Practice Questions (Organic Chemistry)
1(a) A and B have molecular formula C8H10. When A and B are treated with hot concentrated potassium manganate(VII), A forms C7H6O2 while B forms C8H6O4. B reacts with thionyl chloride, SOCl2 to form C and C reacts with methanol, CH3OH in the presence of a catalyst to form D, C10H10O4. Deduce the formula of A to D. [4]
(b)(i) Draw the structural formulae of three alcohols with the molecular formula C5H12O. Label your structures E, F and G. [3]
(ii) Label the chiral carbons on E to G with an asterisk on each carbon. [2]
(iii) E, F and G are treated with an excess of acidified potassium dichromate(VI). E has no reaction with the reagent while F and G gave different products. Suggest simple chemical tests to identify the functional group on each of the products of F and G. Write chemical equations of all the reactions. [8]
(iv) F is treated with PCl3 then with NaCN. After that, it is treated with LiAlH4 to give H. Draw the structural formula of H. [3]
Answers :
1(a) A is C6H5CH2CH3.
B is (CH3)C6H4(CH3). The two methyl groups on the phenyl ring can be in any positions.
When A and B are treated with KMnO4, A becomes C6H5COOH. B becomes HOOCC6H4COOH. B reacts with thionyl chloride to form C, which is ClOCC6H4COCl. C then reacts with methanol to form a diester, D, CH3OOCC6H4COOCH3.
(b) E is a tertiary alcohol, CH3CH2COH(CH3)2
F could be a primary alcohol, CH3CH2CH2CH2CH2OH
G could be a secondary alcohol, CH3CHOHCH2CH2CH3
(iii) E has no reaction with K2Cr2O7 because tertiary alcohols cannot be oxidised.
F is oxidised to a carboxylic acid, CH3CH2CH2CH2COOH. I can add sodium carbonate to the acid and carbon dioxide that is liberated gives a white precipitate with calcium hydroxide.
G is oxidised to a ketone, CH3COCH2CH2CH3. The ketone gives an orange precipitate with 2,4-DNPH.
(iv) F reacts with PCl3 to form CH3CH2CH2CH2CH2Cl. It then reacts with NaCN to form CH3CH2CH2CH2CH2CN. Then it reacts with LiAlH4, a reducing agent, to form an amine, H, which is CH3CH2CH2CH2CH2CH2NH2.
(b)(i) Draw the structural formulae of three alcohols with the molecular formula C5H12O. Label your structures E, F and G. [3]
(ii) Label the chiral carbons on E to G with an asterisk on each carbon. [2]
(iii) E, F and G are treated with an excess of acidified potassium dichromate(VI). E has no reaction with the reagent while F and G gave different products. Suggest simple chemical tests to identify the functional group on each of the products of F and G. Write chemical equations of all the reactions. [8]
(iv) F is treated with PCl3 then with NaCN. After that, it is treated with LiAlH4 to give H. Draw the structural formula of H. [3]
Answers :
1(a) A is C6H5CH2CH3.
B is (CH3)C6H4(CH3). The two methyl groups on the phenyl ring can be in any positions.
When A and B are treated with KMnO4, A becomes C6H5COOH. B becomes HOOCC6H4COOH. B reacts with thionyl chloride to form C, which is ClOCC6H4COCl. C then reacts with methanol to form a diester, D, CH3OOCC6H4COOCH3.
(b) E is a tertiary alcohol, CH3CH2COH(CH3)2
F could be a primary alcohol, CH3CH2CH2CH2CH2OH
G could be a secondary alcohol, CH3CHOHCH2CH2CH3
(iii) E has no reaction with K2Cr2O7 because tertiary alcohols cannot be oxidised.
F is oxidised to a carboxylic acid, CH3CH2CH2CH2COOH. I can add sodium carbonate to the acid and carbon dioxide that is liberated gives a white precipitate with calcium hydroxide.
G is oxidised to a ketone, CH3COCH2CH2CH3. The ketone gives an orange precipitate with 2,4-DNPH.
(iv) F reacts with PCl3 to form CH3CH2CH2CH2CH2Cl. It then reacts with NaCN to form CH3CH2CH2CH2CH2CN. Then it reacts with LiAlH4, a reducing agent, to form an amine, H, which is CH3CH2CH2CH2CH2CH2NH2.
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