Organic Chemistry Practice Questions — JAMB UTME — Page 3

Question 41Question →

Consider the cumulated diene compound penta-1,2-diene, represented by the condensed structure $\text{CH}_2=\text{C}=\text{CH}-\text{CH}_2-\text{CH}_3$ . What are the hybridization states of the central carbon atom (C2) and the methylene carbon atom (C4), respectively?

sp

and

sp^3

sp^2

and

sp^3

sp

and

sp^2

sp^2

and

sp^2

Show answer & explanation

Answer: $sp$ and $sp^3$

Answer

The central carbon atom (C2) is

sp

hybridized and the methylene carbon atom (C4) is

sp^3

hybridized.

In penta-1,2-diene, the central allene carbon atom at position 2 (C2) forms two double bonds (

2\ \sigma

bonds and

2\ \pi

bonds), which necessitates

sp

hybridization. The methylene carbon atom at position 4 (C4) forms four single

\sigma

bonds with two hydrogen atoms and two carbon atoms, which corresponds to

sp^3

tetrahedral hybridization.

Step-by-Step Solution

1

Determine the number of

\sigma

and

\pi

bonds on carbon-2 (C2).

C2 forms two double bonds, which consists of

2\ \sigma

bonds and

2\ \pi

bonds.

Carbon atoms involved in two double bonds (cumulated dienes/allenes) use two

sp

hybrid orbitals to form

\sigma

bonds at an angle of

180^\circ

.

2

Determine the hybridization state of C2.

C2 is

sp

hybridized with linear geometry.

Two

\sigma

bonding domains correlate to

sp

hybridization.

3

Determine the bonding domains and hybridization of carbon-4 (C4).

C4 is bonded to two hydrogen atoms, C3, and C5 via single covalent bonds, giving

4\ \sigma

bonds.

Four single

\sigma

bonding domains require

sp^3

hybridization with tetrahedral geometry.

Key Concept

Hybridization in Cumulated Dienes and Saturated Carbons

Estimated Time:1m 30s

Question 42Question →

In a saturated acyclic alkane such as propane ( $\text{C}_3\text{H}_8$ ), each carbon atom exhibits tetrahedral geometry through $sp^3$ hybridization. What is the percentage of $s$ -orbital character present in each of these hybrid orbitals?

25\%

33.3\%

50\%

75\%

Show answer & explanation

Answer: $25\%$

Answer

The percentage of

s

-orbital character in each

sp^3

hybrid orbital of a tetrahedral carbon atom is

25\%

.

In tetrahedral carbon compounds,

sp^3

hybridization involves mixing one

s

orbital and three

p

orbitals to produce four degenerate orbitals. Therefore, the proportion of

s

-character in each hybrid orbital is

\frac{1}{4}

, which equals

25\%

.

Step-by-Step Solution

1

Determine the composition of orbitals in

sp^3

hybridization.

An

sp^3

hybrid orbital is produced by combining one

s

atomic orbital and three

p

atomic orbitals, yielding a total of 4 equivalent hybrid orbitals.

Hybridization mixes pure atomic orbitals to form equivalent hybrid orbitals around a central tetrahedral carbon atom.

2

Calculate the fractional contribution of the

s

orbital.

The fraction of

s

-character is

\frac{1}{1 + 3} = \frac{1}{4} = 0.25

.

One out of the four constituent atomic orbitals is an

s

orbital.

3

Convert the fraction into a percentage.

0.25 \times 100\% = 25\%

.

Multiplying the fractional contribution by 100 gives the percentage of

s

-character.

Key Concept

Orbital Hybridization and s/p Character in Tetrahedral Carbon

Question 43Question →

Match each structural feature or carbon center of 2-methylbut-1-en-3-yne ( $HC\equiv C-C(CH_3)=CH_2$ ) on the left with its corresponding hybridization state, geometric descriptor, or orbital overlap description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are: (1) The

C_2-C_3

single bond matches with the

\sigma

-bond formed by head-on

sp-sp^2

hybrid orbital overlap. (2) The methyl carbon center matches with

sp^3

hybridization, tetrahedral geometry, and

\approx 109.5^\circ

bond angles. (3) The alkene double bond matches with one

\sigma

-bond (

sp^2-sp^2

) and one

\pi

-bond (

2p-2p

). (4) The terminal acetylenic carbon matches with

sp

hybridization, linear geometry, and

180^\circ

bond angle.

Each carbon atom in 2-methylbut-1-en-3-yne adopts a hybridization state determined by its steric number (number of attached atoms and lone pairs).

