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1 CRBONYLS CRBONYLS Outline mechanism for reaction methanol with propanoyl chloride. Clearly start curly arrow from middle lone pair or middle a bond. CRBONYLS CRBONYLS display formula butanoic anhydride. - - Each chain must contain 4 carbon atoms which includes C in functional group acid anhydride.

2 CRBONYLS CRBONYLS Outline mechanism for following reaction: Clearly start curly arrow from middle lone pair or middle a bond. CRBONYLS CRBONYLS Complete following equation: 3CH 3 OH HOCH CH(OH)CH OH - - The equation shows how biodiesel is produced (a transester). The by-product is glycerol (Propane-1,,3-triol).

3 CRBONYLS CRBONYLS Outline mechanism for reaction ammonia with 1-chloropropane. Clearly start curly arrow from middle lone pair or middle a bond. CRBONYLS CRBONYLS Write an equation for preparation butyl ethanoate from an acid anhydride an alcohol. CH 3 CH CH CH OH + (CH 3 CO) O CH 3 COOH + CH 3 COOCH CH CH CH Butyl indicates butanol Ethanoate indicates ethanoic anhydride

4 Give a use for se compounds which are formed when palm oil is heated with an excess NaOH (aq). CRBONYLS Soaps. Sodium salts fatty acids are used as soaps. CRBONYLS CRBONYLS CRBONYLS Write an equation to show how biodiesel is formed from this triester: + 3 CH 3 OH - - The second product (byproduct) is glycerol, used in cosmetics, food colourings medicines.

5 Explain how a racemic mixture is formed in lab. The nucleophile attacks carbocation in planar C=O bond from above or below plane with equal likelihood. Without terms planar equal likelihood (owtte) you cannot score full marks. What is formed when this compound is dehydrated? - - This question draws on your S knowledge: dehydration alcohols produces alkenes.

6 polymer formed from this monomer (show one repeat unit): This is a condensation polymer as water is removed in process: OH from COOH group H from OH group. MINO CID MINO CID What is zwitterion formed from this amino acid? - - The COOH group is deprotonated; NH group acts as a base in amino acids.

7 n ester formed from methanol with molecular formula C 6 H 1 O is optically active. ester. The information provided indicates that you need to find a carboxylic acid with 5 carbon atoms that is branched to create a chiral carbon atom. a branched, non-cyclic pair geometric isomers with molecular formula C 6 H pieces information need to be incorporated into structure: branched, noncyclic, E/Z forms. different pairs exist that match this description.

8 structural formula a tertiary amine which is an isomer pentylamine. pentylamine Diethylmethylamine (3 o amine) In a tertiary amine re are 3 hydrogen substituents on nitrogen atom (3 H atoms have been replaced with 3 R groups). The molecular formula remains same). a branched, non-cyclic optical isomer with molecular formula C 6 H 1. * pieces information need to be incorporated into structure: branched, noncyclic, optical. The central carbon atom is chiral.

9 two stereoisomers formed when following compound is dehydrated: This question is drawing on your S knowledge: dehydration alcohols produces alkenes. all structural isomers that have a molecular formula C 3 H 6 O contain an OH group as well as a C=O group (broad peak at cm -1 in IR spectrum) - - The broad peak at given wavenumber indicates that this is not a carboxylic acid.

10 structure geometric isomer C 4 H 7 NO which is formed when ammonia reacts with an acyl chloride. Find acyl chloride that shows E/Z isomerism replace Cl with NH group from ammonia. HCl will be an inorganic product. all branched chain isomers with molecular formula C 5 H branched alkenes. If helps if you isomers as you draw m so you avoid repeating isomers.

11 two structures an ester that contains a benzene ring has an M r 136. The benzene ring has a formula mass C 6 H 5 = 77; an ester contains xo = 3; = 7; 7 = xc + 3xH needed to complete structure. MINES MINES a primary, secondary tertiary isomer CH 3 CH CH NH - - In a secondary amine re are H substituents; in a tertiary amine re are 3 H substituents in a primary amine re is one H substituent.

