Hydrocarbons: The Building Blocks of Organic Chemistry
Hydrocarbons are made up of just carbon and hydrogen. They form the basis of many other organic molecules and are the primary constituents of fossil fuels like coal, petroleum, and natural gas.
Classification of Hydrocarbons
Hydrocarbons are classified into two main categories based on their structure:
Aliphatic Hydrocarbons: These have a straight chain or branched chain structure.
Alkanes: Saturated hydrocarbons with single covalent bonds between carbon atoms. Example: Methane (CH₄).
Alkenes: Unsaturated hydrocarbons with one or more double bonds between carbon atoms. Example: Ethene (C₂H₄).
Alkynes: Alkynes are unsaturated hydrocarbons characterized by the presence of one or more triple bonds between carbon atoms. An example of an alkyne is ethyne, with the chemical formula C₂H₂.
Aromatic Hydrocarbons: These have a ring-like structure with alternating single and double bonds. Example: Benzene (C₆H₆).
Isomerism in Hydrocarbons
Compounds with identical molecular formulas but differing structural arrangements are known as isomers.Hydrocarbons can exhibit isomerism, leading to compounds with different properties despite having the same composition.
Sources and Uses of Hydrocarbons
Fossil Fuels: Coal, petroleum, and natural gas are rich sources of hydrocarbons.
Petrochemicals: Hydrocarbons from fossil fuels are used to produce various petrochemicals, such as plastics, synthetic fibers, and detergents.
Fuels: Hydrocarbons are used as fuels for transportation, heating, and cooking.
Reactions of Hydrocarbons
Hydrocarbons undergo various chemical reactions, including:
Combustion: Burning in oxygen to produce carbon dioxide and water, releasing heat and light.
Substitution reactions: Substitution reactions substitute hydrogen atoms.
Addition reactions: Adding atoms or groups to the double or triple bonds in unsaturated hydrocarbons.
Environmental Concerns
The excessive use of fossil fuel-derived hydrocarbons has led to environmental concerns, such as:
Greenhouse gas emissions: Contributing to global warming.
Air pollution: Causing respiratory problems and acid rain.
Efforts are being made to develop sustainable alternatives to fossil fuels and reduce their environmental impact.
1. How do you account for the formation of ethane during chlorination of methane ?
Ans :
The mechanism shows that in the propagation step, CH₃ free radicals are produced. These radicals can undergo three possible reactions, labeled (i), (ii), and (iii). In the chain termination step, two CH₃ free radicals combine to form an ethane (CH₃-CH₃) molecule.
2. Write IUPAC names of the following compounds :
Ans :
3. For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated :
(a) C4H8 (one double bond)
(b) C5H8 (one triple bond)
Ans : (a)
4. Write IUPAC names of the products obtained by the ozonolysis of the following compounds :
(i) Pent-2-ene
(ii) 3,4-Dimethylhept-3-ene
(iii) 2-Ethylbut-1-ene
(iv) 1-Phenylbut-1-ene
Ans :
5. An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of ‘A’.
Ans : Step 1:The structures of the products should be arranged side by side with their oxygen atoms facing each other.
Step 2. To form the alkene “A” from the given structure, we need to remove the oxygen atoms and create a double bond between the two carbon atoms that were previously connected to the oxygen atoms.
6. An alkene ‘A’ contains three C – C, eight C – H σ bonds and one C – C π bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of ‘A’.
Ans : (i) The aldehyde with a molar mass of 44 u is ethanal, which has the chemical formula CH₃CHO.
(ii) The structures of the products should be arranged side by side with their oxygen atoms facing each other.
(iii) To form the alkene “A” from the given structure, we need to remove the oxygen atoms and create a double bond between the two carbon atoms that were previously connected to the oxygen atoms.
The molecule but-2-ene contains three carbon-carbon sigma bonds (C-C σ-bonds), eight carbon-hydrogen sigma bonds (C-H σ-bonds), and one carbon-carbon pi bond (C-C π-bond).
7. Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene?
Ans : (i) The structures of propanal and pentan-3-ene with their oxygen atoms facing each other are:
propanal: CH₃CH₂CHO
pentan-3-ene: CH₃CH₂CH=CHCH₃
In both molecules, the oxygen atom is located on the third carbon atom from the end.
