Hydrocarbons Class 11 Chemistry Notes PDF Download
ALKENES
Alkenes are unsaturated hydrocarbons having a double bond between two carbon atoms. Alkenes have the general formula${{C}_{n}}{{H}_{2n}}$.
Isomerism
1. Structural isomerism
Example, Butene has 3structural isomers,
\[\underset{But-1-ene}{\mathop{C{{H}_{3}}-C{{H}_{2}}-CH=C{{H}_{2}}}}\,\]
\[\underset{But-2-ene}{\mathop{C{{H}_{3}}-CH=CH-C{{H}_{3}}}}\,\]
Isobutene
2. Geometrical isomers
The hindered rotation around the C – C bond gives rise to stereoisomers having different spatial arrangements. Two isomers exist as follows:
Cis and Trans Isomers of but-2-ene
Methods of Preparation
1. By dehydration of alcohol: Dehydration of alcohol in the presence of acids forms alkene. The reaction is an elimination reaction.
\[R-C{{H}_{2}}-C{{H}_{2}}-OH\xrightarrow[\Delta ]{{{H}^{+}}}R-CH=C{{H}_{2}}+{{H}_{2}}O\]
2. By the dehydrohalogenation of alkyl halides: it involves an alkyl halide in the presence of alcoholic KOH to yield alkene. \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br\xrightarrow{alc.KOH}C{{H}_{3}}CH=C{{H}_{2}}+HBr\]
If dehydrogenation of alkyl halide gives two products, the major product will be according to Saytzeff’s rule that states that the most substituted alkene will be the major product.
Dehydrohalogenation of Alkyl Halides
The ease of dehydrohalogenation has the order:
Tertiary alkyl halide > secondary alkyl halide > primary alkyl halide.
Alkyl halides follow the order:
alkyl iodide > alkyl bromide > alkyl chloride.
3. By the dehalogenation of vicinal dihalides: Dehydrohalogenation of vicinal dihalides in the presence of Zinc dust along with alcoholic solution yields pure alkene.
Dehydrohalogenation of Vicinal Dihalides
4. Kolbe’s electrolysis method: The electrolysis of sodium or potassium salts of dicarboxylic acid produces alkene at anode.
Kolbe’s Electrolysis Method
If, $Na/Liq.N{{H}_{3}}$ is used, trans alkene is formed, and in presence of Ni cis alkene is formed.
Physical Properties
1. Melting point: the trans isomers have a high melting point than cis isomer due to symmetry and crystal lattice.
2. Boiling point: Cis isomer has a more dipole moment as it is more polar and therefore has a high boiling point than trans isomer of alkene.
Chemical Properties
Alkenes are reactive due to the presence of double bonds. Due to THE presence of $\pi $ bonds alkenes are able to react towards electrophilic addition reaction. They also give free radical addition reactions.
1. Addition reactions:
(i) Addition of hydrogen (catalytic hydrogenation) \[C{{H}_{2}}=C{{H}_{2}}+{{H}_{2}}\xrightarrow[200-{{300}^{o}}C]{Ni}C{{H}_{3}}-C{{H}_{3}}\]
(ii)Addition of halogens (Chlorine and bromine)
\[C{{H}_{2}}=C{{H}_{2}}+B{{r}_{2}}\to \underset{ethylene\,dibromide\,(colorless)}{\mathop{BrC{{H}_{2}}-C{{H}_{2}}Br}}\,\]
Addition of bromine is used as a test for detecting the presence of unsaturation (C – C double bond or triple bond).
(ii) Addition of hydrogen halides
\[C{{H}_{2}}=C{{H}_{2}}+HX\to C{{H}_{3}}C{{H}_{2}}X\]
The order of reactivity among hydrogen halides is
HI > HBr > HCl > HF
In case of unsymmetrical alkenes, addition occurs according to Markonikov’s rule. This reaction takes place through an ionic mechanism. Electrophilic addition to a carbon–carbon double bond involves the formation of an intermediate that is the more stable carbocation.
