Chemistry Paper 2 June 2012

1. Aproduces hydroxide ions in water
2. Bdonates a proton
3. Caccepts a proton
4. Ddonates a pair of electrons

1. A$[CH_2ClCOO^-][H^+]$
2. B$\frac{[CH_2ClCOO^-][H^+]}{[CH_2ClCOOH]}$
3. C$\frac{[CH_2ClCOO^-]}{[CH_2ClCOOH]}$
4. D$\frac{[CH_2ClCOOH]}{[CH_2ClCOO^-][H^+]}$

1. Awithdraws electrons and stabilises the anion formed
2. Bincreases the relative molecular mass
3. Cforms hydrogen bonds with water
4. Dreleases electrons towards the $-COOH$ group

Answer this when you sit the paper.

Answer this when you sit the paper.

Answer this when you sit the paper.

Answer this when you sit the paper.

1. AGroup VII
2. BGroup II
3. CGroup IV
4. DGroup VI

1. Atetrahedral
2. Bpyramidal
3. CT-shaped
4. Dtrigonal planar

1. Athe nuclear charge decreases
2. Bthe outer electrons are more strongly held
3. Cthe atoms become more electronegative
4. Dthe atomic radius and the shielding both increase

1. Aremove one mole of electrons from one mole of gaseous molecules, breaking every covalent bond within them.
2. Bremove one mole of electrons from one mole of gaseous atoms, forming one mole of gaseous 1+ ions.
3. Cremove one electron from a single gaseous atom, forming a stable solid ionic lattice structure.
4. Dadd one mole of electrons to one mole of gaseous atoms, forming one mole of gaseous 1- ions.

1. Athe atomic radius decreases across the period, but shielding also decreases at exactly the same rate, so the two effects should cancel each other out.
2. Bthe nuclear charge increases while shielding from inner electrons stays roughly constant, so the outer electrons are pulled in more strongly.
3. Cthe number of electron shells increases by one for each successive element across the period.
4. Dthe outer electrons pair up within the same orbital, making each one easier to remove than the last.

Answer this when you sit the paper.

Answer this when you sit the paper.

Answer this when you sit the paper.

1. A$1s^22s^22p^63s^2$
2. B$1s^22s^22p^63s^23p^1$
3. C$1s^22s^22p^6$
4. D$1s^22s^22p^63s^1$

1. A$Mg + 2HCl \rightarrow MgCl_2 + H_2$
2. B$MgCO_3 + HCl \rightarrow MgCl_2 + CO_2$
3. C$Mg(OH)_2 + 2HCl \rightarrow MgCl_2 + 2H_2O$
4. D$MgO + HCl \rightarrow MgCl + H_2O$

1. Ano precipitate forms, since the ionic product is $2.5 \times 10^{-5}$
2. Ba precipitate forms, since the ionic product is $1.0 \times 10^{-4}$
3. Cno precipitate forms, since the ionic product is $5.0 \times 10^{-6}$
4. Da precipitate forms, since the ionic product is $2.5 \times 10^{-5}$

Answer this when you sit the paper.

Answer this when you sit the paper.

Answer this when you sit the paper.

Answer this when you sit the paper.

1. A$C_5H_{12}O$
2. B$C_3H_4O_2$
3. C$C_4H_8O_2$
4. D$C_4H_8O$

1. Abutan-2-ol
2. Bbutanoic acid
3. Cbutanone
4. Dbutanal

1. Abutan-1-ol
2. Bbutanal
3. Cbutane
4. Dbutan-2-ol

1. Abutanal, since aldehydes reduce Fehling's solution but Q gives no reaction with alkaline iodine.
2. Bbutan-2-ol, since secondary alcohols are readily oxidised by the copper(II) ions present in Fehling's solution.
3. Cbutanoic acid, since carboxylic acids also reduce Fehling's solution by donating a proton to it.
4. Dbutanone, since ketones give a positive Fehling's test result in the same way that aldehydes do.

1. A$CH_3COCH_2CH_3 + 3I_2 + 4NaOH \rightarrow CHI_3 + CH_3CH_2CH_2OH + 3NaI + 3H_2O$
2. B$CH_3COCH_2CH_3 + 3Cl_2 + 4NaOH \rightarrow CHCl_3 + CH_3CH_2COONa + 3NaCl + 3H_2O$
3. C$CH_3COCH_2CH_3 + 3I_2 + 4NaOH \rightarrow CHI_3 + CH_3CH_2COONa + 3NaI + 3H_2O$
4. D$CH_3COCH_2CH_3 + I_2 + NaOH \rightarrow CHI_3 + CH_3CH_2COONa + NaI + H_2O$

Answer this when you sit the paper.

1. Ahydroxyl and carboxylic acid
2. Bketone and ester
3. Caldehyde and hydroxyl
4. Dcarboxylic acid and amide

1. Aan alkene
2. Ban aldehyde
3. Ca ketone
4. Dan ester

1. Asodium metal
2. Baqueous sodium carbonate
3. Caqueous bromine
4. D2,4-dinitrophenylhydrazine

1. Areduction of the $-COOH$ group to an aldehyde, leaving the ring $-OH$ groups completely untouched.
2. Bformation of an ester between the $-COOH$ group and the methanol solvent.
3. Cno visible reaction, because cholic acid has no ketone or aldehyde group for $NaBH_4$ to reduce.
4. Dreduction of the $-OH$ groups all the way to alkanes, removing every oxygen atom from the ring system entirely.

Tap to enlarge

1. Aoxidation of the secondary $-OH$ groups to ketone ($C=O$) groups on the ring system.
2. Bno visible reaction, since all the $-OH$ groups in cholic acid are tertiary, not secondary.
3. Creduction of the $-COOH$ group to a primary alcohol.
4. Dfurther oxidation of the $-COOH$ group all the way to $CO_2$ gas, releasing it from the molecule.

Answer this when you sit the paper.

1. Aoxidation
2. Bcondensation
3. Caddition
4. Dhydrolysis

1. A$C_6H_5CO_2C_6H_5$
2. B$C_6H_5NHC_6H_5$
3. C$C_6H_5CH_2NHC_6H_5$
4. D$C_6H_5CONHC_6H_5$

1. Aester
2. Bamine
3. Camide
4. Dnitrile