C_1

and

C_2

are

sp

-hybridized (linear,

180^\circ

),

C_3

and

C_4

are

sp^2

-hybridized (trigonal planar,

120^\circ

), and the methyl group carbon is

sp^3

-hybridized (tetrahedral,

109.5^\circ

). Consequently, single bonds between differently hybridized carbons utilize hybrid orbitals corresponding to each carbon (

sp-sp^2

for

C_2-C_3

), and double bonds consist of one

\sigma

bond (

sp^2-sp^2

) plus one

\pi

bond (

2p-2p

).

Step-by-Step Solution

1

Analyze the expanded structural formula of 2-methylbut-1-en-3-yne

H-C(1)\equiv C(2)-C(3)(CH_3)=C(4)H_2

Determining the bonding domains around each carbon atom establishes its hybridization state and structural role.

2

Determine hybridization state for each carbon center

C_1

(

sp

),

C_2

(

sp

),

C_3

(

sp^2

),

C_4

(

sp^2

), and methyl carbon (

sp^3

)

Carbons with 2 electron domains are

sp

(linear), 3 domains are

sp^2

(trigonal planar), and 4 domains are

sp^3

(tetrahedral).

3

Map orbital overlap types to specific bonds

C_2-C_3

is an

sp-sp^2

\sigma

-bond; double bond

C_3=C_4

comprises an

sp^2-sp^2

\sigma

-bond and a

2p-2p

\pi

-bond.

Single bonds are formed by head-on overlap of hybrid orbitals, while double bonds consist of one coaxial

\sigma

bond and one collateral

\pi

bond.

4

Match left items with their corresponding right item descriptions based on hybridization and geometry principles

All 4 items are accurately matched to their structural characteristics.

Ensures complete alignment between structural features and underlying orbital hybridization properties.

Key Concept

Orbital Hybridization and Overlap Types in Hydrocarbon Frameworks

Estimated Time:2m 0s

Question 44Question →

But-1-ene and but-2-ene share the molecular formula $C_4H_8$ but differ in the location of their carbon-carbon double bond. Which type of structural isomerism do these two compounds exhibit?

Chain isomerism

Positional isomerism

Functional group isomerism

Geometric isomerism

Show answer & explanation

Answer: Positional isomerism

Answer

Positional isomerism

Positional isomerism occurs when compounds with the same carbon skeleton and the same functional group differ only in the location of that functional group on the chain. But-1-ene and but-2-ene both have a straight four-carbon chain, but the double bond starts at position 1 and position 2 respectively.

Step-by-Step Solution

1

Examine the structures of both molecules.

Both but-1-ene (

CH_2=CH-CH_2-CH_3

) and but-2-ene (

CH_3-CH=CH-CH_3

) have a straight four-carbon chain and an alkene functional group.

Comparing carbon chain structure and functional group identity helps determine isomer type.

2

Identify the structural difference between the two molecules.

The double bond is located between carbon-1 and carbon-2 in but-1-ene, but between carbon-2 and carbon-3 in but-2-ene.

Compounds with the same carbon framework that differ only in the location of the functional group are classified as positional isomers.

Key Concept

Positional Isomerism in Alkenes

Question 45Question →

Three isomeric alkanes have the molecular formula $C_5H_{12}$ : pentane, 2-methylbutane, and 2,2-dimethylpropane. Which of the following statements correctly accounts for the trend in their boiling points?

Pentane has the highest boiling point because its straight-chain structure provides a larger surface area for intermolecular van der Waals forces.

2,2-Dimethylpropane has the highest boiling point because its compact spherical shape maximizes molecular interactions.

2-Methylbutane has a higher boiling point than pentane because branching increases the strength of intermolecular forces.

All three isomers have identical boiling points because they possess the same molecular formula and molar mass.

Show answer & explanation

Answer: Pentane has the highest boiling point because its straight-chain structure provides a larger surface area for intermolecular van der Waals forces.

Answer

Pentane has the highest boiling point because its straight-chain structure provides a larger surface area for intermolecular van der Waals forces.

Pentane possesses an unbranched, straight-chain hydrocarbon structure. This spatial arrangement allows adjacent molecules to lie close together with maximum surface contact. As a result, intermolecular van der Waals forces are strongest in pentane, requiring the highest temperature to transition from liquid to gas.

Step-by-Step Solution

1

Analyze the molecular structures of the three

C_5H_{12}

isomers.

Pentane is unbranched (straight-chain), 2-methylbutane is monobranched, and 2,2-dimethylpropane is highly branched (spherical).

Structural branching determines molecular shape and the overall contact area available between molecules.

2

Relate molecular shape to intermolecular forces.

Alkanes are non-polar and held together by weak London dispersion (van der Waals) forces, which scale with molecular surface contact area.

Greater surface contact area leads to stronger attractive forces that require more thermal energy to overcome.

3

Determine the boiling point trend based on surface area.

Pentane has the largest surface area of contact, giving it the strongest intermolecular forces and the highest boiling point, while 2,2-dimethylpropane has the lowest.

