Organic Chemistry

102 questions

Question 61Question

Match each class of alkanol listed on the left with its characteristic oxidation behavior on the right when reacted with acidified potassium heptaoxodichromate(VI) solution.

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Items

Primary alkanol
Secondary alkanol
Tertiary alkanol

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Answer

Primary alkanols pair with oxidation to an alkanal and then an alkanoic acid; secondary alkanols pair with oxidation to an alkanone; tertiary alkanols pair with resistance to oxidation under mild conditions.
Primary alkanols possess two alpha-hydrogens and oxidize in two steps to form alkanals and then alkanoic acids. Secondary alkanols possess one alpha-hydrogen and oxidize to form alkanones. Tertiary alkanols lack alpha-hydrogens entirely, rendering them resistant to oxidation under mild conditions.

Step-by-Step Solution

1
Examine the structural environment of primary alkanols (RCH2OHR-CH_2OH)
Primary alkanols have two α\alpha-hydrogen atoms attached to the carbon holding the OH-OH group, permitting two sequential oxidation steps.
Oxidation requires the removal of hydrogen from the hydroxyl-bearing carbon atom.
2
Examine the structural environment of secondary alkanols (R2CHOHR_2CHOH)
Secondary alkanols have only one α\alpha-hydrogen atom, yielding an alkanone (R2C=OR_2C=O).
Alkanones resist further oxidation under mild conditions because no additional α\alpha-hydrogens are available.
3
Examine the structural environment of tertiary alkanols (R3COHR_3COH)
Tertiary alkanols possess zero α\alpha-hydrogen atoms on the hydroxyl-bearing carbon atom.
Without an α\alpha-hydrogen, oxidation cannot proceed without breaking carbon-carbon bonds.

Key Concept

Classification and oxidation products of alkanols
Question 62Question

Which of the following structural isomers of hexane, C6H14C_6H_{14}, contains exactly one tertiary carbon atom?

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Answer: 2-Methylpentane

Answer

2-Methylpentane contains exactly one tertiary carbon atom.
A tertiary carbon atom is defined as a carbon atom bonded to three other carbon atoms. In 2-methylpentane, the carbon atom at position 2 is bonded to C-1, C-3, and the branch methyl group, making it the only tertiary carbon atom in the molecule.

Step-by-Step Solution

1
Define a tertiary carbon atom
A tertiary carbon atom (33^\circ) is directly bonded to three other carbon atoms.
Carbon classification depends on the number of attached alkyl/carbon groups.
2
Analyze the carbon skeleton of 2-Methylpentane
In CH3CH(CH3)CH2CH2CH3CH_3-CH(CH_3)-CH_2-CH_2-CH_3, C-2 is bonded to C-1, C-3, and the methyl group.
This structural arrangement yields exactly one tertiary carbon atom.
3
Evaluate the remaining options
2,3-Dimethylbutane has two tertiary carbons, 2,2-Dimethylbutane has one quaternary carbon (zero tertiary), and unbranched hexane has zero tertiary carbons.
Only 2-Methylpentane satisfies the condition of having exactly one tertiary carbon atom.

Key Concept

Classification of carbon atoms in structural isomers
Question 63Question

But-2-ene exhibits geometric (cis-trans) isomerism because each carbon atom involved in the double bond is bonded to two non-identical groups, whereas but-1-ene does not exhibit geometric isomerism.

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Answer: True

Answer

The statement is true because geometric isomerism requires restricted rotation about the C=CC=C bond together with two distinct groups attached to each double-bonded carbon atom—a condition satisfied by but-2-ene but not by but-1-ene.
The statement correctly describes the structural rule for geometric isomerism in alkenes. But-2-ene meets the condition because both double-bonded carbons carry two non-identical groups (H-H and CH3-CH_3), whereas but-1-ene fails the condition because its terminal carbon carries two identical hydrogen atoms.

Step-by-Step Solution

1
Identify the structural requirements for geometric (cis-trans) isomerism.
Geometric isomerism in alkenes requires a rigid C=CC=C double bond where each of the two unsaturated carbon atoms is attached to two non-identical substituents.
If either carbon atom of the double bond carries two identical groups, swapping those groups produces an identical molecule rather than a distinct stereoisomer.
2
Analyze the substituent groups attached to the double-bonded carbons in but-2-ene (CH3CH=CHCH3CH_3-CH=CH-CH_3).
Carbon-2 is attached to H-H and CH3-CH_3, and Carbon-3 is also attached to H-H and CH3-CH_3.
Since both double-bonded carbons have two different groups attached, but-2-ene exists as two stereoisomers: cis-but-2-ene and trans-but-2-ene.
3
Analyze the substituent groups attached to the double-bonded carbons in but-1-ene (CH2=CHCH2CH3CH_2=CH-CH_2-CH_3).
Carbon-1 is attached to two identical hydrogen atoms (H-H and H-H).
The presence of two identical hydrogen atoms on Carbon-1 prevents the formation of cis-trans isomers for but-1-ene.

Key Concept

Structural Criteria for Geometric (Cis-Trans) Isomerism
Question 64Question

Match each chemical process or reaction involving alkanes and petroleum refining in Column A with its corresponding chemical description or primary purpose in Column B.

