IB DP Chemistry Topic 10.2 Functional group chemistry HL Paper 2

Question

(a) Oxidation of ethanol with potassium dichromate, K2Cr2O7, can form two different organic products. Determine the names of the organic products and the methods used to isolate them.    [2]

(b) Write the equation and name the organic product when ethanol reacts with methanoic acid.    [2]

Equation:

Product name:

Answer/Explanation

Ans

a.

ethanal AND distillation 

ethanoic acid AND reflux «followed by distillation» 

Award [1] for both products OR both methods.

b.

Equation: CH3CH2OH + HCOOH HCOOCH2CH3 + H2

Product name: ethyl methanoate 

Accept equation without equilibrium arrows. Accept equation with molecular formulas (C2H6O + CH2O2 C3H6O2 + H2O) only if product name is correct.

Question

Consider the following sequence of reactions.

\[{\text{RC}}{{\text{H}}_3}\xrightarrow{{reaction 1}}{\text{RC}}{{\text{H}}_2}{\text{Br}}\xrightarrow{{reaction 2}}{\text{RC}}{{\text{H}}_2}{\text{OH}}\]

\({\text{RC}}{{\text{H}}_{\text{3}}}\) is an unknown alkane in which R represents an alkyl group.

All the isomers can by hydrolysed with aqueous sodium hydroxide solution. When the reaction of one of these isomers, X, was investigated the following kinetic data were obtained.

N10/4/CHEMI/HP2/ENG/TZ0/05.g

The alkane contains 82.6% by mass of carbon. Determine its empirical formula, showing your working.[3]

a.

A 1.00 g gaseous sample of the alkane has a volume of 385 cm3 at standard temperature and pressure. Deduce its molecular formula.[2]

b.

State the reagent and conditions needed for reaction 1.[2]

c.

Reaction 1 involves a free-radical mechanism. Describe the stepwise mechanism, by giving equations to represent the initiation, propagation and termination steps.[4]

d.

The mechanism in reaction 2 is described as SN2. Explain the mechanism of this reaction using curly arrows to show the movement of electron pairs, and draw the structure of the transition state.[3]

e.

There are four structural isomers with the molecular formula \({{\text{C}}_{\text{4}}}{{\text{H}}_{\text{9}}}{\text{Br}}\). One of these structural isomers exists as two optical isomers. Draw diagrams to represent the three-dimensional structures of the two optical isomers.[2]

f.

(i)     Deduce the rate expression for the reaction.

(ii)     Determine the value of the rate constant for the reaction and state its units.

(iii)     State the name of isomer X and explain your choice.

(iv)     State equations for the steps that take place in the mechanism of this reaction and state which of the steps is slow and which is fast.[9]

g.
Answer/Explanation

Markscheme

\({n_{\text{C}}} = \frac{{82.6}}{{12.01}} = 6.88\) and \({n_{\text{H}}} = \frac{{17.4}}{{1.01}} = 17.2\);

ratio is 1:2.5;

\({{\text{C}}_2}{{\text{H}}_5}\);

No penalty for using 12 and 1.

a.

\(\left( {M = \frac{{22400}}{{385}}} \right) = 58.2/\left( {M = \frac{{mRT}}{{PV}}} \right) = 58.3\);

\({{\text{C}}_4}{{\text{H}}_{10}}\);

b.

Br2/bromine ;

UV/ultraviolet light;

Accept hf/hv/sunlight.

c.

initiation:

\({\text{B}}{{\text{r}}_2} \to {\text{2Br}} \bullet \);

propagation:

\({\text{Br}} \bullet + {\text{RC}}{{\text{H}}_3} \to {\text{HBr}} + {\text{RC}}{{\text{H}}_2} \bullet \);

\({\text{RC}}{{\text{H}}_2} \bullet + {\text{B}}{{\text{r}}_2} \to {\text{RC}}{{\text{H}}_2}{\text{Br}} + {\text{Br}} \bullet \);

termination: [1 max]

\({\text{Br}} \bullet + {\text{Br}} \bullet \to {\text{B}}{{\text{r}}_2}\);

\({\text{RC}}{{\text{H}}_2} \bullet + {\text{Br}} \bullet \to {\text{RC}}{{\text{H}}_2}{\text{Br}}\);

\({\text{RC}}{{\text{H}}_2} \bullet + {\text{RC}}{{\text{H}}_2} \bullet \to {\text{RC}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{R}}\);

Award [1] for any termination step.

Accept radical with or without throughout.

Do not penalise the use of an incorrect alkane in the mechanism.

d.

N10/4/CHEMI/HP2/ENG/TZ0/05.e/M

curly arrow going from lone pair/negative charge on O in OHto C;

Do not allow curly arrow originating on H in OH.

curly arrow showing Br leaving;

Accept curly arrow either going from bond between C and Br to Br in bromoethane or in the transition state.

representation of transition state showing negative charge, square brackets and partial bonds;

Do not penalize if HO and Br are not at 180° to each other.

