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Q9.18:
Arrange benzene, $n$-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour.
Solution :
1. Analysis of Molecular Structures & Hybridization
The acidic character of hydrocarbons is primarily determined by the stability of the conjugate base formed after the removal of a proton ($H^+$). This stability is directly proportional to the electronegativity of the carbon atom carrying the negative charge, which is a function of the $s$-character in the hybrid orbital.
| Compound | Carbon Hybridization | s-Character (%) | Electronegativity |
|---|---|---|---|
| Ethyne ($HC \equiv CH$) | $sp$ | 50% | Highest |
| Benzene ($C_6H_6$) | $sp^2$ | 33.3% | Intermediate |
| $n$-Hexane ($CH_3CH_2CH_2CH_2CH_2CH_3$) | $sp^3$ | 25% | Lowest |
2. Step-by-Step Theoretical Justification
Step 1: Analyzing Ethyne ($sp$ hybridization): The carbon atoms in ethyne are $sp$ hybridized [Per VSEPR Theory and Valence Bond Theory]. The $50\% \ s$-character makes the $sp$ hybrid orbital closer to the nucleus, exerting a greater pull on the electron pair of the $C-H$ bond. Consequently, the conjugate base (acetylide ion, $HC \equiv C^-$) is relatively stable because the negative charge resides in an orbital with high $s$-character.
Step 2: Analyzing Benzene ($sp^2$ hybridization): The carbons in the benzene ring are $sp^2$ hybridized. With $33.3\% \ s$-character, the electronegativity of the carbon is lower than that of $sp$ carbon but higher than $sp^3$ carbon. Thus, the acidity of benzene is intermediate [Per Inductive Effect and Resonance Theory].
Step 3: Analyzing $n$-Hexane ($sp^3$ hybridization): The carbons in $n$-hexane are $sp^3$ hybridized. The $25\% \ s$-character results in the lowest electronegativity among the three. The conjugate base (alkyl carbanion) is highly unstable, making $n$-hexane essentially non-acidic.
3. Visual Representation of Orbital Electronegativity
4. Establishing the Sequence
Based on the principle that acidity increases with the $s$-character of the hybrid orbital involved in the $C-H$ bond [Per Electronegativity Trends], the decreasing order of acidity is:
Ethyne ($HC \equiv CH$) > Benzene ($C_6H_6$) > $n$-Hexane ($CH_3CH_2CH_2CH_2CH_2CH_3$)
Final Solution: The decreasing order of acidic behaviour is Ethyne > Benzene > $n$-Hexane. This is due to the decreasing percentage of $s$-character in the hybrid orbitals of the carbon atoms ($sp > sp^2 > sp^3$), which leads to a decrease in the electronegativity of the carbon atom and a subsequent decrease in the stability of the resulting conjugate base.
More Questions from Class 11 Chemistry Hydrocarbons EXERCISES
- Q9.1: How do you account for the formation of ethane during chlorination of methane ?
- Q9.10: Why is benzene extra ordinarily stable though it contains three double bonds?
- Q9.11: What are the necessary conditions for any system to be aromatic?
- Q9.12(i): Explain why the following systems are not aromatic? (i)
- Q9.12(ii): Explain why the following systems are not aromatic? (ii)
- Q9.12(iii): Explain why the following systems are not aromatic? (iii)
- Q9.13(i): How will you convert benzene into (i) $p$-nitrobromobenzene
- Q9.13(ii): How will you convert benzene into (ii) $m$-nitrochlorobenzene
- Q9.13(iii): How will you convert benzene into (iii) $p$-nitrotoluene
- Q9.13(iv): How will you convert benzene into (iv) acetophenone?
- Q9.14: In the alkane $H_3C-CH_2-C(CH_3)_2-CH_2-CH(CH_3)_2$, identify $1^{\circ},2^{\circ},3^{\circ}$ carbon atoms and give the number of H atoms bonded to each one of these.
- Q9.15: What effect does branching of an alkane chain has on its boiling point?
- Q9.16: 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.
- Q9.17: Write down the products of ozonolysis of 1,2-dimethylbenzene ($o$-xylene). How does the result support Kekulé structure for benzene?
- Q9.19: Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?
- Q9.2(a): Write IUPAC names of the following compounds : (a) $CH_3CH=C(CH_3)_2$
- Q9.2(b): Write IUPAC names of the following compounds : (b) $CH_2=CH-C \equiv C-CH_3$
- Q9.2(c): Write IUPAC names of the following compounds : (c)
- Q9.2(d): Write IUPAC names of the following compounds : (d)
- Q9.2(e): Write IUPAC names of the following compounds : (e)
- Q9.2(f): Write IUPAC names of the following compounds : (f)
- Q9.2(g): Write IUPAC names of the following compounds : (g)
- Q9.20(i): How would you convert the following compounds into benzene? (i) Ethyne
- Q9.20(ii): How would you convert the following compounds into benzene? (ii) Ethene
- Q9.20(iii): How would you convert the following compounds into benzene? (iii) Hexane
- Q9.21: Write structures of all the alkenes which on hydrogenation give 2-methylbutane.
- Q9.22(a): Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, $E^+$ (a) Chlorobenzene, 2,4-dinitrochlorobenzene, $p$-nitrochlorobenzene
- Q9.22(b): Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, $E^+$ (b) Toluene, $p-H_3C-C_6H_4-NO_2$, $p-O_2N-C_6H_4-NO_2$.
- Q9.23: Out of benzene, $m$-dinitrobenzene and toluene which will undergo nitration most easily and why?
- Q9.24: Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.
- Q9.25: Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example.
- Q9.3(a): 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) $C_4H_8$ (one double bond)
- Q9.3(b): For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated : (b) $C_5H_8$ (one triple bond)
- Q9.4(i): Write IUPAC names of the products obtained by the ozonolysis of the following compounds : (i) Pent-2-ene
- Q9.4(ii): Write IUPAC names of the products obtained by the ozonolysis of the following compounds : (ii) 3,4-Dimethylhept-3-ene
- Q9.4(iii): Write IUPAC names of the products obtained by the ozonolysis of the following compounds : (iii) 2-Ethylbut-1-ene
- Q9.4(iv): Write IUPAC names of the products obtained by the ozonolysis of the following compounds : (iv) 1-Phenylbut-1-ene
- Q9.5: An alkene ‘$A$’ on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of ‘$A$’.
- Q9.6: An alkene ‘$A$’ contains three $C-C$, eight $C-H$ $\sigma$ bonds and one $C-C$ $\pi$ bond. ‘$A$’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of ‘$A$’.
- Q9.7: Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene?
- Q9.8(i): Write chemical equations for combustion reaction of the following hydrocarbons: (i) Butane
- Q9.8(ii): Write chemical equations for combustion reaction of the following hydrocarbons: (ii) Pentene
- Q9.8(iii): Write chemical equations for combustion reaction of the following hydrocarbons: (iii) Hexyne
- Q9.8(iv): Write chemical equations for combustion reaction of the following hydrocarbons: (iv) Toluene
- Q9.9: Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?
CBSE Solutions for Class 11 Chemistry Hydrocarbons
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