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Items

Catalytic Cracking
Reforming
Complete Combustion
Free-Radical Substitution

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Answer

Catalytic Cracking matches with the thermal breakdown of long-chain hydrocarbons into shorter alkanes and alkenes. Reforming matches with converting straight-chain alkanes into branched or aromatic hydrocarbons to boost octane rating. Complete Combustion matches with reacting alkanes in excess oxygen to produce CO2CO_2 and H2OH_2O. Free-Radical Substitution matches with replacing hydrogen atoms with halogens under UV light.
Catalytic Cracking breaks larger hydrocarbon molecules into smaller, more useful molecules (alkanes and alkenes). Reforming increases fuel quality (octane rating) by isomerizing straight chains to branched chains or aromatics. Complete Combustion converts alkanes in excess oxygen to carbon dioxide and water. Free-Radical Substitution halogenates alkanes in the presence of UV light.

Step-by-Step Solution

1
Identify the primary function of Catalytic Cracking.
Cracking involves breaking heavy petroleum fractions into smaller alkanes and alkenes.
Heavy oils have low demand, whereas lighter fractions like petrol and gases have high industrial demand.
2
Identify the structural transformation involved in Reforming.
Reforming converts straight-chain alkanes into branched-chain alkanes and aromatic compounds.
Straight-chain alkanes cause engine knocking; branched and aromatic structures improve fuel efficiency by increasing the octane rating.
3
Determine the products of Complete Combustion of alkanes.
Alkanes react completely with excess oxygen to yield CO2(g)CO_2(g) and H2O(g)H_2O(g).
Hydrocarbon oxidation in excess O2O_2 yields fully oxidized carbon dioxide and water.
4
Determine the mechanism for alkane halogenation.
Halogenation of alkanes requires ultraviolet light to generate free radicals for substitution.
Alkanes are unreactive saturated hydrocarbons (paraffins) and require UV light to initiate homeolytic fission of chlorine or bromine molecules.

Key Concept

Chemical reactions of alkanes and industrial petroleum refining processes
Question 65Question

Match each chemical transformation involving alkanols listed on the left with the appropriate reagent, enzyme, or catalyst required on the right.

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Items

Conversion of glucose into ethanol and carbon dioxide
Dehydration of ethanol to produce ethene gas
Complete oxidation of ethanol to ethanoic acid
Industrial hydration of ethene to ethanol

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Answer

Glucose is fermented to ethanol using the enzyme zymase; dehydration of ethanol to ethene uses excess concentrated H2SO4\text{H}_2\text{SO}_4 at 170C170^\circ\text{C}; ethanol is oxidized to ethanoic acid using acidified K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7 under reflux; and industrial synthesis of ethanol from ethene uses steam with a phosphoric acid (H3PO4\text{H}_3\text{PO}_4) catalyst at high temperature and pressure.
Each chemical process matches its unique catalyst or reaction conditions: zymase catalyzes glucose fermentation to ethanol, excess concentrated H2SO4\text{H}_2\text{SO}_4 at 170C170^\circ\text{C} dehydrates ethanol to ethene, acidified K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7 under reflux oxidizes ethanol to ethanoic acid, and phosphoric acid (H3PO4\text{H}_3\text{PO}_4) on silica catalyzes the industrial hydration of ethene to ethanol.

Step-by-Step Solution

1
Identify the biological catalyst for sugar fermentation
Fermentation of glucose (C6H12O62C2H5OH+2CO2\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2) is catalyzed specifically by the enzyme zymase.
Yeast produces zymase, which converts simple hexose sugars directly into ethanol.
2
Identify the reagent and temperature for elimination/dehydration
Heating ethanol with excess concentrated H2SO4\text{H}_2\text{SO}_4 at 170C170^\circ\text{C} yields ethene via removal of a water molecule.
Concentrated tetraoxosulfate(VI) acid acts as a dehydrating agent; high temperature (170C170^\circ\text{C}) favors ethene formation over ethoxyethane formation.
3
Identify the oxidizing conditions for full alkanol oxidation
Primary alkanols undergo two-stage oxidation: first to an alkanal, then under reflux with acidified K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7 to an alkanoic acid.
Acidified potassium heptaoxodichromate(VI) is a strong oxidizing agent capable of carrying the oxidation of ethanol fully to ethanoic acid.
4
Identify the industrial catalytic addition reaction conditions
Direct hydration of ethene (C2H4+H2OC2H5OH\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}) uses a phosphoric acid catalyst.
The reversible addition of steam across the double bond of ethene requires a solid phosphoric acid catalyst at 300C300^\circ\text{C} and high pressure.

Key Concept

Reagents, enzymes, and conditions for alkanol preparation and reactions
Question 66Question

What is the IUPAC name of the organic compound formed by the reduction of butan-2-one using lithium tetrahydridoaluminate(III) (LiAlH4LiAlH_4)?

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Answer: butan-2-ol; 2-butanol; butan 2 ol; 2-Butanol; Butan-2-ol

Answer

butan-2-ol
The reduction of alkanones (ketones) with reducing agents such as LiAlH4LiAlH_4 converts the carbonyl group (C=OC=O) into a secondary alcohol group (CH(OH)-CH(OH)-). Since the carbonyl carbon in butan-2-one is at position 2, the resulting hydroxyl group is also at position 2, yielding the secondary alcohol butan-2-ol.

