(a) (i) What is a structural isomer?
(ii) Write all the structural isomeric alkanols with the molecular formula C\(_4\)H\(_{10}\)O.
(iii) Which of the isomers from (a)(ii) above does not react easily on heating with acidified K\(_2\)Cr\(_2\)O\(_7\)?
(b) Chlorine reacted with excessentane in the presence of light. Chloropentane and a gas which fumes on cont 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 is heated, which compounc would distil off first? Give a reason for your answer.
(v) Write the formula of the main product that would have be formed if but -1-ene 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 followinc compounds:
(i) CH\(_3\)CH(CH\(_3\))CH=CH\(_2\); (ii) CH\(_2\)=CHCOOH.
(d) (i) Write the structure of the amino acid, CH\(_3\)CH(NH\(_2\))COOH in: I. acidic medium; II. alkaline medium.
(ii) On analysis, an ammonium salt cf an alkanoic acid gave 60.5% carbon and 6.5% hydrogen. If 0.309 g of the salt yielded 0.0313 g of nitrogen, determine the empirical formula cf the salt. [H = 1.00; C =12.0; N =14.0; O = 16.0]
Explanation
(a)(i) Structural isomers are compounds with the same molecular formula but different arrangement of constituent at / or different structures
(ii) CH\(_3\) CH\(_2\) CH\(_2\) CH\(_2\) OH
(b)(i) C\(_5\)H\(_{12}\) + C\(_2\) \(to\) C\(_5\) H\(_{11}\)CI + HCI
(ii)
(iii) Light act as catalyst
(iv) Pentane C\(_2\)H\(_{12}\) would come out first Pentane has a lower molar mass (72) as compared to chloropentane (C\(_5\)H\(_{12}\)CI) (86.5) Hence weaker inter- molecular forces — van der waals' forces
(V) C\(_4\)H\(_{8}\)CI\(_{12}\)
(c)(i)
(ii)
(d)(i)
(ii) %N = \(\frac{0.0313}{ 0.309}\) x 100
= 10.1%
%0 = 100 — (60.5 + 6.5 + 10.1)
= 22.9
C; \(\frac{60.5}{12}\) H; \(\frac{6.5}{1}\) N; \(\frac{10.1}{14}\) O; \(\frac{22.9}{16}\)
C; \(\frac{5.04}{0.72}\) H; \(\frac{6.5}{0.72}\) N; \(\frac{0.72}{0.72}\) O; \(\frac{1.43}{0.72}\)
Empirical formula — C\(_7\)H\(_9\)NO\(_2\)