Q11. C5H11 has eight isomeric alkyl groups. Name the eight alcohols C5H11OH obtained by attaching –OH to each (NCERT Problem 9.2).
From pentane: pentan-1-ol, pentan-2-ol, pentan-3-ol. From 2-methylbutane: 2-methylbutan-1-ol, 2-methylbutan-2-ol, 3-methylbutan-2-ol, 3-methylbutan-1-ol. From 2,2-dimethylpropane: 2,2-dimethylpropan-1-ol. The central carbon of 2,2-dimethylpropane carries no hydrogen, so it gives no group.
Q12. Write what happens at each electrode in the Kolbe electrolysis of aqueous sodium acetate.
Anode: 2CH3COO− → 2CH3COO• + 2e−; each acetate radical loses CO2 to give •CH3; 2•CH3 → CH3–CH3. Cathode: water takes the electrons, 2H2O + 2e− → H2 + 2OH−, so NaOH remains in solution. Overall: 2CH3COONa + 2H2O → C2H6 + 2CO2 + H2 + 2NaOH.
Q13. Why is the iodination of methane carried out with HIO3 or HNO3?
CH4 + I2 ⇌ CH3I + HI is slow and reversible: HI reduces CH3I back to methane. An oxidising agent destroys the HI (HIO3 + 5HI → 3I2 + 3H2O), so the equilibrium moves forward.
Q14. An alkane C5H12 gives only one monochloro product. Identify it.
2,2-Dimethylpropane, C(CH3)4: all twelve hydrogens are equivalent. Pentane gives three monochloro products and 2-methylbutane four.
Q15. There is an energy barrier to rotation about the C–C bond in ethane, yet the rotation is called almost free. Explain.
The barrier, about 12.5 kJ mol−1 between the eclipsed and staggered forms, is small. Molecules gain this much energy from collisions even at ordinary temperatures, so they turn freely from one conformation to another and the conformers cannot be isolated.
Q16. Which isomer of C6H14 gives exactly two monochloro products?
2,3-Dimethylbutane, (CH3)2CHCH(CH3)2: its four methyl groups are equivalent and so are its two CH groups. The other isomers give three (hexane, 2,2-dimethylbutane), four (3-methylpentane) or five (2-methylpentane).
Q17. In the alkane H3C–CH2–C(CH3)2–CH2–CH(CH3)2, identify the 1°, 2° and 3° carbon atoms and give the number of H atoms on each (NCERT Exercise 9.14).
The compound is 2,4,4-trimethylhexane. Primary: C1, C6 and the three methyl branches, 3 H each (15 H). Secondary: C3 and C5, 2 H each. Tertiary: C2, 1 H. C4, the C(CH3)2 carbon, is quaternary and has no H. Total 15 + 4 + 1 = 20 H, as C9H20 requires.
Q18. Why is the Wurtz reaction not preferred for preparing alkanes with an odd number of carbon atoms? Illustrate with an example (NCERT Exercise 9.25).
An odd alkane needs two different alkyl halides, and these give a mixture. For propane, CH3Br and C2H5Br with sodium give ethane (CH3–CH3) and butane (C2H5–C2H5) as well as propane (CH3–C2H5), and the three are hard to separate.
Q19. Write the IUPAC name of CH3(CH2)4CH(CH2CH(CH3)2)(CH2)3CH3 (NCERT Exercise 9.2 (f)).
The longest chain runs through the pentyl and butyl arms: 5 + 1 + 4 = 10 carbons, decane (through the isobutyl arm it is only 5 + 1 + 3 = 9). The CH2CH(CH3)2 group (2-methylpropyl, or isobutyl) is on carbon 5 counted from the butyl end (6 from the other end). Name: 5-(2-methylpropyl)decane, also written 5-isobutyldecane.
Q20. Write the equation for the complete combustion of butane (NCERT Exercise 9.8 (i)).
C4H10 + 13/2 O2 → 4CO2 + 5H2O, or 2C4H10 + 13O2 → 8CO2 + 10H2O (NCERT equation 9.18).