Benzedrine is a pharmaceutical which stimulates the central nervous system in a similar manner to adrenalin. Benzedrine Adrenalin
|
|
- Clyde Hicks
- 6 years ago
- Views:
Transcription
1 1. Adrenalin is a hormone which raises blood pressure, increases the depth of breathing and delays fatigue in muscles, thus allowing people to show great strength under stress. Benzedrine is a pharmaceutical which stimulates the central nervous system in a similar manner to adrenalin. HO H ( ) NH HO (OH) N Benzedrine Adrenalin (a) (i) On the structure for benzedrine mark with a ( * ) any asymmetric carbon atom that causes chirality. Suggest why adrenalin is more soluble in water than is benzedrine () (b) Give the structural formulae of the organic products obtained when benzedrine reacts with: (i) an aqueous acid such as dilute hydrochloric acid; ethanoyl chloride in the absence of a catalyst; (iii) excess ethanoyl chloride in the presence of the catalyst anhydrous aluminium chloride. () Maltby Academy 1
2 (d) It is possible to eliminate a molecule of water from adrenalin which for the purpose of this question may be represented as R (OH) NH. Draw the structural formulae of the two stereoisomers produced. () (e) The mass spectra of both benzedrine and adrenalin have a peak at a mass/charge ratio of 44. Draw the structure of the species which give these peaks. (i) in benzedrine; in adrenalin. (Total 11 marks). This question concerns the three isomers A, B and C, each of which has a relative molecular mass of 14. O C C O H A B OH C Maltby Academy
3 (a) The mass spectrum of substance A is shown below. Identify the species responsible for the peaks labelled 1, and. Peak 1... Peak... Peak... () (b) The infra-red spectra of two of these substances were also measured. (i) Use the table and the spectra below to identify which spectrum is that of substance C. Bond Wavenumber / cm 1 Bond Wavenumber / cm 1 C H (arenes) O H (hydrogen bonded) C H (alkanes) O H (not hydrogen bonded) C==O C==C (arenes) The spectrum of substance C is spectrum number... Give one reason for your choice.... (iii) Give one other reason why the other spectrum could not be that of substance C Maltby Academy
4 Transmittance Transmittance Relative intensity m/e Wavenumber/cm Wavenumber/cm 1 (c) State which of the substances A, B and C will react with the following reagents and state what would be observed. Maltby Academy 4
5 (i) Bromine dissolved in hexane. Substance(s)... Observation... () A warm ammoniacal solution of silver nitrate. Substance(s)... Observation... () (iii),4-dinitrophenylhydrazine solution. Substance(s)... Observation... () (iv) Give the structural formula of the organic product(s) obtained in (c)(i). Maltby Academy 5
6 (v) Give the structural formula of the organic product(s) obtained in (c). (Total 18 marks). The drug ibuprofen can be synthesised from benzene by the route shown below. H C H C H C H C Step A Step B Step C CO/Pd catalyst C O H C OH H C CO H Ibuprofen (a) Name the type and mechanism of the reaction in Step A, and suggest a suitable reagent and catalyst. Type and mechanism... Name of the reagent for Step A... Catalyst... () Maltby Academy 6
7 (b) Step C is a reduction. Give ONE reason why lithium tetrahydridoaluminate, LiAlH 4, is preferred to hydrogen as a reducing agent in this reaction () (c) A sample of the final product was analysed by combustion g was burnt in oxygen. It produced.78 g carbon dioxide and g water. State the molecular formula of ibuprofen and show that these results are consistent with it. (4) Maltby Academy 7
8 (d) Ibuprofen can be analysed by instrumental methods. The infrared spectra of ibuprofen and two other drugs, aspirin and paracetamol, not necessarily in that order, are shown opposite. H C Ibuprofen has the formula H C CO H Aspirin has the formula CO H O C O OH Paracetamol has the formula H N C O (i) Explain, referring to the structure of each molecule, why infrared spectroscopy is not a good technique to distinguish aspirin from ibuprofen. Maltby Academy 8
9 Transmittance (%) Transmittance (%) Transmittance (%) Deduce which of X, Y or Z is the infrared spectrum of paracetamol, giving a piece of evidence from the spectrum you select. Spectrum X Spectrum Y Wavenumber/cm Spectrum Z Wavenumber/cm Wavenumber/cm 1 Maltby Academy 9
10 (iii) Ibuprofen and aspirin can be distinguished using their mass spectra. A line at mass/charge ratio 57 occurs only in the mass spectrum of ibuprofen. Give the formula of the ion which produces this line. Suggest the mass/charge ratio of one line which occurs in the mass spectrum of aspirin but not ibuprofen, and the formula of the species which produces it. () (Total 14 marks) 4. This question concerns the following compounds containing four carbon atoms. A Butanoic acid, COOH B Butanone, CO C Propyl methanoate, HCOO D Butanoyl chloride, COCl Select, from A to D, the compound that (a) can be made by the oxidation of a primary alcohol. A B C D Maltby Academy 10
