chemistry Unit 4.5

Unit 4.5 Organic chemistry II Recognise structural isomers, stereoisomerism as geometrical (cis-trans) or o ptical Explain the existence of geometrical (cis-trans) isomerism resulting from restricted rotation about a carbon-carbon double bond Understand the existence of optical isomerism resulting from a chiral centre in molecules with a single asymmetric carbon atom, Understand optical isomers as object and non- superimposable mirror images Recall optical activity as the ability of a single optical isomer to rotate the plane of polarisation of plane polarised monochromatic light and understand the nature of a racemic mixture Isomer Same molecular formula, different structural formulae

Structural isomerism Occurs when 2 or more different structural formulae can be written for the same molecular formula Chain isomers Different arrangements of carbon skeleton Similar chemical properties, differ in physical properties(Mt)because of change in shape of molecule

Positional isomers Same skeleton and functional group, sid e chains/functional groups are in different positions on the carbon chain Differ in physical properties

Functional group isomers Same atoms arranged into different fu nctional groups Differ in physical & chemical properties

 – 1

 – 1

2 bonds in the C=C double bond are not the same, bond energy of C=C (612kJmol ) is greater than C-C (348kJmol ) but not as twice as hence pi bond is weaker than the sigma bond Stereoisomerism found in any molecule of the type: Stereoisomerism Molecules have same molecular formula, same structural formula, but atoms have a different 3d arrangement(orientation in space). Differ in physical properties there’ s restricted rotation about a bond(C=C double bond where each of the two C ato ms carries 2 Geometric Isomerism Occurs when there’s different atoms/groups)differ in physical properties(different positions of groups, chains affects shape, dipoles, intermolecular forces) Single sigma bond  Free rotation about this bond without any reduction in degree of overlap  Double bond  Restricted rotation about C=C double bond because rotation would lead to a decrease in overlap of p orbitals that give the pi

Optical Isomerism Where molecules(chiral molecules) have mirror-image isomers that are not superimposable on the original compound. Sole criterion for chirality chirality is existence of non-superimposable mirror images • Commonest origin of chirality is a carbon atom having 4 diff groups attached to it(the chiral centre) • Molecules will not be chiral if one chiral centre is the mirror image of the other in a 2 chiral centred molecule • Possible to have chirality in molecules that don’t have chiral chiral centres(molecule is helical) Chiral molecules rotate plane of polarisation if plane-polarised monochromatic light is shone through them(sodium light) must be monochromatic because angle of rotation depends on wavelength wavelength of light used. In some wavelengths it’s 0 Dextrorotatory (+) in front of the name, molecules that rotate the plane to the right(clockwise looking into the sample) Laevorotatory (-) in front of the name, molecules rotating the plane to the left Racemic Mixture No rotation, solution contains equal(molar) amounts of each 2

forms of chiral molecule. Clockwise rotation of one isomer isomer cancelled by anticlockwise rotation from the other Chiral Has 2 isomers that are non superimposable mirror images Chiral molecule A molecule that is non-superimposable on its mirror image Questions How is optical activity detected experimentally? experimentally? Rotation of plane polarised light light bond polarity Electronegativity (EN) Strength(of an atom)to attract(the pair of)electrons in a covalent bond - EN affects bond length with larger differences giving shorter bonds - More electronegative atoms attract shared shared electrons more towards, towards, itself and acquire a partial negative charge • Electronegativity decreases going down a group(most electronegative electronegative element is fluorine) • Electronegativity increases increase s across period 3, elements on the LHS lose electrons and elements on the RHS gain electrons to achieve a stable structure * Ionic bonds are partially covalent when EN is small * Covalent bonds are partially ionic(creation of dipoles)when EN is large Polarisability The ease with which the electron cloud of an anion is distorted by a cation so there’s electron sharing • Smaller and higher the charge(higher the charge density)on the cation, the more polarising it is • Larger and higher the charge on the anion, the more easily it is polarized Features favouring ionic bonding: • Large cation metal, of low charge, having a low IE • Large anion non metal, of low charge, having a high EA - Large anions most most stable with large large cations Small cations most stable with small a nions - A covalent bond is polar if electrons in the bond are unequally shared formation of polyesters and pol yamides