(ii) The structure of the alkene formed by removing oxygen atoms from the given compound and joining the two fragments by a double bond is:
CH₃-CH=CH-CH₃
8. Write chemical equations for combustion reaction of the following hydrocarbons:
(i) Butane
(ii) Pentene
(iii) Hexyne
(iv) Toluene
Ans :
9. Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?
Ans :
The boiling point of a molecule is influenced by the strength of its intermolecular forces. In the case of cis- and trans-isomers, dipole-dipole interactions play a significant role.
Cis-isomers have a higher dipole moment due to the alignment of polar bonds. This results in stronger dipole-dipole interactions between cis-isomer molecules compared to trans-isomers. As a consequence, cis-isomers have higher boiling points than their corresponding trans-isomers.
Therefore, the higher boiling point of cis-isomers compared to trans-isomers can be attributed to the stronger dipole-dipole interactions present in the cis-configuration.
10. Why is benzene extra ordinarily stable though it contains three double bonds?
Ans :
Benzene’s exceptional stability despite having three double bonds is due to its resonance structure.
In benzene, the six pi electrons are delocalized over the entire ring, creating a cloud of electron density above and below the plane of the molecule. Aromaticity is the term used to describe this delocalization.
Key points:
Cyclic Structure: Benzene is a cyclic compound with six carbon atoms, forming a hexagonal ring.
Planar Structure: The benzene ring is planar, meaning all six atoms are in the same plane.
Conjugated System: The six pi electrons are delocalized across all six carbon atoms, forming a conjugated system.
Aromaticity: This delocalization of electrons provides extra stability to the molecule, making it more resistant to chemical reactions compared to other molecules with isolated double bonds.
The stability gained from aromaticity is significant enough to offset the expected instability associated with having three double bonds in a molecule. This is why benzene is considered an aromatic compound and exhibits unique properties compared to non-aromatic compounds.
11. What are the necessary conditions for any system to be aromatic?
Ans : Aromatic compounds are cyclic organic compounds with a delocalized π electron system that follows Hückel’s Rule. According to this rule, a compound is aromatic when it possesses:
Cyclic structure: The molecule must form a closed ring.
Planar structure: All atoms in the ring must be in the same plane.
Conjugated π system: All atoms in the ring must have a p orbital, creating a continuous system of overlapping π orbitals.
4n+2 π electrons: The number of π electrons in the ring must follow the formula 4n+2, where n is an integer (0, 1, 2, …). This is known as Hückel’s Rule.
Examples of aromatic compounds:
Benzene (C₆H₆): Has 6 π electrons (n=1), satisfying Hückel’s Rule.
Naphthalene (C₁₀H₈): Has 10 π electrons (n=2), satisfying Hückel’s Rule.
Pyridine (C₅H₅N): Has 6 π electrons (n=1), satisfying Hückel’s Rule.
Aromatic compounds exhibit unique properties, such as stability, aromaticity, and distinct chemical reactivity.
12. Explain why the following systems are not aromatic?
Ans : The given structures are not aromatic because they do not meet the criteria for aromaticity, which include:
Cyclic structure: The molecules must form a closed ring.
Planar structure: All atoms in the ring must be in the same plane.
Conjugated π system: All atoms in the ring must have a p orbital, creating a continuous system of overlapping π orbitals.
4n+2 π electrons: The number of π electrons in the ring must follow the formula 4n+2, where n is an integer (0, 1, 2, …). This is known as Hückel’s Rule.
Let’s analyze each structure:
This structure is cyclic and planar, but it has 6 π electrons (n=1.5), which does not follow Hückel’s rule. Therefore, it is not aromatic.
(ii)
This structure is cyclic but not planar. The two hydrogen atoms attached to the central carbon atom are out of the plane of the ring. Therefore, it does not have a continuous conjugated π system and is not aromatic.
(iii)
This structure is cyclic and planar, but it has 8 π electrons (n=2), which does not follow Hückel’s rule. Therefore, it is not aromatic.
In conclusion, the given structures are not aromatic because they either violate the Hückel’s rule or lack a planar structure, preventing the formation of a continuous conjugated π system.
13. How will you convert benzene into
(i) p-nitrobromobenzene
(ii) m- nitrochlorobenzene
(iii) p – nitrotoluene
(iv) acetophenone?