11 (b) would be expected to react most rapidly with ethanol. A B C D (c) would have 4 different chemical shifts in its nmr spectrum and a broad absorption between cm 1 in its infrared spectrum. A B C D (Total marks) 5. In moths a pheromone, P, acts as an attractant for the opposite sex. P has the molecular formula C 7 H 1 O. What can be deduced about the structure of P from the following information? (a) (i) 1 mole of P reacts with 1 mole of Br molecules to form a compound with the formula C 7 H 1 OBr. When lithium tetrahydridoaluminate is reacted with P a compound with the formula C 7 H 14 O is formed. Maltby Academy 11
12 (iii) P forms an orange precipitate with,4-dinitrophenylhydrazine. (iv) When P is heated with Fehling s or Benedict s solution, the solution remains blue. (v) P is a Z-isomer. (b) What does the following physical data tell you about the structure of P? Use your Data booklet where necessary. (i) The infrared spectrum of P has the following absorptions at wavenumbers above 1600 cm cm 1 90 cm cm cm 1 () Maltby Academy 1
13 The nmr spectrum does not have a peak corresponding to a chemical shift,, of between 9 and 10. (iii) The mass spectrum showed the presence of peaks at mass/charge ratios of 15 and 9, but no peak at 4. () (c) Given that P has a straight chain of carbon atoms in its formula, use the information you have deduced above to suggest a displayed formula for the pheromone P. () Maltby Academy 1
14 (d) How could you use a purified sample of the orange precipitate in (a)(iii) to confirm the formula of P? () (Total 16 marks) 6. (a) (i) *( ) 1 (the three) OH groups allow adrenalin to form more hydrogen bonds with water (than does benzedrine) (b) (i) ( )NH (Cl ) + 1 Can use R in place of C 6 H 5 ( ) in both (i) and ( )N C ( 1) H O 1 (iii) C ( ) N C O O substitution(s) in ring at any position(s) production of amide H (d) R NH H C C C C H R H H NH Maltby Academy 14
15 (e) (i) (( ) NH ) + 1 ( NH ) + / ((OH) ) + max 1 for (e) if no charges shown must show some structure in answers ie. C H 5 N(0) 1 [11] + 7. (a) Peak 1 m/e 77: C 6 H 5 Peak m/e 105: C 6 H 5 CO + + Peak m/e 14: C 6 H 5 CO if + left off penalize only once (b) (i) IR spectrum number 1 Because of the broad peak at 0cm 1 caused by OH / because peak at 0 is OH / because it does not have a peak at 1750 and is only one without C O 1 (iii) The other has a peak at approx cm 1 caused by C=O 1 (c) (i) C 1 (brown) goes colourless 1 B 1 silver mirror / black precipitate 1 (iii) A and B red/orange/yellow ppt 1 (iv) C 6 H 5 Br Br OH 1 (v) C 6 H 5 COO but allow the carboxylic acid as product consequential 1 [15] Maltby Academy 15
16 8. (a) Electrophilic substitution IGNORE extras eg Friedel Craft, alkylation UNLESS contradictory 1-chloro-()-methylpropane IGNORE punctuation Accept ()-methyl-1-chloropropane Accept ( ) Cl/( ) Cl Accept Bromo / iodo for chloro Reject 1-methyl--chloropropane Reject missing 1 from position of Cl in name Catalyst AlCl /aluminium chloride Accept Al Cl 6, AlBr, FeBr (b) LiAlH 4 is a source of H / hydride ion Hydrogen might reduce/attack benzene ring/ H won t attack region of negative charge/ H can attack (δ + ) C in keto group Reject comments on conditions or safety eg temperature, pressure Reject LiAlH 4 /H is a more powerful reducing agent Reject H is a nucleophile/a stronger nucleophile Reject any mention of attack on carboxylate ion (for nd mark) Maltby Academy 16
17 (c) Note: although many candidates have calculated the empirical formula, this is not required. Molecular formula of ibuprofen = C 1 H 18 O Allow if given at end Allow marks for masses and number of moles if answers are rounded to SF in OR but method is correct. EITHER M r = g = mol = mol 16 mass CO produced from 1 C = =.78 g mass H O from 18 H = = g OR Mass C = Mass H = (.78 1) 44 (0.786) 9 = 0.758g = 0.087g (0.758) Moles C = = Moles H = Ratio C:H = 0.06: = 1:18 4 (d) (i) (Aspirin and ibuprofen) both contain same (types of) bond(s)(so absorb at same frequency/wavenumber) 1 Accept list of at least 4 bonds which are present in both Reject groups for bonds Data is required for mark Y = paracetamol Peak at (N H) IGNORE mention of amine OR (N H or amide) OR ((phenolic) O H) OR Only Y has peaks above 000 cm 1 (so must contain different type of bond to X and Z) 1 Reject C H in arene = 00 as present in both Reject (amide) Maltby Academy 17
18 (iii) 57 in Ibuprofen C 4 H 9 + / ( ) + /( ) + OR C O H + /CCO H + Accept structural or displayed formulae Do not allow lines at 15 ( + ) 76 (C 6 H 4 + ) 4 (C H 7 + or CO + ) 45 (COOH + ) as present in both Aspirin 59 OCO + /C H O + OR 11 C 6 H 4 CO H + OR 180 C 9 H 8 O 4 + (parent ion) OR 17 C 6 H 4 (CO H)O + Penalise no/wrong charges once only [14] 9. (a) A 1 (b) D 1 (c) A 1 [] 10. (a) (i) contains one carbon-carbon double bond Accept alkene 1 is a carbonyl compound / C=O group reduced (to (OH)) Accept aldehyde or ketone 1 (iii) is a carbonyl compound Accept aldehyde or ketone 1 Maltby Academy 18
19 (iv) is a ketone / P is not an aldehyde 1 Reject aldehyde (v) has two groups on the same side of a C=C Accept cis isomer 1 (b) (i) QWC 060 alkene (C H stretching) 90 alkane (C H stretching) 1690 ketones (C=O stretching) 1660 alkene (C=C stretching) 4 Correct marks Correct marks Correct 1 mark not an aldehyde 1 (iii) QWC 15 group 9 C H 5 group 4 no C H 7 group (c) C H 5 C C H H CO ketone and Z rest of molecule Accept Fully displayed (d) Measure its melting temperature And compare with data book values [16] Maltby Academy 19