12 How do you get from ene to (CH 3 CH ) NH in 3 steps? MINES React ene with HBr to produce CH 3 CH Br. React bromoethane with ammonia to produce ethylamine. React ethylamine with bromoethane. The first reaction is electrophilic addition. The second reaction substitutes Br for an NH group. The NH group substitutes Br in bromoethane in step 3. MINES MINES MINES (CH 3 ) NH can be formed by reaction an excess CH 3 NH with CH 3 Br. Outline a mechanism for this reaction. - - The lone pair electrons is on N atom which is not very clear in this diagram. The mechanism is a nucleophilic substitution reaction.

13 MINES MINES structures two compounds formed when CH 3 NH reacts with ethanoic anhydride. There are amine molecules needed; one to form salt with ethanoic acid produced or to form amide. MINES MINES Write equations to show how butylamine is prepared in a two-step synsis starting from 1-bromopropane, CH 3 CH CH Br. CH 3 CH CH Br + KCN CH 3 CH CH CN + KBr CH 3 CH CH CN + 4[H]/H CH 3 CH CH CH NH - - In first reaction HCN is not allowed as actual reagent added is KCN or NaCN. If in doubt, write CN instead produce Br - to balance equation.

14 MINES MINES How do you get from CH 3 Br to C H 3 N, C H 7 N finally (CH 3 CH )(CH 3 )NH? You get from CH 3 Br to C H 3 N (CH 3 CN) with CN. dd H to CH 3 CN to get to C H 7 N (CH 3 CH NH ). React CH 3 CH NH with CH 3 Br to get final product ethylmethylamine. For step one KCN or NaCN in aqueous acid is required. For step H with a Ni catalyst as a reducing agent. Step 3 is a nucleophilic substitution where Br is replaced by amine. MINES MINES Write an equation for reaction water with ethylamine showing how an alkaline solution is formed. CH 3 CH NH + H O OH - + CH 3 CH + NH mines act as bases/proton acceptors.

15 Give IUPC alanine draw zwitterion alanine. MINO CIDS -aminopropanoic acid MINO CIDS zwitterion is an ion with a positive negative charge an overall charge zero. The H from COOH is transferred over to NH group in an amino acid. product reaction alanine with methanol in presence concentrated sulfuric acid. MINO CIDS MINO CIDS - - The COOH group reacts with alcohols to form esters.

16 structure formed by aspartic acid at high ph. MINO CIDS MINO CIDS High ph is alkaline conditions. The amino acid will be deprotonated as a result. There are two COOH groups that will donate a proton each. t low ph NH group would accept a proton. Show how two aspartic acid molecules react to form a dipeptide. MINO CIDS MINO CIDS - - Eir answer is correct depending on which two COOH groups is used to form peptide link.

17 Give IUPC threonine. MINO CIDS MINO CIDS -amino-3-hydroxybutanoic acid First break peptide bond draw out threonine. Remember alphabetical order. The functional group is COOH group. structure formed by lysine at low ph. MINO CIDS MINO CIDS - - First, break peptide bond draw out lysine. Lysine contains two NH groups which both accept a proton at low ph.

18 Give two reasons why sweetener aspartame is not used in foods that need to be cooked. MINO CIDS spartame contains an ester bridge which will be broken during hydrolysis. The peptide link will also be hydrolysed two amino acids will be produced as a result. MINO CIDS Cooking will hydrolyse aspartame. Toxic methanol will be produced as well as amino acids which do not taste sweet. Give IUPC aspartic acid: MINO CIDS -aminobutane-1,4-dioic acid. MINO CIDS - - The amine COOH group are separated by one C atom, making this an α-amino acid, hence number in front amino. The numbers 1 4 are not strictly required here.

19 MINO CIDS structure an alternative dipeptide that could be formed by two amino acids that formed this dipeptide: First, break peptide link n flip amino acids to link alternative NH COOH groups toger. MINO CIDS MINO CIDS What is produced when aspartic acid reacts with dilute NaOH? MINO CIDS - - Eir above structures will score a mark depending on amount NaOH added, one or both COOH groups will be deprotonated.

