paper 4 organic Flashcards
By ref to the formation of sigma and pi bonds, describe and explain the shape of a benzene molecule
- Benzene has a shape of trigonal planar bond angle of
120°, C-C bonds have the same length. - It also contains delocalised e- in the pi system
- All carbon atoms undergoes sp2 hybridisation.
- The sp2 C-H and C-C bonds overlaps to form sigma bond
- The unhybridised p-orbital in carbon atoms overlap sideways with each other above and below to form pi bond
Describe the relative basicities of NH3, ethylamine, phenylamine, (least to most basic) and explain
- phenylamine < ammonia < ethyamine
- In phenylamine, the LP on N can be delocalized into benzene ring, making it less available to accept PROTON
- In ethylamine, the e- donating alkyl grp w +ve inductive effect increases the e- density on N, making N more available to accept PROTON
CH3COOH, CCl3COOH, CH3CONH2
rank in order of acidity (least –> most) and explain
- CH3CONH2 is neutral
- In CCl3COOH, chlorine is an e- withdrawing grp with -ve
inductive effect, and weakens the strength of O-H bond. - The structure of -ve ion is more stableised due to the
delocalization of -ve charge by the C=O, hence more H+
produced.
State and explain the relative acidities of Cl3CCOOH CH2ClCOOH, CH3COOH
- Relative acidities increases in the order CH3COOH,
CH2ClCOOH, Cl3CCOOH - Cl is an e- withdrawing grp with -ve inductive effect and
weakens the strength of O-H bond. - The more -ve ion, the more stable the -ve charge
becomes , more H+ ion is produced. - Therefore Cl3CCOOH is the strongest acid
- In CH3COOH, the e- donating alkyl grp with +ve
inductive effect will ↑ the strength of O-H bond and less
likely to ionise
Compare and Explain the relative acidities of hexanoic acid, hexan-1-ol and phenol
- Relative acidities increases in order hexan-1-ol, phenol, hexanoic acid
- In hexanoic acid, the C=O is an e- withdrawing group
with -ve inductive effect and weakens the strength of O-H bond. More H+ produced - Besides, the -ve ion stabilizes the structure of the -ve structure due to delocalization of -ve charge
- In phenoxide ion, the lone pair e- in oxygen ion can delocalize into the benzene ring and stabilise the structure of -ve ion due to negative inductive effect of ring. O-H bond weakened, more H+ produced
- In hexan-1-ol, the e- donating alkyl group increases the e- density and strengthens the O-H bond
- Hence, it is less likely to ionise, less H+ produced
Describe 2 observations when HOOCCOOH is warmed with acidified manganate (VII) ions
- the purple colour of KMnO4 decolorises
- effervescence takes place
Polyalkenes biodegrade very slowly.
Explain why by referring to the structures of polymers,
- C–C bonds are non-polar.
- No dipoles so cannot be hydrolysed
Some polymers will degrade in the environment.
Describe 2 processes by which this can happen.
- Hydrolysis using acid/base
2. Action of UV light
Chlorobenzene and Phenol both show lack of reactivity towards reactants that cause the breaking of the C–X bond.
Explain why
- Both C–Cl and C–O are stringer in Chlorobenzene and
Phenol - the p-orbital in Cl or lone pair of e- in O or Cl can be
delocalised into the benzene ring
eg of non-biodegradable polymers
- PVC
- polyethene
eg of biodegradable polymers
- polyesters
- polyamides
state and explain the relative rates of hydrolysis of the following compounds : CH3CH2CL , CH3COCL , C6H5CL
- relative rates of hydrolysis ↑ in the order: C6H5CL,
CH3CH2CL, CH3COCL - In C6H5CL, the p-orbital on CL overlaps with pi system,
no hydrolysis takes place - In CH3COCL, the carbon in C-CL is more e- deficient
since it is attached to an O atom, C-CL bond is the
weakest - In CH3CH2CL, the C-CL bond is strengthened by the
+ve inductive effect alkyl grp
Write an eqn to show how compound H, C10H13NH2 behaves as a base
C10H13NH2 + H30+ ⇌ C10H13NH3+ + H20
Phe-tyr, Phe and Tyr
Suggest why the other 2 species give spots R and S that are so close together
mass/charge ratios are about the same for phe-tyr and phe (charge density)
Construct an eqn for the formation of NO2+ in nitration of benzene
HNO3 + H2SO4 –> HSO4- + NO2+ + H2O