Q11. Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour, and give the reason (NCERT Exercise 9.18).
Ethyne > benzene > n-hexane. The hydrogens are on sp, sp2 and sp3 carbons respectively. The more s character (50%, 33%, 25%), the more electronegative the carbon, and the more easily its hydrogen leaves as a proton.
Q12. Why does benzene undergo electrophilic substitution easily but nucleophilic substitution only with difficulty (NCERT Exercise 9.19)?
The ring carries a cloud of six delocalised π electrons above and below its plane. This electron-rich cloud attracts electrophiles and repels nucleophiles. Substitution, rather than addition, keeps the aromatic ring.
Q13. How would you convert (i) ethyne, (ii) ethene and (iii) hexane into benzene (NCERT Exercise 9.20)?
(i) Pass ethyne through a red hot iron tube at 873 K: 3C2H2 → C6H6. (ii) Add Br2 to ethene to get 1,2-dibromoethane. Treat it with alcoholic KOH, then NaNH2, to get ethyne, then proceed as in (i). (iii) Aromatization: heat hexane to 773 K at 10–20 atm over oxides of vanadium, molybdenum or chromium on alumina: C6H14 → C6H6 + 4H2.
Q14. Suggest a Lewis acid other than anhydrous aluminium chloride that can be used in the ethylation of benzene (NCERT Exercise 9.24).
Anhydrous iron(III) chloride, FeCl3. NCERT also names FeBr3 as a Lewis acid for halogenation.
Q15. Arrange toluene, p-CH3C6H4NO2 and p-O2NC6H4NO2 in decreasing order of reactivity with an electrophile (NCERT Exercise 9.22 (b)).
Toluene > p-nitrotoluene > p-dinitrobenzene. The activating –CH3 group raises the reactivity, and each deactivating –NO2 group lowers it.
Q16. How many π electrons does naphthalene have, and which value of n in Hückel’s rule does it fit?
Ten π electrons (five C=C in each Kekulé structure). 4n + 2 = 10 gives n = 2.
Q17. Write the IUPAC name of C6H5–CH2–CH2–CH=CH2 (NCERT Exercise 9.2 (d)).
4-Phenylbut-1-ene. The chain of four carbons contains the double bond, numbered from the CH2= end, and the benzene ring is a phenyl substituent on carbon 4.