Condensation polymers

esters with acids and alkalis in aqueous solution carbonyl compounds with hydrogen cyanide, 2,4-dinitrophenylhydrazine, alkaline ammoniacal silver nitrate solution, Fehling’s solution, ethanoyl chloride with water, alcohols, ammonia and pr imary amines primary amines with aqueous hydrogen ions, acid chlorides nitriles undergoing hydrolysis and undergoing reduction amides with phosphorus(V) oxide and bromine in aqueous alkali amino acids with acids and bases, their z witterion structures. iodine in the presence of alkali (or potassium iodide and sodium chlorate(I)), (sodium borohydride) sodium tetrahydridoborate(III) (lithium aluminium aluminium hydride) lithium tetrahydridoaluminate(III) tetrahydridoaluminate(III) Halogeno-compounds with magnesium to form Grignard reagents. Grignard reagents reactions with water, carbon dioxide and carbonyl compounds.

Recall that Grignard reagents act as nucleophiles Carboxylic acids with alcohols, (lithium aluminium hydride), phosphorus pentachloride, sodium carbonate and sodium hydrogencarbonate rules for nomenclature Functional group compound Compound type Halogenoalkanes Grignard reagent Aldehydes Ketones

An atom/group of atoms in an organic compound that determines all the possible chemical reactions for that Functional group R-X (X is Cl, Br, I) R-MgX (X is a halogen ato m) 1 1 R -CHO R can be a H atom 1 2 R -CO-R 1 2 R & R must contain at least one C atom

Names based on longest continuous C chain Meth = 1 Eth = 2 Prop = 3 But = 4 Pent = 5 Hex = 6 Hept = 7 Oct = 8 Carbonyl(C=O) compounds aldehydes & ketones

Compound type Acid chlorides Amines Amides Nitriles Amino acids

Functional group R-COCl R-NH2 R-CONH2 R-CN RCH(NH3+)COO-

 primary (1°), secondary (2°) and tertiary (3°) alcohols is based upon the number of carbon atoms the C-OH group's carbon is bonded to. Ene –  Ene – means means C=C bond Alkyl group names come before name of longest C chain preceded, by a number to indicate C atom at which substitution occurred Alkyl group number comes from shortest C chain

3-ethyl-2-methylhexane

3-methylhex-2-ene

Alkyl groups A methyl group is CH 3 An ethyl group is CH3CH2 2-methylpentane

2,3-dimethylbutane

2,2-dimethylbutane

1,1,1-trichloroethane

pentan-3-one

2-bromo-2-methylpropane

1-iodo-3-methylpent-2-ene

2-methylpropan-1-ol

ethane-1,2-diol

HCHO methanol CH3CHO ethanol 2-methylpentanal(aldehyde)

CH3COCH3 butanone propanone(ketone)

ethylamine 2-hydroxypropanoic acid(carboxylic acid) The hydroxy part of the name shows the presence of an -OH group. Normally, you would show that by the ending ol, but this time you can't because you've already got another ending.

Propanamide

Dimethylamine

Trimethylamine

esters 2-aminopropane

Ethanenitrile 2-hydroxypropanenitrile

ethanoyl chloride(acid chloride)

Alanine (2-aminopropanoic acid) An amino acid contains both an amino group, -NH2, and a carboxylic acid 1 group, -COOH Glycine(amino acid) where R=R =H

Grignard reagent

recognise oxidation, reduction, condensation, nucleophilic substitution or nucleophilic addition Redox (reduction oxidation reaction) describes /  describes / all all chemical reactions in which atoms have their oxidation number/state changed Oxidation state is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic Oxidation describes the loss of electrons of  electrons by a molecule, atom or ion (From +2 to +4) Reduction describes the gain of electrons of electrons by a molecule, atom or ion (From +4 to +2) Elimination reaction Elements of a simple molecule(H2O)are removed from the organic molecule and not replaced by any other atom/group of atoms Addition reaction 2 molecules react together forming a single product Electrophilic addition Addition reaction where, electrophile attacks a molecule at a region of high electron density Substitution reaction(atom/group reaction(atom/group of atoms in a molecule replaced by another atom/group of atoms) Nucleophilic substitution δ+ C atom can be attacked by a nucleophile  –  OH – , CN – , NH3 nucleophiles which react with haloalkanes C – Br Br bond breaks :OH provides a pair of electrons for C  –   –  heterolytically, both electrons from the bond taken by Br then OH bonds to C Nucleophile (literally nucleus lover as in nucleus and phile) is a reagent that forms a chemical bond to its reaction partner (the electrophile) by donating both bonding electrons Electrophile (literally electron-lover) is a reagent attracted to electrons that participates in a chemical reaction b y accepting an electron pair in order to bond to a nucleophile Condensation reaction is a chemical reaction in which two molecules or moieties combine to form one single molecule, with the loss of a small molecule. When this small molecule is water, it is known as a dehydration reaction; other possible small molecules lost are hydrogen chloride, methanol, or acetic acid. Compound Alkane Alkene