1. Study the reaction scheme below, then answer the questions that follow.
1. Study the reaction scheme below, then answer the questions that follow. (a) (i) Butanal contains a carbonyl group. State a chemical test for a carbonyl group and describe the result of the test. Test...
More information(a) (i) Give the equation representing the overall reaction. (1) Give the equation representing the formation of the electrophile.
1. Benzene reacts with concentrated nitric acid in the presence of concentrated sulphuric acid at about 50 º in an electrophilic substitution reaction to give nitrobenzene. (a) Give the equation representing
More informationPMT. This question is about the reaction sequence shown below
1. The quality of written communication will be assessed in this question. To gain full marks you must explain your ideas clearly using equations and diagrams where appropriate. This question is about
More informationAQA A2 CHEMISTRY TOPIC 4.10 ORGANIC SYNTHESIS AND ANALYSIS TOPIC 4.11 STRUCTURE DETERMINATION BOOKLET OF PAST EXAMINATION QUESTIONS
AQA A2 CHEMISTRY TOPIC 4.10 ORGANIC SYNTHESIS AND ANALYSIS TOPIC 4.11 STRUCTURE DETERMINATION BOOKLET OF PAST EXAMINATION QUESTIONS 1 1. Consider the following reaction sequence. CH 3 CH 3 CH 3 Step 1
More informationPaper Reference. Paper Reference(s) 6256/01 Edexcel GCE Chemistry (Nuffield) Advanced Level Unit Test 6 (Synoptic)
Centre No. Candidate No. Paper Reference(s) 6256/01 Edexcel GCE Chemistry (Nuffield) Advanced Level Unit Test 6 (Synoptic) Thursday 19 June 2008 Morning Time: 2 hours Materials required for examination
More information(a) Which test always gives a positive result with carbonyl compounds? (b) Which test would give a positive result with ethane-1,2-diol?
1 Some chemical tests are described below. Warm with Fehling s (or enedict s) solution Warm with acidified potassium dichromate(vi) solution dd sodium carbonate solution dd 2,4-dinitrophenylhydrazine solution
More informationQ1. The following pairs of compounds can be distinguished by simple test tube reactions.
Q1. The following pairs of compounds can be distinguished by simple test tube reactions. For each pair of compounds, give a reagent (or combination of reagents) that, when added separately to each compound,
More informationName this organic reagent and state the conditions for the preparation. Reagent... Conditions (3)
1. (a) Propanoic acid, C 3 C 2 COO, can be prepared from carbon dioxide and an organic reagent. Name this organic reagent and state the conditions for the preparation. Reagent... Conditions...... (3) (b)
More informationInternational Advanced Level Chemistry Advanced Subsidiary Unit 2: Application of Core Principles of Chemistry
Write your name here Surname Other names Pearson Edexcel International Advanced Level Centre Number Candidate Number Chemistry Advanced Subsidiary Unit 2: Application of Core Principles of Chemistry Friday
More informationName/CG: 2012 Term 2 Organic Chemistry Revision (Session II) Deductive Question
Name/G: 2012 Term 2 rganic hemistry Revision (Session II) Deductive Question 1(a) A yellow liquid A, 7 7 N 2, reacts with alkaline potassium manganate (VII) and on acidification gives a yellow solid B,
More informationCorrect Answer D Titration with standard acid solution 1. A Collecting and measuring the volume of gas 1
Unit 4: General Principles of hemistry I Section A (a) The hydrolysis of -bromobutane using hydroxide ions 4 9 Br(l) + - (aq) 4 9 (l) + Br - (aq) D Titration with standard acid solution (b) The decomposition
More informationCHEM 3.2 (AS91388) 3 credits. Demonstrate understanding of spectroscopic data in chemistry
CHEM 3.2 (AS91388) 3 credits Demonstrate understanding of spectroscopic data in chemistry Spectroscopic data is limited to mass, infrared (IR) and 13 C nuclear magnetic resonance (NMR) spectroscopy. Organic
More informationChapter 16. Answers to examination-style questions. Answers Marks Examiner s tips. 1 (a) (i) C 6 H 12 O 6 2C 2 H 5 OH + 2CO 2 (ii) fermentation
Chapter 6 (a) (i) C 6 O 6 C 5 O + CO (ii) fermentation (b) (i) C 5 O + 3O CO + 3 O (ii) CO or carbon monoxide or C or carbon (a) (i) potassium (or sodium) dichromate(vi) or correct formula or potassium
More information3.8 Aldehydes and ketones
3.8 Aldehydes and ketones Introduction: p's to p's Like the alkenes, the carbonyl group consists of a s bond and a p bond between the carbon and oxygen: Oxygen is more electronegative than carbon meaning
More informationTOK: The relationship between a reaction mechanism and the experimental evidence to support it could be discussed. See
Option G: Further organic chemistry (15/22 hours) SL students study the core of these options and HL students study the whole option (the core and the extension material). TOK: The relationship between
More informationPearson Edexcel Level 3 GCE Chemistry Advanced Paper 2: Advanced Organic and Physical Chemistry
Write your name here Surname Other names Pearson Edexcel Level 3 GCE Centre Number Candidate Number Chemistry Advanced Paper 2: Advanced Organic and Physical Chemistry Specimen Paper for first teaching
More informationAlcohols. Nomenclature. 57 minutes. 57 marks. Page 1 of 9
..0 Alcohols Nomenclature 57 minutes 57 marks Page of 9 M. (a) % O =.6 % () If % O not calculated only M available C O () = 5. =.5 =.5 Ratio: : 0: ( C 0 O) () If arithmetic error in any result lose M If
More informationChapter 1 Reactions of Organic Compounds. Reactions Involving Hydrocarbons
Chapter 1 Reactions of Organic Compounds Reactions Involving Hydrocarbons Reactions of Alkanes Single bonds (C-C) are strong and very hard to break, therefore these compounds are relatively unreactive
More information91391 Demonstrate understanding of the properties of organic compounds
(2017:2) 91391 Demonstrate understanding of the properties of organic compounds Collated Identification Questions (ii) Explain how Benedict s solution can be used to distinguish between propanone and propanal.
