(a) (i) Draw the structures of the isomers of the alkene with molecular formurat C(_4)H(_8)
(ii) State the class of alkanols to which each of the following compounds belongs:
I. CH(_3)C(CH(_3))(_2)OH;
II. CH(_3)CH(CH(_3))CH(_2)OH;
III. CH(_3)CH(_2)CH(CH(_3))OH.
(b) (i) Write the formulae of the products formed in the following reactions:
I. CH(_3)CH(_2)COOH (frac{K_{(s)}}{})
II. CH(_3)CH(_2)COOH. (frac{C_4H_6OH, heat}{Conc.H_2SO_4})
III. CH(_3)CH(_2)CH(_2)CH(_2)OH (frac{H^+/KMnO_4}{(excess)})
(ii) Name the major product(s) of each of the reactions in (b)(i).
(c) A gaseous hydrocarbon R of mass 7.0 g occupies a volume of 2.24 dm(^3) at s. t.p. If the percentage composition by mass of hydrogen is 14.3, determine its:
(i) empirical formula;
(ii) molecular formula. [ H = 1.00, C = 12.0, Molar volume of gas at s.t.p, = 22.4 dm(^3) ]
(d) Define structural isomerism.
Explanation
(a)(i) Structural Isomers of C(_4)H(_4)
(ii) (I)Tertiary(3°)alkanol.
(II) Primary (1°)alkanol.
(III) Secondary (2°) alkanol
(b)(i)
(I) CH(_3)CH(_2)COOK, H(_2)
(II) CH(_3)CH(_2)COOC(_4)H(_9) H(_2)O
(III) CH(_3)CH(_2)CH(_2)COOH
(ii) Major products in (b)(i) (1) — Potassuim propanoate (II) — Butyl propanoate (III) — Butanoic acid
(C)(i) % carbon by mass = 100 – 14.3 (l) = 85.7%
Ratio of atoms = (frac{C}{frac{85.7}{12.0}}) = 7.1415; (frac{H}{frac{14.3}{1.0}}) = 14.3
Divide by smaller number = (frac{7.1415}{7.1415}) = 1; (frac{14.3015}{7.1415}) = 2
Empirical Formular = CH(_2)
(i) 2.24dm(^3) = 7.0g of R
22.4dm(^3) = (frac{7.0}{2.24}) x 22.4 = 70gmol(^1)
OR
Molar volume of gas = 22.4 dm(^3) at s.t.p 22.4
22.4 dm(^3) gas = 1 mole
22.4 dm(^3) = (frac{2.24}{22.4} times 1) = 0.1 mol
But numbers of moles = 0.1
Mr = (frac{7}{0.1}) = 70gmol(^1)
Molecular formular = C(_5)H(_{10})
(CH_2)n = 70
14n = 70
n = (frac{70}{4}) = 5
(d) The occurence of / existence of / condition whereby compounds with the same molecular formula have different structural formulae.