Haloalkane

Alcohol

Primary alcohol Secondary alcohol Tertiary alcohol Grignard reagent

Reagent Halogen Acidic(purple)KMnO4 Alkaline(purple)KMnO4 HBr H2SO4 Bromine water

NaOH(aq) or KOH(aq)

Product Haloalkane Alcohol (colourless) Alcohol (green) haloalkane Alcohol Decolourised from orange to colourless Dihaloalkane Alcohol

NaOH(ethanol) or KOH(ethanol) KCN(ethanol)

Alkene Nitrile

Dilute nitric acid, then silver nitrate

ppt of silver halide, white- chloride, yellow –  yellow – iodide iodide Carbon dioxide and water RCl + POCl3 + HCl Misty fumes of HCl which turn blue litmus red Haloalkane (green) aldehyde that will react with Tollens reagent to give a silver mirror (green) ketone will not react with Tollens reagent (orange) no reaction

Combustion PCl5

HX Acidified potassium dichromate (orange)

Reaction type Substitution Reduction Electrophilic addition Electrophilic addition Addition

Nucleophilic substitution Elimination Nucleophilic substitution

• Metal carbon bond has ionic character because electronegativities (ability to attract electrons) of metals are less than that of  carbon Ether must be perfectly dr y since water destroys resulting Grignard reagent dry ether, heat(reflux) C2H5I + Mg  C2H5MgI Halogenalkane ethyl magnesium iodide(Grignard reagent) Compound Reagent Product Reaction type Grignard reagent Water Alkane RH Nucleophilic substitution RMgX Carbon dioxide Carboxylic acid RCOOH Methanal HCHO Primary alcohol RCH2OH 1 Aldehydes R CHO Secondary alcohol 1 RCH(OH)R 1 2 1 2 Ketones R COR Tertiary alcohol RR R COH 1 1 Carboxylic acids RCOOH Alcohol R OH Ester RCOOR Nucleophilic substitution followed by elimination LiAlH4 Alcohol RCH2OH Reduction PCl5 Acid chloride RCOCl Nucleophilic substitution + Na2CO3 and NaHCO3 Sodium salt RCOO Na Acid-base CO2 gas(gives white ppt with limewater) 1 1 Esters RCOOR Aqueous mineral acid eg HCl(aq) Alcohol R OH and acid Hydrolysis (equil) RCOOH 1 NaOH(aq) Alcohol R OH and Hydrolysis (equil) + salt RCOO Na Aldehydes RCHO or Hydrogen Cyanohydrin RCH(OH)CN or Nucleophilic substitution 1 1 ketones RCOR cyanide(HCN(covalent)) and RR C(OH)CN potassium cyanide 2, 4-dinitrophenylhydrazine 2, 4-dinitrophenylhydrazine Nucleophilic substitution Test for carbonyl(C=O) group (Orange ppt) followed by elimination Sodium borohydride NaBH 4 or Primary alcohol RCH2OH or Reduction lithium aluminium hydride secondary alcohol 1 LiAlH4 RCH(OH)R Aldehydes RCHO (not Ammonical silver nitrate solution Silver mirror Reduction of the silver ion ketones) (Tollens reagent) Test for CHO group  Fehling’s solution/Benedicts Copper(I) oxide ppt (Red) Reduction of the copper(II) solution(Blue) ion potassium (green) dichromate(VI)(orange) Aldehydes RCHO acidic conditions Carboxylic acid RCOOH Oxidation alkaline conditions salt RCOO X ketone, 2° alcohol NaOH + I2 RCOONa + CHI3 Haloform (iodoform/yellow ppt) Acid chlorides ROCl Water Acid RCOOH Nucleophilic substitution Ammonia Amide RCONH2 1 1 Alcohol R OH Ester RCOOR 1 Amine R NH2 N- substituted amide 1 R CONHR + Amines RNH2 Aqueous acid eg HCl(aq) RNH3 Cl Acid-base 1 Acid chloride R OCl N-substituted amide Nucleophilic substitution 1 R CONHR Amides RCONH 2 Phosphorus(V) oxide P 4O10 Nitrile RCN Dehydration Bromine followed by NaOH(aq) Amine RNH2 Substitution followed by rearrangement and elimination Nitriles RCN Aqueous acid eg HCl(aq) Acid RCOOH Hydrolysis + NaOH(aq) Salt RCOO Na lithium aluminium hydride Amine RCH2NH2 Reduction LiAlH4 + Amino acids Aqueous acid eg HCl(aq) Salt RCH(NH3 )COOH Acid-base + + RCH(NH3 )COO NaOH(aq) Salt RCH(NH2)COO Na Questions State the reagents and conditions necessary to convert CH 3CH2CH(CONH2)CH3 to 2-aminobutane • heat • with bromine • and sodium hydroxide