Increased branching compacts the molecule into a sphere, minimizing contact area and lowering the boiling point.

Key Concept

Effect of structural branching on alkane boiling points and intermolecular forces

Estimated Time:1m 0s

Question 46Question →

How many acyclic structural isomers exist for the haloalkane with the molecular formula $C_4H_9Cl$ ?

2

3

4

5

Show answer & explanation

Answer: 4

Answer

There are 4 acyclic structural isomers for

C_4H_9Cl

.

The molecular formula

C_4H_9Cl

allows for both chain isomerism (butane vs. methylpropane carbon backbones) and positional isomerism (placement of the chlorine atom). This results in four distinct IUPAC structures: 1-chlorobutane, 2-chlorobutane, 1-chloro-2-methylpropane, and 2-chloro-2-methylpropane.

Step-by-Step Solution

1

Identify structural isomers based on the straight-chain butyl skeleton (

CH_3-CH_2-CH_2-CH_3

).

Two positional isomers are obtained: 1-chlorobutane (

CH_3CH_2CH_2CH_2Cl

) and 2-chlorobutane (

CH_3CH_2CH(Cl)CH_3

).

Varying the position of the chlorine atom on a four-carbon unbranched chain yields two distinct structures.

2

Identify structural isomers based on the branched methylpropane skeleton (

(CH_3)_2CH-CH_3

).

Two chain/positional isomers are obtained: 1-chloro-2-methylpropane (

(CH_3)_2CHCH_2Cl

) and 2-chloro-2-methylpropane (

(CH_3)_3CCl

).

Attaching the chlorine atom to a primary carbon versus the tertiary carbon of the methylpropane backbone yields two additional unique isomers.

3

Sum the distinct structural isomers found.

Total number of isomers =

2 + 2 = 4

.

All possible constitutional connectivities for

C_4H_9Cl

have been evaluated without double counting.

Key Concept

Structural Isomerism in Haloalkanes

Estimated Time:1m 15s

Question 47Question →

What is the IUPAC name of the major organic product formed when one mole of hydrogen bromide ( $HBr$ ) reacts with prop-1-ene in the absence of peroxides?

Show answer & explanation

Answer: 2-bromopropane; 2-Bromopropane

Answer

2-bromopropane

Electrophilic addition of hydrogen bromide to prop-1-ene without peroxides follows Markovnikov's rule. The hydrogen atom attaches to the terminal carbon (

CH_2

), leading to the formation of a secondary carbocation intermediate. Subsequent nucleophilic attack by the bromide ion yields 2-bromopropane as the major product.

Step-by-Step Solution

1

Identify the reactants and the reaction type

The reaction takes place between an unsymmetrical alkene, prop-1-ene (

CH_3CH=CH_2

), and hydrogen bromide (

HBr

). This is an electrophilic addition reaction.

Alkenes readily undergo electrophilic addition across the unsaturated carbon-carbon double bond.

2

Apply Markovnikov's rule to determine carbocation stability

In the absence of peroxides, the reaction follows Markovnikov's rule. The proton (

H^+

) adds to the double-bonded carbon holding more hydrogen atoms (

C1

,

CH_2

), generating a more stable secondary carbocation (

CH_3CH^+CH_3

).

A secondary carbocation is more stable than a primary carbocation due to alkyl group electron-donating inductive effects.

3

Attach the bromide ion and name the resulting haloalkane

The bromide ion (

Br^-

) attacks the secondary carbocation to form

CH_3CH(Br)CH_3

. The systematic IUPAC name for this compound is 2-bromopropane.

Numbering the three-carbon parent chain gives the bromine substituent locant position 2.

Key Concept

Markovnikov's rule in electrophilic addition reactions of alkenes

Question 48Question →

A molecule possessing two identical chiral carbon atoms, such as $2,3$ -dichlorobutane, yields a total of four optically active stereoisomers.

True

False

Show answer & explanation

Answer: False

Answer

The statement is false. A molecule with two identical chiral carbon atoms generates three stereoisomers in total: one pair of optically active enantiomers and one optically inactive meso compound.

The statement is false because

2,3

-dichlorobutane contains two identical chiral carbon atoms. Its internal plane of symmetry in the meso configuration causes internal compensation of optical rotation, yielding only two optically active enantiomers and one optically inactive meso form (three stereoisomers in total).

Step-by-Step Solution

1

Identify the chiral carbon atoms and their bonded groups in

2,3

-dichlorobutane.

Carbon-2 and Carbon-3 are both chiral centers, each attached to four groups:

-H

,

-Cl

,

-CH_3

, and

-CH(Cl)CH_3

. Because both carbons have identical sets of substituents, the chiral centers are identical.

Recognizing identical chiral centers is essential for identifying internal planes of symmetry.

