Edexcel Jun 2010 Paper 2 Q20 (with explained solutions) |

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Edexcel Jun 2010 Paper 2 Q20

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PMT 20 (a) Propene, C3H6, reacts with hydrogen bromide, HBr, in an electrophilic addition reaction. 2-bromopropane is formed as the major product. H3CCH=CH2 + HBrH3CCH(Br)CH3 (i) Complete the mechanism for the reaction, using curly arrows where appropriate. Show clearly the structure of the intermediate carbocation formed. (3) Mechanism H3C H C C Br 14 *N36390A01424*<br />
PMT (1) (2) (b) Four isomers, each with the molecular formula C4H10O, are shown below. Isomer A: Isomer B: Isomer C: Isomer D: CH3CH2CH2CH2OH CH3CH2CH(OH)CH3 (CH3)3COH CH3CH(CH3)CH2OH (ii) Draw the structure of the alternative carbocation that can be formed in the reaction between propene and hydrogen bromide. (i) Which isomer is a secondary alcohol? Justify your answer.(ii) Which isomer is resistant to oxidation when heated with acidified potassium dichromate(VI)? Justify your answer in terms of the structure of the isomer. (2)*N36390A01524* 15 Turn over<br />
PMT (iii) Which isomer can be oxidized to a ketone? Draw the displayed formula of the ketone produced. (1)(iv) Which isomers can be oxidized to an aldehyde? (1)(v) Phosphorus(V) chloride (phosphorus pentachloride), PCl5, is used to test for the presence of an OH group. What would you expect to see when any of the above four isomers, A, B, C or D, are reacted with phosphorus(V) chloride? (1)(vi) Complete the equation for the reaction shown below. State symbols are not required. (2) C4H9OH + PCl5(Total for Question 20 = 13 marks) 16 *N36390A01624*<br />

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Question Number 20 (a)(i) Acceptable Answers H3C Br (1) for both arrows H3C (:) Br H3C Br Mark (1) for carbocation (1) for arrow 1st mark:top arrow must start from the double bond / close to the double bond and not from either of the C atoms of the C=C bond top arrow can end on, or close to, the H in HBr lower arrow must start from the bond and not the H atom in HBr REJECT full charges on the HBr 2nd mark: the carbocation must have a full + and not + 3rd mark: the bromide ion must have a fulland notthe lone pair need not be shown on the Brarrow from bromide ion can start anywhere on the Br or from the minus sign or the lone pair (if shown) on Br and can go to the C or the + sign on the intermediate 3rd mark available even if an incorrect intermediate has been drawn Acceptable Answers Reject Mark H3C OR + CH3 CH2 CH2 6CH0201 1006 Question Number 20(a)(ii)<br />
Question Number 20(b)(i) Question Number 20(b)(ii) Question Number 20(b)(iii) Acceptable Answers Reject Mark B /CH3CH2CH(OH)CH3 /butan-2-ol (1) Because the C atom bearing the OH is attached to two other C atoms / C with OH group attached to one H (atom) (1) ALLOW Because the C atom bearing the OH is attached to two alkyl groups These marks are stand alone Just OH is on the second C atom / OH is in the chain, not on the end OR OH attached to two methyl / two CH3 groups OH- (instead of OH) Acceptable Answers Reject Mark C /(CH3)3COH /(2-)methylpropan-2-ol (1) Because it is a tertiary (alcohol)/no C-H bonds to break (1) ACCEPT a description of a tertiary alcohol These marks are stand alone tertiary structure / tertiary carbon / tertiary carbocation Acceptable Answers Reject Mark BOTH B / CH3CH2CH(OH)CH3 / butan-2-ol AND H H O H C C C C HH H BOTH required for the one mark Structural / skeletal formula 6CH0201 1006<br />
Question Number 20(b)(iv) Question Number 20(b)(v) Question Number 20(b)(vi) Acceptable Answers Reject Mark A / CH3CH2CH2CH2OH / butan-1-ol and D / CH3CH(CH3)CH2OH / (2-)methylpropan-1-ol BOTH needed for one mark Acceptable Answers Reject Mark Steamy fumes / misty fumes / white mist White smoke Acceptable Answers Reject Mark (C4H9OH + PCl5 (cid:198)) C4H9Cl + POCl3 + HCl (1) for HCl (1) for rest of the equation correct NOTE: Equation must be completely correct for the second mark. ACCEPT PCl3O instead of POCl3 6CH0201 1006<br />

Question 19 20 21 22

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