2019 OCR Chemistry (A) - Paper 2: Organic Synthesis [Unofficial Mark Scheme]

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Total votes: 185

georgew9828

Multi-choice:

A – Longest chain from all

D – Radical Propagation

C - E isomer

B – Pi bond breaking and sigma bond making in electrophilic addition

B - Benzene with KOH (only substituted onto haloalkane section)

B - NMR with 4 environments

D (I think) - NMR prediction with 13C trace (couldn’t have been either alkane or the alcohol with 3 CH3s attached to same carbon atom

B – 6 Chiral Centres on molecule

Think B -Which carbonyl compound would form (couldn’t have been the 2-methyl propan-2-ol, but could be other 2)

A - Reactions with Phenol, reacted with all solutions given

Think C – Had secondary amide and ester (no ketone) Functional groups in the molecule

Cant remember the other multichoices

Question 16 – Alkenes and Isomers:

Mechanism with H-Br (standard electrophilic addition mechanism), with H accepting electrons from pi bond, and then Br donating electrons to carbocation (could do this with either the secondary or tertiary intermediate)

Why one compound forms more than other - Tertiary intermediate is more stable, so forms more readily in the reaction mixture, hence more of this organic product is made

Identifying the difference between 2 long skeletal compounds (can’t remember names). Add Tollens to both, only one would form silver mirror (as only 1 had aldehyde functional group)

Structural Isomers – Same molecular formula, different structural formula

Stereoisomers – Same structural formula, different arrangement of atoms in space

Stereoisomers of both compounds – The alcohol compound had E/Z isomers due to different priority groups either side of the carbon pi bond (only one double bond had this as the other had 2 CH3 groups attached). Draw E/Z isomers of this. The aldehyde had optical isomers as it had a chiral carbon atom. Draw mirror images of each other using tetrahedral shape

Question 17 – Amino Acids:

Flow diagram of compound

Using excess Ethanoyl chloride – formed an amide on one side of the molecule (with NH2 group) and formed ester at the bottom of the molecules (with OH group)

Using

Titration calculation – Calc moles of acid (using titre and con), use 1:1 ratio for amino acid. Then scale up the moles using the dilution factor to find the moles in 250cm³. Then calculate the Mr of the amino acid. Subtract the Mr of the known parts of an amino acid (NH2 and COOH and CH groups) to find the value for the group R. Then look at table to see which one it is. Answer was Leucine (R = 57gmol^-1)

TLC Plate – Can’t remember the value, think it was 0.33. Distance of Leucine / Distance of solvent

Amino acid was Glycine, in solvent W Glycine and Leucine moved the same distance, and in solvent X Alanine and Glycine moves same distance. Therefore it must be glycine as the Rf in both solvents is the same

Question 19 – 6 Marker Synthesis and other stuff

Stage 1, oxidise the aldehyde to a carboxylic acid under reflux, with acidified potassium dichromate and sulphuric acid. Include equations (using [O] – colour of oxidising agent goes from orange to green because Cr⁶⁺ goes to Cr³⁺)

Stage 2, esterification with methanol and concentrated sulphuric acid to form the ester required

Calculation – mass needed 22.5g

NMR Table

Include the range from the data sheet for each one

Each peak was triplet, triplet, quartet, triplet

Question 19 – Benzene & Related stuff:

Kekule and Delocalisation stuff – Kekule and new model suggests sigma bonds between atoms from the direct overlap of electron orbitals. Kekule suggested localised pi bonds from overlap of p orbitals, in an alternating pairs. Delocalised structure suggests sideways overlap of p orbitals above and below plane of molecule, forming a ring shaped structure above and below the plane, which is a delocalised ring of electrons

Evidence – Enthalpy of hydrogenation is less exothermic, bond lengths are all equal (evidence is diffraction), doesn’t react by electrophilic addition like alkenes (cannot have localised pi bonds as doesn’t cause the decolourising of bromine water on addition)

Flow diagram

Reduction from NaBH₄makes alcohol

NaCN/H+ makes nitrile on the substituted part

Draw intermediate – Br substituted for NH2 with ethanolic NH3

Forming carboxylic acid – Use any aqueous acid (I went for H2O and HCl)

Question ? – Mechanisms:

Curly arrow – Movement of a pair of electrons

Partial mechanism drawn – Formed water and 3-methyl pentane with a positive charge on carbon 3 due to its bond being broken

Heterolytic fission – The bond was broken as the oxygen atom received both electrons from the shared pair, and neutralised its own positive charge forming a tertiary carbocation and water products

Carbonyl Mechanism – OH donated to carbonyl carbon (arrow towards carbon), O atom broke one of its bonds (arrow from double bond to oxygen). Then, Cl falls off from compound, and the double bond O reforms (arrow from C-Cl bond to Cl atom and from oxygen lone pair back to C-O bond)

OH is a nucleophile as it is donating a pair of electrons.