2

Evaluate theoretical stereoisomer combinations using the

2^n

rule.

The theoretical maximum without symmetry considerations is

2^2 = 4

configurations:

(2R, 3R)

,

(2S, 3S)

,

(2R, 3S)

, and

(2S, 3R)

.

The formula

2^n

determines the upper limit of stereoisomers for

n

chiral centers.

3

Examine internal symmetry and optical activity of the configurations.

The

(2R, 3R)

and

(2S, 3S)

forms are non-superimposable mirror images (enantiomers) and are optically active. However, the

(2R, 3S)

and

(2S, 3R)

structures are identical to each other due to an internal plane of symmetry. This single achiral meso compound is optically inactive via internal compensation.

Internal symmetry cancels optical rotation, leaving only two optically active stereoisomers and one optically inactive meso stereoisomer (3 total).

Key Concept

Meso compounds and internal optical compensation in molecules with identical chiral centers.

Question 49Question →

Match each type of organic isomerism on the left with its corresponding example pair of compounds on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Chain isomerism pairs with butane and

2

-methylpropane; Positional isomerism pairs with drop-in pair butan-

1

-ol and butan-

2

-ol; Functional group isomerism pairs with ethanoic acid and methyl methanoate; Geometric isomerism pairs with

cis

-but-

2

-ene and

trans

-but-

2

-ene.

Each type of isomerism is matched to its definitive structural characteristic: chain isomerism involves skeleton branching changes (butane and

2

-methylpropane), positional isomerism involves relocations of the same functional group (butan-

1

-ol and butan-

2

-ol), functional group isomerism involves different organic families sharing a formula (ethanoic acid and methyl methanoate), and geometric isomerism involves spatial orientations across a double bond (

cis

-but-

2

-ene and

trans

-but-

2

-ene).

Step-by-Step Solution

1

Analyze the carbon skeletons of butane and

2

-methylpropane.

Both have formula

C_4H_{10}

, but one is unbranched while the other is branched, defining chain isomerism.

Chain isomers possess identical molecular formulas but different carbon chain arrangements.

2

Examine the functional group positions in butan-

1

-ol and butan-

2

-ol.

The

-\text{OH}

group is on carbon-

1

in butan-

1

-ol and carbon-

2

in butan-

2

-ol, defining positional isomerism.

Positional isomers have the same functional group located on different carbon atoms along the same parent chain.

3

Compare the functional groups of ethanoic acid and methyl methanoate.

Ethanoic acid is a carboxylic acid (

CH_3COOH

) while methyl methanoate is an ester (

HCOOCH_3

). Both have the formula

C_2H_4O_2

, defining functional group isomerism.

Functional group isomers share a molecular formula but belong to different organic homologous families.

4

Evaluate the spatial arrangement in

cis

-but-

2

-ene and

trans

-but-

2

-ene.

The double bond restricts rotation, placing methyl groups on the same side (

cis

) or opposite sides (

trans

), defining geometric isomerism.

Geometric isomerism is a stereoisomerism type arising from restricted double-bond rotation.

Key Concept

Classification of Structural Isomerism and Stereoisomerism

Question 50Question →

Match each organic reagent setup or chemical process in Column A with its corresponding chemical reaction product or characteristic diagnostic observation in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Bubbling

but-1-yne

into ammoniacal silver nitrate forms a white precipitate of silver but-1-ynide due to acidic acetylenic hydrogen replacement; passing

propene

into cold dilute alkaline

KMnO_4

decolourizes purple

MnO_4^-

forming a diol and brown

MnO_2

; heating ethanol with excess concentrated

H_2SO_4

at

170^\circ\text{C}

produces ethene via intramolecular dehydration; adding water to

CaC_2

yields ethyne gas via hydrolysis.

The matches correctly pair each chemical reaction with its distinctive mechanism or diagnostic test outcome: terminal alkyne acidity forming silver salts, alkene hydroxylation via Baeyer's reagent, alcohol elimination yielding ethene at elevated temperature, and carbide hydrolysis generating ethyne.

Step-by-Step Solution

1

Analyze the reaction of terminal alkynes with ammoniacal silver nitrate

but-1-yne

is a 1-alkyne containing a hydrogen atom bonded to an

sp

-hybridized carbon (

\text{CH}_3\text{CH}_2\text{C}\equiv\text{CH}

). This hydrogen is slightly acidic and is readily displaced by

Ag^+

to form a insoluble white precipitate of silver but-1-ynide.

To distinguish terminal alkynes from internal alkynes and alkenes.

2

Analyze the reaction of alkenes with Baeyer's reagent

propene

(

\text{CH}_3\text{CH}=\text{CH}_2

) undergoes syn-hydroxylation with cold, dilute alkaline

KMnO_4

to form

propane-1,2-diol

. The purple trioxomanganate(VII) is reduced to brown manganese(IV) oxide (

MnO_2

).