Ans : Conversion of Benzene to Derivatives
Benzene is a versatile aromatic hydrocarbon that can undergo various reactions to form different derivatives. Here’s how to convert benzene into the specified compounds:
(i) p-Nitrobromobenzene
Nitration: React benzene with a mixture of concentrated nitric acid and concentrated sulfuric acid to introduce a nitro group (-NO₂) at the para position. This forms nitrobenzene.
Bromination: React nitrobenzene with bromine in the presence of a catalyst (e.g., iron or aluminum bromide) to introduce a bromine atom at the para position relative to the nitro group. This forms p-nitrobromobenzene.
(ii) m-Nitrochlorobenzene
Nitration: React benzene with a mixture of concentrated nitric acid and concentrated sulfuric acid to introduce a nitro group (-NO₂) at any position on the benzene ring.
Chlorination: React the nitrobenzene with chlorine in the presence of a catalyst (e.g., iron or aluminum chloride) to introduce a chlorine atom at the meta position relative to the nitro group. This forms m-nitrochlorobenzene.
(iii) p-Nitrotoluene
Nitration: React toluene (methylbenzene) with a mixture of concentrated nitric acid and concentrated sulfuric acid to introduce a nitro group (-NO₂) at the para position relative to the methyl group. This forms p-nitrotoluene.
(iv) Acetophenone
Friedel-Crafts Acylation: React benzene with acetyl chloride (CH₃COCl) in the presence of a Lewis acid catalyst (e.g., anhydrous aluminum chloride) to introduce an acetyl group (-COCH₃) at any position on the benzene ring. This forms acetophenone.
14. In the alkane H3C – CH2 – C(CH3) 2 – CH2 – CH(CH3) 2, identify 1°,2°,3° carbon atoms and give the number of H atoms bonded to each one of these.
Ans :
To identify the primary, secondary, and tertiary carbon atoms in the alkane H₃C-CH₂-C(CH₃)₂-CH₂-CH(CH₃)₂, let’s break it down:
Primary (1°) carbon atoms: Carbon atoms that are directly bonded to only one other carbon atom.
C1: CH₃-
C5: -CH₂-CH(CH₃)₂
Secondary (2°) carbon atoms: Carbon atoms that are directly bonded to two other carbon atoms.
C2: -CH₂-
C4: -CH₂-CH(CH₃)₂
C6: -CH(CH₃)₂
Tertiary (3°) carbon atoms: A tertiary carbon atom is one that is connected to three other carbon atoms.
C3: -C(CH₃)₂-
The hydrogen content of each carbon atom
1° carbons (C1 and C5): 3 hydrogen atoms each (CH₃-)
2° carbons (C2, C4, and C6): 2 hydrogen atoms each (-CH₂-)
3° carbon (C3): 1 hydrogen atom (-CH-)
In summary, there are 2 primary carbon atoms with 3 hydrogen atoms each, 3 secondary carbon atoms with 2 hydrogen atoms each, and 1 tertiary carbon atom with 1 hydrogen atom.
15. What effect does branching of an alkane chain has on its boiling point?
Ans : Branching of an alkane chain decreases its boiling point.
Here’s why:
The boiling point of a substance is influenced by the strength of its intermolecular attractions. For alkanes, London dispersion forces are the primary type of intermolecular attraction.
Branching and Surface Area: Branched alkanes have a more compact shape than their unbranched counterparts. This diminishes the surface area available for intermolecular interactions.
Weakened Intermolecular Forces: With a smaller surface area, the London dispersion forces between branched alkanes are weaker compared to unbranched alkanes.
Lower Boiling Point: Weaker intermolecular forces require less energy to overcome, resulting in a lower boiling point for branched alkanes.
In summary, the branching of an alkane chain reduces its surface area, weakens its intermolecular forces, and consequently lowers its boiling point.
16. Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.
Ans : The addition of HBr to propene is an electrophilic addition reaction. However, the mechanism of this reaction can be influenced by the presence of a catalyst like benzoyl peroxide.
In the absence of a catalyst, the electrophile H⁺ adds to the double bond of propene to form a secondary carbocation intermediate, which is then attacked by the nucleophile Br⁻ to form 2-bromopropane.
In the presence of benzoyl peroxide, the reaction remains electrophilic, but the electrophile becomes a bromine free radical (Br•). This free radical adds to the propene molecule to form a more stable secondary free radical. The secondary free radical then abstracts a hydrogen atom from HBr to give 1-bromopropane.