Mg

CH 3 CH 2 CHBrCH 3

Grignard A

i) CO 2 ii) HCl(aq)

B

PCL5 CONH 2 CH 3 CH 2

C

C

CH 3

H D CH3CH2CH(MgBr)CH3 CH3CH2CH(COOH)CH3 CH3CH2CH(COCl)CH3 B C Reagents and conditions for CD • ammonia • and room temperature Butanone can be made from 2-bromobutane by a synthetic route involving two steps, the first using aqueous sodium hydroxide and the second potassium dichromate(VI) solution acidified with dilute sulphuric acid. (i) Give the structural formula of the intermediate compound in this synthetic route. A

CH3

CH

CH 2 CH3

OH (ii) Butanone reacts with 2,4-dinitrophenylhydrazine 2,4-dinitrophenylhydrazine solution but not with Fehling’s solution. Why is this? • contains C=O so reacts with 2,4 dnp • but cannot be oxidised so no reaction with Fehlings’ solution (iii)

Butanone also reacts with iodine in sodium hydroxide solution. What structural feature of butanone butanone is shown by this reaction?

CH3

C

CH 3 CH CH

if

O

OH

included then zero

(1) (iv) Give the structural formulae of both the organic products from the reaction in (iii). CHI3 and CH3CH2COONa (a) Write the structural formulae of the organic products obtained when ethanoyl chloride reacts with the following compounds. Give the names of these products. (i) Ammonia, NH3 (ii) Methanol, CH3OH. CH3CONH2 (b)

(i)

CH3COOCH3

Bromoethane reacts with magnesium to form the Grignard reagent CH3CH2MgBr. This Grignard Grignard reagent reacts with: CO2, followed by hydrochloric acid, to form compound A; water to form compound B; methanal, followed by hydrochloric acid, to form compound C. Compounds A and C react together, in the presence of a suitable catalyst, to form compound D. Write the structural formulae of compounds A, B, and C. A = CH3CH2COOH or C2H5COOH B = CH3CH3 C = CH3CH2CH2OH or C2H5CH2OH

(ii)

Draw the full structural formula of compound D

H

H

H

O

C

C

C

H

H

O

H

H

H

C

C

C

H

H

H

H

(iii)

Give the names of compounds C and D C = propan-1-ol D = propyl propanoate

(iv)

Identify a catalyst for the reaction between compounds A and C

(a)

sulphuric acid / phosphoric acid / hydrochloric acid Write equations to show the reactions of the amino acid alanine, CH 3CH(NH2)COOH, with:



CH3CH(NH3+Cl)COOH

(i)

HCl

CH3CH(NH2)COOH + HCl

(ii)

NaOH

CH3CH(NH2)COOH + NaOH

(b)

Explain why alanine has a relatively high melting temperature (290 °C)  Exists as zwitterion  Strong attraction between oppositely charged ions

(c)

Explain why alanine exists as two optical isomers Draw diagrams to show the structures of the two optical isomers.

COOH C CH 3



CH3CH(NH2)COONa+ + H2O

COOH

H

C

H

NH 2

CH 3

NH 2

(ii)

Explain how separate pure samples of each optical isomer can be distinguished from each other.  Rotates the plane of (plane) polarised (monochromatic) light in opposite directions  measure rotation (of plane of polarised light) in opposite directions

(d) (i)

A mixture of isomeric alkenes is obtained when butan-2-ol is dehydrated. Draw diagrams to show the two structural isomers obtained when butan-2-ol is dehydrated.