To identify mild oxidation of double bonds in unsaturation testing.

3

Analyze the acid-catalyzed dehydration condition of ethanol

Heating ethanol with concentrated

H_2SO_4

at a high temperature (

170^\circ\text{C}

) favours intramolecular elimination of water yielding ethene (

C_2H_4

).

Lower temperatures (

140^\circ\text{C}

) yield ethoxyethane instead, so temperature controls product selectivity.

4

Analyze the laboratory preparation of ethyne

Calcium dicarbide (

CaC_2

) reacts directly with water to yield ethyne gas (

C_2H_2

) and calcium hydroxide (

Ca(OH)_2

).

This is the primary laboratory synthesis method for ethyne.

Key Concept

Chemical reactions, laboratory preparation methods, and diagnostic unsaturation tests for alkenes and alkynes.

Question 51Question →

During the fractional distillation of crude oil in a refining column, components separate based on differences in their boiling point ranges. Which of the following petroleum fractions is collected at the very top of the fractionating column?

Refinery gas

Bitumen

Kerosene

Diesel oil

Show answer & explanation

Answer: Refinery gas

Answer

Refinery gas is collected at the very top of the fractionating column.

Refinery gas comprises short-chain alkanes containing 1 to 4 carbon atoms. Because these small molecules experience weak intermolecular forces, they possess the lowest boiling points among all petroleum fractions and rise to the coolest region at the very top of the fractionating tower.

Step-by-Step Solution

1

Analyze how fractional distillation separates petroleum fractions.

Fractions separate according to their boiling point ranges, which depend on hydrocarbon chain length and molar mass.

Smaller alkane molecules have weaker van der Waals forces and lower boiling points.

2

Determine the temperature gradient in a fractionating column.

The bottom of the column is the hottest, while the top is the coolest.

Vapors rise through the column and condense when the temperature drops below their respective boiling points.

3

Identify the fraction with the lowest boiling point.

Refinery gas (

C_1

–

C_4

alkanes like methane, ethane, propane, and butane) has the lowest boiling point range (

< 20^\circ\text{C}

) and remains gaseous at the top.

Components with the lowest boiling points travel to the coolest section at the top before exiting.

Key Concept

Fractional Distillation of Crude Oil and Boiling Point Trends

Estimated Time:45s

Question 52Question →

Which of the following reagents forms a characteristic precipitate when reacted with propyne, but produces no precipitate when reacted with propene?

Ammoniacal silver nitrate solution

Bromine water

Acidified potassium tetraoxomanganate(VII) solution

Concentrated tetraoxosulfate(VI) acid

Show answer & explanation

Answer: Ammoniacal silver nitrate solution

Answer

Ammoniacal silver nitrate solution

Ammoniacal silver nitrate solution contains complexed silver ions

[Ag(NH_3)_2]^+

which react selectively with the acidic hydrogen attached to the triply bonded carbon in terminal alkynes like propyne (

CH_3C\equiv CH

), forming a insoluble white precipitate of silver propynide. Propene lacks this acidic acetylenic hydrogen and gives no precipitate.

Step-by-Step Solution

1

Identify the structural difference between propyne and propene

Propyne (

CH_3C\equiv CH

) is a terminal alkyne containing a weakly acidic hydrogen attached to a triply-bonded carbon atom, whereas propene (

CH_3CH=CH_2

) is an alkene.

Terminal alkynes possess acidic acetylenic hydrogens (

R-C\equiv C-H

), unlike alkenes.

2

Evaluate the chemical reagents for specific reactivity with terminal acetylenic hydrogen

Ammoniacal silver nitrate solution (

[Ag(NH_3)_2]^+

) reacts with terminal alkynes to precipitate silver dicarbide/alkynide (a white precipitate). Alkenes do not undergo this substitution reaction.

Reagents like bromine water and acidified

KMnO_4

test for general double/triple bond unsaturation and react with both compounds.

Key Concept

Distinction between terminal alkynes and alkenes using ammoniacal silver nitrate solution

Estimated Time:45s

Question 53Question →

Arrange the following crude oil (petroleum) fractions in order of increasing boiling point range, starting from the fraction with the lowest boiling point to the one with the highest boiling point.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from lowest to highest boiling point range is Refinery gas, followed by Petrol (Gasoline), Kerosene (Paraffin), and Diesel oil (Gas oil).

In the fractional distillation of petroleum, fractions condense and exit the fractionating column at different levels depending on their boiling point ranges. Smaller alkane molecules have lower molecular masses, weaker intermolecular forces, and lower boiling points. Thus, Refinery gas (

C_1-C_4

) has the lowest boiling point, followed by Petrol (

C_5-C_{10}

), Kerosene (

C_{10}-C_{16}

), and Diesel oil (

C_{14}-C_{20}

) with the highest boiling point among the listed options.

