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Hydrocarbons - Alkenes and Alkynes

Grade 11ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond (C=CC=C). Their general formula is CnH2nC_nH_{2n}.

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Alkynes are unsaturated hydrocarbons containing at least one carbon-carbon triple bond (−C≡C−-C \equiv C-). Their general formula is CnH2n−2C_nH_{2n-2}.

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In alkenes, the double-bonded carbon atoms are sp2sp^2 hybridized, resulting in a trigonal planar geometry with bond angles of approximately 120∘120^\circ.

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In alkynes, the triple-bonded carbon atoms are spsp hybridized, resulting in a linear geometry with bond angles of 180∘180^\circ.

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Geometrical Isomerism (ciscis-transtrans isomerism) occurs in alkenes due to restricted rotation around the C=CC=C bond. It requires each carbon of the double bond to be attached to two different groups.

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Markovnikov's Rule: During the addition of a polar reagent (like HBrHBr) to an unsymmetrical alkene, the negative part of the reagent (nucleophile) attaches to the carbon atom containing the lesser number of hydrogen atoms.

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Kharasch Effect (Peroxide Effect): In the presence of organic peroxides, the addition of HBrHBr (only HBrHBr) to unsymmetrical alkenes follows Anti-Markovnikov orientation.

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Acidity of Alkynes: Terminal alkynes (e.g., HC≡CHHC \equiv CH) are weakly acidic because the spsp hybridized carbon has 50%50\% ss-character, making it highly electronegative and able to stabilize the conjugate base.

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Ozonolysis: The process where alkenes or alkynes react with O3O_3 followed by reductive cleavage (with Zn/H2OZn/H_2O) to form carbonyl compounds (aldehydes or ketones). This is used to locate the position of the multiple bond.

📐Formulae

CnH2n (General formula for Alkenes)C_nH_{2n} \text{ (General formula for Alkenes)}

CnH2n−2 (General formula for Alkynes)C_nH_{2n-2} \text{ (General formula for Alkynes)}

CH3−CH=CH2+HBr→CH3−CH(Br)−CH3 (Markovnikov Addition)CH_3-CH=CH_2 + HBr \rightarrow CH_3-CH(Br)-CH_3 \text{ (Markovnikov Addition)}

CH3−CH=CH2+HBr→PeroxideCH3−CH2−CH2Br (Anti-Markovnikov Addition)CH_3-CH=CH_2 + HBr \xrightarrow{\text{Peroxide}} CH_3-CH_2-CH_2Br \text{ (Anti-Markovnikov Addition)}

CaC2+2H2O→Ca(OH)2+C2H2 (Preparation of Ethyne)CaC_2 + 2H_2O \rightarrow Ca(OH)_2 + C_2H_2 \text{ (Preparation of Ethyne)}

n(CH2=CH2)→High P, T—(CH2−CH2)—n (Polymerization)n(CH_2=CH_2) \xrightarrow{\text{High P, T}} \text{---}(CH_2-CH_2)\text{---}_n \text{ (Polymerization)}

💡Examples

Problem 1:

An alkene 'A' on ozonolysis gives a mixture of ethanol (CH3CHOCH_3CHO) and propanone (CH3COCH3CH_3COCH_3). Identify the structure and IUPAC name of 'A'.

Solution:

CH3−CH=C(CH3)2CH_3-CH=C(CH_3)_2 (2-Methylbut-2-ene)

Explanation:

To find the alkene, remove the oxygen atoms from the two carbonyl products and join the carbon atoms with a double bond. CH3−CH=OCH_3-CH=O and O=C(CH3)2O=C(CH_3)_2 become CH3−CH=C(CH3)2CH_3-CH=C(CH_3)_2.

Problem 2:

How can you distinguish between Ethene (C2H4C_2H_4) and Ethyne (C2H2C_2H_2) using a chemical test?

Solution:

Use Ammoniacal Silver Nitrate (Tollen′sReagentTollen's Reagent).

Explanation:

Ethyne, being a terminal alkyne, reacts with ammoniacal silver nitrate to form a white precipitate of silver acetylide (Ag−C≡C−AgAg-C \equiv C-Ag). Ethene does not react as it lacks acidic hydrogens.

Problem 3:

Convert 1-Bromopropane to Propene.

Solution:

CH3CH2CH2Br+alc. KOH→ΔCH3CH=CH2+KBr+H2OCH_3CH_2CH_2Br + \text{alc. } KOH \xrightarrow{\Delta} CH_3CH=CH_2 + KBr + H_2O

Explanation:

This is a dehydrohalogenation reaction (β\beta-elimination) where an alcoholic solution of potassium hydroxide acts as a strong base to remove HBrHBr.