3.14
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Q1: Why do alkanes have low reactivity?
Alkanes exhibit low reactivity due to strong non-polar C–C and C–H σ bonds. These bonds are difficult to break, making alkanes unreactive under normal conditions. This characteristic led to alkanes being termed 'paraffins,' derived from Latin words meaning 'little affinity.'
Q2: How does heat of combustion indicate alkane stability?
Heat of combustion (−ΔH°) is the energy released when alkanes burn in excess oxygen at high temperature, producing carbon dioxide and water. More negative heat values indicate greater energy release and lower potential energy, revealing which isomers are more stable. Straight-chain alkanes release more heat than branched isomers.
Q3: Why do branched alkanes have different combustion energies than straight-chain isomers?
Branched isomers have lower potential energies and greater stability than straight-chain alkanes, releasing less heat during combustion. The increased branching reduces overall potential energy, making branched structures thermodynamically more stable. This difference in heat of combustion helps predict relative isomer stabilities.
Q4: What is strain energy in cycloalkanes?
Strain energy is the difference between actual and predicted heats of combustion in cycloalkanes. It represents combined angular, torsional, and steric strains arising from non-ideal bond angles and conformations. Smaller rings experience greater strain due to compressed bond angles deviating from the ideal 109.5°.
Q5: Why does cyclopropane have maximum strain among cycloalkanes?
Cyclopropane exhibits maximum strain because its three-membered ring compresses bond angles excessively from the ideal 109.5° to 60°. This extreme angular compression creates severe angular strain, making cyclopropane highly unstable. As ring size increases, bond angles approach ideal values, reducing strain significantly.
Q6: Is cyclohexane strain-free, and why?
Cyclohexane is virtually strain-free because its six-membered ring allows bond angles to approach the ideal tetrahedral value of 109°. The ring adopts a stable conformation that minimizes angular, torsional, and steric strains. This makes cyclohexane one of the most stable cycloalkanes.
Q7: What causes strain in larger cycloalkanes like C7 to C9?
Cycloalkanes with 7 to 9 carbons experience moderate strain primarily from torsional and steric strains rather than angular strain. Their non-ideal bond angles in various conformations create unfavorable interactions between atoms. These strains increase the actual heat of combustion above predicted values.
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