Deviation from Markonikov’s rule:
It has been observed that addition of HBr to unsymmetrical alkenes like propene in presence of air, peroxide or light yields n-propyl bromide by anti-Markovnikov's rule. The effect is also called the peroxide effect or Kharasch effect.
Markonikov’s and Anti-Markovnikov's Rule for Halogenation of Propene
Addition of hypochlorous acid
Reaction of Hypochlorous Acid of Ethylene
Addition of sulphuric acid
Reaction of Sulphuric Acid of Propene
Alkyl hydrogen sulphates are water soluble, when heated at about${{160}^{o}}C$ , they give olefins. In reaction with water they give alcohol.
Formation of Alcohol from Alkyl Hydrogen Sulphates
Addition of water
It is also in accordance with Markovnikov’s rule.
Addition of Water to Alkene
Addition of alkanes (alkylation)
Alkylation of Alkanes in Alkene
Addition of diborane (hydroboration)
Hydroboration of Alkanes in Alkene
In case of unsymmetrical alkenes, addition follows Anti Markovnikov's rule.
\[6C{{H}_{3}}C{{H}_{2}}CH=C{{H}_{2}}+{{B}_{2}}{{H}_{6}}\to \underset{tributyl\,borane}{\mathop{2{{\left( C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}} \right)}_{3}}~B}}\,\]
Trialkyl borane on oxidation $({{H}_{2}}{{O}_{2}}/O{{H}^{-}})$ gives alcohol and on reduction $(LiAl{{H}_{4}})$ gives alkane.
Oxymercuration – demercuration
Oxymercuration – Demercuration of Alkene
Addition of oxygen
Addition of Oxygen in Alkene
1. Oxidation:
Oxidation by cold alkaline $KMn{{O}_{4}}$ (Bayer’s reagent)
Oxidation of Alkene
It is a test for detecting double bonds in alkene. Hydroxylation by $KMn{{O}_{4}}$ is always a syn addition. The cis alkene on hydroxylation gives meso compound and trans alkene gives racemic mixture. Like Bayer’s reagent $Os{{O}_{4}}$ also gives glycol and the hydroxylation is a syn addition.
Hydroxylation of Alkene Using OsO4
Oxidation by per acids $(RC{{O}_{3}}H)$
Oxidation of Alkene by per Acids $(RC{{O}_{3}}H)$
Reaction of Acetic Acid with Alkene
This addition occurs in trans manners. The cis alkene gives racemic mixture and trans alkenes give meso compound.
Ozonolysis
Ozonolysis of Alkene
Oxidation by hot concentrated alkaline $KMn{{O}_{4}}$
\[RCH=C{{H}_{2}}+KMn{{O}_{4}}(conc.)\to RCOOH+C{{O}_{2}}+{{H}_{2}}O\]
1. Substitution reaction:
Chlorination is done by treating the alkene with carbon tetrachloride in liquid phase or with chlorine gas:
Chlorination of Alkene
Allylic bromination (bromination at allylic carbon atom) is very easily achieved by treating the alkene having hydrogen atom at the allylic carbon atom with N-bromosuccinimde (NBS).
Allylic Bromination of Alkene
Illustration 3 : Which of the following alkene has the lowest heat of hydrogenation?
Alkene with Lowest Heat of Hydrogenation
Solution: (B). Higher the stability of alkene, lower the heat of hydrogenation.
Illustration 4 :
Reaction of Alkene with Alkaline KMnO4
Which is true about this reaction?
(A) A is meso 1, 2-butan-di-ol formed by syn addition.
(B) A is meso 1, 2-butan-di-ol formed by anti addition.
(C) A is a racemic mixture of d and l, 1, 2-butan-di-ol formed by anti addition.
(D) A is a racemic mixture of d and l, 1, 2-butan-di-ol formed by syn addition.
