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TPJC H2 Chemistry / Page 1 of 10

TPJC JC2 H2 Chemistry Answers for H2 Topical Revision Oxygen-containing Organic Compounds

1 2 3 4 5 6 7 8 9 10

D B B B B D D B A D

11 12 13 14 15 16 17 18 19 20

B A C C D D D A D D

Question 7

For oxidation with KMnO4, cold conditions would reduce the oxidising ability of KMnO4, enabling the diol to form. Question 13

The silver nitrate precipitates out the iodide ions (NaI or HI) that was formed as a SIDE PRODUCT in the Iodoform test. The mole ratio given shows that there were 3 moles of NaI (or HI) formed with 1 mole of the organic compound, indicating 1 methyl carbonyl functional group present. Note that the CHI3 would have been removed upon filtration at the beginning and would NO longer have been present in the later steps.

Structured/Essay Section

1 (a) Suggest the role of CH3CH2O- in step 1 of the mechanism.

It is acting as a base to abstract H+ from the ester.

(b) Suggest the type of reaction in step 2 of the mechanism.

Nucleophilic substitution

(c) Suggest the structural formulae of the final organic products formed when two molecules of CH3CH2CH2CO2CH2CH2CH3 undergo Claisen condensation with CH3O

- in a similar process as above.

C

CH(CH2CH3)CO2CH2CH2CH3

O

CH3CH2CH2

CH3OH [2]

[Total:4]

2 (a) Primary alcohol, secondary alcohol, ester (b) (i) Step II: Condensation (Nucleophilic Substitution)

Step III: Elimination / Oxidation

(ii)

Intermediate J


(c) Reagent Structural formula of organic compounds

Solid Na2CO3

OH

OH

O

O

OH

HO

COO-Na+HO

aq NaOH, heat

O-Na+

O-Na+OH

O

OH

HO

COO-Na+HO

+Na-O

PCl5

OH

OH

O

O

Cl

Cl

COClCl

aq Br2

OH

OH

O

O

OH

HO

COOHHO

Br

Br

Br

OH

Br

OR

OH

OH

O

O

OH

HO

COOHHO

Br

Br

Br

Br

OH


(d) (i) H2C CH2

OH

CH

Br

CH2 CH2

OH

H2C CH CH

Br

CH CH2Step I

G H

Step II

H2C CH CH

OH

CH CH2Step III

OH

C

O C

O

C

OH

O

Mesoxalic acid Intermediate J

Step Reagents and conditions

I excess concentrated H2SO4, 180 C or Al2O3, 400 oC

II NaOH(aq), heat/reflux

III KMnO4(aq), dilute H2SO4, heat/reflux

(ii) Add Br2 dissolved in CCl4 under room conditions each to compound G and H separately. With compound H, decolourisation of Br2 and no decolourisation for G

H2C CH CH

Br

CH CH2+ 2Br2 (CCl4)

H2C CH

Br

CH

Br

CH CH2

BrBr Br (other acceptable answers include: Br2(aq), PCl5, Na metal)

3 (i) Glucose, brick red ppt of Cu2O.

HOCH2 C C C C C O

O

OH

H

H

OH

OH

H

OH

H

(ii) Step 1: HCN, NaCN catalyst

Step 2: dilute H2SO4, heat under reflux Intermediate Y: RCH(OH)CH(OH)CN

Intermediate Z: RCH(OH)CH(OH)CH2OH (iii)

CH2OH

C

C

OHH

C

H OH

C

HHO

C

H OH

C

OHH

OH

(iv) Type of reaction: nucleophilic addition


4(a) (i)

(ii) (iii)

(iv)

(v)

Suggest the role of anhydrous magnesium sulfate in the procedure above. The anhydrous magnesium sulfate is a drying agent; it removes any remaining traces of water from the product. Calculate the number of moles of ester used and hence calculate the Mr of liquid J. n(ester) = 11.6/116 = 0.100 mol n(liquid J) = 0.100 mol Mr (liquid J) = 7.4/0.100 = 74.0 Propose a possible structural formula of the ester.

What would be the structure of an isomeric ester that would produce an optically active aqueous layer after hydrolysis?

(The acid formed from the ester, undergoes neutralisation with the alkaline medium to form an ionic salt which could be present in the aqueous layer.)