20 What is produced when serine reacts with an excess bromoethane? MINO CIDS With an excess haloalkane, any o amine produced will continue substitution reaction until a quarternary amine is formed. MINO CIDS MINO CIDS What is produced when this amino acid reacts with ethanoyl chloride? MINO CIDS - - cyl chlorides can react with amines to form an amide link or with carboxylic acids to form an ester. In second product all COOH NH groups have reacted with ethanoyl chloride.

21 MINO CIDS Which route to an amine, nucleophilic substitution with a haloalkane or reduction a nitrile, will give a less pure product? Why? With an excess haloalkane, any o amine produced will continue substitution reaction until a quarternary amine is formed. MINO CIDS Nucleophilic substitution. Furr substitution reactions can occur with amines that are formed. MINO CIDS Which major organic product is formed when this compound is used in excess reacted with bromomethane? MINO CIDS - - n excess primary amine will make furr substitution reactions less likely.

22 MINO CIDS What is amino acid formed when this dipeptide is hydrolysed? -aminobutanoic acid Two same amino acid were joined toger to make this dipeptide. MINO CIDS What type polymer is shown? Condensation polymer. - - Polyester is an allowed alternative answer.

23 What type polymer is shown which alkenes make up this polymer? ddition polymer. dditional polymer is not allowed does not score. Which diol was used to make this polymer? Pentane-1,5-diol. - - The e at end pentane is needed, as are both numbers.

24 What is common this polymer? Terylene or Polyester. You are required to know three condensation polymers Nylon-6,6, Terylene Kevlar, including ir monomers. What is common this polymer? Kevlar or Polyamide. - - You are required to know three condensation polymers Nylon-6,6, Terylene Kevlar, including ir monomers.

25 one repeat unit polymer formed when propanedioic acid reacts with hexane-1,6-diamine. Hexane-1,6-diamine is one monomers in Nylon-6,6 refore you are required to know its structure. What type polymer is this what is type polymerisation by which it was formed? Polyamide polymer or nylon-,4; Condensation polymerisation. - - When naming nylon polymers, first number refers to number carbon atoms in diamine, second one to number carbon atoms in dicarboxylic acid.

26 one repeating unit formed by polymer H C=CHCN. common mistake is to use C in CN group end up with three C atoms in a chain. What is empirical formula this repeating unit? C 3 H 4 O - - Empirical formula = simplest whole number ratio.

27 What reducing agent is needed for this reaction? ROMTICS Sn HCl or Fe HCl or H Ni or Pt LilH 4 or NaBH 4 are incorrect reducing agents here. ROMTICS What reducing agent do you need for this reaction? ROMTICS H Ni or H Pt ROMTICS - - Essentially converting alkene to an alkane, hence use hydrogen a nickel catalyst.

28 ROMTICS Describe how ethanoyl chloride can react with benzene. Include a mechanism. lcl 3 catalyst/carrier needed to form electrophile: lcl 3 + CH 3 COCl lcl CH 3+ CO The overall equation for this reaction is CH 3 COCl + C 6 H 6 C 6 H 5 COCH 3 + HCl. The mechanism is electrophilic substitution. ROMTICS Outline a mechanism to show how 4-methylnitrobenzene is produced from methylbenzene. ROMTICS ROMTICS - - Make sure that nitro group methyl group are in opposite positions so 4- methylnitrobenzene is produced.

29 Write equations to show how electrophile + NO is produced. ROMTICS H SO 4 + HNO 3 HSO H NO 3 + H NO 3 + H O + + NO Both acids need to be concentrated. lternative equations are: H SO 4 + HNO 3 HSO NO + + H 3 O + or H SO 4 + HNO 3 NO + + HSO H O ROMTICS Use data provided to state explain stability benzene compared with hypotical cyclohexatriene. ROMTICS ROMTICS Benzene is more stable than cyclohexatriene. From data given, adding 3H molecules should release 360kJmol -1. The actual energy change is 15 kjmol -1 different/less due to electrons being delocalised in benzene. - - Remember that question asks you to compare cyclohexatriene to benzene. This means commenting on 3 double bonds, not 1.