Solution: (A). On cis alkene there is syn addition of two –OH groups forming meso compound
Addition of hypochlorous acid
Addition of sulphuric acid
Addition of water
Addition of alkanes (alkylation)
Addition of diborane (hydroboration)
Oxymercuration – demercuration
Addition of oxygen
Oxidation by cold alkaline $KMn{{O}_{4}}$ (Bayer’s reagent)
Oxidation by per acids $(RC{{O}_{3}}H)$
Ozonolysis
Oxidation by hot concentrated alkaline $KMn{{O}_{4}}$
Exercise 2
Halogenation of Alkene
Predominant A is:
Predominant Product A
Product X and Y from Alkene
X and Y are:
$\text{C}{{\text{H}}_{3}}-\text{C}{{\text{H}}_{2}}-\text{CH}=\text{C}{{\text{H}}_{2}},\left( \text{C}{{\text{H}}_{3}}\text{C}{{\text{H}}_{2}}\text{COOH}+\text{C}{{\text{O}}_{2}} \right)$
$\text{C}{{\text{H}}_{3}}-\text{CH}=\text{CH}-\text{C}{{\text{H}}_{3}},\text{C}{{\text{H}}_{3}}\text{COOH (2 moles) }$
$\text{C}{{\text{H}}_{3}}-\text{CH}=\text{CH}-\text{C}{{\text{H}}_{3}},\text{C}{{\text{H}}_{3}}\text{CHO (2 moles) }$
$\text{C}{{\text{H}}_{3}}-\text{C}{{\text{H}}_{2}}-\text{CH}=\text{C}{{\text{H}}_{2}},\left( \text{C}{{\text{H}}_{3}}\text{C}{{\text{H}}_{2}}\text{CHO}+\text{HCHO} \right)$
The reaction of propene with HOCl proceeds via the addition of:
${{H}^{+}}$ in the first step
$C{{l}^{+}}$ the first step
$O{{H}^{-}}$ in the first step
$C{{l}^{+}}$ and $O{{H}^{-}}$ in a single step
$\text{C}{{\text{H}}_{3}}-\text{C}{{\text{H}}_{2}}-\text{CH}=\text{C}{{\text{H}}_{2}},\left( \text{C}{{\text{H}}_{3}}\text{C}{{\text{H}}_{2}}\text{COOH}+\text{C}{{\text{O}}_{2}} \right)$
$\text{C}{{\text{H}}_{3}}-\text{CH}=\text{CH}-\text{C}{{\text{H}}_{3}},\text{C}{{\text{H}}_{3}}\text{COOH (2 moles) }$
$\text{C}{{\text{H}}_{3}}-\text{CH}=\text{CH}-\text{C}{{\text{H}}_{3}},\text{C}{{\text{H}}_{3}}\text{CHO (2 moles) }$
$\text{C}{{\text{H}}_{3}}-\text{C}{{\text{H}}_{2}}-\text{CH}=\text{C}{{\text{H}}_{2}},\left( \text{C}{{\text{H}}_{3}}\text{C}{{\text{H}}_{2}}\text{CHO}+\text{HCHO} \right)$
${{H}^{+}}$ in the first step
$C{{l}^{+}}$ the first step
$O{{H}^{-}}$ in the first step
$C{{l}^{+}}$ and $O{{H}^{-}}$ in a single step
ALKYNES
Alkynes are characterized by the presence of a triple bond between two carbon atoms. The general formula of alkyne is${{C}_{n}}{{H}_{2n-2}}$ .
Methods of Preparation
1. By the dehydrohalogenation of vicinal dihalides:
\[C{{H}_{2}}=C{{H}_{2}}\xrightarrow{B{{r}_{2}}}Br-C{{H}_{2}}-C{{H}_{2}}-Br+KOH(alc.)\to Br-CH=C{{H}_{2}}\xrightarrow{NaN{{H}_{2}}}CH\equiv CH\]
\[C{{H}_{3}}-CHB{{r}_{2}}+KOH\text{ }(alc.)\text{ }\to C{{H}_{2}}=CH-Br\xrightarrow{NaN{{H}_{2}}}CH\equiv CH\]
2. By dehalogenation of vicinal tetrahalides:
Reaction with active metals like Zinc, Mg etc. gives acetylene.
