2010 H2 Revision Package for BT2 Part 1 Solutions A. Hybridisation 1 (a) A bond is formed from the end-on/head-on overlapping of orbitals. A bond is formed from the side-on overlapping of p-orbitals. b)i) sp 2 and sp 3 hybridisation ii) 8 bonds and 1 bond 2 Carbon atom C X Type of hybridisati on sp 2 Description of hybridisati on One s and two p orbitals take part in the mixing process Three sp 2 hybridised orbitals are formed. H H H H H H
Transcript
2010 H2 Revision Package for BT2 Part 1 Solutions
A. Hybridisation1 (a) A bond is formed from the end-on/head-on overlapping of orbitals.
A bond is formed from the side-on overlapping of p-orbitals.
b)i) sp 2 and sp 3 hybridisation
ii)
8 bonds and 1 bond
2 Carbon atom CX
Type of hybridisation sp2
Description of hybridisation
One s and two p orbitals take part in the mixing process
Three sp2 hybridised orbitals are formed.
Carbon atom CY
Type of hybridisation sp3
Description of hybridisation
One s and three p orbitals take part in the mixing process
Four sp3 hybridised orbitals are formed.
3 sp2: 5
H
H H
HH
H
sp3: 3
4 i) sp2 hybridisation
ii) 120o
B. Explanation Based QuestionsI Comparison of acidity and basicity
5TJC 2006
CH3 alkyl groups are electron donating groups which increases the strength of the O-H bond hence a proton is lost less easily. Furthermore, the phenoxide ion (conjugate base) is destabilised as the electron density on the O atom is increased.
NO2 groups are electron withdrawing groups which decreases the strength of the O-H bond hence a proton is lost more easily. Furthermore, the phenoxide ion (conjugate base) is stabilised as the electron density on the O atom is effectively dispersed.
6RJC2006
(a) Lone pair of electrons on N atom interacts with the pi electrons of the adjacent C=O bond and the benzene ring. Thus, the lone pair of electrons are not available for coordination to a proton.
(b) The acidity of an acid is determined by the stability of the conjugate base (anion).
In phenoxide ion, the negative charge is delocalised into the benzene ring, hence stabilising the conjugate base.
In CH3COO-, the negative charge on oxygen is delocalised into the two electronegative O atoms resulting in a resonance effect. Thus, the ethanoate ions are more resonance stabilised than the phenoxide ion.
Hence, the position of equilibrium 1 shifts more to the right to produce more H3O+ ions.
7IJC 2007
8NYJC2008
Tertiary amines are less basic than primary amines due to the bulky substituents hindering the formation of dative bond.
The lone pair on N of the heterocyclic aromatic ring is found in sp2 orbital which is planar with the aromatic ring. Hence it is not delocalised into the aromatic ring.
The lone pair on N of phenylamine is delocalised into the aromatic ring, hence
less available to accept protons.9DHS2008 (i)
(ii) is more basic than
Presence of electron–donating alkyl group increases the availability of lone pair of electrons on the nitrogen atom for dative bonding with proton. The positive charge on the conjugate acid is dispersed hence stabilising the cation relative to the base.
In 4–aminobenzoic acid, the lone pair of electrons on the nitrogen atom is delocalised into the electron cloud in benzene. This reduces the availability of lone pair of electrons for dative bonding with proton, making it less basic.
10TPJC2007
< <
The stronger the base, the larger the Kb value. Basicity depends on the availability of the lone pair of electrons on N atom for donation to a proton.
II Comparison of ease of hydrolysis
11RJC 2006
Rate will increase when 2-iodobutane is used. C- I bond is weaker than the C – Cl bond due to less effective atomic orbital overlap in the former. Rate of reaction is dependant on the ease of breaking of the carbon – halogen bond.
Availability of the lone pairs of electrons enhanced by electron-releasing alkyl –CH2- group
Lone pair of electrons is delocalised into the benzene ring, so less available for donation
CH3CO- group is electron withdrawing, enhancing the delocalization of the lone pair of electrons on N atom into the ring, so even less available for donation to a proton.
12HCI2006
13AJC2001
Iodocyclohexane is the most susceptible to hydrolysis as the C – I bond is weaker than the C – Cl bond in cyclohexane due to less effective atomic orbital overlap in the former (Iodine is bigger in atomic radius than chlorine). Hence the C-I bond is easier to break than C-Cl bond.
In Chlorobenzene, the one pair of electrons on Cl is delocalised into the benzene ring as the p orbitals of Cl overlaps with the pi electron system of the benzene ring. Hence, the C-Cl bond is stronger and making the carbon of the C-Cl bond less electron deficient. Hence, substitution is very difficult.
