Q11. Write the structural formulas and IUPAC names of all the alkenes C4H8, counting geometrical isomers (NCERT Exercise 9.3 (a)).
Four: CH2=CH–CH2–CH3, but-1-ene; CH3–CH=CH–CH3 as cis-but-2-ene (both CH3 on one side) and trans-but-2-ene (opposite sides); CH2=C(CH3)–CH3, 2-methylprop-1-ene. Only but-2-ene has two different groups on each doubly bonded carbon.
Q12. Why is the peroxide effect seen with HBr but not with HCl or HI?
The H–Cl bond (430.5 kJ mol−1) is too strong to be broken by the free radical. The H–I bond (296.8 kJ mol−1) breaks easily, but the iodine radicals combine to form I2 instead of adding to the double bond. Only H–Br (363.7 kJ mol−1) lets the radical chain run (NCERT page 312).
Q13. Arrange (a) HCl, HBr and HI in order of reactivity towards alkenes, and (b) the primary, secondary and tertiary alkyl bromides in order of rate of dehydrohalogenation.
(a) HI > HBr > HCl. (b) Tertiary > secondary > primary.
Q14. Which alkene gives only propanone on ozonolysis?
2,3-Dimethylbut-2-ene, (CH3)2C=C(CH3)2: cutting its C=C gives two identical (CH3)2C=O molecules.
Q15. How many σ and π bonds are there in buta-1,3-diene?
CH2=CH–CH=CH2, C4H6: 10 atoms in an open chain, so 9 σ bonds, and 2 π bonds (one in each double bond).
Q16. An alkene C4H8 gives methanal and propanal on ozonolysis. Identify it.
Join the two carbonyl carbons: H2C=O and O=CHCH2CH3 give CH2=CHCH2CH3, but-1-ene.
Q17. Write the IUPAC names of (a) CH3CH=C(CH3)2 and (b) CH3–CH=CH–CH2–CH=CH–CH(C2H5)–CH2–CH=CH2 (NCERT Exercise 9.2 (a) and (g)).
(a) 2-Methylbut-2-ene. (b) The ten-carbon chain holds all three double bonds; numbering from the CH2= end gives them 1, 5 and 8 (the other end gives 2, 5 and 9), with the ethyl group on carbon 4: 4-ethyldeca-1,5,8-triene.
Q18. Write the IUPAC names of the products of ozonolysis of (i) pent-2-ene, (ii) 3,4-dimethylhept-3-ene, (iii) 2-ethylbut-1-ene and (iv) 1-phenylbut-1-ene (NCERT Exercise 9.4).
(i) CH3CH=CHCH2CH3: ethanal and propanal. (ii) CH3CH2C(CH3)=C(CH3)CH2CH2CH3: butan-2-one and pentan-2-one. (iii) CH2=C(C2H5)2: methanal and pentan-3-one. (iv) C6H5CH=CHCH2CH3: benzaldehyde (benzenecarbaldehyde) and propanal.
Q19. An alkene A gives a mixture of ethanal and pentan-3-one on ozonolysis. Write the structure and IUPAC name of A (NCERT Exercise 9.5).
Join the carbonyl carbons: CH3CH=O and O=C(C2H5)2 give CH3CH=C(C2H5)2, 3-ethylpent-2-ene.
Q20. An alkene A contains three C–C σ bonds, eight C–H σ bonds and one C–C π bond. On ozonolysis it gives two moles of an aldehyde of molar mass 44 u. Name A (NCERT Exercise 9.6).
The aldehyde of molar mass 44 u is ethanal, CH3CHO (24 + 4 + 16 = 44). Two moles of it come from CH3CH=CHCH3, but-2-ene, which has three C–C σ bonds, eight C–H bonds and one π bond.
Q21. Propanal and pentan-3-one are the ozonolysis products of an alkene. What is its structural formula (NCERT Exercise 9.7)?
CH3CH2CH=C(CH2CH3)2, 3-ethylhex-3-ene.
Q22. Draw the cis and trans structures of hex-2-ene. Which has the higher boiling point, and why (NCERT Exercise 9.9)?
Draw them like but-2-ene in the cis-trans figure above, with C3H7 in place of one CH3: in cis-hex-2-ene the CH3 and C3H7 groups are on the same side of the double bond; in trans-hex-2-ene, on opposite sides. The cis isomer boils higher: its bond dipoles do not cancel, so the molecule is more polar and its molecules attract each other more strongly (as with cis-but-2-ene, 277 K, against trans, 274 K).
Q23. Write the structures of all the alkenes that give 2-methylbutane on hydrogenation (NCERT Exercise 9.21).
The double bond can sit on any C–C bond of the 2-methylbutane skeleton: CH2=C(CH3)CH2CH3 (2-methylbut-1-ene), (CH3)2C=CHCH3 (2-methylbut-2-ene) and (CH3)2CHCH=CH2 (3-methylbut-1-ene).
Q24. Write the equation for the complete combustion of pentene (NCERT Exercise 9.8 (ii)).
C5H10 + 15/2 O2 → 5CO2 + 5H2O, or 2C5H10 + 15O2 → 10CO2 + 10H2O.