Dehalogenation of Vicinal Tetrahalides
3. By Kolbe electrolysis method:
Kolbe Electrolysis Method
4. By heating iodoform or chloroform with silver powder or zinc: This method can be used for the preparation of only acetylene.
\[CH{{I}_{3}}+6Ag+CH{{I}_{3}}\to CH\equiv CH+6AgI\]
5. From acetylene: Higher alkynes can be prepared from acetylene when treated with sodium metal in liquid ammonia.
\[CH\equiv CH+Na\xrightarrow{Liq.N{{H}_{3}}}CH\equiv C-Na+{}^{1}/{}_{2}{{H}_{2}}\]
\[\underset{sodium\,acetylide}{\mathop{CH\equiv CNa}}\,+C{{H}_{3}}Br\to \underset{propyne}{\mathop{CH\equiv C-C{{H}_{3}}}}\,+NaBr\]
Similarly, $CH\equiv CH\xrightarrow[Liq.N{{H}_{3}}]{2Na}NaC=CNa\xrightarrow{2C{{H}_{3}}Br}C{{H}_{3}}C=CC{{H}_{3}}$
Chemical Properties
Alkyne gives electrophilic addition reactions due to the presence of loosely held $\pi $ electrons, but electrophilic addition reactions in alkynes are slower than that of alkenes.
Terminal hydrogen present in alkynes is acidic in nature. Since s electrons are closer to the nucleus than p electrons, the electrons present in the bond having more s character will be closer to the nucleus. The amount of s character in various types of C – H bond are as follows
Hybridisation of Alkene, Percentage of S-Character
Relative acidities:
\[\text{HOH}=\text{HOR} > \text{CH}\equiv \text{CR} > \overset{..}{\mathop{\text{N}}}\,{{\text{H}}_{3}} > \text{C}{{\text{H}}_{2}}=\text{C}{{\text{H}}_{2}} > \text{C}{{\text{H}}_{3}}-\text{C}{{\text{H}}_{3}}\]
Relative basicities:
\[\text{O}{{\text{H}}^{-}}=\text{O}{{\text{R}}^{-}} < {{\text{C}}^{-}}\equiv \text{C}-\text{R} < \text{NH}_{2}^{-} < \text{C}{{\text{H}}^{-}}=\text{C}{{\text{H}}_{2}} < \text{CH}_{2}^{-}-\text{C}{{\text{H}}_{3}}\]
1. Addition of hydrogen:
\[\underset{acetylene}{\mathop{CH\equiv CH}}\,+{{H}_{2}}\xrightarrow{Ni}\underset{ethylene}{\mathop{C{{H}_{2}}=C{{H}_{2}}}}\,\xrightarrow{{{H}_{2}}}\underset{ethane}{\mathop{C{{H}_{3}}-C{{H}_{3}}}}\,\]
In case of alkynes where triple bond is not present at the end of the chain, on reduction gives cis or trans alkene, which depends upon the choice of reducing agent. With sodium in liquid ammonia the alkene is trans form and on catalytic reduction the alkene is cis form.
Addition of Hydrogen to Alkyne
2. Electrophilic addition:
Addition of halogens
Electrophilic Addition of Chlorine in Alkyne
The order of reactivity of halogens is $C{{l}_{2}} > B{{r}_{2}} > {{I}_{2}}$
Addition of halogen acid
Electrophilic Addition Halogen Acid of Alkyne
The order of reactivity of halogen acids is HI > HBr > HCl.
In the presence of peroxide, anti Markonikov’s product is obtained.