III. Nitrogen Compounds
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15
16 (a) (i)
(ii) Side chains of amino acid A are most likely to be involved. An increase in pH causes –CH2COOH to be deprotonated to form –CH2COOH + OH- à -CH2COO- + H2O
(iii) amino acid B: The side chain in B forms disulhide bridge, -CH2-S-S-CH2-
amino acid D: The side chain in D forms hydrogen bonding between –CH2OH and other suitable R groups that are also capable to form hydrogen bonds (e.g.: -CH2OH or –COOH).
(b)à Protein undergoes denaturation in which there is a major change in the secondary, tertiary and quaternary structure of proteins.
à This is because during cooking, hydrogen bonds, electrostatic forces of attraction and Van der Waals forces are disrupted by heat.
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18
19 (a)
(b) β-pleated sheet consists of adjacent polypeptide strands stabilised by hydrogen
bonds between the backbone C=O group of one strand and the backbone N-H
group of the adjacent strand.
(c) Asparagine : Hydrogen bond; Glutamic acid : Ionic bond
(d) (i) (ii)
(e)
Nucleophilic Substitution
20(a)(i) Students can draw parallel or anti-parallel sheets.
Sketch showing- Mirror image of pleated sheets, with C=O mirroring N-H - Show and label hydrogen bonds
State R groups (side chains) project above or below the sheet and are 90º to the plane of
the pleated sheet.
(ii) In alkali medium, the some of the R groups are deprotonated which affect the electrostatic interactions between them, causing unfolding of the protein chain(OR for mentioning the disruption of tertiary structure of protein)
(iii) leu-ser-pro-ala-asp
(iv)
asp residue ser residue
21 ai)
ii)
b) Ala-His-His-Pro-Ser
ci) Hydrogen Bonding
O
C N
H
O
C N
H
Diagram: [1]
ii) At pH above and below 2, pepsin has denatured as the side chain groups with opposite
charges / the ionic interaction between the side chain residues of the amino acids are
broken which disrupts the tertiary structure / causes unfolding of the polypeptide
chain.
C. Distinguishing Test Questions
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23 (i)
(ii)
(iii)
24 A and BReagent Solid sodium carbonate / PCl3Condition Room TemperatureObservations Effervescence is observed in B which will form a white ppt in
limewater when the gas evolved is bubbled through. No effervescence is observed in A and no white ppt is formed with limewater. / White fumes are evolved in B while A will have not form white fumes.
B and CReagent Solid sodium carbonateCondition Room Temperature
Observations Effervescence is observed in B which will form a white ppt in limewater when the gas evolved is bubbled through. No effervescence is observed in B and no white ppt is formed with limewater.
A and CReagent NaOH (aq), followed by HNO3 (aq) and AgNO3 (aq)Condition HeatObservations White ppt observed in A whilst C has no white ppt formed.
25 (i) Left (A), Right (B)Reagent NaOH (aq)Condition HeatObservations Pungent gas evolved in A that will turn moist red litmus paper
blue while B will has no effect on moist red litmus paper.White ppt formed in A while B has not white ppt formed (Benzoic acid is a white ppt).
(ii) Left (A), Right (B)Reagent I2, NaOH (aq)Condition HeatObservations Yellow ppt formed in A while no yellow ppt formed in B.
(iii) Left (A), Right (B)Reagent NaOH (aq)Condition HeatObservations Pungent gas evolved in A that will turn moist red litmus paper
blue while B will has no effect on moist red litmus paper.
(iv) Left (A), Right (B)Reagent KMnO4 (aq) followed by I2, NaOH (aq) Condition Acidified and Heat followed by heatObservations Both A and B will decolourise purple KMnO4 (aq) (double bond
is cleaved). On reaction with I2 / NaOH, yellow ppt is observed in A but no yellow ppt formed with B.
(v) Left (A), Right (B)Reagent KMnO4 (aq) Condition Acidified and Heat.
Observations B will decolourise purple KMnO4 (aq) but purple KMnO4 (aq) remains with A.
(vi) Left (A), Right (B)Reagent H2O Condition ColdObservations A will form white fumes while B does not form white fumes
26 A: C6H5CH(OH)CH3 and B: phenylethanone
Reagent I2 / NaOH (aq) Condition HeatObservations Both A and B will form yellow ppt. (Iodoform test)
A: C6H5CH(OH)CH3 and B: phenylethanone
Reagent 2,4 dinitrophenylhydrazine Condition HeatObservations Orange ppt formed in phenylethanone whilst no orange ppt in A.
27(a) Phenol and benzoic acid can both react with NaOH (aq) whilst Phenylmethanol and cyclohexanol cant.