H

H

H

H

H

C

C

C

C

H

H

H

H

H

H

H

H

C

C

C

C

H

but-1-ene

H

H but-2-ene

(ii) One of the above structural isomers can itself exist as two different stereoisomers. Draw diagrams to clearly illustrate these two two stereoisomers, and name this type of stereoisomerism.

H3 C

CH 3 C

C

H 

geometric

H+J

H

CH 3 C

NaOH

O

Cl

Reagent 2

HCN NaCN

CH 3

H

K Reagent 1

CH 3

C CH 3

C

trans-but-2-ene

CH 3 propanone CH 3

C

cis- but-2-ene I2

H

H3 C

OH C

C

O

CH 3

COOH

Cl Reagent 3

CH 3

Cl C

CH 3

C

O

Br2 NaOH

M

NH 2 (a)

Give the structural formula of: H J K M

CH3

CH3 H is CHl3 J is CH3 COONa/CH3COO – 

K is

CH3

C CN

(b)

OH

M is

CH 3

C Cl

Identify: Reagent1/2/3  Reagent 1 Named dilute acid e.g. HCl(aq) or NaOH (aq) then add HCl  Reagent 2 PCl5 / SOCl2 / PCl3  Reagent 3 (Conc) ammonia (solution) / NH3

CH3 NH 2

or

CH3

C NH 2

OH

(1)

(c)

Compounds produced when glucose C6H12O6, is metabolised include: CH2(OH)CH(OH)CHO CH3COCOOH CH3CH(OH)COOH 2,3-dihydroxypropanal 2-oxopropanoic acid 2-hydroxypropanoic acid

(i)

Draw the full structural formula for 2,3-dihydroxypropanal.

C

H

H

H

C

C

C

O

OH OH

(1) (ii) Suggest two of these compounds which would give a positive test with 2,4-dinitrophenylhydrazine solution. State what would see for a positive te st result. yellow

2,3-dihydroxypropanal and 2-oxopropanoic acid (iii)

/ orange / orange – red red ppt / solid / crystals

Describe a test which would enable you to distinguish between the two compounds identified in part (ii). Add Fehlings’ solution/ Benedicts’ solution red/orange ppt for2,3-dihydroxypropanal and no result for 2-oxopropanoic acid Add ammoniacal silver nitrate silver mirror for2,3dihydroxypropanal and no result for 2-oxopropanoic acid Add named carbonate effervescence/ bubbling for 2-oxopropanoic acid and no result for2,3-dihydroxypropanal Add iodine + sodium hydroxide solution / Kl + NaClO yellow ppt for 2-oxopropanoic acid and no result for 2,3 -dihydroxypropanal

Add dilute sulphuric acid + po tassium dichromate dichromate goes green for 2,3-dihydroxypropanal and no result for 2-oxopropanoic acid Draw two diagrams to clearly represent the optical isomers that result from the chirality of this alcohol C4H9OH

C2 H5 C CH 3 (b) (ii)

(iii) (a)

C2 H 5

OH

C

HO

H

Alcohols react with carboxylic acids to form esters. Write an equation for a typical esterification reaction. CH3COOH + C2H5OH CH3COOC2H5 + H2O Suggest how this type of reaction could be used to form polyesters Alcohol group at one end and acid group at the other React at each end Give another type of reagent that could be used to make an ester from an alcohol Acyl chlorides / acid chlorides / acid halides / RCOCl / acid anhydride (i) Give the structural formula of a nitrile, C4H7N, that has an unbranched chain.

H (ii) (iii)

H

H

H

C

C

C

H

H

H

C

N

(1)

Primary amines can be made by reducing nitriles. Suggest a reagent that could be used for this purpose. LiAlH4 Draw the structural formula of the amine produced by reducing the nitrile given in (a)(i).