Step-by-Step Solution

1
Identify the functional group and carbon chain length of the reactant.
Butan-2-one is a 4-carbon alkanone (ketone) with the carbonyl group at carbon-2 (CH3COCH2CH3CH_3-CO-CH_2-CH_3).
Determining the reactant structure establishes the expected reduction product.
2
Apply the reduction reaction mechanism for alkanones.
Reducing agents such as LiAlH4LiAlH_4 or NaBH4NaBH_4 reduce alkanones to secondary alcohols by adding hydrogen across the C=OC=O double bond.
The carbonyl group (C=OC=O) is converted to a secondary alcohol group (CH(OH)-CH(OH)-).
3
Name the resulting alcohol using IUPAC nomenclature.
CH3CH(OH)CH2CH3CH_3-CH(OH)-CH_2-CH_3 is named butan-2-ol.
The hydroxyl group (OH-OH) remains on carbon-2 of the 4-carbon parent chain.

Key Concept

Reduction of alkanones to secondary alcohols
Estimated Time:1m 0s
Question 67Question

Compound WW is a four-carbon carbonyl compound that yields a negative result when tested with ammoniacal silver nitrate solution. Complete reduction of compound WW using lithium aluminium hydride (LiAlH4\text{LiAlH}_4) produces an alcohol, compound ZZ. What is the IUPAC name of compound ZZ?

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Answer: butan-2-ol; 2-butanol; Butan-2-ol; 2-Butanol

Answer

The IUPAC name of compound ZZ is butan-2-ol.
Compound WW does not react with Tollen's reagent (ammoniacal silver nitrate), which confirms it is an alkanone rather than an alkanal. The only four-carbon alkanone is butan-2-one. Reducing butan-2-one with LiAlH4\text{LiAlH}_4 yields the secondary alcohol butan-2-ol.

Step-by-Step Solution

1
Determine the functional group of compound WW based on the distinction test.
Compound WW is an alkanone (ketone).
Alkanals (aldehydes) reduce ammoniacal silver nitrate (Tollen's reagent) to silver metal (silver mirror), whereas alkanones (ketones) give a negative result.
2
Identify the specific structure and IUPAC name of compound WW.
Compound WW is butan-2-one (CH3COCH2CH3\text{CH}_3\text{COCH}_2\text{CH}_3).
Since compound WW is a four-carbon alkanone, its only structural isomer is butan-2-one.
3
Determine the product of the reduction reaction.
Compound ZZ is a secondary alcohol, butan-2-ol (CH3CH(OH)CH2CH3\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3).
Reduction of alkanones with LiAlH4\text{LiAlH}_4 converts the carbonyl group (C=O\text{C=O}) into a secondary alcohol group (CH-OH\text{CH-OH}).

Key Concept

Reduction of alkanones to secondary alcohols and distinction between alkanals and alkanones using Tollen's reagent
Question 68Question

In electrophilic aromatic substitution, benzene reacts with strong electrophiles generated by specific catalyst-reagent combinations. Match each benzene reaction system on the left with its corresponding active electrophile species generated during the reaction mechanism on the right.

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Items

Nitration using concentrated HNO3\text{HNO}_3 and concentrated H2SO4\text{H}_2\text{SO}_4
Friedel-Crafts acylation using ethanoyl chloride (CH3COCl\text{CH}_3\text{COCl}) and anhydrous AlCl3\text{AlCl}_3
Catalytic bromination using Br2\text{Br}_2 and FeBr3\text{FeBr}_3
Sulfonation using fuming or concentrated H2SO4\text{H}_2\text{SO}_4

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Answer

Nitration produces the nitronium ion (NO2+\text{NO}_2^+); Friedel-Crafts acylation generates the acylium ion (CH3C+=O\text{CH}_3\text{C}^+=\text{O}); Catalytic bromination produces the bromonium ion (Br+\text{Br}^+); Sulfonation generates neutral sulfur trioxide (SO3\text{SO}_3).
Each benzene electrophilic substitution reaction relies on a specific reagent and catalyst mechanism to create a powerful electrophile capable of disrupting benzene's stable aromatic system. Nitration generates NO2+\text{NO}_2^+ via acid-base protonation of nitric acid by sulfuric acid. Friedel-Crafts acylation forms the acylium ion CH3C+=O\text{CH}_3\text{C}^+=\text{O} through chloride abstraction by the Lewis acid AlCl3\text{AlCl}_3. Bromination generates a polarized Br+\text{Br}^+ complex using FeBr3\text{FeBr}_3. Sulfonation relies on SO3\text{SO}_3, which features an electron-deficient sulfur atom due to polar S=O bonds.

Step-by-Step Solution

1
Identify the electrophile in nitration
Concentrated H2SO4\text{H}_2\text{SO}_4 acts as an acid to protonate HNO3\text{HNO}_3. Loss of H2O\text{H}_2\text{O} yields NO2+\text{NO}_2^+ (nitronium ion).
H2SO4\text{H}_2\text{SO}_4 is a stronger acid than HNO3\text{HNO}_3 and forces HNO3\text{HNO}_3 to act as a base.
2
Identify the electrophile in Friedel-Crafts acylation
The catalyst AlCl3\text{AlCl}_3 abstracts Cl\text{Cl}^- from CH3COCl\text{CH}_3\text{COCl}, leaving the resonance-stabilized cations CH3C+=O\text{CH}_3\text{C}^+=\text{O}.
AlCl3\text{AlCl}_3 is an electron-deficient Lewis acid capable of coordinating chloride.
3
Identify the electrophile in bromination
FeBr3\text{FeBr}_3 coordinates with a bromine atom of Br2\text{Br}_2, polarising the bond to create an effective Br+\text{Br}^+ electrophile.
Benzene requires a Lewis acid catalyst to polarize halogen molecules sufficiently for reaction.
4
Identify the electrophile in sulfonation
Equilibrium in concentrated/fuming H2SO4\text{H}_2\text{SO}_4 produces neutral SO3\text{SO}_3, which has a highly electron-deficient sulfur atom.
The three electronegative oxygen atoms in SO3\text{SO}_3 withdraw electron density from the central sulfur atom.

