(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 isome...
Question 1 Report
(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]
Answer Details
(a)(i) Structural isomer: one of two or more compounds having the same molecular formula but different structural arrangements (different order of bonding) of their atoms.
(a)(ii) Alkanols of C4H10O (the four alcohols):
- Butan-1-ol: CH3CH2CH2CH2OH
- Butan-2-ol: CH3CH2CH(OH)CH3
- 2-methylpropan-1-ol: (CH3)2CHCH2OH
- 2-methylpropan-2-ol: (CH3)3COH
(a)(iii) 2-methylpropan-2-ol (the tertiary alcohol) does not react easily with acidified K2Cr2O7 because a tertiary alcohol has no hydrogen atom on the carbon bearing the OH group and so resists oxidation.
(b) Chlorine with excess pentane in light
(i) \(\text{C}_5\text{H}_{12} + \text{Cl}_2 \xrightarrow{\text{light}} \text{C}_5\text{H}_{11}\text{Cl} + \text{HCl}\)
(ii) Major product (1-chloropentane): CH3CH2CH2CH2CH2Cl.
(iii) Light supplies the energy that homolytically splits the Cl2 molecule into chlorine free radicals, initiating the substitution (free-radical) reaction.
(iv) Pentane would distil off first because it has the lower boiling point; 1-chloropentane has a higher relative molecular mass and stronger dipole/van der Waals forces, hence a higher boiling point.
(v) With but-1-ene the reaction is addition, giving 1,2-dichlorobutane, CH2ClCHClCH2CH3.
(c) Hydration products
- (i) CH3CH(CH3)CH=CH2 adds water (Markovnikov) to give 3-methylbutan-2-ol, CH3CH(CH3)CH(OH)CH3.
- (ii) CH2=CHCOOH gives 3-hydroxypropanoic acid, HOCH2CH2COOH.
(d)(i) Amino acid CH3CH(NH2)COOH:
- I. Acidic medium (amino group protonated): CH3CH(NH3+)COOH.
- II. Alkaline medium (acid deprotonated): CH3CH(NH2)COO-.
(d)(ii) Empirical formula of the ammonium salt
%N \(= \dfrac{0.0313}{0.309}\times100 = 10.1\%\). Then %O \(= 100 - 60.5 - 6.5 - 10.1 = 22.9\%\).
| Element | % | moles (%/Ar) | ratio (/0.72) |
|---|---|---|---|
| C | 60.5 | 5.04 | 7 |
| H | 6.5 | 6.50 | 9 |
| N | 10.1 | 0.72 | 1 |
| O | 22.9 | 1.43 | 2 |
Empirical formula: C7H9NO2 (consistent with an ammonium salt, RCOONH4).
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