Answer 7 |
A, C2H4, is evidently ethene CH2=CH2.. This reacts with hydrogen bromide in the gas phase to give bromoethane, B, CH3CH2Br. With aqueous sodium hydroxide solution bromoethane gives ethanol, C, CH3CH2OH.
Ethanol is oxidised by acidified potassium dichromate solution to give successively D, which since it reacts with both 2,4-dinitrophenylhydrazine and ammoniacal silver nitrate is a carbonyl compound and an aldehyde. D is ethanal, CH3CHO. Ethanal is oxidised further to E, ethanoic acid, CH3COOH.
Magnesium, dry ether and a halogenoalkane signifies the Grignard reaction. Bromoethane B gives F, CH3CH2MgBr, ethylmagnesium bromide. Grignard reagents react with carbonyl compounds in a nucleophilic addition reaction to give alcohols. With aldehydes apart from methanal (which yields a primary alcohol) the product is a secondary alcohol. G, C4H10O, is therefore butan-2-ol, CH3CH2CH(OH)CH3. Oxidation of secondary alcohols with acidified potassium dichromate solution gives ketones, so H is CH3CH2COCH3, butanone. This will also react with the Grignard reagent F; ketones give tertiary alcohols in this reaction, so that I is 3-methylpentan-3-ol, (CH3CH2)2C(OH)CH3. I, like all tertiary alcohols, is not oxidised by acidified potassium dichromate solution.
The reaction of ethene, A, with bromine in the presence of sodium chloride is a straightforward electrophilic addition. The electrophile is Brd+ from the polarisation of the bromine molecule by the C=C bond. The resulting carbocation is then attacked by the negative ions present in the mixture. There are two; attack by Br - gives J, BrCH2CH2Br, 1,2-dibromoethane. Attack by Cl - from the sodium chloride gives K, BrCH2CH2Cl, 1-bromo-2-chloroethane. No C2H4Cl2 is formed because there is no possibility of a Cld+ species being generated.
Problem 7 Organic problems contents Home page