Key Concept

Generation of Electrophiles in Benzene Electrophilic Substitution
Question 69Question

An unknown organic compound YY forms a brick-red precipitate when warmed with Fehling's solution and also gives a yellow precipitate of triiodomethane (CHI3CHI_3) when treated with iodine in sodium hydroxide solution. Which of the following compounds is YY?

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Answer: Ethanal

Answer

Ethanal
Ethanal (CH3CHOCH_3CHO) is an alkanal, so it reduces copper(II) ions in Fehling's solution to copper(I) oxide (Cu2OCu_2O), producing a brick-red precipitate. Furthermore, because its carbonyl group is bonded directly to a methyl group (CH3C=OCH_3C=O), it undergoes halogenation and cleavage in alkaline iodine to produce a yellow precipitate of triiodomethane (CHI3CHI_3).

Step-by-Step Solution

1
Analyze the Fehling's solution reaction.
Fehling's test distinguishes alkanals (aldehydes) from alkanones (ketones). A positive test (brick-red Cu2OCu_2O precipitate) indicates YY must be an alkanal.
Alkanals are easily oxidized to alkanoic acids, whereas alkanones resist mild oxidation.
2
Analyze the triiodomethane (iodoform) reaction.
A positive triiodomethane test (yellow CHI3CHI_3 precipitate) requires a methyl carbonyl group (CH3C=OCH_3C=O) or a secondary alcohol structure (CH3CH(OH)CH_3CH(OH)-).
Iodine in aqueous alkali oxidizes and iodinates compounds containing the methyl carbonyl structural unit.
3
Combine the structural requirements.
The compound must be both an alkanal (CHO-CHO) and contain a methyl carbonyl group (CH3C=OCH_3C=O). Ethanal (CH3CHOCH_3CHO) is the only alkanal that possesses a CH3C=OCH_3C=O group.
Other alkanals like methanal (HCHOHCHO) or propanal (CH3CH2CHOCH_3CH_2CHO) lack the CH3C=OCH_3C=O structural unit.

Key Concept

Distinction tests for carbonyl compounds: Fehling's test and Iodoform test
Estimated Time:1m 0s
Question 70Question

Which of the following alkenes with the molecular formula C5H10C_5H_{10} exhibits geometric (cis-trans) isomerism?

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Answer: Pent-2-ene

Answer

Pent-2-ene is the only isomer listed that exhibits geometric (cis-trans) isomerism because each carbon of the double bond is attached to two distinct groups.
Pent-2-ene has a double bond between carbon-2 and carbon-3. Carbon-2 is attached to a hydrogen atom and a methyl group (CH3CH_3), while carbon-3 is attached to a hydrogen atom and an ethyl group (CH2CH3CH_2CH_3). Because neither carbon atom of the double bond holds two identical groups, spatial restriction gives rise to distinct cis and trans stereoisomers.

Step-by-Step Solution

1
Recall the necessary structural condition for geometric (cis-trans) isomerism in alkenes.
For a molecule to show cis-trans isomerism around a double bond C=CC=C, each of the two carbon atoms in the double bond must be attached to two different atoms or groups.
If either carbon in the double bond has two identical groups attached, rotating spatial arrangements results in identical molecules.
2
Examine the connectivity of each option at the C=CC=C double bond.
Pent-2-ene has C2C_2 bonded to H-H and CH3-CH_3, and C3C_3 bonded to H-H and CH2CH3-CH_2CH_3. Both carbons have two different groups.
This satisfies the criteria for both cis and trans geometric arrangements.
3
Check the remaining options for duplicate attached groups on double-bonded carbons.
Pent-1-ene and 3-methylbut-1-ene both have a terminal =CH2=CH_2 (two H atoms on C1C_1). 2-Methylbut-2-ene has two methyl groups on C2C_2.
None of these three options satisfy the non-identical substituent requirement on both double-bonded carbons.

Key Concept

Geometric Isomerism Requirements in Alkenes
Estimated Time:1m 0s
Question 71Question

An organic compound with the molecular formula C4H10OC_4H_{10}O forms four isomeric alkanols. Which of these alkanol isomers contains an asymmetric carbon atom and exhibits optical isomerism?

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Answer: Butan-2-ol

Answer

Butan-2-ol is the only isomeric alkanol of formula C4H10OC_4H_{10}O containing a chiral carbon atom.
Butan-2-ol has a chiral carbon at position 2 (CH3CH(OH)CH2CH3CH_3-CH(OH)-CH_2-CH_3), which is bonded to four different substituent groups: H-H, OH-OH, CH3-CH_3, and CH2CH3-CH_2CH_3. The presence of this asymmetric center causes optical activity.

Step-by-Step Solution

1
Identify the requirement for optical isomerism.
A molecule exhibits optical isomerism if it contains at least one chiral (asymmetric) carbon atom—a carbon atom bonded to four different groups or atoms.
Chirality causes non-superimposable mirror images (enantiomers) capable of rotating plane-polarized light.
2
Analyze the structural formulas of the four isomeric alkanols of C4H10OC_4H_{10}O.
1) Butan-1-ol: CH3CH2CH2CH2OHCH_3-CH_2-CH_2-CH_2OH
2) Butan-2-ol: CH3CH(OH)CH2CH3CH_3-CH(OH)-CH_2-CH_3
3) 2-Methylpropan-1-ol: (CH3)2CHCH2OH(CH_3)_2CH-CH_2OH
4) 2-Methylpropan-2-ol: (CH3)3COH(CH_3)_3C-OH
Writing structural formulas reveals the substituent groups attached to each carbon atom.
3
Examine carbon-2 in butan-2-ol.
Carbon-2 is attached to H-H, OH-OH, CH3-CH_3, and CH2CH3-CH_2CH_3. All four substituents are distinct.
Since carbon-2 has four different attached groups, it is an asymmetric (chiral) carbon atom.