More informationOption G: Further organic chemistry (15/22 hours)
Option G: Further organic chemistry (15/) TOK: The relationship between a reaction mechanism and the experimental evidence to support it could be discussed. See 16... Core material: G1 G8 are core material
More informationCarbonyls. Aldehydes and Ketones N Goalby chemrevise.org. chemrevise.org
arbonyls Aldehydes and Ketones N Goalby chemrevise.org arbonyls are compounds with a = bond, they can be either aldehydes or ketones. 3 ethanal 3 3 propanone If the = is on the end of the chain with an
More informationMark Scheme Page 1 of 8 Unit ode 2814 Session Jan Year 2004 Qu. Expected answers: Marks: 1 (a) (i) (relative) molecular mass / M r (ii) right / highest m /e / highest mass / second highest mass etc AW
More informationChemistry 2.5 AS WORKBOOK. Working to Excellence Working to Excellence
Chemistry 2.5 AS 91165 Demonstrate understanding of the properties of selected organic compounds WORKBOOK Working to Excellence Working to Excellence CONTENTS 1. Writing Excellence answers to Cis-Trans
More informationChapter 19: Amines. Introduction
Chapter 19: Amines Chap 19 HW: (be able to name amines); 37, 39, 41, 42, 44, 46, 47, 48, 53-55, 57, 58 Introduction Organic derivatives of ammonia. Many are biologically active. Chap 19: Amines Slide 19-2
More information1 Which of the following cannot be used to detect alcohol in a breathalyser test? Fractional distillation. Fuel cell. Infrared spectroscopy
1 Which of the following cannot be used to detect alcohol in a breathalyser test? Fractional distillation Fuel cell Infrared spectroscopy Reduction of dichromate(vi) ions 2 Propanal, H 3 H 2 HO, and propanone,
More informationMonday 19 June 2017 Morning Time allowed: 2 hours 15 minutes
Oxford ambridge and RSA A Level hemistry A H432/02 Synthesis and analytical techniques Monday 19 June 2017 Morning Time allowed: 2 hours 15 minutes *6826116453* You must have: the Data Sheet for hemistry
More informationIB Topics 10, 20 & 21 MC Practice
IB Topics 10, 20 & 21 MC Practice 1. What is the major product of the reaction between HCl and but-2-ene? 1,2-dichlorobutane 2,3-dichlorobutane 1-chlorobutane 2-chlorobutane 2. Which compound can be oxidized
More informationMechanisms. . CCl2 F + Cl.
Mechanisms 1) Free radical substitution Alkane à halogenoalkane Initiation: Propagation: Termination: Overall: 2) Ozone depletion UV light breaks the C Cl bond releasing chlorine radical CFCl 3 F à. CCl2
More informationA mass spectrometer can be used to distinguish between samples of butane and propanal. The table shows some precise relative atomic mass values.
Butane and propanal are compounds with M r = 58.0, calculated using data from your Periodic Table. (a) A mass spectrometer can be used to distinguish between samples of butane and propanal. The table shows
More informationChapter 10: Carboxylic Acids and Their Derivatives
Chapter 10: Carboxylic Acids and Their Derivatives The back of the white willow tree (Salix alba) is a source of salicylic acid which is used to make aspirin (acetylsalicylic acid) The functional group
More information10.2 ALCOHOLS EXTRA QUESTIONS. Reaction excess conc, H SO 180 C
10.2 ALOOLS EXTRA QUESTIONS 1. onsider the reaction scheme below which starts from butanone. N Reaction 1 3 2 3 3 2 3 O Reaction 2 O A 3 O B 2 3 excess conc, SO 180 2 4 but 1 ene and but 2 ene (a) When
More information2 Set up an apparatus for simple distillation using this flask.
The following instructions are from an experimental procedure for the preparation of cyclohexene from cyclohexanol and concentrated phosphoric acid. Read these instructions and answer the questions that
More informationORGANIC - BROWN 8E CH INFRARED SPECTROSCOPY.
