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Unit 4.5 Organic chemistry II
Recognise structural isomers, stereoisomerism as geometrical (cis-trans) or optical
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, side chains/functional
groups are in different positions on the carbon chain
Differ in physical properties
Functional group isomers Same atoms arranged into different functional groups
Differ in physical & chemical properties
2 bonds in the C=C double bond are not the same, bond energy of C=C (612kJmol –1) is greater than C-C (348kJmol –1) but not as twice as b
hence pi bond is weaker than the sigma bond
Stereoisomerism Molecules have same molecular formula, same structural
Stereoisomerism found in any molecule of the type:
formula, but atoms have a different 3d arrangement(orientation in space).
Differ in physical properties
Geometric Isomerism Occurs when there’s restricted rotation about a bond(C=C double bond where each of the two C atoms carries 2
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 b
Optical Isomerism Where molecules(chiral molecules) have mirror-image isomers that are not superimposable on the original
compound. Sole criterion for 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 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 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 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? Rotation of plane polarised 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 electrons more towards, itself and acquire a partial negative charge
• Electronegativity decreases going down a group(most electronegative element is fluorine)
• Electronegativity increases 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 stable with large cations
Small cations most stable with small anions
- A covalent bond is polar if electrons in the bond are unequally shared
formation of polyesters and polyamides
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 primary 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 zwitterion structures.
iodine in the presence of alkali (or potassium iodide and sodium chlorate(I)),
(sodium borohydride) sodium tetrahydridoborate(III)
(lithium aluminium hydride) lithium 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 atom)
R1-CHO R1 can be a H atom
R1-CO-R2
R1 & R2 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 – 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 CH3
An ethyl group is CH3CH2
2-methylpentane
2,3-dimethylbutane
2,2-dimethylbutane
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
1,1,1-trichloroethane
pentan-3-one
ethylamine
Propanamide
propanone(ketone)
Dimethylamine
Trimethylamine
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.
2-aminopropane
esters
Ethanenitrile
2-hydroxypropanenitrile
ethanoyl chloride(acid chloride)
Alanine (2-aminopropanoic acid)
An amino acid contains both an amino group, -NH2, and a carboxylic acid
group, -COOH
Glycine(amino acid) where R=R1=H
Grignard reagent
recognise oxidation, reduction, condensation, nucleophilic substitution or nucleophilic addition
Redox (reduction oxidation reaction) describes / 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 by a molecule, atom or ion (From +2 to +4)
Reduction describes the gain 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 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
:OH– provides a pair of electrons for C
C–Br bond breaks
–
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 by 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
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 – iodide
Carbon dioxide and water
RCl + POCl3 + HCl
Misty fumes of HCl which turn blue
litmus red
Haloalkane
Combustion
PCl5
HX
Reaction type
Substitution
Reduction
Electrophilic addition
Electrophilic addition
Addition
Nucleophilic
substitution
Elimination
Nucleophilic
substitution
Primary alcohol
Secondary alcohol
Tertiary alcohol
Acidified potassium dichromate
(orange)
(green) aldehyde that will react with
Tollens reagent to give a silver mirror
(green) ketone will not react with
Tollens reagent
(orange) no reaction
Grignard reagent
• Metal carbon bond has ionic character because electronegativities(ability to attract electrons) of metals are less than that of
carbon
Ether must be perfectly dry 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
Aldehydes R1CHO
Secondary alcohol
RCH(OH)R1
1
2
Ketones R COR
Tertiary alcohol RR1R2COH
1
Carboxylic acids RCOOH
Alcohol R OH
Ester RCOOR1
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)
Esters RCOOR1
Aqueous mineral acid eg HCl(aq) Alcohol R1OH and acid
Hydrolysis (equil)
RCOOH
NaOH(aq)
Alcohol R1OH and
Hydrolysis (equil)
salt RCOO-Na+
Aldehydes RCHO or
Hydrogen
Cyanohydrin RCH(OH)CN or Nucleophilic substitution
ketones RCOR1
cyanide(HCN(covalent)) and
RR1C(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 NaBH4 or
Primary alcohol RCH2OH or
Reduction
lithium aluminium hydride
secondary alcohol
LiAlH4
RCH(OH)R1
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
Alcohol R1OH
Ester RCOOR1
Amine R1NH2
N- substituted amide
R1CONHR
Amines RNH2
Aqueous acid eg HCl(aq)
RNH3+ClAcid-base
1
Acid chloride R OCl
N-substituted amide
Nucleophilic substitution
R1CONHR
Amides RCONH2
Phosphorus(V) oxide P4O10
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
RCH(NH3+)COO-
Aqueous acid eg HCl(aq)
NaOH(aq)
Salt RCH(NH3+)COOH
Salt RCH(NH2)COO- Na+
Acid-base
Questions
State the reagents and conditions necessary to convert CH3CH2CH(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
PCL 5
CONH 2
CH 3 CH 2
C
CH 3
C
H
D
A
CH3CH2CH(MgBr)CH3
B
CH3CH2CH(COOH)CH3
C
CH3CH2CH(COCl)CH3
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.