(b) A: phenol and B: benzoic acid
Reagent Solid sodium carbonate / aqueous bromineCondition Room temperatureObservations Effervescence is observed in B which will form a white ppt in
limewater when the gas evolved is bubbled through. A will not form effervescene and no white ppt formed in limewater. / Brown bromine decolourises and white ppt formed in A while brown colour remains in B.
A: phenylmethanol and B: cyclohexanol
Reagent K2Cr2O7 or KMnO4
Condition Acidified and HeatObservations Both A and B will turn orange dichromate to green. However, A
will give a white ppt but B does not.
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29
30
31
32 (a)
(b)
(c) Add aqeous bromine at room temperature. Compound A decolourises brown bromine water with the formation of a white ppt whilst Compound A does not decolourise brown bromine water and white ppt is not formed.
(d)
33 Test:Add KMnO4 in H2SO4(aq) to both compounds separately and heat. To the oxidised products, add 2,4-DNPH and warm. A: orange ppt B: no ppt
Alternative: (1)Use alkaline aqueous iodine, warm.(2)Cold alkaline KMnO4, followed by Cr2O7
2-
(3)H3PO4, 300oC, 70atm, followed by Cr2O72-
(4)KMnO4 in H2SO4, passed CO2 gas produced into Ca(OH)2(aq), white ppt for A.34 Reagent(s) and Condition(s): ………… Na metal …………………...
Observations for :
Compound Y: …… no effervescence observed.Compound X: …… effervescence of hydrogen gas observed and it will extinguished burning splint with a ‘pop’ sound.…………
35 (i) Use aq I2, aq NaOH, warm. CCl3CH2COCH3: Yellow ppt formed. CCl3COCH2CH3: No ppt formed.
(ii) Use alcoholic AgNO3 or silver ethanoate, warm. (Aqueous AgNO3 not allowed)
No ppt formed. : White ppt (of AgCl) formed
36 Add aq Br2 to both compounds separately.salicylic acid: reddish brown aq Br2 will decolourise.aspirin: reddish brown aq Br2 will not decolourise.
37 Add dilute HCl(aq) and heat each mixture. Add excess NaOH(aq) followed by I2(aq) to each mixture and heat .
For G, a yellow ppt of CHI3 is formed.For H, no yellow ppt of CHI3 is formed.
38 (i) Add Tollens’ reagent to both compounds separately with warming . For the 1st compound (aldehyde), there would be a silver mirror formed . For the 2nd compound, there would be no silver mirror formed .
(ii) First, add dil H2SO4 (aq) to both compounds separately, and heat . Then, cool and add aqueous bromine to both samples.
For the 1st compound, the reddish-brown bromine remains, with the absence of the formation of any ppt. For the 2nd compound, there would be a decolourisation of reddish-brown bromine , with the formation of a white ppt.
Or Add K2Cr2O7 (aq), dil H2SO4, heatFor the 1st compound, orange solution turns green. For the 2nd compound, orange solution does not turn green.
39 (a)To each compound in separate test tubes, add NaOH(aq) and heat. Add Br2(aq) after cooling the mixture.
The test-tube containing A will decolourise reddish brown aqueous bromine with the formation of a white precipitate. Reddish brown aqueous bromine remains for test tube containing B.
(b)To each compound in separate test tubes, add 2,4-dinitrophenylhydrazine.
An orange precipitate will form in test-tube containing D and no precipitate observed for test tube containing C.
D. Structure Elucidation Questions
40 a) A is
b)
(Note: The positions of the side chains does not matter.)Since C reacts with methanol and hot conc. H2SO4 to give D which has sweet smell, D must be an ester and C must contain COOH.
c) E is CH3CONH2
Since E is neutral and has no reaction with aq alkali, it is not amine.
d) F is
Since upon hydrolysis, the compound C6H7N gives a white ppt with aq Br2, it must be phenylamine.
e) G is cyclohexane
Since G does not react with Br2 in CCl4, it must not contain any C=C.
41 i) A solution of H turns damp blue litmus paper red H is acidic, either RCOOH or phenol is present.H gives a yellow ppt. with iodoform test H contains methyl carbonyl.
H is
ii) When PCl5 is added to J, dense white fumes were observed J contains -COOH or
–OH, but not phenol
J forms a silver mirror with ammonical silver nitrate J contains aldehye J does not form any ppt. when Fehling’s solution J is a benzaldehydeJ does not form a yellow ppt. with aq. alkaline iodine J does not contain methyl carbonyl gp or methyl alcohol gp.