!! www.clutchprep.com CONCEPT: PURPOSE OF ANALYTICAL TECHNIQUES Classical Methods (Wet Chemistry): Chemists needed to run dozens of chemical reactions to determine the type of molecules in a compound.
More information4. Carbonyl chemistry
4. Carbonyl chemistry 4.1. Oxidation of alcohols 4.2 Tests for aldehydes and ketones 4.3 Carbonyl functional groups 4.4 Reactions of carboxylic acids 4.5 Reductions of carbonyl groups 4.6 Esters 4.7 Preparing
More informationChapter 25: The Chemistry of Life: Organic and Biological Chemistry
Chemistry: The Central Science Chapter 25: The Chemistry of Life: Organic and Biological Chemistry The study of carbon compounds constitutes a separate branch of chemistry known as organic chemistry The
More informationScholarship 2015 Chemistry
93102 931020 S SUPERVISOR S Scholarship 2015 Chemistry 9.30 a.m. Friday 27 November 2015 Time allowed: Three hours Total marks: 32 Check that the National Student Number (NSN) on your admission slip is
More informationOCR (A) Chemistry A-level. Module 6: Organic Chemistry and Analysis
OCR (A) Chemistry A-level Module 6: Organic Chemistry and Analysis Organic Synthesis Notes by Adam Robertson DEFINITIONS Heterolytic fission: The breaking of a covalent bond when one of the bonded atoms
More informationM1. (a) Functional group (isomerism) 1
M. (a) Functional group (isomerism) (b) M Tollens (reagent) (Credit ammoniacal silver nitrate a description of making Tollens ) (Ignore either AgNO or [Ag(NH ) 2 + ] or the silver mirror test on their
More informationIt belongs to a homologous series with general formula C n H 2n+1 O
1 Propene can be made by the dehydration of propan-2-ol. What is the percentage yield when 30 g of propene (M r = 42.0) are formed from 50 g of propan-2-ol (M r = 60.0)? 60% 67% 81% 86% 2 Which statement
More informationMASS and INFRA RED SPECTROSCOPY
MASS and INFRA RED SPECTRSCPY Mass Spectroscopy The mass spectrometer was looked at in Unit 1. It was noted there that compounds produce fragmentation patterns when passes through a mass spectrometer.
More informationCARBONYL COMPOUNDS. Section A. Q1 Acrylic acid is produced from propene, a gaseous product of oil refineries.
MCQs Section A Q1 Acrylic acid is produced from propene, a gaseous product of oil refineries. Which statement about acrylic acid is not correct? A Both bond angles x and y are approximately 120. B It decolourises
More informationLecture 11. IR Theory. Next Class: Lecture Problem 4 due Thin-Layer Chromatography
Lecture 11 IR Theory Next Class: Lecture Problem 4 due Thin-Layer Chromatography This Week In Lab: Ch 6: Procedures 2 & 3 Procedure 4 (outside of lab) Next Week in Lab: Ch 7: PreLab Due Quiz 4 Ch 5 Final
More information*SCH12* *20SCH1201* Chemistry. Assessment Unit AS 1 [SCH12] FRIDAY 26 MAY, MORNING. Specification. New
New Specification Centre Number ADVANCED SUBSIDIARY (AS) General Certificate of Education 2017 Candidate Number Chemistry Assessment Unit AS 1 assessing Basic Concepts in Physical and Inorganic Chemistry
More informationPage (Extra space) (4) Benzene can be converted into amine U by the two-step synthesis shown below.
Q1. The hydrocarbons benzene and cyclohexene are both unsaturated compounds. Benzene normally undergoes substitution reactions, but cyclohexene normally undergoes addition reactions. (a) The molecule cyclohexatriene
More informationUnit 4: General Principles of Chemistry I Rates, Equilibria and Further Organic Chemistry (including synoptic assessment)
Write your name here Surname Other names Edexcel GCE Centre Number Candidate Number Chemistry Advanced Unit 4: General Principles of Chemistry I Rates, Equilibria and Further Organic Chemistry (including
More informationUnit 4: General Principles of Chemistry I Rates, Equilibria and Further Organic Chemistry (including synoptic assessment)
Write your name here Surname Other names Pearson Edexcel GCE Centre Number Candidate Number Chemistry Advanced Unit 4: General Principles of Chemistry I Rates, Equilibria and Further Organic Chemistry
More informationF322: Chains, Energy and Resources Basic Concepts
F322: hains, Energy and Resources Basic oncepts 1. Some of the hydrocarbons in kerosene have the formula 10 22. (i) What is the name of the straight chain hydrocarbon with the formula 10 22? (ii) Draw
More informationAdvanced Unit 6B: Chemistry Laboratory Skills II Alternative
Write your name here Surname Other names Edexcel GCE Centre Number Candidate Number Chemistry Advanced Unit 6B: Chemistry Laboratory Skills II Alternative Wednesday 15 May 2013 Morning Time: 1 hour 15
More informationClass XII - Chemistry Aldehydes, Ketones and Carboxylic Acid Chapter-wise Problems
Class XII - Chemistry Aldehydes, Ketones and Carboxylic Acid Chapter-wise Problems I. Multiple Choice Questions (Type-I) 1. Addition of water to alkynes occurs in acidic medium and in the presence of Hg
More informationIGNORE Just benzene has a delocalised ring Benzene does not have C=C double bonds Any references to shape/ bond angles. Acceptable Answers Reject Mark
1(a) All carbon to carbon bonds same length/ longer C-C and shorter C=C not present 1 IGNORE Just benzene has a delocalised ring Benzene does not have C=C double bonds Any references to shape/ bond angles
More informationH H O C C O H Carboxylic Acids and Derivatives C CH 2 C. N Goalby chemrevise.org. Strength of carboxylic acids.
