Alkenes and Alkynes: Preparation, Reactions, and Unsaturation Tests Practice Questions — JAMB UTME

Question 1Question →

An acyclic hydrocarbon $X$ with the molecular formula $C_5H_8$ rapidly decolourises bromine water. However, when $X$ is treated with ammoniacal silver nitrate solution, no precipitate is observed. Upon complete catalytic hydrogenation in the presence of a nickel catalyst, $X$ is converted into pentane. What is the IUPAC name of hydrocarbon $X$ ?

Show answer & explanation

Answer: pent-2-yne; 2-pentyne; Pent-2-yne; 2-Pentyne

Answer

Pent-2-yne (or 2-pentyne)

Hydrocarbon

X

has the molecular formula

C_5H_8

, corresponding to two degrees of unsaturation. Complete hydrogenation to pentane confirms an unbranched five-carbon chain. Decolourisation of bromine water verifies unsaturation. Because

X

yields no precipitate with ammoniacal silver nitrate solution, it lacks acidic terminal acetylenic hydrogens (

R-C \equiv C-H

). Therefore, the triple bond must be located internally between carbon-2 and carbon-3, making the compound pent-2-yne.

Step-by-Step Solution

1

Determine the degree of unsaturation and carbon skeleton of hydrocarbon

X

.

Degree of unsaturation is 2, and the carbon skeleton is a straight 5-carbon chain.

The molecular formula

C_5H_8

corresponds to

C_nH_{2n-2}

, indicating two degrees of unsaturation (an alkyne or alkadiene). Complete catalytic hydrogenation yields pentane (

C_5H_{12}

), proving an unbranched five-carbon chain.

2

Analyze the reaction with bromine water.

Hydrocarbon

X

contains carbon-carbon multiple bonds.

Decolourisation of bromine water confirms the presence of unsaturation.

3

Evaluate the test with ammoniacal silver nitrate solution.

Hydrocarbon

X

is an internal (non-terminal) alkyne.

Terminal alkynes possess acidic acetylenic hydrogen atoms (

R-C \equiv C-H

) that react with ammoniacal silver nitrate to form a characteristic white silver acetylide precipitate. The absence of a precipitate rules out pent-1-yne and confirms that the triple bond is located internally at C-2.

4

Deduce the final IUPAC name of hydrocarbon

X

.

pent-2-yne

Combining a straight 5-carbon chain with an internal triple bond between carbon-2 and carbon-3 gives pent-2-yne.

Key Concept

Distinction between terminal and non-terminal alkynes using ammoniacal silver nitrate test and carbon skeleton determination via hydrogenation

Estimated Time:2m 0s

Question 2Question →

An unknown gaseous hydrocarbon $X$ decolorizes acidified potassium tetraoxomanganate(VII) solution and produces a reddish-brown precipitate when bubbled into an ammoniacal solution of copper(I) chloride. Which of the following IUPAC structural formulas represents hydrocarbon $X$ ?

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

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

\text{CH}_3\text{CH}=\text{CHCH}_3

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

Show answer & explanation

Answer: $\text{CH}_3\text{CH}_2\text{C}\equiv\text{CH}$

Answer

The correct structural formula is

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

(but-1-yne).

The compound

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

(but-1-yne) contains a carbon-carbon triple bond which decolorizes acidified

\text{KMnO}_4

via oxidation. Furthermore, because it is a terminal alkyne with a hydrogen atom directly bonded to an

sp

-hybridized carbon, it reacts with ammoniacal copper(I) chloride solution to yield a reddish-brown precipitate of copper(I) acetylide.

Step-by-Step Solution

1

Analyze the reaction with acidified potassium tetraoxomanganate(VII) (\text{KMnO}_4) solution.

The decolorization of acidified

\text{KMnO}_4

proves that hydrocarbon

X

is unsaturated (contains carbon-carbon double or triple bonds). This eliminates saturated alkanes like butane.

Unsaturated hydrocarbons undergo oxidation addition across carbon-carbon multiple bonds.

2

Analyze the reaction with ammoniacal copper(I) chloride (\text{Cu}_2\text{Cl}_2) solution.

Formation of a reddish-brown precipitate (copper(I) diallylide/acetylide) confirms the presence of a terminal alkyne carrying an acidic acetylenic hydrogen attached to an

sp

-hybridized carbon atom (

-\text{C}\equiv\text{C}-\text{H}

).

Only terminal alkynes have sufficiently acidic hydrogen atoms to be substituted by copper(I) or silver ions in ammoniacal solutions.

