Paper 2 2019 Flashcards
(25 cards)
Which alkane has the highest boiling point?
1. CH3(CH2)5CH3
2. (CH3)3CCH(CH3)2
3. CCH3(CH2)3CH(CH3)2
4. (CH3)2CHCH2CH(CH3)2
CH3(CH2)5CH3
Butane reacts with chlorine in the presence of ultraviolet radiation to form a mixture of organic products. Which equation shows a propagation step in the mechanism for this reaction?
1. Cl2 –> .Cl+ .Cl
2. .Cl + C4H8Cl –> C4H8Cl2
3. C4H9Cl + .Cl –> C4H8Cl2 + H.
4. .Cl + C4H9Cl –> .C4H8Cl + HCl
.Cl + C4H9Cl –> .C4H8Cl + HCl
Which type of bonds are broken and formed in the reaction of ethene and bromine?
- Bonds broken are sigma and pi
- Bond formed is sigma
A student reacts 4.50 g of C6H5NH2 with excess CH3COCI in the reaction below.
C6H5NH2 + CH3COCl –> C6H5NHCOCH3 + HCl
Mr of C6H5NH2 = 93
Mr of C6H5NHCOCH3 = 135
The reaction produces 3.25 g of C6H5NHCOCH3. What is the percentage yield of C6H5NHCOCH3?
49.8%
A carbonyl compound is reacted with NaBH4. Which compound(s) could be formed?
1. 2-Methylpentan-2-ol
2. 2-Methylpentan-1-ol
3. 3-Methylpentan-2-ol
Only 2 and 3
Which chemical(s) can react with phenol?
1. Potassium hydroxide
2. Ethanoyl chloride
3. Nitric acid
1, 2 & 3
3-Methylbut-2-enal is reacted with hydrogen bromide, forming a mixture of two organic products. i) Suggest the two organic products formed.
- 3-bromo-3-methyl-butanal
- 2-bromo-3-methyl-butanal
3-Methylbut-2-enal is reacted with hydrogen bromide, forming a mixture of two organic products. ii) Explain why one of the organic products forms in a much greater quantity than the other organic product.
- Major product forms from most/more stable intermediate/carbocation
- Carbocation bonded to more C atoms / more alkyl
Describe how the observations from a chemical test would distinguish between geraniol and citronellal.
- Tollen’s reagent
- Silver mirror formed with citronellal
Explain why geraniol and citronellal are structural isomers of each other.
- Same molecular formula
- But different structural formulae
Explain the term stereoisomerism.
- Same structural formula but different arrangement of atoms in space
Serine + H+.
Serine but with and NH3+
Serine + (CH3)2CHOH/H2SO4.
Esterification with the carboxylic acid
Serine + CH3COCl.
Esterification with the amine and alcohol
A student is provided with one of the four amino acids with its R group.
Alanine / CH3-
Serine / HOCH2-
Leucine / (CH3)2CHCH2-
Glycine / H-
A student carries out a titration with a standard solution of hydrochloric acid to identify the amino acid.
The student’s method is outlined below. The student dissolves 5.766 g of the amino acid in water and makes the solution up to 250.0 cm^3 in a volumetric flask.
The student titrates this solution with 25.0 cm^3 of 0.15moldm^-3 hydrochloric acid.
21.3cm^3 of the amino acid solution were required for complete neutralisation of the hydrochloric acid.
Determine which amino acid the student used.
n(HCl) = 0.150 × (25.0/1000) = 3.75 × 10–3 (mol)
n(amino acid) in 250 cm3 = 3.75 × 10–3 × (250.0/21.30) = 0.0440 (mol)
M(amino acid) = 5.766/0.0440 = 131 (g mol–1)
Amino acid = (CH3)2CHCH2CH(NH2)COOH/leucine
AND working to show R = 57 to justify choice
OR evidence to show Mr leucine = 131 to justify choice
Two products of ethyl 3-bromopropanoate + H+.
- 3-bromo-ethanoic acid
- Ethanol
Two products of ethyl 3-bromopropanoate + OH-.
- 3-hydroxy-propanoate
- Ethanol
Plan a two-stage synthesis to prepare 12.75g of ester (CH3)2CHCOOCH3 starting from 2-methylpropanal, (CH3)2CHCHO. Assume the overall percentage yield of ester (CH3)2CHCOOCH3 from 2-methylpropanal is 40%.
In your answer include the mass of 2-methylpropanal required, reagents, conditions and equations where appropriate.
Purification details are not required.
Theoretical mass of ester = 12.75 × 100/40 = 31.875 (g)
Theoretical n((CH3)2CHCHO) = 31.875/102 = 0.3125 (mol)
Mass of (CH3)2CHCHO = 72.0 × 0.3125 = 22.5 g
Step 1: Oxidation of aldehyde (CH3)2CHCHO
Reagents: Cr2O72–/H+
Conditions: reflux
Equation: (CH3)2CHCHO + [O] → (CH3)2CHCOOH
Step 2: Formation of ester C
Reagents: methylpropanoic acid/(CH3)2CHCOOH and methanol/CH3OH
Conditions: acid (catalyst), reflux/heat
Equation: (CH3)2CHCOOH + CH3OH → (CH3)2CHCOOCH3 + H2O
Describe, in terms of orbital overlap, the similarities and differences between the bonding in the Kekulé model and the delocalised model of benzene.
Similarities: Orbital overlap (sideways) overlap of p orbitals & π bond/system/ring above and below (bonding (C) atoms/ring/plane)
Differences: Kekulé has: alternating π bonds AND Delocalised has: π ring (system) with all p orbitals overlapped, the π electrons spread around ring
Describe two (or more) pieces of evidence to support the delocalised model of benzene.
- All (C–C) bond lengths are the same
- ∆H hydrogenation less (exothermic) than
expected - Benzene is less reactive than alkenes
- Bromination of benzene requires a catalyst/halogen carrier
- Benzene does not react with/decolourise
bromine
Write the equation for the formation for the electrophile of ethanoyl chloride and AlCl3 and the regeneration of the catalyst.
- CH3COCl + AlCl3 → CH3–C+=O + AlCl4–
- H+ + AlCl4– → AlCl3 + HCl
i) Phenylethanone + NaBH4 = ?1
ii) ?1 + ?2 = phenylethene
ii) Phenylethanone + NaCN/H+= ?3
iv) ?3 + ?4 = phenyl-2-bromo-2-nitril-ethane
v) phenyl-2-bromo-2-nitril-ethane + ?5 = ?6
vi) ?6 + ?7 = C6H5C(CH3)(COOH)(NH2)
- ?1 = phenyl-2-hydroxy-ethane
- ?2 = H3PO4
- ?3 = phenyl-2-hydroxy-2-nitril-ethane
- ?4 = NaBr and H2SO4
- ?5 = H2SO4
- ?6 = phenyl-2-bromo-ethanoic acid
- ?7 = NH3 and ethanol or excess NH3
What does a curly arrow show in a reaction mechanism?
Movement of an electron pair
What is meant by heterolytic fission?
Breaking of a covalent bond
