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Organic Chemistry Revision Notes and Questions 2014

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Organic Chemistry Revision Notes and Questions 2014 Catholic Junior College H2 Chemistry 9647 Organic Chemistry Revision Notes and Questions 2014 Contents Page 1. Synthesis/Conversion questions 2 2. Structural Elucidation questions 11 3. Distinguishing tests 17 4. Explanation based questions 20 5. Data based questions 22 6. Answers 29 Name: _________________________________________ Class: 2T____ 1 1. Synthesis/Conversion type Questions A. To synthesize a compound with a change in functional group. Method 1: Change the functional group by substitution. If the functional group cannot be substituted to give the desired compound in a single step, substitute the functional group into another functional group that can undergo substitution readily (e.g. halogenoalkane) Examples: 1) CH3CH2Br → CH3CH2OH 2) CH3CH2OH → CH3CH2NH2 Method 2: Convert the functional group by oxidation/reduction if the starting compound cannot undergo substitution. Example: 1) CH3COCH3 → CH3CH(Cl)CH3 B. To synthesize a compound with a change in position of functional group (that may or may not be the same as the starting reagent). Step 1: Convert the starting reagent into an alkene by elimination of the functional group. Step 2: Add the new functional group (same or different) back to the alkene. Examples: 1) CH3CH2CH2Cl → CH3CHClCH3 2) CH3 CH3 CH3CHCH2Cl CH3CCH3 OH 2 C. To synthesize a compound with an increase in carbon chain length. Method 1: Substitute the nitrile group into the compound. (i.e. The compound needs to have a halogen group already) The nitrile group can be further converted into amine or carboxylic acid group. [This method does not increase the number of functional groups in the compound.] Examples: 1) CH3CH2Br → CH3CH2CO2H 2) CH3CH2Br → CH3CH2CH2NH2 3) CH3CH2OH → CH3CH2CH2NH2 Method 2: Add the nitrile group into the compound using HCN. (i.e. The compound needs to have a carbonyl group already) The product formed will contain a hydroxyl group adjacent to the nitrile group. The nitrile group can be further converted into amine or carboxylic acid functional group. [This method increases the number of functional groups in the compound by 1.] Examples: 1) CH3CHO → CH3CH(OH)CN 2) CH3CHO → CH3CH(OH)CO2H 3) O OH CH3 C CH3 CH3CCH3 CH2NH2 3 Method 3: Employ other reactions that will increase in number of carbon atoms by more than one. (e.g. Reaction of alcohol with carboxylic acid, phenol with acyl chloride, amine with acyl chloride) Example: 1) CH3CHO → CH3CH(OH)CN D. To synthesize a compound with a decrease in carbon chain length. Method 1: Oxidise the alkene to rupture the C=C double bond. Example: 1) CH3CH=CHCH3 → CH3CH2OH Method 2: Oxidise the compound that contain CH3CH(OH)- group or the CH3CO- group using alkaline aqueous iodine. This reduces the number of carbon atoms by one. Example: 1) CH3CH(OH)CH2CH3 → CH3COCl Method 3: Hydrolysis of esters and amides. 4 Exercises (Synthesis) Below are some two step syntheses. In each case, identify the reagents and conditions for each step and also identify the intermediate compound. (a) CH2CH2Cl A CHCH3 Cl (b) CH3 CH3 C O B CH C Br 3 CH3 H (c) CH3CHCH2CH3 C CH3CHCHCH3 Cl HO OH (d) CH CH3 CH3 3 C C D CH3 C OH CH CH 3 3 H (e) O H NO2 E CH3 C N (f) CH3 CH3 O C C F CH3 C CH CH OH 3 3 5 PRACTICE QUESTIONS- Conversion/synthesis based questions 1. a) Suggest a 2-step synthesis of cyclohexanol, starting from cyclohexane, writing equations for all reactions. OH X Step 1 Step 2 cyclohexanol cyclohexane b) Suggest reagents and conditions for the dehydration of cyclohexanol to cyclohexene. c) Why is there only one isomer of cyclohexene, whereas there are two isomers of hex-3-ene, CH3CH2CH=CHCH2CH3? d) What reagents and conditions are needed to convert 1-bromopropane into the following? (i) CH3CH2CH2NH2; (ii) CH3CH2CH2CN; (iii) CH3CH=CH2 “A” Level Specimen paper/III/1(c) & (d) 2. Coumarin is a naturally occurring organic compound with a grassy odour. A series of reactions is shown below. CH=CHCOCl Q HCl(g) OH Step 1 O O heat coumarin NaOH(aq), heat D2O(l) P R S T H2SO4 (aq) KMnO4(aq) heat a) Name the type of reaction in Step 1. [1] b) Draw the structures of organic compounds P - T. [5] NJC Prelim 06/III/6a,b *3. Propanone reacts with hydrogen cyanide to form a compound with molecular formula C4H7ON. a) Draw a displayed formula for the compound obtained. [1] b) The compound in a) can be used in the synthesis of methyl-2- CH3 methylpropanoate shown on the right. H2C C Outline a reaction scheme, stating reagents and conditions for the C O conversion. [3] NYJC Prelim 09/III/3c H3C O *4. Below is a reaction scheme. Step 1 Step 2 Step 3 (CH3)2C=CH2 K L CH3CH=CH2 J M Give the reagents and conditions required for each of the steps as well as the structures of the intermediate K and L. [5] IJC Prelims 09/II/3b 6 5. The following represents part of a synthetic pathway to obtain the amino acid, phenylalanine, and one of its other derivatives, D. Suggest the reagents and conditions required to carry out Steps 1-5 as well as the structure of intermediate A formed. [6] MJC Prelim 06/II/5 (modified) 6. One type of hydroxyacid E, can be synthesized from compound D in three steps. Give the reagents, conditions and intermediates for this synthesis. IJC Prelim 09/III/4c 7. 