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Question 18M.2.hl.TZ2.9

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Date May 2018 Marks available [Maximum mark: 12] Reference code 18M.2.hl.TZ2.9
Level hl Paper 2 Time zone TZ2
Command term Deduce, Draw, Explain, Formulate, Justify, State, Suggest Question number 9 Adapted from N/A
9.
[Maximum mark: 12]
18M.2.hl.TZ2.9

Organic compounds often have isomers.

A straight chain molecule of formula C5H10O contains a carbonyl group. The compound cannot be oxidized by acidified potassium dichromate(VI) solution.

(a.i)

Deduce the structural formulas of the two possible isomers.

[2]

Markscheme

M18/4/CHEMI/HP2/ENG/TZ2/09.a.i/M

 

Accept condensed formulas.

[2 marks]

(a.ii)

Mass spectra A and B of the two isomers are given.

M18/4/CHEMI/HP2/ENG/TZ2/09.a.ii_01

Explain which spectrum is produced by each compound using section 28 of the data booklet.

 

[2]

Markscheme

A:

CH3CH2COCH2CH3 AND «peak at» 29 due to

(CH3CH2)+/(C2H5)+/(M – CH3CH2CO)+

OR

CH3CH2COCH2CH3 AND «peak at» 57 due to

(CH3CH2CO)+/(M – CH3CH2)+/(M – C2H5)+

 

B:

CH3COCH2CH2CH3 AND «peak at» 43 due to

(CH3CH2CH2)+/(CH3CO)+/(C2H3O)+/(M – CH3CO)+

 

Penalize missing “+” sign once only.

Accept “CH3COCH2CH2CH3 by elimination since fragment CH3CO is not listed” for M2.

[2 marks]

A tertiary halogenoalkane with three different alkyl groups, (R1R2R3)C−X, undergoes a SN1 reaction and forms two isomers.

(b.i)

State the type of bond fission that takes place in a SN1 reaction.

[1]

Markscheme

heterolytic/heterolysis

 

[1 mark]

(b.ii)

State the type of solvent most suitable for the reaction.

[1]

Markscheme

polar protic

 

[1 mark]

(b.iii)

Draw the structure of the intermediate formed stating its shape.

[2]

Markscheme

M18/4/CHEMI/HP2/ENG/TZ2/09.b.ii/M

Shape: triangular/trigonal planar

[2 marks]

(b.iv)

Suggest, giving a reason, the percentage of each isomer from the SN1 reaction.

[2]

Markscheme

«around» 50% «each»

OR

similar/equal percentages

 

nucleophile can attack from either side «of the planar carbocation»

 

Accept “racemic mixture/racemate” for M1.

[2 marks]

(c)

Nitrobenzene, C6H5NO2, can be converted to phenylamine via a two-stage reaction.

In the first stage, nitrobenzene is reduced with tin in an acidic solution to form an intermediate ion and tin(II) ions. In the second stage, the intermediate ion is converted to phenylamine in the presence of hydroxide ions.

Formulate the equation for each stage of the reaction.

 

[2]

Markscheme

Stage one:

C6H5NO2(l) + 3Sn(s) + 7H+(aq) → C6H5NH3+(aq) + 3Sn2+(aq) + 2H2O(l)

 

Stage two:

C6H5NH3+(aq) + OH–(aq) → C6H5NH2(l) + H2O(l)

 

[2 marks]

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