Step-by-Step Solution

1

Determine the relationship between carbon chain length and boiling point in alkane fractions.

Smaller hydrocarbon molecules have weaker intermolecular van der Waals forces and therefore lower boiling points.

Boiling point increases as the number of carbon atoms per molecule increases.

2

Identify the carbon chain lengths for each given fraction.

Refinery gas (

C_1-C_4

), Petrol (

C_5-C_{10}

), Kerosene (

C_{10}-C_{16}

), and Diesel oil (

C_{14}-C_{20}

).

Fractional distillation separates crude oil based on boiling point ranges governed by molecular sizes.

3

Arrange the fractions from smallest carbon number to largest carbon number.

Refinery gas

\rightarrow

Petrol

\rightarrow

Kerosene

\rightarrow

Diesel oil.

This sequence directly corresponds to increasing boiling point range.

Key Concept

Fractional Distillation and Boiling Point Trends of Petroleum Fractions

Estimated Time:45s

Question 54Question →

In the laboratory preparation of ethyne gas, water is added dropwise to solid calcium carbide ( $CaC_2$ ). What is the IUPAC name of the inorganic compound produced as a byproduct in this reaction?

Show answer & explanation

Answer: Calcium hydroxide; calcium hydroxide; Ca(OH)2; Ca(OH)₂

Answer

Calcium hydroxide

Hydrolysis of calcium carbide (

CaC_2

) produces ethyne (

C_2H_2

) as the desired hydrocarbon gas along with calcium hydroxide (

Ca(OH)_2

) as the inorganic byproduct.

Step-by-Step Solution

1

Write the balanced chemical equation for the hydrolysis of calcium carbide.

CaC_2(s) + 2H_2O(l) \rightarrow C_2H_2(g) + Ca(OH)_2(s)

Calcium carbide reacts exothemically with water to liberate ethyne gas and leave behind a solid residue of calcium hydroxide.

2

Identify the chemical name of the inorganic byproduct

Ca(OH)_2

.

Calcium hydroxide

The inorganic salt contains calcium ions (

Ca^{2+}

) and hydroxide ions (

OH^-

), giving the IUPAC name calcium hydroxide.

Key Concept

Laboratory preparation of ethyne by hydrolysis of calcium carbide

Question 55Question →

How many of the acyclic structural isomers with the molecular formula $C_4H_8Cl_2$ possess at least one chiral carbon atom?

3

2

4

5

Show answer & explanation

Answer: 3

Answer

3 acyclic structural isomers of

C_4H_8Cl_2

possess at least one chiral carbon atom.

The correct response is 3. A chiral carbon atom is defined as a carbon atom bonded to four different atoms or functional groups. Among the nine constitutional acyclic isomers of dichlorobutane (

C_4H_8Cl_2

), only 1,2-dichlorobutane (at carbon-2), 1,3-dichlorobutane (at carbon-2), and 2,3-dichlorobutane (at carbon-2 and carbon-3) possess carbon atoms meeting this criterion.

Step-by-Step Solution

1

Identify all acyclic structural (constitutional) isomers of

C_4H_8Cl_2

.

There are 9 total acyclic structural isomers: 1,1-dichlorobutane, 1,2-dichlorobutane, 1,3-dichlorobutane, 1,4-dichlorobutane, 2,2-dichlorobutane, 2,3-dichlorobutane, 1,1-dichloro-2-methylpropane, 1,2-dichloro-2-methylpropane, and 1,3-dichloro-2-methylpropane.

Structural isomers must be systematically generated based on butane and methylpropane carbon skeletons.

2

Examine each structural isomer for the presence of a chiral (asymmetric) carbon atom.

A chiral carbon must be bonded to four completely different atoms or groups.

Chirality requires tetrahedral asymmetry around a carbon atom.

3

Determine which specific isomers contain chiral carbons.

1. 1,2-dichlorobutane: Carbon-2 is bonded to

-H

,

-Cl

,

-CH_2Cl

, and

-CH_2CH_3

(Chiral).
2. 1,3-dichlorobutane: Carbon-2 is bonded to

-H

,

-Cl

,

-CH_3

, and

-CH_2CH_2Cl

(Chiral).
3. 2,3-dichlorobutane: Carbon-2 and Carbon-3 are both bonded to

-H

,

-Cl

,

-CH_3

, and

-CH(Cl)CH_3

(Chiral).

All other structural isomers have identical groups attached to each carbon (e.g., two

-Cl

atoms on C-1 in 1,1-dichlorobutane, or two methyl groups on C-2 in methylpropane derivatives).