J is
42 i) Alcoholic sodium hydroxide and reflux/heat.ii) D B
Or
D B
iii)
any twoOr
any two
43 a) A has chiral carbon as it can rotate plane-polarised light. A contains alcohol / carboxylic acid as it reacts with PCl5. Since there are two Cl atoms in B, there must be two OH groups in A that are substituted. Since B reacts with water to give C, losing one Cl atom, there must be an acid chloride in B, hence A must contain one carboxylic acid and one alcohol.
Conversion of C to DNucleophilic substitution
b) Optical isomerism: Non superimposable mirror images which rotate plane-polarised light.
44 a)Atom C H O
Mole ratio35.8/12 4.5/1 59.7/16
2.98 4.5 3.731 1.5 1.254 6 5
E.F. = C4H6O5
b) The formula of B shows it contains five O. B besides being is a diacid, must contain ether or alcohol. Since B is a diacid, it reacts with ethanol to form diester with eight carbon atoms, B must have started with 4 carbon.
Since compound C reacts with sodium metal producing H2, it must contain –OH group. Hence B is a diacid + alcohol.
D must be a ketone since it gives positive test with 2,4-DNPH but negative test with Fehling’s solution or Tollens’ reagent.
C must contain secondary alcohol which is oxidized to ketone in D.
B C
D
45 ai)
K and L are interchangeableaii) K & L exhibit geometric/cis-trans isomerism.
K , L and M exhibit structural/positional isomerism.
b) Electrophilic addition As bromine molecule approaches the electrons in the C = C bond, it becomes polarised. The Br+ atom acts as an electrophile and adds to one of the C atoms forming a carbocation intermediate which then attracts the Br- ion forming the product
c) Diol product is formed. Purple manganate(VII) solution is decolourised, brown MnO2 ppt is
formed.
46 ai) Carbonyl compoundaii) Aliphatic aldehydeaiii) –OH group in alcohol and carboxylic acid (not phenol)
bi) Q is
Displayed formula Full structural formula
ci) Orange or brown Br2(aq) is decolourised, steamy fumes evolved and white ppt. formed.
cii)
ciii) Electrophilic substitution
47 ai)Atom C H O
Mole ratio 40.0/12 6.65/1 53.3/161 2 1
E.F. = CH2O
b) i) A contains chiral carbon presentii) A is acidic, contains carboxylic acidiii) A contains
group
c)
d)
e) Ester functional group present
f) Ethanamide, CH3CONH2(s) and ammonium ethanoate, CH3CO2NH4(s)
Dissolve the unknown solids in deionised water respectively.Add neutral FeCl3 solution to each solution in a test tube followed by boiling. The test tube which shows red colouration followed by brown precipitate on boiling is ammonium ethanoate, CH3CO2NH4. The test tube which shows no visible reaction is ethanamide, CH3CONH2.
48 ai)
aii) F would be more acidic than phenol because the two electron-withdrawing Cl atoms stabilise the phenoxide ion formed. They do this by drawing the electrons from the oxygen atom of the phenoxide ion into the delocalised electron system in the benzene ring. This makes the O—H bond more polar (weaken O- H bond) and easier to donate a proton and hence the acid strength increases.
bi) Since G reacts with NaOH(aq) losing both Cl atoms to form H, it must have hydrolysed and hence the two Cl atoms must be on aliphatic chain (not on benzene ring). J is insoluble in water but it dissolves in NaOH(aq) J must be acidic, still contains phenol. J reacts with 2,4-dinitrophenylhydrazine and with alkaline aqueous iodine, but not with
Fehling’s solution J is a methyl ketone or contain
bii) Cl2, u.v. light
49 F is insoluble in water, but dissolves in NaOH(aq) F contains phenolF reacts with 2,4-dinitrophenylhydrazine but not with Fehling’s solution F contains ketone or benzaldehydeWith bromine water, F gives G, C8H6O2Br2 (disubstitution) F contains phenol and has one of the positions 2 or 4 occupied. F reacts positively with iodoform test to give H, C7H6O3 F has methyl cabonyl group, CH3CO-H dissolves in both NaOH(aq) and Na2CO3(aq) H contains carboxylic acid
F G H
50 K is not very soluble in water, but dissolves in HCl(aq) K is basic, contains an amine. K dissolves in NaOH(aq), but not in Na2CO3(aq) K is acidic, it contains phenol but not carboxylic acid.L is no longer soluble in HCl(aq), but is still soluble in NaOH(aq) L is no longer basic, but is still acidic, the phenolic group is still present. N reacts with bromine water with 2 Br substituted L is phenolic and has its 2 or 4 position occupied. N is not soluble in either HCl(aq) or NaOH(aq) N is neutral, amine and phenol is no longer present.