H (b)

CH3

H

H

H

H

H

C

C

C

C

H

H

H

H

H (1)

N H

Draw the structure of an isomer of C4H11N which has a chiral centre in the molecule and identify the chiral centre

H H H

H C H

H

C

C

C* N

H

H

H

H H

you

(c) (i)

What feature feature of an amine molecule makes it both a base and a nucleophile? lone pair of electrons on the N atom

(ii)

Give, by writing an equation, an example of an amine acting as a base. C4H9NH2 + H+  C4H9NH3+

(d) (i)

Ethanoyl chloride, CH3COCl, reacts with both amines and alcohols. Give the name of the type of compound produced when ethanoyl chloride reacts with ethylamine, C 2H5NH2 Amide

(ii) State one of the advantages of reacting ethanoyl chloride with ethanol to make an ester rather than reacting ethanoic acid with ethanol. faster / more control / better yield / not equilibrium / no need to heat (e)

Ethanoyl chloride can be made from ethanoic acid.Suggest a reagent suitable for this conversion. PCl5 / PCl3 SOCl2

(ii)

Suggest how chloromethane can be converted into ethanoic acid via a Grignard reagent.

CH 3 Cl

Mg(1)/dry ether(1)

CH 3 MgCl

CO 2 (1) acid (1)

CH 3 COOH

 HCHO

acid (1) CH 3 CH 2 OH K 2 Cr 2 O 7 H 2 SO 4

(1) for all of  the steps and reagents apart from acid which is stand alone

CH 3 CO2 H (a) Three isomers A, B and B have the molecular formula C4H8O. All three compounds give an orange precipitate with 2,4-dinitrophenylhydrazine reagent. B and C also give a silver mirror when warmed with ammoniacal silver nitrate solution. Write the structural formulae of A, B and C . A is CH3CH2COCH3 B and C are CH 3CH2CH2CHO and CH3CH(CH3)CHO (b) (ii) (iii) (c)

Substance A reacts with the Grignard reagent, C 2H5MgBr. Give the equation for the preparation of this Grignard reagent. C2H5Br + Mg  C2H5MgBr State the conditions for this preparation dry ether Write the structural formula of the product obtained when this Grignard reagent reacts with substance A. CH3CH2C(OH)(CH3) C2H5 C2H5MgBr reacts with carbon dioxide to form t he acid C 2H5COOH, converted to propanamide in a t wo step process.

step 1 C2H5COOH step 1 (d) (i)

CH 3

step 2

C2H5COCl PCl5 /PCl3 SOCl2

C2H5CONH2 step 2 NH3

State the reagent required for each step

C2H5MgBr also reacts with ethanal to for m substance D, which exists as a pair of optical isomers Draw the structural formulae of these two isomers and use them to explain why these isomers exist

C2 H5

C2H5

C

C OH

H

HO H

CH 3 Has asymmetric carbon atom (4 diff groups on a carbon) mirror image non-superimposable

(ii) What is the difference in property between these isomers? rotate the plane of plane polarised light in opposite directions (e)(i) Write down down the name name and the structural structural formula of the organic compound compound formed when substance D is heated under reflux with a solution of potassium dichromate(VI) in dilute sulphuric acid. Name and Structural formula. State the colour of the solution remaining after this reaction Butanone

CH3CH2COCH3

green

(b)

(i)

H

Draw the structural formula of the secondary alcohol, C5H12O, which does NOT exist as optical isomers.

H

H

H

H

H

C

C

C

C

C

H

H

OH

H

H

H

(ii) X is obtained by oxidising this secondary alcohol with potassium dichromate(VI) acidified with dil ute sulphuric acid. Draw the structural formula of X.

H (v)

H

H

C

C

H

H

H

H

C

C

C

O

H

H

H

precipitate when treated with iodine in the presence of sodium hydroxide solution. solution. Explain why not X does not give a yellow precipitate no CH3CO / CH3CH(OH) / methyl ketone / methyl secondary alcohol present

C 3H 8O A

K 2 Cr 2O 7 in dilute sulphuric acid

C3 H6 O B

 A Propan-2-ol acid

conc sulphuric acid

C

•None of the compounds in the scheme shows cis -trans isomerism •D reacts with KCN to form 2-methylpropanonitrile •An isomer of A of  A will form C by the same route but will not p roduce B by reaction with potassium dichromate(VI) acidified with (dil) H 2SO4 Instead it makes E, C 3H6O2 Identify using a name or structural formulae:A,B,C,D,E  B Propanone

C  Propene

 D 2-bromopropane

 E  Propanoic

D

HBr Ethanol can be converted into ethylamine by two different routes.

(a) Identify organic compounds V and W by writing their full structural formulae showing all bonds.