Key Concept

Chirality and Structural Isomerism in Haloalkanes

Question 56Question →

Which organic gas is produced when ethanol ( $C_2H_5OH$ ) is heated with excess concentrated tetraoxosulfate(VI) acid ( $H_2SO_4$ ) at $170^\circ\text{C}$ ?

Ethene (

C_2H_4

)

Ethane (

C_2H_6

)

Ethyne (

C_2H_2

)

Ethoxyethane (

(C_2H_5)_2O

)

Show answer & explanation

Answer: Ethene ( $C_2H_4$ )

Answer

Ethene (

C_2H_4

)

When ethanol is heated with excess concentrated tetraoxosulfate(VI) acid at a high temperature (

170^\circ\text{C}

), the acid acts as a dehydrating agent, removing a molecule of water from ethanol to form ethene (

C_2H_4

).

Step-by-Step Solution

1

Identify the functional group and reagent conditions given in the reaction stem.

Ethanol (

C_2H_5OH

) is reacted with excess concentrated

H_2SO_4

at

170^\circ\text{C}

.

Concentrated tetraoxosulfate(VI) acid acts as a powerful dehydrating agent at high temperatures.

2

Determine the type of dehydration taking place under these specific temperature conditions.

Intramolecular dehydration occurs, removing one water molecule (

H_2O

) from a single ethanol molecule.

At

170^\circ\text{C}

with excess acid, the removal of

-OH

and a neighboring

-H

atom creates a carbon-carbon double bond.

3

Write the balanced chemical equation to confirm the product formula.

C_2H_5OH \xrightarrow{\text{conc. } H_2SO_4, 170^\circ\text{C}} C_2H_4 + H_2O

The organic gas evolved is ethene (

C_2H_4

), an alkene.

Key Concept

Laboratory preparation of alkenes via dehydration of alkanols

Estimated Time:45s

Question 57Question →

According to Hückel's rule, a planar cyclic conjugated compound exhibits aromatic stability if the number of delocalized $\pi$ -electrons in its conjugated ring system is equal to which expression, where $n$ is a non-negative integer?

4n + 2

4n

2n + 2

4n - 2

Show answer & explanation

Answer: $4n + 2$

Answer

The expression for the number of delocalized

\pi

-electrons in an aromatic compound is

4n + 2

.

Hückel's rule states that a cyclic, planar, fully conjugated molecule possesses aromatic stability when it contains

(4n + 2)

delocalized

\pi

-electrons, where

n

is an integer (

0, 1, 2, 3, \dots

). For example, benzene has

6

\pi

-electrons, satisfying the rule for

n = 1

.

Step-by-Step Solution

1

Recall the structural criteria for aromaticity.

An aromatic molecule must be cyclic, planar, fully conjugated, and possess a specific number of delocalized

\pi

-electrons.

These structural conditions allow complete ring delocalization of

\pi

-electrons.

2

Apply Hückel's rule to determine the required

\pi

-electron count.

The total number of delocalized

\pi

-electrons must equal

(4n + 2)

, where

n = 0, 1, 2, 3, \dots

This formula predicts closed-shell electronic stability for planar conjugated rings.

Key Concept

Hückel's Rule of Aromaticity

Estimated Time:45s

Question 58Question →

An organic compound $X$ with the molecular formula $C_4H_6$ rapidly decolorizes bromine water in tetrachloromethane, but produces no precipitate when treated with ammoniacal silver nitrate solution. What is the IUPAC name of compound $X$ ?

But-2-yne

But-1-yne

But-1-ene

But-2-ene

Show answer & explanation

Answer: But-2-yne

Answer

But-2-yne

The compound But-2-yne (

CH_3-C\equiv C-CH_3

) has the molecular formula

C_4H_6

and contains an internal carbon-carbon triple bond. As an unsaturated hydrocarbon, it readily decolorizes bromine water. However, because its triple bond is located between carbon-2 and carbon-3, it lacks a terminal acetylenic hydrogen atom (

-C\equiv C-H

). Therefore, it cannot react with ammoniacal silver nitrate to form a precipitate, matching all given experimental observations.

Step-by-Step Solution

1

Determine the structural class using the given molecular formula

C_4H_6

.

The formula

C_4H_6

fits the general formula

C_nH_{2n-2}

(alkynes or alkadienes), indicating two degrees of unsaturation.

Alkynes with 4 carbon atoms have the formula

C_4H_6

.

2

Analyze the response to bromine water in tetrachloromethane.

Decolorization confirms the presence of carbon-carbon multiple bonds (unsaturation).

Electrophilic addition of bromine occurs across the triple bond.

3

Analyze the reaction with ammoniacal silver nitrate solution.

The absence of a precipitate indicates that the alkyne is non-terminal (internal).