51 H, J, K and L all show positive reactions with Na All four contain OH gpK reacts with NaOH: K is a phenol or is acidicK reacts with Br2(aq): K contains a phenol. K & L have no reaction with acidified K2Cr2O7: K & L are neither 10 ROH nor 20ROH, could be 30
ROHH reacts with acidified K2Cr2O7 to give M (C9H10O2): 10 ROH oxidised to acid J reacts with acidified K2Cr2O7 to give N (C9H10O): 20 ROH oxidised to ketone +ve reaction of J & I2/OH-: J has CH3CH(OH)- gp.H, J, & L have side-chain alkyl gp oxidised by hot alkaline MnO4
-
52 bi) No. of moles of NaOH = 0.1 x 6.85/1000 = 6.85 x 10-4 molZ is a monobasic acid.No. of moles of Z = 6.85 x 10-4 molMr of Z = (1.0 x 25/250) / 6.85 x 10-4 = 146If Z is RR’C(OH)CO2HC(OH)CO2H has Mr of 74, so R + R’ has Mr of 146 – 74 = 72R and R’ could be C5H11/H or C4H9/CH3 or C3H7/C2H5
bii) [H+] = 10-2.73 = 0.00186 mol dm-3
[Z] = 6.85 x 10-4 x 1000/25 = 0.0274 mol dm-3
Ka = = = 1.26 x 10-4 mol dm-3
biii) X is chiral X contains a C atom joined to 4 different groups/atomsX has no reaction with Na X has no –OH group (or not alcohol or acid)X has no reaction with Fehling’s solution X is not an aldehydeX forms ppt. with 2, 4-DNPH X must be a ketoneX forms yellow ppt. with I2/OH- X contains CH3CO- groupSince X is chiral and it contains CH3CO-, X must be
and
53 i) Since D is basic, D must contain an amine group.
Since D forms a white ppt. With Br2(aq), D must be phenylamine,
G + OH- à D
G must be an ammonium salt:
ii)
H must be CH3CH2COO-
E must be CH3CH2COCH3
iii) No. of moles of NaOH = 0.50 x 20/1000 = 0.01 mol
No. of moles of F = 0.005 mol
No. of moles of NaOH = 0.01 = 2 No. of moles of F 0.005 1
F must be a dibasic acid.
Since B à E + F + GB = C13H16NOCl, E = CH3CH2COCH3, G = C6H5-NH3
+, (E + G: 9 carbon atoms)F must contains the 4 carbon atoms with 2 COOH groups According to the Mr of F = 138.5.
F must be
iv) B à CH3CH2COCH3 + HOOC – CHCl – COOH + C6H5-NH3+
B is
v)
B undergoes alkaline hydrolysis to form C and
C is
54 a) Since mole ratio of C:H in Z 1:1, Z contains benzene ring.
Since Z is soluble in HCl Z contains
Since Z reacts with Br2(aq) C6H5NH2 is confirmed.
Z is either
Z + fuming H2SO4 + H2O X + Y optically active optically inactive +ve iodoform test
XY + MnO4
-/H+ A, C7H8O2NZ heat
X, Y & Z must have the same C number.
A is
i)
ii)
iii)
55 a)
P is an alkyl bromide that undergoes alkaline hydrolysis to give an alcohol group in Q.
Q contains an aldehyde group or 10 or 20 alcohol which undergoes oxidation by Cr2O72-.
R reacts with alkaline iodine to form yellow precipitate.R must contain CH3CO - group which gives a positive test with iodoform test.
R + PCl5 à dense white fumesR must have a 30 alcohol since it has already been oxidised.Q must contain 20 and 30 alcohols.
i)
ii)
56 a) Since Q has a positive iodoform test and negative test with Tollen’s reagent, Q must be contain
Since Q can be oxidised, Q is not a ketone and thus a methyl alcohol.
d) Electrophilic addition As bromine molecule approaches the electrons in the C = C bond, it becomes polarised. The Br+ atom acts as an electrophile and adds to one of the C atoms forming a carbocation intermediate which then attracts the Br- ion forming the product
57 i) Since J reacts with aqueous Cl2 forming K with 3 H substituted by 3 Cl, J must be contain a phenol with 2,4 and 6 positions unoccupied.J undergoes monosubstitution with Cl2 in CCl4 forming L, Since J contains 7 carbon and besides phenolic group, J must contain an aldehyde substituted on 3 position.J contains aldehyde which undergoes reduction with LiAlH4 in dry ether to give a primary alcohol which will react with PCl5 to give an alkyl halide. Since L does not undergo substitution with NH3 while M does, Cl in L will be on the ring while Cl in M is on aliphatic chain.
ii) K is acidic as the conjugate base formed is stabilized when the lone pair electrons on oxygen delocalised into the benzene ring. Furthermore, the 4 substituted groups on the ring are all electron withdrawing, further enhancing the stability of the conjugate base.
iii) The Cl in M pulls electrons away from the C bonded making the C electron deficient and hence attracts the lone pair electrons on NH3. Cl then leaves and nucleophilic substitution occurs.