Although both reactions involve the addition of H and Br atoms, the different order of addition due to the change in the electrophile leads to the formation of different products. This demonstrates the importance of the stability of the intermediate formed during the reaction.
17. Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support Kekulé structure for benzene?
Ans : The molecule o-xylene can be represented by two resonance structures, known as Kekulé structures. Ozonolysis of each of these structures gives two different products.
The three possible products formed in the bromination of o-xylene are evidence that o-xylene is a resonance hybrid of the two Kekulé structures. If o-xylene were a single structure, only one product would be formed. The fact that three products are observed indicates that o-xylene exists as a hybrid of the two contributing structures, allowing for bromination at different positions.
18. Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour.
Ans :
The decreasing order of acidic behavior among benzene, n-hexane, and ethyne is:
Ethyne > Benzene > n-Hexane
Reasoning:
s-character: The acidic character of a carbon-hydrogen bond increases with the s-character of the carbon atom. The s-character of the carbon atom increases in the order: sp³ < sp² < sp.
Ethyne: In ethyne, the carbon atoms involved in the C-H bond are sp-hybridized, which has the highest s-character. This means the carbon atom in ethyne has a higher electronegativity, making it more acidic.
Benzene: The carbon atoms involved in the C-H bond in benzene are sp²-hybridized, which has a lower s-character compared to ethyne. This makes benzene less acidic than ethyne.
n-Hexane: In n-hexane, the carbon atoms involved in the C-H bond are sp³-hybridized, which has the lowest s-character. This makes n-hexane the least acidic among the three compounds.
In summary, the acidic behavior of these compounds increases with the s-character of the carbon atom involved in the C-H bond.
19. Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?
Ans : Benzene undergoes electrophilic substitution reactions easily but nucleophilic substitutions with difficulty due to its aromatic nature.
Here’s a breakdown of why:
Aromaticity: Benzene has a delocalized π electron system that contributes to its stability. This delocalization creates a cloud of electron density above and below the plane of the molecule.
Electron-Rich Environment: The delocalized π electrons make benzene an electron-rich molecule. This electron-rich nature makes it susceptible to attack by electron-deficient species (electrophiles).
Electrophilic Substitution: When an electrophile (such as a halogen or a proton) approaches the benzene ring, it is attracted to the electron-rich π cloud. The electrophile attacks one of the carbon atoms, disrupting the aromaticity. To regain aromaticity, a proton is lost from the adjacent carbon, resulting in the substitution of the electrophile.
Nucleophilic Substitution: Nucleophilic substitution reactions involve the attack of a nucleophile (electron-rich species) on an electrophilic carbon. In benzene, the carbon atoms are already electron-rich due to the delocalized π electrons. This makes it difficult for a nucleophile to attack and replace a hydrogen atom in the benzene ring without disrupting the aromaticity.
In summary, benzene’s aromatic nature, characterized by its delocalized π electron system, makes it highly susceptible to electrophilic substitution reactions but resistant to nucleophilic substitution reactions.
20. How would you convert the following compounds into benzene?
(i) Ethyne
(ii) Ethene
(iii) Hexane
Ans : Conversion of Compounds to Benzene
(i) Ethyne (C₂H₂) to Benzene (C₆H₆)
Step 1: Trimerization: Ethyne can undergo trimerization under appropriate conditions (e.g., using a catalyst like cobalt or nickel) to form benzene.
(ii) Ethene (C₂H₄) to Benzene
Step 1: Dimerization: Ethene can dimerize to form cyclobutene.
Step 2: Ring Opening and Cyclization: Cyclobutene can undergo ring-opening followed by cyclization to form benzene. This reaction often requires a catalyst.
(iii) Hexane (C₆H₁₄) to Benzene
Step 1: Dehydrogenation: Hexane can be dehydrogenated to form cyclohexane. This reaction requires a catalyst and high temperatures.
Step 2: Aromatization: Cyclohexane can be aromatized to benzene under appropriate conditions, such as heating with a catalyst like platinum or palladium.
21. Write structures of all the alkenes which on hydrogenation give 2-methylbutane.
Ans :
22. Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E+
(a) Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene
(b) Toluene, p-H3C – C6H4 – NO2, p-O2N – C6H4 – NO2.