Only terminal alkynes containing an acidic acetylenic hydrogen atom (

-C\equiv C-H

) react with Tollens' reagent (

[Ag(NH_3)_2]^+

) to yield a insoluble silver alkynide precipitate.

4

Identify the correct IUPAC name among the

C_4H_6

isomers.

But-2-yne (

CH_3-C\equiv C-CH_3

) is an internal alkyne, whereas But-1-yne (

CH_3-CH_2-C\equiv CH

) is a terminal alkyne.

Since compound

X

does not form a precipitate, it must be the internal alkyne, But-2-yne.

Key Concept

Distinction between terminal and internal alkynes using Tollens' reagent (ammoniacal silver nitrate)

Estimated Time:1m 30s

Question 59Question →

When an unknown alkanol is warmed with acidified potassium heptaoxodichromate(VI) ( $K_2Cr_2O_7$ ) solution, the orange solution turns green and an alkanone is produced. Which of the following compounds undergoes this reaction?

Propan-2-ol

Propan-1-ol

2-Methylpropan-2-ol

Ethanol

Show answer & explanation

Answer: Propan-2-ol

Answer

Propan-2-ol is a secondary alkanol that undergoes oxidation to form an alkanone (propanone), changing the color of acidified potassium heptaoxodichromate(VI) from orange to green.

Propan-2-ol is a secondary alkanol (

CH_3-CH(OH)-CH_3

). Upon oxidation with acidified

K_2Cr_2O_7

, the orange dichromate(VI) ions (

Cr_2O_7^{2-}

) are reduced to green chromium(III) ions (

Cr^{3+}

), and the secondary alcohol is converted into propanone (

CH_3COCH_3

), which belongs to the alkanone family.

Step-by-Step Solution

1

Classify the given alkanols by structural type (primary, secondary, or tertiary).

Propan-1-ol and ethanol are primary alkanols; propan-2-ol is a secondary alkanol; 2-methylpropan-2-ol is a tertiary alkanol.

The reaction outcome of alkanol oxidation depends strictly on the classification of the hydroxyl-bearing carbon atom.

2

Determine the oxidation product for each classification type.

Primary alkanols oxidize to alkanals (and further to alkanoic acids); secondary alkanols oxidize to alkanones; tertiary alkanols resist mild oxidation.

Secondary alkanols have one hydrogen atom on the hydroxyl carbon, which allows dehydrogenation to form a carbonyl double bond (

C=O

) bounded by two alkyl groups (ketone/alkanone).

3

Match the specified product (alkanone) to the correct compound.

Propan-2-ol oxidizes to propanone (

CH_3COCH_3

), which is an alkanone.

Only the secondary alkanol propan-2-ol yields an alkanone.

Key Concept

Oxidation of Alkanols (Primary, Secondary, and Tertiary Classification)

Estimated Time:1m 0s

Question 60Question →

In petroleum refining, catalytic cracking breaks down long-chain hydrocarbons into smaller, economically valuable fractions. If one mole of dodecane ( $C_{12}H_{26}$ ) undergoes thermal-catalytic cracking to yield one mole of hexane ( $C_6H_{14}$ ) and two moles of an alkene product $X$ , what is the molecular formula of $X$ ?

C_3H_6

C_3H_8

C_6H_{12}

C_6H_{14}

Show answer & explanation

Answer: $C_3H_6$

Answer

The molecular formula of alkene

X

is

C_3H_6

(propene).

In catalytic cracking, atomic mass is conserved. Subtracting the atoms of hexane (

C_6H_{14}

) from dodecane (

C_{12}H_{26}

) leaves 6 carbon atoms and 12 hydrogen atoms. Since 2 moles of product

X

are formed (

2X = C_6H_{12}

), dividing by 2 yields

C_3H_6

, which is propene, a valid alkene.

Step-by-Step Solution

1

Write the balanced stoichiometric equation for the cracking reaction.

C_{12}H_{26} \rightarrow C_6H_{14} + 2X

Cracking conserves the total number of carbon and hydrogen atoms between reactants and products.

2

Determine the remaining number of carbon and hydrogen atoms allocated to

2X

.

Carbons:

12 - 6 = 6

, Hydrogens:

26 - 14 = 12

. Total fragment formula is

C_6H_{12}

.

Subtract the atoms present in one mole of hexane from dodecane.

3

Divide the carbon and hydrogen count by 2 to find the formula of 1 mole of product

X

.

Carbon count for

X = \frac{6}{2} = 3

, Hydrogen count for

X = \frac{12}{2} = 6

. Formula of

X = C_3H_6

.

Since two moles of alkene

X

are produced, each mole contains half the remaining atoms, fitting the general formula for alkenes (

C_nH_{2n}

).

Key Concept

Catalytic Cracking Stoichiometry of Alkanes

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