Key Concept

Optical isomerism and chirality in alkanols
Question 72Question

An organic compound ZZ with the molecular formula C5H10OC_5H_{10}O gives a positive orange precipitate when treated with 2,4-dinitrophenylhydrazine. When warmed with acidified potassium heptaoxodichromate(VI) (K2Cr2O7K_2Cr_2O_7), compound ZZ is readily oxidized to a carboxylic acid containing five carbon atoms. Upon reduction with lithium tetrahydridoaluminate(III) (LiAlH4LiAlH_4), compound ZZ yields a primary alcohol that exhibits optical activity (chirality). What is the IUPAC name of compound ZZ?

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Answer: 2-methylbutanal

Answer

2-methylbutanal
2-methylbutanal has the molecular formula C5H10OC_5H_{10}O and contains an alkanal functional group. It reacts with 2,4-dinitrophenylhydrazine to form a precipitate. Because it is an alkanal, it undergoes oxidation with acidified potassium heptaoxodichromate(VI) to produce 2-methylbutanoic acid (a 5-carbon carboxylic acid). Upon reduction with lithium tetrahydridoaluminate(III), it forms 2-methylbutan-1-ol, which has an asymmetric carbon atom at position 2 bonded to four different substituent groups (H-H, CH3-CH_3, CH2CH3-CH_2CH_3, CH2OH-CH_2OH), rendering the molecule chiral and optically active.

Step-by-Step Solution

1
Identify the functional group class from the 2,4-DNPH test.
A positive test with 2,4-dinitrophenylhydrazine confirms that compound ZZ is a carbonyl compound (either an alkanal or an alkanone).
Both alkanals and alkanones form colored hydrazone precipitates with 2,4-DNPH.
2
Distinguish between an alkanal and an alkanone using oxidation behavior.
Compound ZZ readily oxidizes with acidified K2Cr2O7K_2Cr_2O_7 to a carboxylic acid with 5 carbons, proving ZZ is an alkanal (pentanal isomer).
Alkanals are easily oxidized to carboxylic acids with the same number of carbon atoms, whereas alkanones resist oxidation under mild conditions.
3
Analyze the reduction product for chirality.
Reduction of an alkanal with LiAlH4LiAlH_4 yields a primary alcohol. Among 5-carbon alkanals (pentanal, 2-methylbutanal, 3-methylbutanal, 2,2-dimethylpropanal), only 2-methylbutanal reduces to 2-methylbutan-1-ol, CH3CH2CH(CH3)CH2OHCH_3CH_2CH(CH_3)CH_2OH.
In 2-methylbutan-1-ol, C-2 is bonded to four distinct groups: H-H, CH3-CH_3, CH2CH3-CH_2CH_3, and CH2OH-CH_2OH, making it a chiral molecule capable of optical isomerism.

Key Concept

Distinction between alkanals and alkanones via oxidation, reduction of carbonyls to alcohols, and optical isomerism in branched primary alcohols.
Question 73Question

An organic compound ZZ with the molecular formula C5H10OC_5H_{10}O gives a negative result with Tollen's reagent and does not form a yellow precipitate when warmed with iodine in sodium hydroxide solution. Upon reduction with lithium tetrahydridoaluminate(III) (LiAlH4LiAlH_4), compound ZZ yields a secondary alkanol. Which of the following is the correct IUPAC name of compound ZZ?

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Answer: pentan-3-one

Answer

pentan-3-one
Pentan-3-one is an alkanone with the structure CH3CH2COCH2CH3CH_3CH_2COCH_2CH_3. Because it is a ketone, it cannot be oxidized by mild oxidizing agents like Tollen's reagent. Furthermore, because its carbonyl carbon is bonded to two ethyl groups rather than a methyl group, it gives a negative triiodomethane (iodoform) test. Reduction of pentan-3-one using LiAlH4LiAlH_4 yields pentan-3-ol, which is a secondary alkanol.

Step-by-Step Solution

1
Analyze the functional group class using Tollen's reagent test.
Compound ZZ gives a negative Tollen's test, confirming it is an alkanone (ketone) rather than an alkanal (aldehyde).
Alkanals are easily oxidized to alkanoic acids and reduce Tollen's reagent to metallic silver, whereas alkanones resist mild oxidation.
2
Evaluate the iodoform (triiodomethane) test requirement.
The absence of a yellow precipitate (CHI3CHI_3) rules out compounds containing a methyl ketone (CH3COCH_3CO-) group.
Only methyl ketones (RCOCH3R-COCH_3) or ethanal (CH3CHOCH_3CHO) undergo triiodomethane formation with iodine in sodium hydroxide.
3
Examine the reduction reaction and identify the specific isomer.
Among the C5H10OC_5H_{10}O ketone isomers (pentan-2-one, pentan-3-one, and 3-methylbutan-2-one), only pentan-3-one lacks a methyl ketone group and reduces with LiAlH4LiAlH_4 to give pentan-3-ol (a secondary alkanol).
Pentan-3-one has the structural formula CH3CH2COCH2CH3CH_3CH_2COCH_2CH_3, fulfilling both qualitative test observations and structural reduction requirements.