Ans : The reactivity of aromatic compounds towards electrophilic substitution reactions is influenced by the electron-withdrawing or electron-donating effects of substituents on the benzene ring.
Electron-withdrawing groups (e.g., nitro, cyano, carbonyl) decrease the electron density of the benzene ring, making it less reactive towards electrophiles.
Electron-donating groups (e.g., alkyl, alkoxy) increase the electron density of the benzene ring, making it more reactive towards electrophiles.
Here’s the arrangement of the compounds in decreasing order of relative reactivity with an electrophile (E+):
(a) Chlorobenzene > p-nitrochlorobenzene > 2,4-dinitrochlorobenzene
Chlorobenzene has one electron-withdrawing chlorine substituent, making it moderately less reactive than benzene.
p-Nitrochlorobenzene has two electron-withdrawing nitro substituents, making it less reactive than chlorobenzene.
2,4-Dinitrochlorobenzene has three electron-withdrawing nitro substituents, making it the least reactive among the three.
(b) Toluene > p-H₃C-C₆H₄-NO₂ > p-O₂N-C₆H₄-NO₂
The methyl group on toluene increases its reactivity relative to benzene due to its electron-donating nature.
p-H₃C-C₆H₄-NO₂ has one electron-donating methyl group and one electron-withdrawing nitro group. The electron-donating effect of the methyl group slightly outweighs the electron-withdrawing effect of the nitro group, making it slightly more reactive than benzene.
p-O₂N-C₆H₄-NO₂ has two electron-withdrawing nitro groups, making it less reactive than benzene.
In summary, the reactivity of aromatic compounds towards electrophiles decreases with an increase in the number and strength of electron-withdrawing substituents.
23. Out of benzene, m–dinitrobenzene and toluene which will undergo nitration most easily and why?
Ans : Benzene will undergo nitration most easily among the three compounds.
Here’s why:
Electron-donating groups: Toluene has a methyl group, which is an electron-donating group. This increases the electron density of the benzene ring, making it more susceptible to electrophilic attack by the nitronium ion (NO₂⁺).
Electron-withdrawing groups: The nitro group (-NO₂) is an electron-withdrawing group. This decreases the electron density of the benzene ring, making it less susceptible to electrophilic attack.
Therefore, m-dinitrobenzene, with two electron-withdrawing nitro groups, is the least reactive towards nitration. Benzene, with no substituents, is more reactive than m-dinitrobenzene. Toluene, with an electron-donating methyl group, is even more reactive towards nitration.
In summary, the reactivity of benzene towards nitration increases in the order: m-dinitrobenzene < benzene < toluene.
24. Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.
Ans : Here are a few Lewis acids other than anhydrous aluminum chloride that can be used for the ethylation of benzene:
Anhydrous ferric chloride (FeCl₃): Ferric chloride is a common Lewis acid used in Friedel-Crafts reactions, including ethylation.
Boron trifluoride (BF₃): This Lewis acid is a strong electrophile and can be used for ethylation reactions, especially with activated aromatic compounds.
Stannic chloride (SnCl₄): Stannic chloride is another Lewis acid that can be used for ethylation of benzene.
Titanium tetrachloride (TiCl₄): This Lewis acid is also effective for ethylation reactions.
The choice of Lewis acid depends on various factors, such as the reactivity of the aromatic compound, the desired reaction conditions, and the cost and availability of the Lewis acid.
25. Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example
Ans : The Wurtz reaction is not preferred for the preparation of alkanes containing an odd number of carbon atoms because it produces alkanes with an even number of carbon atoms.
The coupling of two alkyl halides using sodium metal is known as the Wurtz reaction. The reaction mechanism involves the formation of alkyl radicals, which then combine to form the alkane product. The number of carbon atoms in the product is always twice the number of carbon atoms in the starting alkyl halide.
Example:
If we react two molecules of ethyl bromide (CH₃CH₂Br) using the Wurtz reaction, we get butane (CH₃CH₂CH₂CH₃), which has an even number of carbon atoms.
CH₃CH₂Br + CH₃CH₂Br + 2Na → CH₃CH₂CH₂CH₃ + 2NaBr
To prepare alkanes with an odd number of carbon atoms, alternative methods like the Grignard reaction or the Corey-House synthesis are more suitable. These methods allow for the formation of carbon-carbon bonds with greater flexibility and control over the product structure.