The lone pair electrons on Cl in L delocalized into the benzeme ring, hence strengthening the C – Cl bond. Cl does not leave easily and hence substitution by NH3 is difficult and only occurs under harsh conditions (high P)
58 i)Atom C H O
Mole ratio32/12 4/1 64/16
2 3 3Empirical formula. = C2H3O3
Since molar mass of P is 150 g mol-1, let molecular formula of P be (C2H3O3)n
(12x2+3+16x3) n = 150n = 2Molecular formula = C4H6O6
ii)
Since P with 4 carbon, undergoes esterification with ethanol forming Q with 8 carbon, there should be two COOH groups in P.
Since Q is oxidisable by Cr2O72- to give R, which reacts with 2,4 DNPH to give orange ppt,
Q should contain 2o alcohol to give carbonyl. Since R has no reaction with Tollen’s reagent, R should not contain aldehyde.
Since Q evolves 2 moles of HCl with PCl5 to form S, Q contains 2 alcohol groups.Since S reacts with ethanolic KOH to give product which reacts with 2 moles of H2 during hydrogenation, the product must have two C=C double bonds and hence S must contain 2 Cl groups which can undergo elimination.
iii)
59
60
Compound A
Each structure carries 1 mark; total [9] ; max [8]
A must be an ester which undergoes alkaline hydrolysis with aqq NaOH to give carboxylate salt and alcohol.
B must be the alcohol and since it is positive towards iodoform test, it must contain methyl alcohol CH3CHOH-
B is oxidized by Cr2O72- producing C. B must be a 2o alcohol CH3RCHOH which is oxidized to
C which contains ketone which reacts with 2,4 DNPH is can be oxidized to a carbonyl, B is a secondary alcohol with CH3RCHOH D is ethanedioic acid since it is oxidized to CO2 and H2O by MnO4
-. D contains COOH which is converted to acid chloride with PCl5, forming HCl fumes E. B undergoes dehydration to form F which contains C=C. Vigorous oxidation of F brings about the double bond cleavage to form acid G. Reduction of acid G by LiAlH4 produce a primary alcohol, which undergoes substitution with
SOCl2 to form alkyl halide H. Alkyl halide H undergoes nucelophilic substitution with alcoholic KCN to form a nitrile, which is
reduced to give an amine I.
Compound BCompound C
Compound D
HClCompound E
Compound FCompound G
Compound H Compound I
61 ● mole ratio of C:H ratio 1:1 in X => X probably contains benzene ring
● X contains 3 O atoms but is neutral => X does not contain phenol or carboxylic acid groups
● X does not react with 2,4 DNPH => X is not a carbonyl compound
● X does not react with Br2 dissolved in CCl4 => it does not contain an alkene group
● Since X gives two products when it is oxidized under acidic conditions, but is not an alkene, it must have an ester linkage (heating with acid produces benzoic acid and alcohol, but the alcohol is further oxidized to a carboxylic acid)
● Y has positive tri-iodoform test and is a product of oxidation => Y has a methyl carbonyl group,
● Y reacts with Na2CO3 to give CO2 gas => Y has –COOH group
● Y is
● Since one of the products is benzoic acid, X must be
62C H Cl O
% mass 44 6.6 26 23.4Ar 12 1 35.5 16No. of mol 3.67 6.6 0.732 1.46Mole ratio 5 9 1 2
Information given Deduction1 % composition of each
given elements in GG has empirical formula of C5H9ClO2
G has molecular formula C5H9ClO2
2 G is soluble in aq NaOH, G contains –COOH group. (Phenol is not possible as only 5 C is present)Note: When no heating is involved for addition of acid/alkali, its purpose is to test for presence of basic/acidic/neutral compound)
3 When G is heated with aq NaOH followed by addition of aq HCl, H, C5H10O3 is obtained.
G, contains alkyl halide, undergoes nucleophilic substitution with aq NaOH to give H.with alcohol.
4 H gives a positive iodoform test.
H has a CH3CH(OH)- structure.
5 H when refluxed with conc H2SO4, give J which is a sweet smelling liquid.
J is an ester / H undergoes internal esterification, hence H must contain both –COOH and –OH group.
