Question

8. (a)(i) Members of the same homologous series have a general molecular formula. State two other characteristics of a homologous series.
(ii) The compound CH3(CH2)2CH3 belongs to the alkane family (butane). I. Which homologous series does the compound belong to? II. Write the structures of the three possible isomers of the compound. III. Name the three possible isomers in 8(a)(ii)II.
(b) Write the structure of the major product formed in each of the following reactions:
(i) ethanol with excess acidified potassium tetraoxomanganate (VII);
(ii) excess ethane with chlorine in the presence of sunlight;
(iii) ethanol with propanoic acid with few drops of concentrated tetraoxosulphate (VI) acid.
(c) Name the major product formed in each reaction in 8(b).
(d) Consider the following organic compounds:
Y - (CH3)3COH
X - C6H4COOH (a benzene ring with a –COOH substituent, plus –CH3, –CH3, –CH3 groups as shown in the source figure)
(i) Give the IUPAC name of each compound. (ii) State a chemical test for the functional group in each compound.
(e) An organic compound with a relative molecular mass of 136 contains 70.57% carbon, 5.90% hydrogen and 23.53% oxygen. Determine its: (i) empirical formula; (ii) molecular formula. [H=1.00, C=12.0, O=16.0]

Accepted Answer

(a)(i) Members of the same homologous series: Consecutive members differ in their molecular formula by a –CH2 group and in their relative molecular mass by 14.
(ii) I. The compound CH3(CH2)2CH3 (i.e. CH3CH2CH2CH3, butane) belongs to the alkane family.
II./III. The possible isomers of the compound are: CH3–CH2–CH2–CH3 (Normal butane); and CH3–CH(CH3)–CH3 (2-methylpropane / isobutane).

(b)(i) The major product when ethanol reacts with excess acidified potassium tetraoxomanganate (VII) is ethanoic acid: CH3–C(=O)–OH.
(ii) The major product is chloroethane: H–C(H)(H)–C(H)(Cl)–H, i.e. CH3CH2Cl.
(iii) The major product is ethylpropanoate: CH3CH2–C(=O)–O–CH2CH3.

(c) (i) Ethanoic acid. (ii) Chloroethane. (iii) Ethylpropanoate.

(d)(i) Y is 2-methylpropan-2-ol; X is benzoic acid (benzenecarboxylic acid).
(ii) Test for Y: Make use of Lucas reagent and add a few drops. A cloudy or milky appearance indicates Y (a tertiary alcohol). Test for X: Evolution of colourless, odourless gas (CO2) alongside vigorous effervescence upon adding a hydrogen carbonate indicates X (a carboxylic acid).

(e) C=70.57%, H=5.90%, O=23.53%. Dividing by atomic mass: C=70.57/12=5.88, H=5.90/1=5.9, O=23.53/16=1.4725. Dividing by smallest (1.4725): C=4, H=4, O=1. Thus, the empirical formula is C4H4O.
(ii) (C4H4O)n=136. (48+4+16)n=136. 68n=136; n=136/68=2. Molecular formula = (C4H4O)2 = C8H8O2.

WAEC Chemistry 2013 Theory — Question 6

Sign up free to unlock