19 arboxylic Acids and Derivatives Naming arboxylic acids These have the ending -oic acid but no number is necessary for the acid group as it must always be at the end of the chain. The numbering always
More informationGCE AS/A level 1092/01 CHEMISTRY CH2
Surname Centre Number Candidate Number Other Names 2 GCE AS/A level 1092/01 CHEMISTRY CH2 S15-1092-01 P.M. TUESDAY, 2 June 2015 1 hour 30 minutes For s use Question Maximum Mark Mark Awarded Section A
More informationPaper Reference. Tuesday 23 January 2007 Morning Time: 1 hour 30 minutes
Centre No. Candidate No. Paper Reference(s) 6244/01 Edexcel GCE Chemistry Advanced Unit Test 4 Tuesday 23 January 2007 Morning Time: 1 hour 30 minutes Materials required for examination Nil A calculator
More informationHaloalkanes. Isomers: Draw and name the possible isomers for C 5 H 11 Br
Haloalkanes The basics: The functional group is a halogen atom: F, Cl, Br or I General formula C n H 2n+1 X Use the prefixes: fluoro, chloro, bromo and iodo. Isomers: Draw and name the possible isomers
More informationA drug is designed to simulate one of the following molecules that adsorbs onto the active site of an enzyme.
1 drug is designed to simulate one of the following molecules that adsorbs onto the active site of an enzyme. Which molecule requires the design of an optically active drug? 2 Which one of the following
More informationTopic 4.10 ORGANIC SYNTHESIS AND ANALYSIS. Organic analysis Organic synthesis
Topic 4.10 ORGANIC SYNTHESIS AND ANALYSIS Organic analysis Organic synthesis DISTINGUISHING BETWEEN DIFFERENT ORGANIC COMPOUNDS Many of the organic compounds prepared in AS Unit 2 and in A2 Unit 4 can
More informationOrganic Chemistry. A. Introduction
Organic Chemistry A. Introduction 1. Organic chemistry is defined as the chemistry of CARBON compounds. There are a huge number of organic compounds. This results from the fact that carbon forms chains
More information(2) Read each statement carefully and pick the one that is incorrect in its information.
Organic Chemistry - Problem Drill 17: IR and Mass Spectra No. 1 of 10 1. Which statement about infrared spectroscopy is incorrect? (A) IR spectroscopy is a method of structure determination based on the
More informationChem 1075 Chapter 19 Organic Chemistry Lecture Outline
Chem 1075 Chapter 19 Organic Chemistry Lecture Outline Slide 2 Introduction Organic chemistry is the study of and its compounds. The major sources of carbon are the fossil fuels: petroleum, natural gas,
More informationGCE A level 1094/01 CHEMISTRY CH4
Surname ther Names Centre 2 Candidate GCE A level 1094/01 CHEMISTRY CH4 P.M. MNDAY, 14 January 2013 1¾ hours ADDITINAL MATERIALS In addition to this examination paper, you will need: Data Sheet Periodic
More informationALLOW CO 2 and CO 2 H CH 3
CERRY ILL TUITI CR A CEMISTRY A PAPER 1 MARK SCEME 1 (a) (i) The p R point at which the zwitterion exists 1 ALLW p/point at which there is no overall/net charge IGRE p/point at which there is no charge/
More informationModule 4 revision guide: Compounds with C=O group
opyright N Goalby Bancroft's School Module 4 revision guide: ompounds with = group arbonyls: Aldehydes and Ketones arbonyls are compounds with a = bond, they can be either aldehydes or ketones. 3 ethanal
More informationSection A. 1 at a given temperature. The rate was found to be first order with respect to the ester and first order with respect to hydroxide ions.
2 Section A Answer all questions in the spaces provided. Question 1:The N/Arate of hydrolysis of an ester X (HCOOCH2CH2CH3) was studied in alkaline 1 at a given temperature. The rate was found to be first
More information(07) 3 (e) Calculate the ph of this buffer solution at 298 K. Give your answer to 2 decimal places
7 3 (e) An acidic buffer solution is formed when 10.0 cm3 of 0.125 mol dm 3 aqueous sodium hydroxide are added to 15.0 cm3 of 0.174 mol dm 3 aqueous HX. The value of Ka for the weak acid HX is 3.01 10
More informationChapter 20: Carboxylic Acids and Nitriles شیمی آلی 2
Chapter 20: Carboxylic Acids and Nitriles شیمی آلی 2 Dr M. Mehrdad University of Guilan, Department of Chemistry, Rasht, Iran m-mehrdad@guilan.ac.ir Based on McMurry s Organic Chemistry, 7 th edition The
More information18.1 Arenes benzene compounds Answers to Exam practice questions
Pages 230 232 1 a) Benzene has a planar molecule ; with six carbon atoms in a regular hexagon. Each carbon atom forms a normal covalent ( ) bond with its two adjacent carbons atoms and a hydrogen atom.