30 What reagent is needed for this reaction? ROMTICS Concentrated sulfuric acid or concentrated phosphoric acid or dilute acid with l O 3. This reaction is an elimination reaction tests S knowledge; benzene ring does not take part in this reaction. ROMTICS How do you get from benzene to final product in steps? ROMTICS First, react benzene with propanoyl chloride in presence an lcl 3 catalyst. Then reduce resulting phenylpropanone with aqueous NaBH 4. ROMTICS - - The first step is a Friedel- Crafts acylation reaction (electrophilic substitution) followed by reduction an aldehyde to an alcohol.

31 What type reaction is shown what reagent do you need for this reaction? ROMTICS Water alone is insufficient to score reagent mark. ROMTICS Hydrolysis reaction ( peptide/amide bond is broken); aqueous NaOH or aqueous HCl is required. ROMTICS Which or two substances are required for this reaction? HBr followed by lbr 3 /FeBr 3 or HCl followed by FeCl 3 /lcl 3. ROMTICS - - HBr or HCl will convert alkene into a haloalkane first. The halogen carrier is n required to produce electrophile CH 3+ CH which will react with ring.

32 STRUCTURE How many peaks will re be in a 1 H NMR J? What is splitting pattern for protons labelled a? 4 peaks: The splitting pattern for protons labelled a will be a triplet. STRUCTURE lways check for symmetry before deciding on number peaks/ number environments. STRUCTURE How many peaks will re be in a 13 C NMR this compound? 3 peaks: STRUCTURE - - lways check for symmetry before deciding on number peaks/ number environments.

33 STRUCTURE STRUCTURE structure an optically active carboxylic acid with 6 carbon atoms five peaks in its 13 C NMR. spectrum. Check that re are 5 environments a chiral carbon atom. STRUCTURE How many peaks will re be in a 13 C NMR this compound? 11 peaks: STRUCTURE - - lways check for symmetry before deciding on number peaks/ number environments.

34 STRUCTURE Compounds F G have molecular formula C 6 H 4 N O 4 both are dinitrobenzenes. F has two peaks in its 13 C NMR. spectrum. G has three peaks in its 13 C NMR spectrum. Identify F G. F G STRUCTURE lways check for symmetry before deciding on number peaks/ number environments. STRUCTURE Compounds H J have molecular formula C 6 H 1. Both have only one peak in ir 1 H NMR spectra. H reacts with aqueous bromine but J does not. H J STRUCTURE - - lways check for symmetry before deciding on number peaks/ number environments.

35 STRUCTURE Which bond is responsible for a major absorption at 174cm -1 in an IR spectrum? C=O STRUCTURE lways use data sheet provided to identify bonds. When asked to suggest a range, always stay within data sheet range. STRUCTURE STRUCTURE Suggest structure fragment responsible for major peak in mass spectrum CH 3 COOCH 3 state its m/z value. Write an equation showing formation this fragment from molecular ion. major peak [CH 3 CO] + m / z 43 CH 3 COOCH 3 +. CH 3 CO + + OCH Carefully check positive charges radical dots are correctly assigned.

36 STRUCTURE DETERMINTION Use 1 H NMR information provided to identify compound which has a molecular formula C 6 H 1 O. STRUCTURE DETERMINTION Mark scheme Use splitting pattern to confirm identity each peak. The yellow H group has hydrogen neighbours is split into a triplet whereas red H group has 3 hydrogen neighbours is split into a quadruplet. Comment

37 STRUCTURE DETERMINTION Use information provided to identify compound which has a molecular formula C 6 H 1 O. (Each peak in 1 H NMR is a singlet) STRUCTURE DETERMINTION Mark scheme The IR reveals that re is an alcoholic OH group (broad peak at ) as well as a C=O peak at Comment

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