A2 23 Ans 2020. 9. 8.آ  A2 Level TOPIC 23 ANSWERS & MARK SCHEMES QUESTIONSHEET 5 STRUCTURES OF...

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Transcript of A2 23 Ans 2020. 9. 8.آ  A2 Level TOPIC 23 ANSWERS & MARK SCHEMES QUESTIONSHEET 5 STRUCTURES OF...

  • TOPIC 23 ANSWERS & MARK SCHEMESA2 Level

    QUESTIONSHEET 1

    E/Z (CIS/TRANS) ISOMERISM

    a) Explanation Restricted rotation (1) of atoms or groups about a C=C bond / when C atoms are joined together by a double bond (1) (Do not allow restricted rotation of molecules.) Due to the π- bond locking atoms in position (1)

    Structures and names

    Z-pent-2-ene or (cis pent-2-ene) (1) E-pent-2-ene or (trans pent-2-ene) (1)

    b) Suggestion Alicyclic ring prevents rotation / locks the molecule in position (1)

    Structures

    c) CH 2 =CHCH

    2 CH

    3 No (1)

    (CH 3 )

    2 C=CH

    2 No (1)

    CHCl=CHCl Yes (1) CHCl=CHCH

    3 Yes (1)

    CHCl=CClCH 3

    Yes (1) CH

    3 CH=CClCH

    2 CH

    3 Yes (1)

    Structural feature Both C atoms of the C=C bond must be joined to different atoms or groups (1)

    C C C CCH3

    CH 3 CH

    2

    H H

    CH 3 CH

    2

    CH 3

    H

    H (1)

    (1)

    (1)

    Cl Cl

    Cl

    Cl

    (1)

  • TOPIC 23 ANSWERS & MARK SCHEMESA2 Level

    QUESTIONSHEET 2

    OPTICAL ISOMERISM

    a) (i) Structures

    (Tapered bonds (or similar) to show tetrahedral distribution about C are essential)

    Means of distinguish between them Equal rotation (1) of the plane of polarised light (1) but in opposite directions (1)

    (ii) Butan-2-ol has a chiral centre / asymmetric carbon atom (1) but butanone does not (1)

    b) (i) Isomers whose molecules bear an object to mirror image relationship (1)

    (ii) No effect on the plane of polarised light / optically inactive (1)

    (iii) Racemic mixture / racemate (1)

    (iv) Ethanal is trigonal planar about the carbonyl C atom (1) There is an equal probability / 50:50 chance of CN- attacking from above and below the plane (1) so that equal amounts of the mirror image ions are formed (1)

    This leads to (±)-CH 3 CH(OH)CN / dl-CH

    3 CH(OH)CN (1)

    Or to (±)-lactic acid / dl-lactic acid (1) Maximum 4 marks

    C

    CH 3

    C 2 H

    5

    OHH C

    CH 3

    C 2 H

    5

    HO H

    (1) (1)

    C

    CH 3

    CN

    O - H

    C

    CH 3

    NC

    - O H (1)

    c)

    H C HC C C C

    H

    H

    HH

    H H

    O

    H

    C

    H

    HH

    CH 3

    CH 3

    CH 2

    CH C

    O

    O H

    NH 2

    C

    H

    C N H

    OH

    H H

    CH 3

    HO

    HO

    CH 2C

    CH 3

    CH 3 CH

    2

    CH 2

    C CH 3

    H

    NH 2 H C C

    CH 2 OH

    N

    CH 2 Cl

    CH 3CH3 C O CH2 CH2 CH CH3

    O CH3

    CHCl

    CH 2

    C O

    CH 2

    CH 2

    CH 2

    * *

    * * *

    *

    1 mark per correctly placed * and - ½ per wrongly placed *.

    d) Only one of the isomers is beneficial (1) So half the dose needed if single isomer used (1) The other isomer may have unpleasant side effects (1)

  • TOPIC 23 ANSWERS & MARK SCHEMESA2 Level

    QUESTIONSHEET 3

    TEST QUESTION ON ISOMERISM

    a) (i) Skeletal isomerism CH 3 CH

    2 CH

    2 CH

    2 OH / CH

    3 CH(OH)CH

    2 CH

    3 (1)

    and (CH 3 )

    2 CHCH

    2 OH / (CH

    3 )

    3 COH (1)

    Positional isomerism CH 3 CH

    2 CH

    2 CH

    2 OH (1) and CH

    3 CH(OH)CH

    2 CH

    3 (1)

    Or (CH 3 )

    2 CHCH

    2 OH (1) and (CH

    3 )

    3 COH (1)

    Functional group isomerism Accept any of the C 4 alcohols (1) and any of the C

    4 ethers (1)

    Stereoisomerism

    (ii) Butan-2-ol has a chiral centre / asymmetric C atom (1) ∴ can exhibit optical isomerism / exist in two non-superimposable forms (1) but none of the compounds has a C = C bond (or other structural feature) which can cause restricted rotation of atoms (1) ∴there is no possibility of geometrical isomerism / existence of cis and trans isomers (1)

    b) (i) Geometrical / cis-trans isomerism (1)

    (ii)

    C

    CH 3

    C 2 H

    5

    OHH C

    CH 3

    C 2 H

    5

    HO H (1) and (1)

    (iii) Melting point / solubility (1)

    CH3 CH2

    C=N

    CH3

    . N

    H

    C6H5 (1)

    CH3 CH2

    C=N

    CH3

    . .

    N H

    C6H5

    (1) .