Write down an equation for the alkaline hydrolysis of one of the esters with sodium hydroxide solution.

OR

[6] (b) 1. Reflux phenylethanal with KMnO4/H

+ to get benzoic acid. 2. Reduce phenylethanal with LiAlH4 in dry ether to get 2phenylethanol, C6H5(CH2)2OH. Reflux

benzoic acid and 2 phenylethanol with conc. H2SO4 to get the ester.

(c) Describe the mechanism in steps 1 and 2 using only equations. Include relevant charges and curly arrows to show the movement of electrons.

[3]

*

*

+


5(a) (i) Suggest the identity of the acidic gas. CO2 / carbon dioxide

(ii) Both D and E but not F decolourise potassium manganate(VII). Both D and F but not E gives a yellow precipitate with alkaline aqueous iodine. Suggest the structural formulae of D, E and F.

HCH3

O

C

H H

O

C

CH3 CH3

O

C

D E F

(iii) There is no organic product obtained when E is reacted with hot acidified potassium

manganate(VII). Account for this observation. E, which is methanal, undergoes strong oxidation to form carbon dioxide gas and water.

[5]

(b) When D the organic product, K, is formed. K is then reacted

with 2 chloro 2 methylpropane, (CH3)3CCl, forming a sweet smelling liquid.

(i) Suggest the structural formula of K.

CH3 O-

O

C

K

(ii) In the formation of the sweet smelling liquid, K reacts with (CH3)3CCl. Suggest the type of reaction here. Nucleophilic substitution.

(iii) Propose, with the aid of appropriate equations, the reaction mechanism for the organic synthesis in part (a)(iv).

Clslow

CH3

CH3

ClCCH3

CH3

CH3

CCH3


CH3

CH3

CCH3

CH3 O-

O

C

CH3

C

O

CH3

CH3

CH3

C

O

6

Information/ Type of reaction Deductions: functional group

L rotates plane-polarised light.

L exhibits optical isomerism and contains at least 1 chiral carbon

No reaction occurs between L and 2,4-dinitrophenylhydrazine.

L has no aldehyde or ketone functional group.

L produces M, C10H22O, when it undergoes reduction [] with hydrogen with nickel catalyst

L is unsaturated. Addition of 4 H atoms indicate presence of 2 C=C bonds .

L does not undergo oxidation with acidified potassium dichromate(VI).

L contains a 3o alcohol group.

L undergoes elimination when heated with aluminium oxide to form N as major product.

L contains alcohol that undergoes elimination. N contains an additional C=C bond upon elimination.

N undergoes oxidation with hot concentrated acidified KMnO4 to produce propanone and P, C5H6O5. CO2 gas which gives a white precipitate with limewater is liberated.

Presence of 2 terminal C=C bond due to 2 moles of CO2 produced during oxidative cleavage. (loss of 2 C from N to propanone & P).

P can also be produced by oxidation of the compound shown above with concentrated acidified KMnO4 and heating.

-

L:

CH CHCH2 CH2 CH2

CCH3

CH3

OH

CH CH2 CH CCH2 CH2 CH2

CCH3

CH3

H

CH CH2

OH

or

(Note: The 2 structures below are not accepted as they will lead to formation of a minor product

(not in syllabus))

C CH

CH3

CH3

C

CH3

CH2 CH2CH CH2

OH

or

C CH

CH3

CH3

CH2 CH2 C CH CH2

OH

CH3


M:

CH2 CHCH3 CH2 CH2

CCH3

CH3

OH

CH2 CH3 CH2 CCH3 CH2 CH2

CCH3

CH3

H

CH2 CH3

OH

or

N:

CH CCH2 CH2CH2

C

CH3 CH3

CH CH2 P:

C CH2 CH2 C C

O

OH

O

OH

O

7 (i) Y liberates white fumes with PCl5, therefore it contains alcohol or carboxylic acid group.

Y Y ketone.

Y forms a yellow ppt in the tri-iodomethane test, hence it contains either CH3CO-R or CH3CH(OH)-R where R is H or an alkyl chain.

Y is oxidised to form a carboxylic acid, hence it must contain a primary alcohol. Therefore, Y must be

CH3COCH2CH2OH

(ii) His choice of chemical test was inappropriate because the PCl5 would have undergone hydrolysis to form phosphoric acid and hydrochloric acid.