Question ? – NMR:

Empirical formula C5H6O

Molecular formula 2 x Empirical formula – C10H12O2

Infrared – C=O and O-H (carboxylic acid broad peak) were present

NMR – D2O removed the carboxylic acid proton, 4 protons in a Benzene environment, CH3 in a benzene environment attached. 1 Proton in both a benzene –CH and HC-C=O environment, with a quartet splitting pattern (CH3 was next door). 3 Protons in a HC-R environment. Overall compound had a Ch3 at any position on benzene ring (No carbon-13 NMR to confirm which position) and then CH(CH3)COOH attached at any position on the benzene ring.

Don’t know question number:

Flow diagram of hydrolysis of acids

Acid Hydrolysis – Forms carboxylic acid and alcohol (the Nh2 becomes NH3+ as it accepts a proton from the acid

Excess OH- (alkaline hydrolysis) – Forms carboxylate salt and alcohol. Also, the

Polymerisation question

1^ST was 2 repeat units of an addition polymer (Alkene double bond breaks)

2^nd was 2 repeat units of a condensation polymer (lose H from NH2 and OH from carboxylic acid)

Naming the type of polymer (unlikely to actually name the polymer made)

Addition

Condensation (could be polyamide)

Repeat Units – Calculate the Mr of the repeat unit, then multiply by 100 (half of the 200 units wanted as the Mr of the double repeat unit is twice the single unit)

Acylation mechanism

Draw curly arrows and dipoles to carbocation

Draw arrow from H bond back into benzene ring, to reform the benzene ring electron system

Reform the halogen carrier by combining H+ with AlCl4-, forming AlCl3 and HCl.

(edited 4 years ago)

Scroll to see replies

Reply 1

JamesOHara

Question 16 tollen's reagent is for aldehydes. 2,4-DNP (Brady's reagent) identifies carbonyl compounds/ketones with an orange precipitate

Reply 2

georgew9828

Original post by JamesOHara

Question 16 tollen's reagent is for aldehydes. 2,4-DNP (Brady's reagent) identifies carbonyl compounds/ketones with an orange precipitate

Didn't mean to type out ketone, changed it now

Reply 3

Chocolate Hobnob

Yasss pretty much got em all xx Messed up the last 6-marker but other than that I think it was Okay.

Reply 4

Yazberry

in the multiple choice you break both a pi and sigma and reform a sigma

Reply 5

georgew9828

Original post by Yazberry

in the multiple choice you break both a pi and sigma and reform a sigma

Pretty sure the sigma bond stays formed as the 2 carbons stay bonded together, or else they would just fall apart. Only the pi bond is broken

Reply 6

Yazberry

also ethanoyl chloride reacts with the carboxylic group to form an acid anhydride

Reply 7

georgew9828

Original post by Yazberry

also ethanoyl chloride reacts with the carboxylic group to form an acid anhydride

Also pretty sure that they only react with amides and alcohols and water because of steric hindrance...

Reply 8

ConV1ct__

4 is wrong you gotta remember in the bromine molecule i think it was the bonds break as well so its sigma and pi bond

Reply 9

winteriscoming99

Sigma bond is broken in Bromine you're forgetting.

Original post by georgew9828

Pretty sure the sigma bond stays formed as the 2 carbons stay bonded together, or else they would just fall apart. Only the pi bond is broken

Reply 10

georgew9828

Original post by ConV1ct__

4 is wrong you gotta remember in the bromine molecule i think it was the bonds break as well so its sigma and pi bond

Ah right I get you now, but didn't the question specify about the alkene (ethene) molecule and not the reagent?

Reply 11

Wzfghfhbvvv

Original post by georgew9828

Ah right I get you now, but didn't the question specify about the alkene (ethene) molecule and not the reagent?

It was for the entire mechanism am sure

Reply 12

Hello1234567abvd

Could you say 2,4-DNP instead of tollens? With the aldehyde you would get an orange precipitate and you would not with the alcohol

Reply 13

sanchit.loi

For the heterolytic fission partial mechanism did anyone put H3O+ is formed and a negative charge is on the alkane formed?

I got probs 40%

For Q16 could you use Cr2O7^2- with H2SO4 and say that the alcohol will get oxidised and turn orange to green, whereas there will be no colour change (staying orange) involved in the carboxylic acid?

Reply 16

Steve West

I talked about Brady's reagent forming an orange ppt, as one was aldehyde and the other was a aldehyde. I think any test which only reacts with1 of the 2 compuonds would be acceptable

Original post by KimchiCupRamen

For Q16 could you use Cr2O7^2- with H2SO4 and say that the alcohol will get oxidised and turn orange to green, whereas there will be no colour change (staying orange) involved in the carboxylic acid?

Reply 17

bluemiche19

Surely it’s pi and sigma that break - the pi bond in the alkene and sigma bond in the bromine then one sigma forms

Reply 18

sherzsy

jus wanna say if anyone is feeling upset of this mark scheme, dw its not UNOFFICIAL

Reply 19

Frah1

Original post by tesconyc

I got probs 40%

omg same I dont understand how people found it easy. It was such a weird paper like none of the questions were like before. 2018 paper was much nicer.