Halogenation of Alkyne
Addition of hypohalous acids
Addition of Hypohalous Acids of Alkyne
3. Nucleophilic addition reaction: In these reactions, the addition is initiated by a nucleophile and are generally catalysed by salt of heavy metals (e.g. $H{{g}^{2+}},P{{b}^{2+}},B{{a}^{2+}}$ )which are found to form $\pi $ compound with multiple bonds.
\[\text{HC}\equiv \text{CH}+\text{H}{{\text{g}}^{2+}}\to \underset{\underset{H{{g}^{2+}}}{\mathop{\downarrow }}\,}{\mathop{\text{CH}=\text{CH}}}\,\]
Addition of water
Addition of Water to Alkyne
Addition of hydrogen cyanide
\[HC\equiv CH+HCN\xrightarrow{Ba{{(CN)}_{2}}}\underset{vinyl\,cyanide}{\mathop{C{{H}_{2}}=CHCN}}\,\]
Addition of acetic acid
Addition of Acetic Acid to Alkyne
Addition of alcohol
\[HC\equiv CH+{{C}_{2}}{{H}_{5}}OH\xrightarrow[{{130}^{o}}C]{{{H}_{2}}S{{O}_{4}}}C{{H}_{2}}=CHO{{C}_{2}}{{H}_{5}}\xrightarrow{{{H}_{2}}O}C{{H}_{3}}CHO+{{C}_{2}}{{H}_{3}}OH\]
Addition of ozone and ozonolysis
Ozonolysis of alkyne
4. Oxidation:
Oxidation with alkaline $KMn{{O}_{4}}$
\[\underset{acetylene}{\mathop{HC\equiv CH}}\,+4[O]\xrightarrow{alk.KMn{{O}_{4}}}\underset{oxalic\,acid}{\mathop{HOOC-COOH}}\,\]
Oxidation of Alkyne
Oxidation with acidic ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ $KMn{{O}_{4}}$
\[\underset{acetylene}{\mathop{HC\equiv CH}}\,+\xrightarrow[{{H}^{+}}]{{{K}_{2}}C{{r}_{2}}{{O}_{7}}}HOOC-COOH\to \underset{acetic\,acid}{\mathop{C{{H}_{3}}COOH}}\,\]
Oxidation of Acidic ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ / $KMn{{O}_{4}}$
5. Formation of metallic derivatives:
The group –C≡C–H in alkynes is slightly acidic in nature and hence its hydrogen atom can be easily replaced by certain metals to give metallic derivatives called acetylides or alkynides.
Acidic Hydrogen in Alkenes
Formation of sodium acetylides
\[HC\equiv CH+Na\xrightarrow{Liq.N{{H}_{3}}}\underset{monosodium\,acetylide}{\mathop{HC\equiv C-Na}}\,+Na\xrightarrow[{{120}^{o}}C]{NaN{{H}_{2}}}\underset{disodium\,acetylide}{\mathop{NaC=CNa}}\,\]
Formation of Sodium Propynide
Formation of copper and silver acetylides
\[HC\equiv CH+\underset{ammonical\,cuprous\,chloride}{\mathop{2CuC{{l}_{2}}+2N{{H}_{4}}OH}}\,\to \underset{copper\,acetylide\,(red\,ppt.)}{\mathop{Cu-C\equiv C-Cu\downarrow }}\,+2N{{H}_{4}}Cl+2{{H}_{2}}O\]
\[HC\equiv CH+\underset{ammonical\,silver\,nitrate}{\mathop{2AgN{{O}_{3}}+2N{{H}_{4}}OH}}\,\to \underset{silver\,acetylide\,(white\,ppt)}{\mathop{AgC\equiv CAg\downarrow }}\,+2N{{H}_{4}}N{{O}_{3}}+2{{H}_{2}}O\]
These reactions are used for detecting the presence of acetylenic hydrogen atoms.
Illustration 4 : The products obtained via oxymercuration ($HgS{{O}_{4}}+{{H}_{2}}S{{O}_{4}}$) of 1–butyne would be
Oxymercuration Product of 1–Butyne
Solution: (A)
Oxymercuration reaction of 1–butyne
Addition of halogens
Addition of halogen acid
Addition of hypohalous acids
Addition of water
Addition of hydrogen cyanide
Addition of acetic acid
Addition of alcohol
Addition of ozone and ozonolysis
Oxidation with alkaline $KMn{{O}_{4}}$
Oxidation with acidic ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ $KMn{{O}_{4}}$
Formation of sodium acetylides
Formation of copper and silver acetylides
Class 11 Chemistry Chapter 9 Hydrocarbons Important Topics and Subtopics Covered
Class 11 Chemistry Chapters 9 Details, and Formulas and Concepts.