Key Concept

Distinction tests for carbonyl compounds (Tollen's test and iodoform test) and reduction outcomes of alkanones
Estimated Time:2m 0s
Question 74Question

Match each chemical reaction or test involving carbonyl compounds in Column A with its corresponding characteristic visual outcome in Column B.

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Items

Warming ethanal with Fehling's solution
Warming propanone with Tollen's reagent
Warming propanal with acidified potassium tetraoxomanganate(VII) solution
Warming propanone with iodine in sodium hydroxide solution

Matches

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Answer

Warming ethanal with Fehling's solution yields a brick-red precipitate; warming propanone with Tollen's reagent yields no visible reaction; warming propanal with acidified KMnO4KMnO_4 causes decolorization of the purple solution; and warming propanone with iodine in sodium hydroxide solution produces a pale yellow precipitate.
Alkanals (ethanal and propanal) are reducing agents due to the carbonyl hydrogen atom; thus, ethanal reduces Fehling's solution to a brick-red copper(I) oxide precipitate, and propanal reduces purple acidified KMnO4KMnO_4 to a colorless Mn2+Mn^{2+} solution. Alkanones (propanone) lack this hydrogen atom, so propanone shows no reaction with Tollen's reagent. However, because propanone has a CH3COCH_3CO- group, it responds to the triiodomethane test by forming a pale yellow precipitate.

Step-by-Step Solution

1
Analyze the oxidation reactions of alkanals.
Ethanal reduces Fehling's solution to form brick-red Cu2OCu_2O, while propanal reduces acidified KMnO4KMnO_4, turning the purple solution colorless.
Alkanals possess a hydrogen atom bonded to the carbonyl carbon, enabling easy oxidation by mild and strong oxidizing agents.
2
Analyze the oxidation behavior of alkanones.
Propanone yields no visible reaction with Tollen's reagent.
Alkanones lack a hydrogen atom on the carbonyl carbon and are resistant to oxidation by mild oxidizing agents.
3
Identify the triiodomethane (iodoform) test reaction.
Propanone forms a pale yellow precipitate of triiodomethane (CHI3CHI_3).
Propanone contains the CH3C=OCH_3C=O group required for a positive triiodomethane test.

Key Concept

Distinction tests and oxidation properties of alkanals and alkanones
Estimated Time:1m 0s
Question 75Question

Which of the following organic compounds acts as a weak base in an aqueous solution due to the presence of an unshared pair of electrons on its nitrogen atom?

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Answer: Methylamine (CH3NH2CH_3NH_2)

Answer

Methylamine (CH3NH2CH_3NH_2)
Methylamine (CH3NH2CH_3NH_2) is an aliphatic amine. The nitrogen atom retains a localized lone pair of electrons that readily accepts a proton from water to form a methylammonium ion and a hydroxide ion, demonstrating basic behavior in aqueous solution.

Step-by-Step Solution

1
Identify the functional groups present in each compound.
Methylamine (CH3NH2CH_3NH_2) is an amine; ethanamide (CH3CONH2CH_3CONH_2) is an amide; ethanoic acid (CH3COOHCH_3COOH) is a carboxylic acid; ethanol (CH3CH2OHCH_3CH_2OH) is an alcohol.
Basic properties in organic nitrogen compounds depend on the availability of the nitrogen lone pair.
2
Evaluate the availability of the unshared pair of electrons on the nitrogen atom.
In primary aliphatic amines like methylamine, the lone pair on nitrogen is readily available to accept a proton (H+H^+). In amides like ethanamide, resonance delocalizes the nitrogen lone pair toward the carbonyl oxygen, removing its basic character.
Proton acceptance (Lewis/Brønsted-Lowry basicity) requires an available lone pair.

Key Concept

Basicity of Amines versus Amides
Question 76Question

Match each nitrogen-containing organic compound on the left with its correct structural classification on the right.

Click a left item, then click its matching right item

Items

Methylamine (CH3NH2CH_3NH_2)
Ethanamide (CH3CONH2CH_3CONH_2)
Phenylamine (C6H5NH2C_6H_5NH_2)
Dimethylamine ((CH3)2NH(CH_3)_2NH)

Matches

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Answer

Methylamine pairs with Primary aliphatic amine; Ethanamide pairs with Neutral organic amide; Phenylamine pairs with Primary aromatic amine; Dimethylamine pairs with Secondary aliphatic amine.
Each compound matches its unique chemical definition: Methylamine is a 11^\circ aliphatic amine, Ethanamide is a neutral amide, Phenylamine is a 11^\circ aromatic amine, and Dimethylamine is a 22^\circ aliphatic amine.

Step-by-Step Solution

1
Examine the functional groups and substituents attached to nitrogen in each compound.
Methylamine (CH3NH2CH_3NH_2) and Phenylamine (C6H5NH2C_6H_5NH_2) each have one organic group attached (11^\circ). Dimethylamine ((CH3)2NH(CH_3)_2NH) has two organic groups attached (22^\circ). Ethanamide (CH3CONH2CH_3CONH_2) has a carbonyl group (C=OC=O) linked directly to nitrogen.
The number of alkyl/aryl groups determines amine degree (1,2,31^\circ, 2^\circ, 3^\circ), while a carbonyl-nitrogen bond defines an amide.
2
Classify by aliphatic, aromatic, or neutral amide characteristics.
Methylamine contains an alkyl group (11^\circ aliphatic amine). Phenylamine contains a benzene ring (11^\circ aromatic amine). Dimethylamine has two alkyl groups (22^\circ aliphatic amine). Ethanamide is an amide and exhibits neutral aqueous behavior due to lone pair resonance delocalization.
Structure and electronic delocalization determine both classification and relative basicity.

