CAns Organic chem general

Why is reflux necessary for the preparation of many covalent organic compounds?

• Organic compounds contain strong covalent bonds

• Heating is needed to provide energy to break these bonds as Ea is high and reaction would be slow -> time is needed for reaction to go to completion

• Reflux is necessary to avoid losing volatile reactants and products due to vapourisation during heating

Recrystallisation: explain why a mixture of ethanol and warm water is used to dissolve the aspirin 

• Mixture of ethanol and warm water can dissolve aspirin and the impurities

• When the solution cools, only aspirin will form crystals due to its low solubility in cold water -> this is recrystallisation 

1. Check WHAT attacks (dnt just anyhow pls) 

• Eg. Free radical sub: CH3CH2CH2CH3 radical is a trigonal planar about electron-deficient C atoms, so the chlorine can attack it from top or bottom with equal probability…  

2. KMnO4: BOTH CO2 and decolourisation! 

2. If the question gives R1 and asks you to deduce structure -> give some random R (Eg. CH3)

Benzene 

• Liquid at room temperature 

• C trigonal planar (Bond angle: 120) 

Draw H2SO4

State condition for this reaction

K2Cr2O2 (aq) (NOT KMnO4 -> entire side-chain will become benzoic acid!), H2SO4 (aq), heat 

Why phenoxide ion needs to be generated 

• OH- ion is needed to generate phenoxide ion which is a stronger nucleophile as it is -ve charged

• A lone pair of electrons on O is more easily delocalised into the benzene ring -> richer in electrons -> ring can attract electrophiles more easily

1. Suggest structure of product = include SIDE PRODUCTS

2. Name the function group 

• Primary amine, chloroalkane (-CI)

State what H2, Ni + LiAIH4 + NaBH4 can reduce

• H2, Ni can reduce C=C (+2H), CN (+4H), RCOR (+2H)

• LiAlH4 can reduce CN (+4H), RCOR (+2H), CONH2 (+2H), COOH

• NaBH4 can reduce RCOR (+2H)

Draw mechanism

Draw benzene attack CO2 nucleophile

Nucleophilic attack on C atom by another Cl− ion AND simultaneous P=O bond formation and P−Cl bond cleavage to form the products

Starting molecule: CICHROPCI3

Protonation

Starting molecule: OH(CH2)2CHO

1. B does not contain aldehyde as there is no oxidation with Fehling’s solution

2. Compound C:H = 1:1 -> benzene ring

State deduction for “B forms two possible mono-nitro compounds”

B forms two possible mono-nitro compounds -> B is a symmetrical molecule and is 1,4-disubstituted -> D is formed according to the orientating effect of groups already present on the benzene ring