Do not award M3 if OH —- C bond is represented unless already penalised in M1.

Do not penalise the use of an incorrect alkyl chain in the mechanism.

e.

N10/4/CHEMI/HP2/ENG/TZ0/05.f_1/M ;

N10/4/CHEMI/HP2/ENG/TZ0/05.f_2/M ;

First and second structures should be mirror images. Tetrahedral arrangement around carbon must be shown.

f.

(i)     order with respect to \({\text{O}}{{\text{H}}^ – } = {\text{0}}\);

order with respect to \({\text{X}} = 1\);

rate \( = k{\text{[X]}}\);

Award [3] for final correct answer.

(ii)     0.2(0);

\({\text{mi}}{{\text{n}}^{ – 1}}\);

(iii)     2-bromo-2-methyl-propane;

Do not penalize missing hyphens or added spaces.

Accept 2-methyl-2-bromopropane.

tertiary structure;

(iv)     \({{\text{C}}_4}{{\text{H}}_9}{\text{Br}} \to {{\text{C}}_4}{\text{H}}_9^ +  + {\text{B}}{{\text{r}}^ – }\) / in equation with curly arrows and slow;

\({{\text{C}}_4}{\text{H}}_9^ +  + {\text{O}}{{\text{H}}^ – } \to {{\text{C}}_4}{{\text{H}}_9}{\text{OH}}\) / in equation with curly arrows and fast;

No penalty if primary structure is shown.

No credit for SN2 mechanism, except by ECF.

g.

Examiners report

Although this was the least popular question in Section B there was generally a good level of performance. In (a) most candidates scored at least 2 out of 3 marks for calculating the empirical formula.

a.

Many managed to give a correct molecular formula based on their background knowledge once they had determined the molar mass from the density calculation.

b.

The conditions of free radical substitution were well known.

c.

The mechanism of free radical substitution was well known.

d.

The conditions and mechanism of free radical substitution were well known but the SN2 mechanism in (e) caused more problems.

e.

Again the use of curly arrows proved to be difficult. In some case they originated from the H not the lone pair on O of the nucelophile, or were missing from the C – Br bond. Another common mistake was the omission of a negative charge from the transition state. As the attack of the nucleophile is on the opposite side of the carbon atom to the halogen leaving, the partial bonds in the transition state should be drawn at 180 degrees. Candidates were not penalised however if they failed to do this.

f.

Most candidates were able to draw accurate 3D diagrams for the stereoisomers of 2-bromobutane, to deduce the rate expression from the experimental data presented in (g), and correctly identify X as having a tertiary structure. It was also pleasing to see that most were able to describe the SN1 mechanism.

g.

Question

In an experiment conducted at 25.0 °C, the initial concentration of propanoic acid and methanol were \({\text{1.6 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) and \({\text{2.0 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) respectively. Once equilibrium was established, a sample of the mixture was removed and analysed. It was found to contain \({\text{0.80 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) of compound X.

Two compounds, A and D, each have the formula \({{\text{C}}_{\text{4}}}{{\text{H}}_{\text{9}}}{\text{Cl}}\).

Compound A is reacted with dilute aqueous sodium hydroxide to produce compound B with a formula of \({{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}{\text{O}}\). Compound B is then oxidized with acidified potassium

manganate(VII) to produce compound C with a formula of \({{\text{C}}_{\text{4}}}{{\text{H}}_{\text{8}}}{\text{O}}\). Compound C resists further oxidation by acidified potassium manganate(VII).

Compound D is reacted with dilute aqueous sodium hydroxide to produce compound E with a formula of \({{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}{\text{O}}\). Compound E does not react with acidified potassium manganate(VII).

Deduce the structural formulas for compounds A, B, C, D and E.

A:

B:

C:

D:

E:[5]

a.

Deduce an equation for the reaction between propanoic acid and methanol. Identify the catalyst and state the name of the organic compound, X, formed.[4]

b.

Calculate the concentrations of the other three species present at equilibrium.[3]

c.i.

State the equilibrium constant expression, \({K_{\text{c}}}\), and calculate the equilibrium constant for this reaction at 25.0 °C.[2]

c.ii.

2-chloro-3-methylbutane reacts with sodium hydroxide via an \({{\text{S}}_{\text{N}}}{\text{2}}\) mechanism. Explain the mechanism by using curly arrows to represent the movement of electron pairs.[4]

d.i.

Explain why the hydroxide ion is a better nucleophile than water.[2]

d.ii.