PCl5 + 4H2O H3PO4 + 5HCl 8 (i) Which molecule P or Q will have a higher boiling point? Explain your answer. [2] P will have a higher boiling point as more energy will be required to overcome the more

extensive intermolecular hydrogen bonding in P. On the other hand, due to the proximity of the OH groups in molecule Q, intramolecular hydrogen bonding reduces the extent of intermolecular hydrogen bonding. As such, less energy is required to break the weaker intermolecular forces of attraction in Q.

(ii) Draw the structures of all the possible organic products when P and Q react separately with chlorine

in trichloromethane. [2] Products formed from P: Products formed from Q:


9 (a) (i)

Add Fehlings solution and warm. The 1st compound gives a brick red ppt while the 2nd compound gives no ppt.

(ii)

Add aqueous I2 and NaOH. Warm.

2nd compound gives a pale yellow ppt while the 1st compound gives no ppt. (iii)CH3CH2CH=CHCH2C(CH2CH3)=CH2 and CH3CH2CH2CH=CHCH=CHCH2CH3 Add acidified KMnO4(aq) to each compound and heat

Add 2, 4dinitrophenylhydrazine at room temperature to the resulting mixture. 1st compound gives an orange ppt while the 2nd compound gives no ppt.

[7] (b)

Evidence Deduction

D (C10H8O2) High C:H ratio shows that D has a benzene ring.

D + aq NaOH E D is hydrolysed. Since only E is formed, D is a cyclic ester.

E soluble in Na2CO3. E undergoes acid-base reaction and contains CO2H.

E displays cis-trans isomerism whereas F does not.

E has C=C with 2 different groups attached to each C of the alkene group. Alkene in E is reduced to alkane in F.

1 mol F reacts with 2 mol aq Br2

F contains phenol group which activates the benzene ring. Hence, it undergoes electrophilic substitution readily. 2 mol of Br2 are required since a cyclic phenolic ester forms a 1, 2-disubstituted ring.

F + LiAlH4 G G not soluble in Na2CO3

G does not contain CO2H. Hence CO2H in F is reduced to give primary alcohol in G.

G + Al2O3 H Elimination / dehydration takes place. H is an alkene.

H + HCl I Electrophilic addition takes place.

I optical isomers I possesses a chiral centre.

I + aq AgNO3 White ppt is AgCl. I is confirmed to be a halogenoalkane.


white ppt + org product I undergoes nucleophilic substitution with water to form an alcohol. Since the organic product is not oxidized, alcohol formed is tertiary OR I is a tertiary halogenoalkane.

Structures [1m each]

[Total 15, max 13]

10 Summary of the reaction scheme:

Observations

Deductions

C can exist as a pair of enantiomers but D and E does not show any optical activity.

C contains a chiral carbon whereas D and E does not contain chiral carbon (or are symmetrical).

1 mol of C and D reacts with 2 mol of liquid bromine. 1 mol of E reacts with 3 mol of liquid bromine.

Electrophilic addition. C and D have 2 C=C, R has 3 C=C.

C, D and E react with hydrogen in the presence of nickel to give the same compound.

(Catalytic) reduction/hydrogenation occurs.


C undergoes oxidative cleavage with acidified KMnO4 to give ethanoic acid and a compound containing 10C.

C contains a terminal alkene as there is loss of CO2. D does not contain terminal alkene. C and D contains the CH3CH=C structure.

D undergoes oxidative cleavage with acidified KMnO4 to give ethanoic acid and a compound containing 11C.

F reacts with 1 mole of Na2CO3 to produce CO2.

Acid-base reaction occurs and F is a dibasic acid (dicarboxylic acid).

F reacts with 2 moles of HCN under cold condition with base as the catalyst.

Nucleophilic addition occurs and F contains 2 carbonyl groups.

Structures of unknown:

OO O

OH

O

OO O

OH

P Q R S

C D E F 11(i) LiAlH4 in dry ether, room temperature OR NaBH4 in (alkaline) methanol, room

temperature

(ii)

O O

O O or

O O C

O

(CH2)2 C

O

n

(iii) KMnO4,

dilute H2SO4, heat

Step 1

O O

I2(aq), NaOH(aq)

warmStep 2

O O

OO

Dilute H2SO4

Step 3

O O

OHHO

.

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