Classification of Hydrocarbons:
Hydrocarbons are organic compounds consisting of only carbon and hydrogen atoms.
They are classified into two main groups: aliphatic hydrocarbons (open-chain or cyclic) and aromatic hydrocarbons (containing benzene ring or its derivatives).
Structural Isomerism:
Structural isomers have the same molecular formula but different structural arrangements.
IUPAC Nomenclature:
The International Union of Pure and Applied Chemistry (IUPAC) system is used to systematically name hydrocarbons based on their structure and functional groups.
Rules include identifying the longest continuous carbon chain (parent chain), numbering the chain to give substituents the lowest possible locants, and naming substituent groups based on prefixes.
Types of Hydrocarbons:
Alkanes: Saturated hydrocarbons containing only single bonds between carbon atoms.
Alkenes: Unsaturated hydrocarbons containing at least one carbon-carbon double bond.
Alkynes: Unsaturated hydrocarbons containing at least one carbon-carbon triple bond.
Aromatic Hydrocarbons: Contain one or more benzene rings or derivatives.
Reactions of Hydrocarbons:
Combustion: Hydrocarbons burn in the presence of oxygen to produce carbon dioxide and water.
Substitution: Alkanes undergo substitution reactions, where one or more hydrogen atoms are replaced by other atoms or groups.
Addition: Alkenes and alkynes undergo addition reactions, where atoms or groups are added to the carbon-carbon double or triple bonds.
Importance of Revision Notes for Class 11 Chemistry Chapter 9 Hydrocarbons
Summarises Key Points: Condenses important concepts for quick review.
Saves Time: Provides a fast way to revise before exams.
Highlights Essentials: Focuses on crucial topics and definitions.
Improves Memory: Helps in better retention of information.
Enhances Exam Prep: Targets weak areas for more effective study.
Clarifies Concepts: Simplifies complex ideas for easier understanding.
Includes Visuals: Uses diagrams and charts for better grasp.
Boosts Confidence: Prepares students thoroughly for exams.
Tips for Learning the Chemistry Class 11 Chapter 9
Focus on core processes with illustrations and examples of Hydrocarbon
Draw and label Nomenclature for clarity.
Create summaries of each process of Types of Isomerism.
Connect concepts to everyday examples of Alkane, Alkynes and Alkenes.
Solve past exam questions to test understanding.
Explain concepts to others to reinforce learning.
Revisit material frequently to retain information.
Conclusion
Class 11 Chemistry Chapter 9 Hydrocarbons is pivotal for understanding the basic building blocks of organic chemistry. By learning about the different types of hydrocarbons, their properties, and their reactions, students gain a solid foundation in organic chemistry. This knowledge is not only essential for further studies but also has practical implications in fields such as petrochemicals, environmental science, and industrial applications. Mastery of this chapter equips students with the skills to analyse and utilise hydrocarbons in various contexts.
Related Study Materials for Class 11 Chapter 9 Hydrocarbons
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FAQs on Hydrocarbons Class 11 Notes: CBSE Chemistry Chapter 9
1. What are the main types of hydrocarbons to focus on for a quick revision of Class 11 Chapter 9?
For a quick revision, focus on the four main categories of hydrocarbons:
- Alkanes: Saturated hydrocarbons with only single carbon-carbon bonds (C-C).
- Alkenes: Unsaturated hydrocarbons containing at least one carbon-carbon double bond (C=C).
- Alkynes: Unsaturated hydrocarbons containing at least one carbon-carbon triple bond (C≡C).
- Aromatic Hydrocarbons: Cyclic compounds containing a stable benzene ring.
2. What is the core concept to remember about isomerism in hydrocarbons?
The key concept is that isomers have the same molecular formula but different structural arrangements. For revision, focus on two main types: Structural Isomerism (different connectivity, like chain and position isomers) and Geometrical Isomerism (different spatial arrangement around a double bond, known as cis-trans isomerism).