(b) Identify the reagents used in Steps A to E

Step A: NH3 Step B: K2Cr2O7 and H2SO4 Step D: P2O5 OR P4O10 Step E: LiAlH4

Step C: PCl5 OR SOCl2 OR PCl3

(c) (i) What type of organic compound would be for med when ethylamine, CH 3CH2NH2 reacts with ethanoyl chloride, CH 3COCl? (N-substituted) amide (ii) A polymer is formed when the two monomers shown below react together under suitable conditions. H2N(CH2)6NH2 ClOC(CH2)4COCl Draw sufficient of the polymer chain to make its structure clear.

An organic compound, A, with molecular formula C5H10O contains a carbonyl group. (a) Compound A is reacted with iodine in the presence of alkali. A pale yellow precipitate forms. (i) What is the formula of this precipitate? CHI3 (ii) What does this reaction indicate about the structure of  A? methyl ketone (iii) Compound A has a branched carbon chain. Draw the structural formula and give the name of  A

methylbutanone pr oduce sodium pentanoate and a red precipitate. (b) Pentanal is a structural isomer of  A. When heated with Fehling’s solution, it reacts to pr oduce (i) Identify the homologous series to which pentanal belongs. aldehyde (ii) Suggest the identity of the red precipitate formed in this reaction. copper(I) oxide (c) State a reagent which could be used to convert the sodium pentanoate made in the reaction above into pentanoic acid. Conc HCl/conc H 2SO4 (d) Solid sodium hydrogencarbonate, NaHCO 3, is reacted with excess concentrated pentanoic acid solution. (i) State what you would see as this reaction proceeds. effervescence/fizzing/bubbles (ii) Write a balanced chemical equation for this reaction CH (CH ) COOH + NaHCO → CH (CH ) COONa + H O + CO 3

2 3

3

3

2 3

2

2

3. Two compounds, A and B, are isomers with molecular formula C4H8O. • A and B give an orange-yellow precipitate with 2,4-dinitrophenylhydrazine. • Both compounds react with sodium borohydride (sodium t etrahydridoborate(III)). • When the compounds are warmed separately with Fehling’s solution, A forms a red-brown precipitate but B does not. • Compound B forms yellow crystals when warmed with aqueous iodine and sodium hydroxide, whereas A does not. (a) Draw full structural formulae for A and B, showing all bonds.

A B Give name and formula for the organic product of the reaction between compound B and sodium borohydride in water butan(-)2(-)ol CH 3CH (OH )C 2 H 5 (e) Compound C has the molecular formula C4H8O. • When phosphorus pentachloride, PCl 5, was added to a dry sample of  C, steamy fumes were observed. • When bromine water was shaken with a sample of  C, the bromine water turned colourless. • Compound C can be oxidised to a carboxylic acid which has a geometric isomer. Use the information above to draw the formulae of the two i somers which could be compound C

Cinnamaldehyde (a) To show the presence of the carbonyl group, a few drops of a solution of 2,4-dinitrophenylhydrazine are added to a sample of  cinnamaldehyde. (i) What observation is made in the reaction above? yellow / orange/red and precipitate / crystals / solid (iii) Suggest a further reaction, including the result, to show that cinnamaldehydecontains an aldehyde group. (warm with) Fehling’s/Benedict’s solution, sol ution, red ppt (iv) Why does the reaction you have given in (iii) not give a positive result with a ketone? ketone cannot be oxidised (b) Cinnamaldehyde can be converted into co mpound A

(i) Give the reagents and conditions which bring about this conversion. HCN+base or KCN+acid (ii) State, with a reason, how many stereoisomers exist for compound A. Four (c) Compound A reacts with lithium tetrahydridoaluminate(III), LiAlH4. The mixture is then treated with dilute acid to give the final organic product. (i) Name the type of reaction occurring occurring between compound A and LiAlH4. reduction (ii) Draw the structural formula of the final organic product. C H CH=CHCH(OH)CH NH 6

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(d) Cinnamaldehyde reacts with the Grignard reagent ethyl magnesium bromide, C 2H5MgBr. (iii) Draw the structural formula of product formed when cinnamaldehyde reacts with C2H5MgBr, and the intermediate is hydrolysed. C H CH=CHCH(OH)CH CH 6

(iv) State the type of alcohol formed in (d)(iii)

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Secondary