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Identifying an unknown chemical

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Introduction

Compound A Compound (A) has no effect on blue litmus paper, and so it is not an acid. When reacted with 2,4-dinitrophenylhydrazine, an orange precipitate is formed. This indicates the presence of a C=O (carbonyl) bond. This suggests that (A) must be an aldehyde or a ketone. There is no reaction with sodium carbonate, so there is no O-H (alcohol) group the lack of observable reaction with sodium confirms this. Because there is no noticeable change when mixed with bromine water, there cannot be a C=C bond (an alkene would decolourise bromine water, forming a halogenoalkane). There is no change when heated with ethanoic acid and sulphuric acid, further indication that (A) is not an alcohol. According to the 2,4-dinitrophenylhydrazine (Brady's reagent) test, (A) must be an aldehyde or a ketone. To determine which of these it is, Tollen's reagent is used. When (A) is heated with Tollen's reagent (ammoniacal silver nitrate), no observable change occurs. This tells us that (A) must be a ketone- had (A) ...read more.

Middle

This suggests the following structure: The compound is a ketone; it has a C=O bond, but not on the end carbon atom. The RMM of this compound (propanone) is = This matches the RMM determined by the mass spectoscopy. This is the only possible structure, taking into account the wet test results and infra-red spectrum, which confirmed (A) to be a ketone, the NMR, which confirmed the structure, and the mass spectrum, which found the RMM of (E). (E) has to be propanone. Compound E Compound (E) does not effect blue litmus paper: it mustn't, therefore, be acidic. There is no reaction with acidified potassium dichromate, so it is not a primary or secondary alcohol. It could, however, be a tertiary alcohol. It does not noticeably react with 2,4-dinitrophenylhydrazine (Brady's reagent), and so there is no C=O carbonyl bond; it is not an aldehyde or a ketone. It is not an alkene, as the bromine water test showed no noticeable change (it would have been decolourised by an alkene), forming a halogenoalkane. ...read more.

Conclusion

Attached to this are three methyl groups, each with three hydrogen atoms (protons). The fourth group is an O-H group (alcohol). I have now determined that (E) is an alcohol. To find out the relative molecular mass of (E), I must use the mass spectrum. The molecular peak on the mass spectrum is at 74 m/z, with a relative intensity of 13. There is a peak at 73 with a relative intensity of 2 The molecular ion had an m/z of 74, and a relative intensity of 13. It had the greatest m/z on the spectrum, as there was no "M+1" peak caused by carbon-13. The homologous series of alcohols has the general formula CnH2n+1OH. According to this formula, the molecular masses will be: The RMM of (E) (74) is exactly the same as the RMM of butanol (74). As the infra-red and NMR spectra and the wet tests suggest that (E) is an alcohol, (E) must be an isomer of butanol. As the acidified potassium dichromate test proves (E) to be a tertiary alcohol, I believe that (E) can only be 2-methyl-propan-2-ol: Identifying an unknown chemical 3/3 ...read more.

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