Key Concept

Classification of amines (primary, secondary, aromatic, aliphatic) and amides.
Question 77Question

An unknown disaccharide XX with the molecular formula C12H22O11\text{C}_{12}\text{H}_{22}\text{O}_{11} does not reduce Fehling's solution. Upon acid-catalyzed hydrolysis, XX yields an equimolar mixture of two isomeric hexoses, YY and ZZ. Compound YY rapidly produces a deep red color when heated with Seliwanoff's reagent, whereas compound ZZ is oxidized by bromine water to form a monocarboxylic acid. Which of the following correctly identifies disaccharide XX and explains why it fails to react with Fehling's solution prior to hydrolysis?

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Answer: Sucrose; because the glycosidic linkage involves the anomeric carbons of both glucose and fructose, eliminating free hemiacetal/hemiketal groups.

Answer

Sucrose; because the glycosidic linkage involves the anomeric carbons of both glucose and fructose, eliminating free hemiacetal/hemiketal groups.
Seliwanoff's test specifically identifies ketohexoses such as fructose through rapid dehydration to hydroxymethylfurfural and reaction with resorcinol. Bromine water selectively oxidizes aldoses like glucose to aldonic acids without oxidizing ketoses. A disaccharide yielding glucose and fructose upon hydrolysis is sucrose. Sucrose is a non-reducing sugar because its glycosidic bond connects C-1 of glucose and C-2 of fructose, locking both anomeric carbon atoms and preventing ring opening to form reactive carbonyl groups.

Step-by-Step Solution

1
Analyze the chemical test results of the hydrolysis products YY and ZZ.
Compound YY gives a positive Seliwanoff's test (rapid red color), which is characteristic of a ketose (fructose). Compound ZZ is oxidized by bromine water (a mild oxidizing agent), which selectively oxidizes aldoses (glucose) to aldonic acids.
Seliwanoff's reagent differentiates ketoses from aldoses, while bromine water differentiates aldoses from ketoses.
2
Identify the disaccharide XX based on its hydrolysis products.
Disaccharide XX hydrolyzes into glucose and fructose, identifying XX as sucrose.
Sucrose (C12H22O11\text{C}_{12}\text{H}_{22}\text{O}_{11}) is composed of one glucose unit and one fructose unit.
3
Determine the structural basis for the non-reducing nature of sucrose.
In sucrose, the glycosidic bond connects C-1\text{C-1} (α\alpha-anomeric carbon of glucose) to C-2\text{C-2} (β\beta-anomeric carbon of fructose).
Because both potential reducing centers (anomeric carbons) are locked in the glycosidic bond, sucrose lacks a free hemiacetal or hemiketal group, rendering it a non-reducing sugar that does not reduce Fehling's reagent.

Key Concept

Glycosidic bond linkages and reducing vs. non-reducing carbohydrate behavior
Question 78Question

Four organic substances—starch, nylon-6,6, polyethene, and natural rubber—were analyzed to compare their chemical structures, reaction behaviors, and environmental properties. Which of the following statements provides a correct scientific comparison among these materials?

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Answer: Starch consists of monomeric glucose units joined by glycosidic linkages that undergo acid hydrolysis, whereas nylon-6,6 contains recurring amide linkages formed via condensation polymerization.

Answer

Starch consists of monomeric glucose units joined by glycosidic linkages that undergo acid hydrolysis, whereas nylon-6,6 contains recurring amide linkages formed via condensation polymerization.
Starch is a carbohydrate (polysaccharide) composed of glucose units connected by glycosidic bonds that can be broken down by acid hydrolysis. Nylon-6,6 is a synthetic condensation polymer (polyamide) made from a dicarboxylic acid and a diamine, containing amide linkages.

Step-by-Step Solution

1
Analyze the structural classification and linkages of starch and nylon-6,6.
Starch is a naturally occurring polysaccharide composed of glucose monomer units linked via α\alpha-glycosidic bonds. Nylon-6,6 is a synthetic polyamide composed of hexanedioic acid and hexane-1,6-diamine monomers joined by amide (CONH-\text{CO}-\text{NH}-) linkages with the elimination of water.
Identifying monomer units and linkage types distinguishes condensation polymers and carbohydrates.
2
Evaluate the biodegradability of synthetic addition polymers like polyethene versus natural polymers like starch.
Starch is readily degraded by enzymatic action and micro-organisms. Polyethene is a synthetic addition polymer containing strong CC\text{C}-\text{C} single bonds that resist microbial enzymatic attack, rendering it non-biodegradable.
Synthetic addition polymers lack functional linkages susceptible to hydrolysis.
3
Assess the chemical action of concentrated H2SO4\text{H}_2\text{SO}_4 on carbohydrates.
Concentrated H2SO4\text{H}_2\text{SO}_4 acts as a dehydrating agent, removing water elements from starch according to (C6H10O5)nconc. H2SO46nC+5nH2O(\text{C}_6\text{H}_{10}\text{O}_5)_n \xrightarrow{\text{conc. H}_2\text{SO}_4} 6n\text{C} + 5n\text{H}_2\text{O}, turning the sample black.
This is a dehydration reaction, not an acid-base neutralization.
4
Analyze the chemical test reagents for unsaturation in rubber versus terminal alkynes.
Bromine water or acidified KMnO4\text{KMnO}_4 is used to test for unsaturation in alkenes and dienes like natural rubber. Ammoniacal silver nitrate selectively reacts with terminal alkynes to give a silver acetylide precipitate.
Different unsaturated functional groups require specific chemical reagents for qualitative identification.