More informationMOSTLY ALCOHOLS. Question 2, 2017 The structure of a molecule of an organic compound, threonine, is shown below.
MOSTLY ALCOHOLS Modified Question 1, 2017 A chemistry class was learning about the chemistry of haloalkanes. They were researching the effect of heat and concentrated potassium hydroxide in ethanol, conc.
More informationChirality, Carbonyls and Carboxylic Acids
hirality, arbonyls and arboxylic Acids Questions on this unit may include material from UNIT 2 see syllabus Isomerism Structural isomerism. Structural isomerism was dealt with in UNIT 2. All isomers are
More informationChemistry Assessment Unit A2 1
Centre Number 71 Candidate Number ADVANCED General Certificate of Education January 2013 Chemistry Assessment Unit A2 1 assessing Periodic Trends and Further Organic, Physical and Inorganic Chemistry AC212
More informationUnit 4: General Principles of Chemistry I Rates, Equilibria and Further Organic Chemistry (including synoptic assessment)
Write your name here Surname Other names Edexcel GCE Centre Number Candidate Number Chemistry Advanced Unit 4: General Principles of Chemistry I Rates, Equilibria and Further Organic Chemistry (including
More informationExperiment [RCH 2 Cl] [OH ] Initial rate/mol dm 3 s
1. The kinetics of the hydrolysis of the halogenoalkane RCH 2 Cl with aqueous sodium hydroxide (where R is an alkyl group) was studied at 50 ºC. The following results were obtained: Experiment [RCH 2 Cl]
More informationI N V E S T I C E D O R O Z V O J E V Z D Ě L Á V Á N Í CARBONYL COMPOUNDS
= substances containing the carbonyl group Aldehydes have the C=O group at the end of the chain, the aldehydic group is then... Naming of aldehydes: In the systematic name there is a suffix after the stem
More informationPearson Edexcel Level 3 GCE Chemistry Advanced Paper 2: Advanced Organic and Physical Chemistry
Write your name here Surname Other names Pearson Edexcel Level 3 GCE Centre Number Candidate Number Chemistry Advanced Paper 2: Advanced Organic and Physical Chemistry Sample Assessment Materials for first
More informationCarboxylic Acids and Nitriles
Carboxylic Acids and Nitriles Why this Chapter? Carboxylic acids present in many industrial processes and most biological processes They are the starting materials from which other acyl derivatives are
More informationInfrared Spectroscopy: How to use the 5 zone approach to identify functional groups
Infrared Spectroscopy: How to use the 5 zone approach to identify functional groups Definition: Infrared Spectroscopy is the study of the Infrared Spectrum. An Infrared Spectrum is the plot of photon energy
More informationMolecular Geometry: VSEPR model stand for valence-shell electron-pair repulsion and predicts the 3D shape of molecules that are formed in bonding.
Molecular Geometry: VSEPR model stand for valence-shell electron-pair repulsion and predicts the 3D shape of molecules that are formed in bonding. Sigma and Pi Bonds: All single bonds are sigma(σ), that
More information3.2.9 Alkenes. Addition Reactions. 271 minutes. 268 marks. Page 1 of 35
..9 Alkenes Addition Reactions 71 minutes 68 marks Page 1 of 5 Q1. Propene reacts with bromine by a mechanism known as electrophilic addition. (a) Explain what is meant by the term electrophile and by
More informationA.M. WEDNESDAY, 3 June hours
Candidate Name Centre Number 2 Candidate Number GCE AS/A level 1092/01 New AS CEMISTRY C2 ADDITIONAL MATERIALS In addition to this examination paper, you will need a: calculator; Data Sheet containing
More informationFurther Synthesis and Analysis
Further Synthesis and Analysis 12 4 In 2008, some food products containing pork were withdrawn from sale because tests showed that they contained amounts of compounds called dioxins many times greater
More informationMCAT Organic Chemistry Problem Drill 10: Aldehydes and Ketones
MCAT rganic Chemistry Problem Drill 10: Aldehydes and Ketones Question No. 1 of 10 Question 1. Which of the following is not a physical property of aldehydes and ketones? Question #01 (A) Hydrogen bonding
More informationLevel 3 Chemistry Demonstrate understanding of the properties of organic compounds
1 ANSWERS Level 3 Chemistry 91391 Demonstrate understanding of the properties of organic compounds Credits: Five Achievement Achievement with Merit Achievement with Excellence Demonstrate understanding
More informationUse your knowledge of organic reaction mechanisms to complete the mechanism for this step by drawing two curly arrows on the following equation.
Q1.The carboxylic acid 3-methylbutanoic acid is used to make esters for perfumes. The following scheme shows some of the reactions in the manufacture of this carboxylic acid. (a) One of the steps in the
More informationChapter 12 Alcohols from Carbonyl Compounds: Oxidation-Reduction and Organometallic Compounds
Chapter 12 Alcohols from Carbonyl Compounds: Oxidation-Reduction and Organometallic Compounds Introduction Several functional groups contain the carbonyl group Carbonyl groups can be converted into alcohols
More informationChemistry *P46941A0128* Pearson Edexcel P46941A
Write your name here Surname ther names Pearson Edexcel International Advanced Level Centre Number Candidate Number Chemistry Advanced Unit 5: General Principles of Chemistry II Transition Metals and rganic
More information2 Answer all the questions. 1 The first commercially useful azo dye was chrysoidine, designed by Otto Witt in H 2 N. chrysoidine.