3. How can I quickly summarise the key preparation methods for alkenes?
For a quick summary, remember these four primary methods for preparing alkenes:
- Dehydration of Alcohols: Removing a water molecule from an alcohol using an acid catalyst.
- Dehydrohalogenation of Alkyl Halides: Removing a hydrogen halide (HX) using an alcoholic base, often following Saytzeff's rule.
- Dehalogenation of Vicinal Dihalides: Removing two halogen atoms from adjacent carbons using zinc dust.
- Kolbe's Electrolytic Method: Electrolysis of sodium or potassium salts of dicarboxylic acids.
4. What is the key difference to remember between Markovnikov's and Anti-Markovnikov's rule?
The key difference lies in where the negative part of the adding reagent (like HBr) attaches to an unsymmetrical alkene.
- Markovnikov's Rule: The negative part adds to the carbon atom with fewer hydrogen atoms. This is the standard mechanism.
- Anti-Markovnikov's Rule (Peroxide Effect): In the presence of peroxide, the negative part (specifically from HBr) adds to the carbon atom with more hydrogen atoms.
5. What is the main takeaway from the ozonolysis of alkenes and alkynes?
The main takeaway is that ozonolysis is a powerful reaction used to determine the position of double or triple bonds in a molecule. The reaction cleaves the molecule at the C=C or C≡C bond, forming smaller carbonyl compounds (aldehydes, ketones, or carboxylic acids), which can then be identified to deduce the original structure.
6. Why are alkenes and alkynes more reactive than alkanes?
Alkenes and alkynes are more reactive than alkanes because of the presence of loosely held pi (π) electrons in their double and triple bonds. These π-electron clouds are exposed and act as a source of electrons for attacking electrophiles, making alkenes and alkynes highly susceptible to electrophilic addition reactions. Alkanes, being saturated with strong sigma (σ) bonds, are much less reactive.
7. How does the choice of reducing agent create different products during the hydrogenation of alkynes?
The choice of reducing agent is crucial as it determines the stereochemistry of the resulting alkene.
- Using Lindlar's catalyst (partially deactivated palladium) results in a cis-alkene because the hydrogen atoms add to the same side of the triple bond (syn-addition).
- Using sodium in liquid ammonia (Birch reduction) results in a more stable trans-alkene because the mechanism involves an intermediate that favours the addition of hydrogen atoms to opposite sides (anti-addition).
8. What is the concept behind the acidic nature of terminal alkynes?
The hydrogen atom attached to a triply bonded carbon in a terminal alkyne (like in propyne) is acidic. This is because the carbon atom is sp-hybridised, which has 50% s-character. The high s-character makes the carbon atom more electronegative, pulling the electron density from the C-H bond and making it easier for the hydrogen to be released as a proton (H⁺). This acidity is why terminal alkynes react with strong bases like NaNH₂.
9. How can you use simple chemical tests to distinguish between an alkane, an alkene, and a terminal alkyne?
You can use these key tests for a quick distinction:
- Alkane: Does not react with bromine water or cold, dilute alkaline KMnO₄ (Baeyer's reagent). The colours of the reagents persist.
- Alkene: Decolourises both bromine water and Baeyer's reagent, indicating the presence of a C=C double bond.
- Terminal Alkyne: Also decolourises both reagents but gives an additional, unique positive test. It forms a white precipitate with ammoniacal silver nitrate (Tollens' reagent) or a red precipitate with ammoniacal cuprous chloride due to its acidic hydrogen.
10. What is the significance of Hückel's rule for revising aromatic hydrocarbons?
Hückel's rule is a quick check to determine if a cyclic compound is aromatic. For a compound to be aromatic, it must be cyclic, planar, have a continuous ring of p-orbitals, and contain a total of (4n+2) π electrons, where 'n' is a whole number (0, 1, 2, etc.). For benzene, n=1, so it has (4*1+2) = 6 π electrons, confirming its exceptional stability.
