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Q9.15:
What effect does branching of an alkane chain has on its boiling point?
Solution :
Theoretical Foundation: Intermolecular Forces in Alkanes
The boiling point of an alkane is primarily governed by the magnitude of London dispersion forces (van der Waals forces) acting between its molecules. These forces are dependent on two critical factors: the total surface area of the molecule and the efficacy of molecular packing.
Step 1: Analysis of Surface Area and Molecular Shape
As the degree of branching in an alkane chain increases, the molecule shifts from a linear, elongated geometry to a more compact, spherical geometry. For isomers with the same molecular formula ($C_nH_{2n+2}$), the following physical logic applies:
- Linear Alkanes: Possess a larger surface area, allowing for greater contact between adjacent molecules [Per the principles of Molecular Geometry and London Dispersion Forces].
- Branched Alkanes: Possess a reduced surface area due to their compact, spherical shape, which minimizes the contact surface available for intermolecular attraction.
Step 2: Graphical Representation of Molecular Interaction
Step 3: Correlation with Boiling Point
The boiling point ($T_b$) is proportional to the strength of intermolecular forces. Since London dispersion forces are weaker in branched structures due to the decreased surface area of contact, less energy (thermal kinetic energy) is required to overcome these forces to transition the substance from a liquid phase to a gas phase.
| Alkane (Isomers of $C_5H_{12}$) | Structure Type | Boiling Point (K) |
|---|---|---|
| n-Pentane | Linear | 309.1 K |
| Isopentane | Slightly Branched | 301.0 K |
| Neopentane | Highly Branched | 282.5 K |
Analytical Conclusion
The branching of an alkane chain leads to a decrease in its boiling point. The structural modification results in a more spherical shape, which decreases the effective surface area for van der Waals interactions between individual molecules [Based on Kinetic Molecular Theory]. Consequently, the intermolecular attractions are weakened, facilitating vaporization at lower temperatures.
Final Solution: The boiling point of an alkane decreases as the degree of branching increases because branching reduces the surface area of the molecule, thereby decreasing the strength of the intermolecular London dispersion forces.
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.
- Q9.16: 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?
- Q9.18: Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour.
- 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) $CH_3(CH_2)_4 CH (CH_2)_3 CH_3$ with a $CH_2–CH(CH_3)_2$ group attached to the CH.
- Q9.2(g): Write IUPAC names of the following compounds : (g) $CH_3 – CH = CH – CH_2 – CH = CH – CH – CH_2 – CH = CH_2$ with a $C_2H_5$ group attached to the third CH from the left.
- 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
Chapters in CBSE - Class 11 Chemistry
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