3

Differentiate between terminal alkynes, internal alkynes, and alkenes based on structural formulas.

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

is but-1-yne (a terminal alkyne) which fulfills both conditions.

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

is an internal alkyne lacking a terminal acidic hydrogen.

Internal alkynes and alkenes fail the ammoniacal copper(I) chloride test despite being unsaturated.

Key Concept

Distinction between general unsaturation tests and terminal alkyne confirmation tests.

Estimated Time:1m 30s

Question 3Question →

Match each chemical process or reagent involving alkenes and alkynes in Column A with its corresponding chemical observation or primary product in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairing links calcium carbide hydrolysis to ethyne gas preparation, propene with bromine in carbon tetrachloride to reddish-brown decolorization, but-1-yne with ammoniacal silver nitrate to white precipitate formation, and but-2-yne with ammoniacal silver nitrate to no visible reaction.

Hydrolysis of calcium carbide is a standard laboratory preparation of ethyne gas. Electrophilic addition of bromine across the double bond of propene decolorizes the reddish-brown bromine solution, demonstrating general unsaturation. Ammoniacal silver trioxonitrate(V) reacts specifically with terminal alkynes like but-1-yne because of their acidic terminal hydrogens to form a white silver alkynide precipitate, whereas internal alkynes like but-2-yne lack a terminal acidic hydrogen and show no reaction.

Step-by-Step Solution

1

Identify the preparation method for ethyne

Calcium carbide reacts with water according to

\text{CaC}_2 + 2\text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_2 + \text{Ca(OH)}_2

, yielding ethyne gas.

This is the primary laboratory method for generating ethyne.

2

Evaluate the reaction testing for general unsaturation in propene

Bromine in

\text{CCl}_4

adds across the carbon-carbon double bond of propene to form 1,2-dibromopropane.

The consumption of molecular bromine causes the reddish-brown color of the solution to discharge.

3

Distinguish between terminal and internal alkynes using ammoniacal silver trioxonitrate(V)

But-1-yne possesses a terminal

\text{C}\equiv\text{C-H}

bond with an acidic proton that forms a white silver alkynide precipitate, whereas internal but-2-yne lacks terminal hydrogen atoms and gives no precipitate.

Ammoniacal

\text{AgNO}_3

selectively reacts only with terminal alkynes based on terminal hydrogen acidity.

Key Concept

Unsaturation testing, ethyne synthesis, and differentiation of terminal from internal alkynes

Estimated Time:1m 30s

Question 4Question →

A gaseous hydrocarbon rapidly decolourizes bromine water in tetrachloromethane, but fails to form a precipitate when bubbled through an ammoniacal solution of silver nitrate. Which of the following hydrocarbons exhibits this chemical behavior?

Propene

Propyne

Ethyne

Propane

Show answer & explanation

Answer: Propene

Answer

Propene

Propene is an alkene (

CH_3CH=CH_2

). Its carbon-carbon double bond allows it to undergo electrophilic addition with bromine water, decolourizing the reddish-brown solution. However, because it is an alkene rather than a terminal alkyne, it lacks an acidic hydrogen atom bonded to a triply-bonded carbon atom, so it does not produce a precipitate when treated with ammoniacal silver nitrate.

Step-by-Step Solution

1

Analyze the reaction with bromine water.

Rapid decolourization of bromine water confirms the presence of unsaturation (a double or triple carbon-carbon bond). Saturated hydrocarbons like propane are eliminated.

Unsaturated hydrocarbons undergo addition reactions across their double or triple bonds to absorb bromine.

2

Analyze the outcome with ammoniacal silver nitrate solution.

The absence of a precipitate indicates that the hydrocarbon is NOT a terminal alkyne (

R-\text{C}\equiv\text{C}-\text{H}

).

Only terminal alkynes possess acidic acetylenic hydrogen atoms capable of being replaced by silver ions to form insoluble metallic acetylides.

3

Identify the hydrocarbon matching both criteria.

Propene (

CH_3CH=CH_2

) is an alkene. It decolourizes bromine water due to unsaturation but yields a negative test with ammoniacal silver nitrate.

Alkenes contain carbon-carbon double bonds but lack acidic acetylenic hydrogen atoms.

Key Concept

Distinguishing alkenes from terminal alkynes using unsaturation and acetylenic hydrogen reagents

Estimated Time:1m 0s

Question 5Question →

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 6Question →

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 7Question →

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 8Question →

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 9Question →

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 10Question →

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

Alkenes and Alkynes: Preparation, Reactions, and Unsaturation Tests

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