  • TOPIC 23 ANSWERS & MARK SCHEMESA2 Level

    QUESTIONSHEET 4

    STRUCTURE OF BENZENE

    a) (i) Platinum catalyst, room temperature / nickel catalyst, heat (1)

    (ii)

    (iii) Since three double bonds are hydrogenated, ∆H = 3(-120) = -360 kJ mol-1 (1)

    (iv) Benzene is more stable than expected (1) by (360 – 208) = 152 kJ mol-1 (1) This is because benzene contains a delocalised system of electrons (1)

    b) (i) Delocalised Electrons are not restricted to “2-electron, 2-centre” bonds (1) but are dispersed / spread / mobile over a greater number of nuclear centres (1)

    Unsaturated One or more (carbon-carbon) bonds are multiple bonds / the compound will undergo an addition reaction with hydrogen (1)

    (ii)

    X

    X

    (1)

    (iii) Benzene does not undergo electrophilic addition reactions (1)

    (1)

    and

    X

    X (1)Benzene does not produce disubstituted isomers of the form

    Electron-diffraction measurements show that all the carbon-carbon bonds are of equal length / intermediate between C-C and C=C (1) The enthalpy of hydrogenation of benzene is less exothermic than expected (1) The enthalpy of formation of benzene is less endothermic than expected (1) Maximum 3 marks

  • TOPIC 23 ANSWERS & MARK SCHEMESA2 Level

    QUESTIONSHEET 5

    STRUCTURES OF CYLOALKENES

    a) Alkene I Cyclohexene (1) Alkene II 1,3-Cyclohexadiene (1) Alkene III 1,4-Cyclohexadiene (1)

    b) + H

    2 → (1)

    + 2H 2 → (1)

    c) If the π-orbitals of II were completely localised / by comparison with cyclohexene (1) enthalpy of hydrogenation of II could be calculated as –119.6 x 2 = -239.2 kJ mol-1 (1) This is 7.5 kJ mol-1 different from the experimental value (1) ∴II is energetically more stable than if it had two completely localised π-bonds (1) A slight degree of delocalisation of π-electrons must be occurring (1) Maximum 4 marks

    d) Shortest bonds will be a and c (1) Longest bonds will be d, e and f (1) b will be slightly shorter than d, e, and f (1) because of the slight degree of delocalisation (1)

    e) Estimate -239.2 kJ mol-1 (1)

    Reasons π-bonds are too far apart to overlap (1) No delocalisation of π-electrons (1) ∴ no increase in stability (1)

  • TOPIC 23 ANSWERS & MARK SCHEMESA2 Level

    QUESTIONSHEET 6

    BOND TYPE AND REACTIVITY

    a) (i) Electrophiles are attracted by electrons of the π-bond (1) The π-bond is weak / easily broken (1)

    (ii) C⎯H bonds are little polarised / C and H have similar electronegativity (1) For an alkane to react, strong σ-bonds would have to be broken (1)

    b) (i) Polarisation of the bond / Cδ+ ⎯Xδ- (1) due to the high electronegativity of halogens compared with carbon (1)

    (ii) RCl < RBr < RI (1)

    (iii) Bond strength decreases from C⎯Cl to C⎯I (1) hence a halide ion can leave more readily (1) This outweighs the decreasing strength of the –I effect / the lower electronegativity of the halogens (1) Maximum 2 marks

    c) (i) Nucleophilic reagents / nucleophiles (1)

    (ii) Bond polarisation Cδ+ Oδ- / oxygen is much more electronegative than carbon (1) The π-bond is easily polarised (1)

    (iii) The π-bond is weak / easily broken (1) but the σ-bond is strong (1)

    (iv) They are repelled by the π-electrons of the C C bond (1)

    (v) HCN would have to undergo heterolytic fission / dissociation (to H+ and CN-) (1) This is difficult because HCN is a very weak acid / because of the strength of the H⎯CN bond (1)

    ⎯⎯

    ⎯⎯

  • TOPIC 23 ANSWERS & MARK SCHEMESA2 Level

    QUESTIONSHEET 7

    (ii) Aluminium chloride / iron (III) chloride (1)

    (iii) Electrophilic (½) substitution (½)

    (iv) Heterolytic fission / heterolysis (1)

    c) Example CH 3 CH

    2 Br CH

    3 CH

    2 OH (1)

    Nucleophilic substitution (1)

    CH 3 CH

    2 Br CH

    2 = CH

    2 (1)

    Elimination (1)

    REACTION CONDITIONS AND PRODUCT

    a) (i) Free radical (½) substitution (½)

    (ii) Homolytic fission / homolysis (1)

    (iii) Stage 1

    Ethanolic / alcoholic NaOH(1)

    Aqueous NaOH (1)

    Cl Cl .. 2Cl

    . (1)

    CH 3

    + Cl .

    + HCl (1)

    Stage 2

    . CH

    2

    + Cl2

    CH 2 Cl

    + HCl (1)

    . CH

    2

    Stage 3

    + Cl .

    (1)

    . CH

    2 CH2Cl

    Or 2Cl .

    Cl2

    Or . CH

    2 CH2 CH2 (1)

    b) (i) CH3

    + Cl2

    CH 3

    + HCl (1) Cl

    Or

    CH 3

    + Cl2

    CH 3

    + HCl (1)

    Cl

    (1)

    2

  • TOPIC 23 ANSWERS & MARK SCHEMESA2 Level

    QUESTIONSHEET 8

    NITRATION OF ARENES

    a) Reagents Concentrated nitric acid (1) and concen