6 H also forms K when refluxed with conc H2SO4.
H undergoes elimination or dehydration to form K.
7 K exhibits cis-trans. K must contain C=C with each C bonded to different atoms or groups of atoms.
8 Since K reacts with bromine to give on a product with Br on adjacent C atoms,.
K must contain C=C which undergo electrophilic addition with Br2.
9 K reacts with acidified KMnO4 forming L, C3H4O4.
K undergoes strong oxidation, C=C is cleaved.
Structure of K (pt no 8) is
Therefore structure of H (pt no 5 and 4) and L (pt no 9) are
Structure of G (pt no 3) is Structure of J (pt no 6) is
Explanations required are: Functional group possible for such reaction or any unique structure is present Type of reaction undergone Whether acid/basic/neutral compound is present Write balanced equations if asked by the question.
63 C H O% mass 49.3 6.85 43.85Ar 12 1 16No. of mol 4.108 6.85 2.741Mole ratio 1.5 2.5 1
3 5 2
Empirical formula of P: C3H5O2
P undergoes basic hydrolysis with hot, aqueous sodium hydroxide to form Q and sodium ethanoate. P is an ester Q is an alcohol
Q undergoes nucleophilic substitution with HCl and ZnCl2 to form R. Q is an alcohol R is a halogenoalkane / chloroalkane
R undergoes nucleophilic substitution with ethanolic KCN to form S. S is a nitrile
S undergoes acid hydrolysis with aq. HCl to form T. T is a carboxylic acid
T is optically inactive T does not contain a chiral carbon.
U undergoes electrophilic addition with aq. bromine. U is an alkene / contains C=C bond
T undergoes oxidation to form V T contains an –OH group
V is a carboxylic acid
Structures (displayed formula not required):
P:
Q: R:
S: T:
U: V:
64 ● Mole ratio of C:H ratio 1:1 in A => A probably contains benzene ring
● A undergoes elimination with ethanolic NaOH forming B which decolourises aq Br2 => A probably contains alkyl halide
=> B probably contains C = C
● A contains alkyl halide which undergoes nucleophilic substitution with aq NaOH forming C, which contains alcohol and undergoes dehydration easily to form D.
● D gives orange ppt with 2,4 DNPH => D contains carbonyl
● A undergoes oxidation with MnO4- forming E with two O, very likely side chain
oxidation occurs to A, forming benzoic acid in E.
A is (CH3 can be anywhere on ring) B is
D is E is
65 R has a C=C. On vig oxidation, 2C must have been converted to CO2
S has a carboxyl fg (reaction with NaHCO3) and a ketone with the group CH3CO R to T is a nucleophilic substitution ( note same no. of H-atoms) T is reduced to U, a saturated amine V is a salt from neutralistion with HCl T is hydrolysed to a diacid.
The Mr of Q is 59.Let Q be (C3H9N)n. Then Mr of Q = (3n)(12.0) + (9n)(1.0) + (n)(14.0) = 59 59n = 59 i.e. n = 1Hence the molecular formula of Q is C3H9N.
(b)
Description Deductions
+ 2NH4+
+ 3OH- + 4[O]
+ 2H+ + 4H2O
P, C15H20O3NCl 6 degrees of unsaturation / high C:H ratio.Benzene ring may be present since P has at least 6 C atoms
P is neutral compound Amines, carboxylic acids, phenols absent OR amides, esters present
Q reacted with excess iodomethane to give T (Mr = 229).T gives a yellow precipitate with aqueous silver nitrate. Q is straightchain.
T is a quarternary ammonium iodide OR quarternary ammonium salt T can be [CH3CH2CH2N(CH3)3]+I T has Mr = (6)(12.0)+(16)(1.0)+(1)(14.0)+127 = 229 which agrees with the given value (show working) ORSince Mr of T is 229, three CH3 groups must be substituted on Q(show working : 22912757 = 45 = 3 x 15) Hence Q is a primary amine Since Q is a straightchain amine, Q can be CH3CH2CH2NH2.
R on acidification yields U, which is also obtained from acidic hydrolysis of P.
R is the sodium carboxylate U is a carboxylic acid
U reacts with a small amount of concentrated sulphuric acid to give γ-butyrolactone
U undergoes esterification (or intra-esterification) U must be a hydroxy acid U must be HO2CCH2CH2CH2OH R must be Na+ O2CCH2CH2CH2OH
S (C8H7O3Na), reacts with acidified potassium dichromate(VI) and heating with immediate distillation, to give V.