2 Answer all the questions. 1 The first commercially useful azo dye was chrysoidine, designed by Otto Witt in 1875. N N NH 2 H 2 N chrysoidine (a) Witt made this dye by first forming the diazonium salt
More information*P51939A0124* Pearson Edexcel WCH04/01. P51939A 2018 Pearson Education Ltd.
Write your name here Surname Other names Pearson Edexcel International Advanced Level Centre Number Candidate Number Chemistry Advanced Unit 4: General Principles of Chemistry I Rates, Equilibria and Further
More informationJUNIOR COLLEGE CHEMISTRY DEPARTMENT EXPERIMENT 14 SECOND YEAR PRACTICAL. Name: Group: Date:
JUNIOR COLLEGE CHEMISTRY DEPARTMENT EXPERIMENT 14 SECOND YEAR PRACTICAL Name: Group: Date: This practical will serve as (i) an introduction to aromatic chemistry and (ii) a revision of some of the reactions
More information18.8 Oxidation. Oxidation by silver ion requires an alkaline medium
18.8 Oxidation Oxidation by silver ion requires an alkaline medium Test for detecting aldehydes Tollens reagent to prevent precipitation of the insoluble silver oxide, a complexing agent is added: ammonia
More information3.2.8 Haloalkanes. Nucleophilic Substitution. 267 minutes. 264 marks. Page 1 of 36
3.2.8 Haloalkanes Nucleophilic Substitution 267 minutes 264 marks Page 1 of 36 Q1. (a) The equation below shows the reaction of 2-bromopropane with an excess of ammonia. CH 3 CHBrCH 3 + 2NH 3 CH 3 CH(NH
More informationSPECTROSCOPY MEASURES THE INTERACTION BETWEEN LIGHT AND MATTER
SPECTROSCOPY MEASURES THE INTERACTION BETWEEN LIGHT AND MATTER c = c: speed of light 3.00 x 10 8 m/s (lamda): wavelength (m) (nu): frequency (Hz) Increasing E (J) Increasing (Hz) E = h h - Planck s constant
More information17 Alcohols H H C C. N Goalby chemrevise.org 1 H H. Bond angles in Alcohols. Boiling points. Different types of alcohols H 2 C CH 2 CH 2
17 Alcohols General formula alcohols n 2n+1 Naming Alcohols These have the ending -ol and if necessary the position number for the group is added between the name stem and the ol If the compound has an
More informationHydrocarbons and their Functional Groups
Hydrocarbons and their Functional Groups Organic chemistry is the study of compounds in which carbon is the principal element. carbon atoms form four bonds long chains, rings, spheres, sheets, and tubes
More informationAcceptable Answers Reject Mark. Acceptable Answers Reject Mark. ALLOW Iron/Fe or Zn/Zinc for tin Conc for concentrated. Acceptable Answers Reject Mark
(a)(i) (a)(ii) Concentrated nitric acid AND concentrated sulfuric acid concentrated nitric and sulfuric acids Concentrated HNO and concentrated H SO 4 Extra reagents To prevent multiple substitutions/
More informationThe mechanism of the nitration of methylbenzene is an electrophilic substitution.
Q1.Many aromatic nitro compounds are used as explosives. One of the most famous is 2-methyl-1,3,5-trinitrobenzene, originally called trinitrotoluene or TNT. This compound, shown below, can be prepared
More informationIsomerism and Carbonyl Compounds
Isomerism and Carbonyl Compounds 18 Section B Answer all questions in the spaces provided. 7 Esters have many important commercial uses such as solvents and artificial flavourings in foods. Esters can
More informationCHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME. ±½ a square
Chemistry Advanced Level Paper 3 (9CH0/03) 1(a)(i) suitable scale and axes labelled including units (1) all points plotted correctly (1) line of best fit (1) Plotted points use at least half the available
More informationChapter 16 Aldehydes and Ketones I. Nucleophilic Addition to the Carbonyl Group
Chapter 16 Aldehydes and Ketones I. Nucleophilic Addition to the Carbonyl Group Nomenclature of Aldehydes and Ketones Aldehydes are named by replacing the -e of the corresponding parent alkane with -al
More informationHow to Interpret an Infrared (IR) Spectrum
How to Interpret an Infrared (IR) Spectrum Infrared (IR) Spectroscopy allows the identification of particular bonds present within molecules. In this class we have simplified IR analysis by only focusing
More information1 The structures of six organic compounds are shown. H H H H H H H C C C H H H H H O H D E F. (a) Give the name of F. ... [1]
1 The structures of six organic compounds are shown. A B O O D E F O (a) Give the name of F.. [1] (b) Identify two of the compounds that are members of the same homologous series. Give the general formula
More informationPhysical Properties. Alcohols can be: CH CH 2 OH CH 2 CH 3 C OH CH 3. Secondary alcohol. Primary alcohol. Tertiary alcohol
Chapter 10: Structure and Synthesis of Alcohols 100 Physical Properties Alcohols can be: CH 3 CH 3 CH CH 2 OH * Primary alcohol CH 3 OH CH * CH 2 CH 3 Secondary alcohol CH 3 CH 3 * C OH CH 3 Tertiary alcohol
More information