Key Concept

Classification of polymers and biomolecules by linkage type (glycosidic vs amide), reaction mode (addition vs condensation), and chemical reactivity.
Estimated Time:2m 0s
Question 79Question

Match each nitrogen-containing organic compound on the left with the statement on the right that accurately accounts for its aqueous basicity and lone-pair electronic behavior.

Click a left item, then click its matching right item

Items

Dimethylamine, (CH3)2NH(CH_3)_2NH
Phenylamine, C6H5NH2C_6H_5NH_2
Ethanamide, CH3CONH2CH_3CONH_2
Triethylamine, (C2H5)3N(C_2H_5)_3N

Matches

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Answer

Dimethylamine matches the statement describing higher aqueous basicity than ammonia due to inductive donation and solvation; Phenylamine matches the statement describing reduced basicity from aromatic resonance delocalization; Ethanamide matches the statement describing neutrality caused by carbonyl resonance; Triethylamine matches the statement describing steric hindrance affecting conjugate acid solvation.
The correct matches reflect fundamental physical-organic chemistry principles governing nitrogen basicity: Dimethylamine combines inductive donation with high conjugate acid solvation stability; Phenylamine suffers basicity loss from aromatic resonance delocalization; Ethanamide lone-pair delocalization into the carbonyl group yields a neutral compound; Triethylamine basicity in water is moderated by steric crowding that interferes with hydration of the ammonium cation.

Step-by-Step Solution

1
Analyze the electronic structure of Dimethylamine ((CH3)2NH(CH_3)_2NH).
Two methyl groups supply electron density via +I+I inductive effects, enhancing nitrogen lone-pair availability, while the secondary cation remains readily solvated by water.
Secondary aliphatic amines are generally the strongest bases in aqueous media.
2
Analyze the resonance interactions in Phenylamine (C6H5NH2C_6H_5NH_2).
The unshared electron pair on nitrogen participates in resonance with the benzene ring, lowering lone-pair availability.
Aromatic amines are significantly weaker bases than ammonia and aliphatic amines.
3
Examine the functional group characteristics of Ethanamide (CH3CONH2CH_3CONH_2).
Resonance delocalization between nitrogen's lone pair and the adjacent C=OC=O double bond (O=CNOC=N+O=C-N \leftrightarrow ^-O-C=N^+) deprives nitrogen of basic character.
Amides behave as neutral organic compounds in aqueous solution.
4
Evaluate steric effects in Triethylamine ((C2H5)3N(C_2H_5)_3N).
Three ethyl groups create steric crowding around the nitrogen cation, hindering stabilization through hydration in water.
In aqueous solution, tertiary aliphatic amines are often weaker bases than secondary aliphatic amines due to solvation factors.

Key Concept

Relative basicity of aliphatic amines, aromatic amines, and amides governed by inductive, resonance, and solvation steric effects.
Estimated Time:2m 0s
Question 80Question

Match each organic nitrogen compound on the left with its corresponding relative basicity or acid-base structural property on the right.

Click a left item, then click its matching right item

Items

Phenylamine (C6H5NH2C_6H_5NH_2)
Methylamine (CH3NH2CH_3NH_2)
Ethanamide (CH3CONH2CH_3CONH_2)
Dimethylamine ((CH3)2NH(CH_3)_2NH)

Matches

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Answer

Phenylamine matches with being weaker than ammonia due to aromatic delocalization; Methylamine matches with being stronger than ammonia due to +I inductive effect; Ethanamide matches with being neutral due to carbonyl resonance; Dimethylamine matches with being stronger than methylamine in aqueous solution.
Phenylamine is less basic than ammonia because its lone pair is delocalized across the aromatic ring. Methylamine is more basic than ammonia due to inductive electron donation by the methyl group. Ethanamide is neutral because its lone pair participates in resonance with the carbonyl group. Dimethylamine is more basic than methylamine in water due to two electron-donating methyl groups.

Step-by-Step Solution

1
Analyze how structural features influence nitrogen lone pair availability.
Electron-donating alkyl groups (+I effect) enhance lone pair availability (increasing basicity), while electron-withdrawing groups or resonance delocalization decrease lone pair availability (decreasing basicity).
Lewis/Brønsted-Lowry basicity of nitrogen compounds depends directly on lone pair availability to accept a proton.
2
Evaluate phenylamine and ethanamide.
Phenylamine delocalizes its lone pair into the benzene ring, making it weaker than NH3NH_3. Ethanamide delocalizes its lone pair into the C=OC=O double bond, making it neutral in aqueous solution.
Resonance delocalization significantly stabilizes the unprotonated state and lowers basicity.
3
Compare methylamine and dimethylamine.
Methylamine has one alkyl group increasing basicity over NH3NH_3. Dimethylamine has two alkyl groups supplying greater electron density, making it more basic than methylamine in aqueous solution.
Inductive electron donation by methyl groups stabilizes the positive conjugate ammonium ion.

Key Concept

Relative basicity of amines and amides based on inductive and resonance effects
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