon, facilitating nucleophilic attack by alcohol molecules. In cyanohydrin formation, CN⁻ is the attacking nucleophile.
- �� Options B, C and D assign incorrect catalytic roles to dry HCl.
- �� Dry HCl is neither a reducing agent nor a neutralizing agent.
Used
- Match the Following
- Dry HCl = Protonates O → Activates C=O
17 Identify the reaction type for the intermediate step that converts the initial ammonia derivative adduct into the final >C=N-Z product:
�� Ammonia derivative first forms an addition product. �� Water is eliminated from the intermediate. �� A C=N bond is formed.
Aldehydes and ketones react with ammonia derivatives through nucleophilic addition followed by dehydration. The dehydration step converts the unstable intermediate into a stable compound containing the >C=N-Z linkage.
- �� Option A: No substitution occurs.
- �� Option C: Decarboxylation is not involved.
- �� Option D: Water is removed, not added.
Used
- Reaction Type Identification
- Ammonia derivative + Carbonyl → Addition → Dehydration → C=N
18 The name of the yellow, orange, or red solid derivative formed when an aldehyde reacts with 2,4-Dinitrophenylhydrazine:
�� 2,4-DNP reagent reacts with carbonyl compounds. �� Colored precipitates are formed. �� Product is called 2,4-dinitrophenylhydrazone.
Aldehydes and ketones react with 2,4-dinitrophenylhydrazine (Brady's reagent) to form yellow, orange, or red crystalline derivatives known as 2,4-dinitrophenylhydrazones. This reaction is widely used for identification of carbonyl compounds.
- �� Option A: Formed with semicarbazide.
- �� Option B: Formed with primary amines.
- �� Option D: Formed with hydroxylamine.
Used
- Naming
- 2,4-DNP Test = Colored Hydrazone
19 Match List-I (Reduction Process) with List-II (Reagents):
| List-I | List-II |
|---|---|
| 1. Clemmensen reduction | a. Hydrazine followed by heating with KOH in ethylene glycol |
| 2. Wolff-Kishner reduction | b. Zinc-amalgam and concentrated hydrochloric acid |
| 3. NaBH₄ reduction | c. Sodium borohydride |
| 4. LiAlH₄ reduction | d. Lithium aluminium hydride |
�� Clemmensen uses Zn-Hg/HCl. �� Wolff-Kishner uses hydrazine and KOH. �� NaBH₄ and LiAlH₄ are reducing agents. �� Both reduce carbonyl compounds to alcohols.
Reduction Process — Reagent Clemmensen Reduction — Zn-Hg / HCl Wolff-Kishner Reduction — NH₂NH₂ / KOH NaBH₄ Reduction — Sodium borohydride LiAlH₄ Reduction — Lithium aluminium hydride
- �� Options B, C and D contain incorrect reagent-process matching.
Used
- Match the Following
- Wolff-Kishner = Basic (NH₂NH₂/KOH)
20 Identify the overall reaction type taking place when an aldehyde is treated with Fehling's reagent (resulting in a reddish brown precipitate):
�� Fehling's reagent oxidizes aldehydes. �� Cu²⁺ is reduced to Cu₂O. �� Reddish-brown precipitate confirms aldehyde.
Fehling's solution oxidizes aliphatic aldehydes to the corresponding carboxylate ions in alkaline medium. During the reaction, Cu²⁺ ions are reduced to cuprous oxide (Cu₂O), which appears as a reddish-brown precipitate.
- �� Option A: No nucleophilic addition occurs.
- �� Option C: Aldehyde is oxidized, not reduced.
- �� Option D: Cannizzaro reaction is disproportionation, not Fehling's test.
Used
- Reaction Type Identification
- Fehling's Test = Brick-red Cu₂O + Aldehyde Oxidation