4-bromobenzoyl chloride can be synthesized from methylbenzene via multi-step transformation. CH3 COCl Br Suggest reagents and conditions for each step and draw the structural formulae of the intermediates formed. [5] NYJC Prelim 08/III/4d 7 *8. Salbutamol, commonly known as Ventolin, is a drug that is used to treat acute asthma and to relieve symptoms associated with other conditions pertaining to reversible airway obstructions. In the laboratory, salbutamol can be made from the starting compound, 2-hydroxybenzoic acid via the following route: a) State the reagents and conditions for Step 1 and Step 2. [2] b) i) Suggest a possible 3-step synthesis for the conversion of compound J to salbutamol. Your answer should include all essential reagents and conditions, as well as the structures of any intermediates formed. [3] ii) A lab technician suggested the use of chlorine gas under UV light to convert compound H to J in Step 3. Suggest why his choice of condition may not be appropriate for the conversion stated. [1] MJC Prelim 06/III/6a,b *9. 5-nitro-2-propoxyphenylamine, T, is an artificial sweetening agent which is 4000 times as sweet as sucrose. It can be made from propoxybenzene, R. Suggest a two-stage synthesis of T starting from propoxybenzene, R, identifying the intermediate compound S. The reactivity of propoxybenzene is similar to that of phenol. [3] “A” Levels/N08/III/5d 8 *10. Compound A can be used to synthesize compounds B and C via a series of steps. CH CH3 3 H N CONH(CH ) COOH O2N CH2Cl H N CH2Cl 2 2 A B C a) Describe the reagents and conditions required to synthesize compounds B and C respectively. b) Compounds B and C can further react under suitable conditions to give compound D. CH3 CH3 H N CH2 N CONH(CH2)2COOH D Give the structural formulae of the organic product(s) for each of the reaction of compound D with (i) cold dilute hydrochloric acid; (ii) hot aq. KOH; (iii) aq. Br2 11. A cyclic amide can be synthesised from compound J as shown below. a) Draw the structures of K and L. [2] b) A student suggests the use of hot, acidified potassium manganate(VII) as the oxidising agent in Step II. Explain why this choice of reagent is inappropriate and suggest a suitable alternative. [2] RI Prelim 13/III/4d 9 *12. A phenyl ester can be converted to a hydroxyl phenyl ketone via a rearrangement reaction known as the Fries rearrangement. It involves migration of an acyl group of phenyl ester to the benzene ring. The reaction is ortho, para–selective and the position of acylation can be regulated by the choice of temperature. A Lewis acid catalyst like AlCl3 is required. R O OH O ortho–product AlCl3 R O > 100 C R O AlCl3 rt para–product OH Using the Fries rearrangement in step II of the reaction scheme below, give the reagents and conditions for steps I – IV and deduce the structures for intermediates A – C. [7] DHS Prelim 13/III/4c 10 2. Structural Elucidation Questions Observations/Data from Qn Deduction From the molecular formulae - C: H ratio of about 1:1 Benzene ring is present. - C: H ratio of about 1:2 Unsaturation is present. (e.g. C=C, C=O) If compound only has C and H, but does not decolourise Br2, cycloalkane is present. - C: H ratio > 1:2 Saturated compound. (No double bond) - Neutral compound with N,O Amide may be present. - Neutral compound with O Alcohols or ester may be present. Rotate plane polarized light/Optically A chiral carbon present in the compound. (Carbon with active compound four different groups attached to it) Can show geometrical isomerism 2 different substituents joined to each carbon in C=C present, and C=C is not in a rring Compound decolourises bromine. Alkene OR phenol OR phenylamine may be present. - If alkene is present, no. of moles of bromine reacted is equal to the number of moles of double bond. The molecular formula will also show an increase in no. of bromine atoms. - If phenol OR phenylamine is present, the molecular formula will show an increase in the number of bromine atoms with a decrease in no.
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