CH3
CH
CH2 CH3
OH
(ii)
Butanone reacts with 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 is shown by this reaction?
CH3
C
CH 3 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 CH 3CH2MgBr.
This 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)
(i)
(ii)
sulphuric acid / phosphoric acid / hydrochloric acid
Write equations to show the reactions of the amino acid alanine, CH 3CH(NH2)COOH, with:
HCl
CH3CH(NH2)COOH + HCl  CH3CH(NH3+Cl)COOH
NaOH
CH3CH(NH2)COOH + NaOH  CH3CH(NH2)COONa+ + H2O
(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
COOH
H
C
H
NH 2
NH 2
CH 3
(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
stereoisomers, and name this type of stereoisomerism.
H3 C
CH 3
C
C
H
 geometric
H+J
H
Cl
CH 3
C
O
HCN
NaCN
Reagent 2
K
Reagent 1
CH 3
C
CH 3
H
OH
C
C
O
CH 3
COOH
Cl
Reagent 3
CH 3
Cl
C
CH 3
C
O
Br2
NaOH
M
NH 2
(a)
C
CH 3
trans-but-2-ene
CH 3
propanone
CH 3
C
cis- but-2-ene
I2
NaOH
H
H3 C
Give the structural formula of: H J K M
CH3
CH3
H is CHl3 J is CH3 COONa/CH3COO–
K is CH
3
C
OH M is CH 3
C
CH3
NH 2
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
(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
C
C
C
CH3
Cl
CN
(b)
H
or
C
OH
(1)
NH 2
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 test result.
yellow / orange / orange–red ppt / solid / crystals
2,3-dihydroxypropanal and 2-oxopropanoic acid
(iii)
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 + potassium 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 C 4H9OH
C2 H5
C
CH3
(b)
(ii)
(iii)
(a)
OH
C2 H5
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)
CH3
H
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).
you
H
(b)
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
(c) (i)
What 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, C2H5NH2
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 CH3CH2CH2CHO and CH3CH(CH3)CHO
(b)
(ii)
(iii)
(c)
Substance A reacts with the Grignard reagent, C2H5MgBr. 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 the acid C2H5COOH, converted to propanamide in a two step process.
step 1
C2H5COOH
step 1
(d)
(i)
C2H5COCl
PCl5 /PCl3 SOCl2
step 2
C2H5CONH2
step 2 NH3
State the reagent required for each step
C2H5MgBr also reacts with ethanal to form 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
CH 3
C2 H5
C2H5
C
C
OH
HO
H
CH 3
H
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 the name and the structural formula of the organic 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
(i)
Draw the structural formula of the secondary alcohol, C5H12O, which does NOT exist as optical isomers.
(b)
H
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 dilute sulphuric acid.
Draw the structural formula of X.
H
H
H
C
C
H
H
H
H
C
C
C
O
H
H
H
(v)
X does not give a yellow precipitate when treated with iodine in the presence of sodium hydroxide solution. Explain why not
no CH3CO / CH3CH(OH) / methyl ketone / methyl secondary alcohol present
•None of the compounds in the scheme shows cis-trans isomerism
K 2 Cr 2O 7 in
•D reacts with KCN to form 2-methylpropanonitrile
dilute sulphuric
C 3 H 8 O acid
C 3 H 6 O •An isomer of A will form C by the same route but will not produce B by reaction with
potassium dichromate(VI) acidified with (dil) H2SO4 Instead it makes E, C3H6O2
A
B
Identify using a name or structural formulae:A,B,C,D,E
A Propan-2-ol
acid
conc sulphuric acid
C
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 formed when ethylamine, CH 3CH2NH2 reacts with ethanoyl chloride, CH3COCl?
(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
(b) Pentanal is a structural isomer of A. When heated with Fehling’s solution, it reacts to produce sodium pentanoate and a red precipitate.
(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 H2SO4
(d) Solid sodium hydrogencarbonate, NaHCO3, 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 tetrahydridoborate(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
CH3CH(OH)C2H5
(e) Compound C has the molecular formula C4H8O.
• When phosphorus pentachloride, PCl5, 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 isomers 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, 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 compound 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), LiAlH 4. The mixture is then treated with dilute acid to give the final
organic product. (i) Name the type of reaction 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
5
2
2
(d) Cinnamaldehyde reacts with the Grignard reagent ethyl magnesium bromide, C2H5MgBr.
(iii) Draw the structural formula of product formed when cinnamaldehyde reacts with C 2H5MgBr, and the intermediate is hydrolysed.
C H CH=CHCH(OH)CH CH
6
(iv) State the type of alcohol formed in (d)(iii)
5
2
3
Secondary