Chemistry Unit 4 Goodie bag

March 28, 2018 | Author: Jacob Salkin | Category: Alkene, Hydrogen Compounds, Hydrogen, Functional Group, Atoms


Comments



Description

A2 Chemistry – UNIT 4.9 Mass Spectrometry and Infra Red 1. Consider the following reaction scheme: ste p 1 C 2H 5B r s te p 2 C 2H 5M gB r C 2H 5C O O H PC l5 r o o m te m p e r a tu re C 3H 5O C l A lC l3 benzene C 9H L iA lH 4 , d ry e th e r B A (a) 10O C 9H 12 O C (4) Give the equation for the reaction between propanoic acid and phosphorus pentachloride, making clear the structure of compound A. (2) (b) Compound B reacts with 2,4-dinitrophenylhydrazine but not with ammoniacal silver nitrate solution. Give the structural formula for compound B and explain why B reacts with 2,4-dinitrophenylhydrazine but not with ammoniacal silver nitrate. (3) (c) (i) Give the structural formula for C. (1) (ii) State why the ether solvent must be dry in the conversion of B to C. (1) (d) The infrared spectra of B and of C are given below, together with a table of infrared absorbencies. What is the evidence from these spectra for the conversion of the functional group in B to that in C? (3) Com pound B: T ra n s m itta n c e (% ) 100 50 0 4000 3000 2000 1500 W avenum ber / cm NT Exampro 1000 500 –1 1 Com pound C : T ra n s m itta n c e (% ) 100 50 0 4000 3000 2000 1500 W avenum ber / cm 1000 500 –1 Infra-red absorbencies. Bond Assignment Wavenumbers/cm–1 CH Alkanes Alkenes, arenes 2850–2950 3000–3100 C==C Alkenes 1610–1680 Benzene ring Arenes 1450–1650 C==O Aldehydes, ketones, amides, esters, carboxylic acids 1680–1750 O H Free 3580–3670 Hydrogen bonded in alcohols or phenols 3230–3550 2500–3300 Hydrogen bonded in carboxylic acids (e) (i) PCl5 is made by the action of excess chlorine gas on white phosphorus, P4. Give an equation to represent this reaction. (2) (ii) PCl5 decomposes on heating just above its sublimation temperature PCl5(g)  PCl3(g) + Cl2(g) Explain whether or not this is a redox reaction. (3) (Total 25 marks) NT Exampro 2 2. This question concerns the three isomers A, B and C, each of which has a relative molecular mass of 134. O O C C H 2 C H C H 3 CH 2 2 C H A B C H C H C H 2O H C (a) The mass spectrum of substance A is shown below. Identify the species responsible for the peaks labelled 1, 2 and 3. Peak 1 ...................................................................................................................... ... Peak 2 ...................................................................................................................... ... Peak 3 ...................................................................................................................... ... (3) 100 2 R e la t iv e in t e n s it y 80 60 1 40 20 3 0 25 (b) m /e 75 100 125 The infra-red spectra of two of these substances were also measured. (i) NT Exampro 50 Use the table and the spectra below to identify which spectrum is that of substance C. Bond Wavenumb er/ cm–1 C—H (arenes) 3000-3100 Bond O––H (hydrogen bonded) Wavenumb er/ cm–1 3200-3570 3 C—H (alkanes) 2850-3000 O––H (not hydrogen bonded) 3580-3650 C==O 1680-1750 C==C (arenes) 1450-1600 The spectrum of substance C is spectrum number .......................................... (1) (ii) Give one reason for your choice. ...................................................................................................................... (1) (iii) Give one other reason why the other spectrum could not be that of substance C. ...................................................................................................................... ...................................................................................................................... (1) 100 2 R e la t iv e in t e n s ity 80 60 1 40 20 3 0 25 50 T ra n s m itta n c e 100 75 100 125 50 0 NT Exampro m /e 4000 3000 2000 1500 W a v e n u m b e r /c m – 1 1000 4 T ra n s m itta n c e 100 50 0 (c) 4000 3000 2000 1500 W a v e n u m b e r /c m – 1 1000 State which of the substances A, B and C will react with the following reagents and state what would be observed. (i) Bromine dissolved in hexane. Substance(s) ................................................................................................... .. Observation .................................................................................................... .. (2) (ii) A warm ammoniacal solution of silver nitrate. Substance(s) ................................................................................................... .. Observation .................................................................................................... .. (2) (iii) 2,4-dinitrophenylhydrazine solution. Substance(s) ................................................................................................... .. Observation .................................................................................................... .. (iv) NT Exampro Give the structural formula of the organic product(s) obtained in (c)(i). 5 (1) (v) Give the structural formula of the organic product(s) obtained in (c)(ii). (1) (Total 15 marks) NT Exampro 6 ANSWERS 1. (a) O O C H 3CH 2C O H + P C l5 C H 3C H 2C C l + P O C l3 + H C l structure of acid chloride showing at least expansion of COCl (1) products (1) 2 O C (b) C 2H 5 (1 ) 2,4–dnp reacts with carbonyl group (1) this carbonyl compound is a ketone and cannot be oxidised by/ cannot reduce ammoniacal silver nitrate (1) This is consequential on B being shown as a ketone 3 O H C 2H C (c) (i) (e) H (1) This is consequential on a ketone being shown in (d) The lithium aluminium hydride would react with any water (1) 1 1 B has absorption peak in region 1680 – 1750 showing carbonyl (1) but peak in region of 3250 – 3550 is present ( not present in B) showing alcohol group / allow broad peak if range omitted (1) comparison between two spectra (1) 3 (ii) (d) 5 (i) (ii) P4 + 10Cl2  4PCl5 Allow P4 + PCl3  PCl5 for species (1) for balancing (1) If start with P2 or P allow one mark for balanced equation 2 Oxidation number of phosphorus changes from 5 to 3 (1) Oxidation number of chlorine changes from –1 to 0 (1) Change of oxidation number is redox (1) 3 [25] 2. (a) (b) Peak 1 m/e 77: C6H5+ (1) Peak 2 m/e 105: C6H5CO+ (1) Peak 3 m/e 134: C6H5COCH2CH3+ (1) if + left off penalize only once 3 (i) IR spectrum number 2 1 (ii) Because of the broad peak at 3320cm–1 caused by OH / because peak at 3320 is OH / because it does not have a peak at 1750 and is only one C without (c) NT Exampro O 1 (iii) The other has a peak at approx. 1700 cm–1 caused by C=O 1 (i) C 1 (brown) goes colourless 1 2 (ii) B 1 silver mirror / black precipitate 1 2 7 NT Exampro (iii) A (1) and B (1) red/orange/yellow ppt 1 3 (iv) C6H5CHBr–CHBr–CH2OH 1 (v) C6H5CH2CH2COO– but allow the carboxylic acid as product consequential 1 8
Copyright © 2024 DOKUMEN.SITE Inc.