Treating V with 2,4-DNPH gives an orange precipitate which has the following structure:
Since V undergoes condensation with 2,4DNPH to give the hydrazone,
V must be an aldehyde with the structure
Since V is obtained from (controlled) oxidation of SS must be a primary alcohol with the structure
When P is refluxed with aqueous sodium hydroxide, three compounds Q, R and S are obtained.
P undergoes alkaline hydrolysis of the ester and amide groups and nucleophilic substitution of the Cl.P must be an ester, amide and chloroalkane.
P :
68 V has a high C:H ratio, it is likely to contain a benzene ring.
V does not dissolve in NaOH, hence it is not acidic (It does not contain a carboxyl group or a phenol group).
V does not give orange precipitate with Brady’s reagent, hence V is not a carbonyl
compound.
V reacts with cold, dilute acidic KMnO4, therefore V is an alkene and W is a diol.
V undergoes reduction with LiAlH4 to give 2 products. V is an ester. X and Y are primary alcohols.
X gives a positive triiodomethane test, it contains – CH(OH)CH3 group.
Y decolourises aqueous bromine to give a whit precipitate Z (C10H10Br4O2). Y is a phenol and only the 3rd carbon with respect to the –OH group is substituted. Y is also an alkene.
V, W, Y and Z have chiral carbons, non-superimposable mirror images and no plane of symmetry.
V: W: X: CH3CH2OH
Y:
(ii) Z:
69
E. Conversion Based Questions
70
(b) C2H5OH + conc. H2SO4, heat
(c) CH3CONH2 + NaOH(aq) CH3COOH + NH3(g)
heat NH3 gas turns moist red litmus paper blue and forms white fumes with a glass rod dipped in concentrated HCl.
CH3CH2NH2 + NaOH(aq) No reaction. HeatMoist red litmus paper remains red and no fumes are formed with concentrated HCl.
71
Step 1 = KOH (ethanolic), heatStep 4 = Sn, conc. HCl, followed by excess NaOH
(iii) Add Br2(aq) at r.t.p. to chloramphenicol and X respectively. X will form a white ppt, steamy fumes evolved and Br2(aq) is decolourised whilst chloramphenicol does not give a white ppt. Type of reaction involved is electrophilic substitution.
(b)
75 1: NaOH(alcoholic), heat2: conc. H2SO4 ( at 0 oC followed by boiling with H2O ) 3: NaOH(aq), heat.
76 (a) To synthesize compound B via 2 steps: Step 1: Sn, Conc. HCl, heat followed by excess NaOH, step 2: CH3Cl (alcoholic), heat.To synthesize compound C via 3 steps: Step 1: acidified KMnO4 solution, heat, step 2: PCl5, r.t.p. step 3: NH2CH2CH2CO2H, r.t.p.
(b)(i)
(ii)
(iii)
77 (a) Let R = CH3O-C6H5
LiAlH4 in dry ether PCl5 at rtp KCN(alc)
R – CHO ============> R – CH2OH =======> R – CH2Cl =====> R –
CH2CN
Rtp heat
(b) (i)
(c) C is a racemic mixture where the d isomers and l isomers are in 50:50 ratio.
2, 4-dinitrophenylhydrazine at rtp.
Condensation reaction.
78 (a)
I CH3CH2Cl, AlCl3, warm V dilute HCl, heatII Cl2, U.V. VI NaOH(alcoholic), heatIII KCN(ethanolic), heat VII NaOH(aq), heatIV NaOH(aq), heat VIII KMnO4, dil. H+, heat
(b) Oxidation with K2Cr2O7, dilute H+, heat. Compound 2 does not turn orange Cr2O7
2-/H+ to green .Compound 2 changes the orange Cr2O72-/H+ to green.
Compound 1 is oxidised to C6H5CH2COOH.
79 (i) Compound J to Salbutamol
(ii)
80 .(a) (i) L: P:
(ii)
(iii) acidified aq. KMnO4, reflux/heat.
81 A is CH2BrCH2Br F is CH3CH2COOCH2CH3
B is CH2OHCH2OH G is CH3CH2COClC is CH2BrCH3 H is CH3CH2CONH2
(c) React both X and Coumarin respectively with: 2, 4 DNPH at r.t.p.X gives an orange ppt. whilst Coumarin has no orange ppt formed.React both Y and Coumarin respectively with: Br2(aq) at r.t.p.Y gives a white ppt with brown Br2(aq) decolourised and evolution of steamy fumes, whilst Coumarin does not decolourise brown bromine to form a white ppt.
Cyclohexene has only 1 isomer as the molecule only exhibits cis isomerism. The trans form would distort the ring with great strain and hence is not stable and non-existent.
Hex-3-ene has geometrical cis-trans isomerism due to the restriction of rotation caused by the C=C double bond.