1-chlorobutane can be converted to a pentylamine via a two stage process. Deduce equations for each step of this conversion including any catalyst required and name the organic product produced at each stage.[5]

d.iii.
Answer/Explanation

Markscheme

M11/4/CHEMI/HP2/ENG/TZ1/07.a/M

Accept condensed formulas.

Award [1 max] if A and D are other way round (and nothing else correct).

Award [2 max] if A and D are other way round but one substitution product B or E is correct based on initial choice of A and D.

Award [3 max] if A and D are other way round but both substitution products B and E are correct based on initial choice of A and D.

M2 (for B) and M5 (for E) may also be scored for substitution product if primary chloroalkane used.

Penalize missing hydrogens once only in Q.7.

a.

\[{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{COOH}} + {\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}} \rightleftharpoons {\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{COOC}}{{\text{H}}_{\text{3}}} + {{\text{H}}_{\text{2}}}{\text{O}}\]

[1] for reactants and [1] for products.

(concentrated) sulfuric acid/\({{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\);

Do not accept just \({H^ + }\) or acid.

methyl propanoate;

b.

[CH3CH2COOH]:

\((1.6 – 0.80 = ){\text{ }}0.8{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ – 3}}{\text{)}}\);

[CH3OH]:

\((2.0 – 0.80 = ){\text{ }}1.2{\text{ (mol}}\,{\text{d}}{{\text{m}}^{ – 3}}{\text{)}}\);

[H2O]:

\({\text{0.80 (mol}}\,{\text{d}}{{\text{m}}^{ – 3}}{\text{)}}\);

c.i.

\(({K_{\text{c}}} = )\frac{{{\text{[C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{COOC}}{{\text{H}}_{\text{3}}}{\text{][}}{{\text{H}}_{\text{2}}}{\text{O]}}}}{{{\text{[C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{COOH][C}}{{\text{H}}_{\text{3}}}{\text{OH]}}}}\);

\(\left( {{K_{\text{c}}} = \frac{{[{{(0.80)}^2}]}}{{\left[ {(1.2 \times 0.8)} \right]}} = } \right){\text{ }}0.7\);

Allow 0.67.

Award [1 max] for 0.83.

c.ii.

curly arrow going from lone pair/negative charge on O in \({\text{H}}{{\text{O}}^ – }\) to C;

Do not allow curly arrow originating on H in \(H{O^ – }\).

curly arrow showing Cl leaving;

Accept curly arrow either going from bond between C and Cl to Cl in 2-chloro-3-methylbutane or in the transition state.

representation of transition state showing negative charge, square brackets and partial bonds;

Do not penalize if HO and Cl are not at 180° to each other.

Do not award M3 if OH —- C bond is represented.

formation of organic product 3-methylbutan-2-ol and \({\text{C}}{{\text{l}}^ – }\);

d.i.

\({\text{O}}{{\text{H}}^ – }\) has a negative charge/higher electron density;

greater attraction to the carbon atom (with the partial positive charge) / OWTTE;

Do not allow just greater attraction.

d.ii.

\({\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{Cl}} + {\text{KCN}} \to {\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CN}} + {\text{KCl}}\);

Accept \(C{N^ – }\) for KCN and \(C{l^ – }\) for KCl.

pentanenitrile;

Allow 1-cyanobutane.

\({\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{CN}} + {\text{2}}{{\text{H}}_2} \to {\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{N}}{{\text{H}}_2}\);

pentan-1-amine / 1-aminopentane / 1-pentylamine / 1-pentanamine;

Catalyst: nickel/Ni / palladium/Pd / platinum/Pt;

Penalise missing hydrogen once only in Q.7.

d.iii.

Examiners report

This was the least popular question in Section B. Most candidates either scored all five marks in (a) or just one.

a.

(b) was usually well done, though it was disappointing that more candidates did not use the equilibrium sign.

b.

In (c), a significant number of candidates omitted water from the equilibrium calculations.

c.i.

In (c), a significant number of candidates omitted water from the equilibrium calculations.

c.ii.

The organic reaction mechanism in (d) (i) was very poorly presented. Many even tried drawing curly arrows from NaOH as an attacking species. The majority could identify the product of the reaction but a mechanism was far beyond them. Transition states were poor or missing completely.

d.i.

In (ii) although many knew that \({\text{O}}{{\text{H}}^ – }\) has a negative charge, few linked this to the greater attraction to the carbon atom.

d.ii.

In (iii) very few candidates did well here and the name of pentan-1-amine was rarely given. Other mistakes included incorrect catalysts. Further common mistakes included some candidates not including all the hydrogens in the structural formulas. In general for this part there was very poor knowledge of organic synthesis amongst candidates. Very few had a good “stab” at this question. The fact that pentylamine was mentioned in the question initially meant that very few candidates accessed the last mark for the name of the product.

d.iii.
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