Question

4. (a)(i) What is a structural isomer? (ii) Write all the structural isomeric alkanols with the molecular formula C4H10O. (iii) Which of the isomers from 4(a) above does not react easily on heating with acidified K2Cr2O7?
(b) Chlorine reacted with excess pentane in the presence of light. Chloropentane and a gas which fumes on contact with air were produced. (i) Write an equation for the reaction. (ii) Draw the structure of the major product. (iii) What is the role of light in the reaction? (iv) If a mixture of pentane and the major product would distil off first? Give a reason for your answer.
(v) Write the formula of the main product that would have been formed if but-1-ene (C4H8) has been used instead of pentane.
(c) Give the name and structural formula of the product which would be formed by hydration of each of the following compounds: (i) CH3CH(CH3)CH=CH2; (ii) CH2=CHCOOH.
(d)(i) Write the structure of the amino acid, CH3CH(NH2)COOH in: I. acidic medium; II. alkaline medium.
(ii) On analysis, an ammonium salt of an alkanoic acid gave 60.5% carbon and 6.5% hydrogen. If 0.309g of the salt yielded 0.0313g of nitrogen, determine the empirical formula of the salt. [H=1.00; C=12.0; N=14.0; O=16.0]

Accepted Answer

(a)(i) Structural isomers are compounds with the same molecular formula but different structures.
(ii) Structural isomers of C4H10O (alkanols): CH3CH2CH2CH2OH (butan-1-ol); CH3CH2CH(OH)CH3 (butan-2-ol); (CH3)2CHCH2OH (2-methylpropan-1-ol); (CH3)3COH (2-methylpropan-2-ol).
(iii) 2-methylpropan-2-ol does not react easily with K2Cr2O7, since it is a tertiary alkanol.

(b)(i) C5H12+Cl2→C5H11Cl+HCl.
(ii) The major product is chloropentane, e.g. CH3CH2CH2CH2CH2Cl (1-chloropentane).
(iii) Light serves as a catalyst for the reaction.
(iv) Pentane would distil off first. This is because chloropentane has a higher relative molecular mass than pentane (C5H11Cl ≈ 106.5, pentane C5H12 ≈ 72), and generally the lower-boiling component distils off first.
(v) If but-1-ene were used instead of pentane, the main product formed would be C4H8Cl2 (1,2-dichlorobutane).

(c)(i) Hydration of CH3CH(CH3)CH=CH2 gives 3-methylbutan-2-ol: CH3CH(CH3)CH(OH)CH3.
(ii) Hydration of CH2=CHCOOH gives 2-hydroxypropanoic acid: CH3CH(OH)COOH.

(d)(i) In acidic medium: CH3CH(NH3⁺)COOH. In alkaline medium: CH3CH(NH2)COO⁻.
(ii) C=60.5%, H=6.5%. N = (0.0313/0.309)×100% = 10.129%. C+H+N+O=100%, so 60.5+6.5+10.129+O=100%, O = 22.871%. Mole ratio: C=60.5/12=5.042, H=6.5/1=6.5, N=10.129/14=0.7235, O=22.871/16=1.4292. Dividing by smallest (0.7235): C=7, H=9, N=1, O=2. Empirical formula = C7H9NO2.

WAEC Chemistry 2012 Theory — Question 4

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