Amines Flashcards
What are the steps in the reduction of nitriles ?
Step 1- Halogenoalkanes react with the cyanide ion in aqueous ethanol. The halide ion is replaced
Step 2-the nitrile group can be reduced to primary amines with a nickel / hydrogen (H2) catalyst
How do you produce phenylamine from benzene ?
Step 1: Benzene is reacted with a mixture of concentrated nitric acid and concentrated sulfuric acid
Step 2: Nitrobenzene is reduced to phenylamine, using tin and and hydrochloric acid as the reducing agent
The tin and hydrochloric acid react to form hydrogen, which reduces the nitrobenzene by removing oxygen atoms of the NO2 group
Water is produced
Amines react with acid chlorides and acid anhydrides in which mechanism ?
Nucleophilic addition elimination
Primary amines can be prepared by the reaction of halogenoalkanes with ammonia or by the reduction of nitriles.
Justify the statement that it is better to prepare primary amines from nitriles rather than from halogenoalkanes
With halogenoalkane:
further reaction (of primary amines)
OR
Impure product/mixture of products/lower atom economy
With nitriles
No further reaction
OR
Single product / higher atom economy
A student dissolves a few drops of propylamine in 1 cm3 of water in a test tube.
Give an equation for the reaction that occurs.
Describe what is observed when Universal Indicator is added to this solution.
CH3CH2CH2NH2 + H2O ⇌ CH3CH2CH2NH3+ + OH–
Green turns blue
Phenylamine can be prepared by a process involving the reduction of nitrobenzene using tin and an excess of hydrochloric acid.
Give an equation for the reduction of nitrobenzene to form phenylamine. Use [H] to represent the reducing agent.
Explain why an aqueous solution is obtained in this reduction even though phenylamine is insoluble in water.
C6H5NO2 + 6[H] → C6H5NH2 + 2H2O
C6H5NH2 present as ionic salt OR C6H5NH3+ (Cl–) OR phenyl ammonium (chloride)
Amines E, F and G are weak bases.
Explain the difference in base strength of the three amines and give the order of increasing base strength.
E= C6H6NH2 F= C6H6CH2CH2NH2 G= C6H6NHCH2CH3
(Strength depends on availability of) lone pair on N (atom)
M1
E N (next to ring): (lp) delocalised into ring
M2
(lp) less available (to donate to or to accept a H+)
M3
F or G: N (next to alkyl): (positive) inductive effect/electrons pushed to N
M4
(lp) more available (to donate to or to accept a H+)
M5
order of increasing base strength E
Primary amines can be prepared by the reaction of halogenoalkanes with ammonia or by the reduction of nitriles.
Justify the statement that it is better to prepare primary amines from nitriles rather than from halogenoalkanes.
With halogenoalkane: further reaction (of primary amines) OR Impure product/mixture of products/lower atom economy Ignore bi-product / yield
With nitriles
No further reaction
OR
Single product / higher atom economy
Aqueous solutions of ammonia, ethylamine and phenylamine are prepared.
Each solution has the same concentration.
Which is the correct order for the pH values of these solutions?
A
ammonia > ethylamine > phenylamine
B
ammonia > phenylamine > ethylamine
C
ethylamine > ammonia > phenylamine
D
ethylamine > phenylamine > ammonia
C
Which compound is the strongest base?
A
Ammonia
B
Ammonium chloride
C
Methylamine
D
Phenylamine
C
What type of reaction is used to convert (CH3)3N into the cationic surfactant [(CH3)3N(CH2)15CH3]Cl?
A
Bronsted–Lowry acid-base reaction
B
Nucleophilic addition
C
Nucleophilic addition-elimination
D
Nucleophilic substitution
D
Suggest why lidocaine hydrochloride is used medically in preference to lidocaine.
Explain your answer.
Salt is ionic
Independent marks
(More) soluble (in blood/body fluids/water)
nitrobenzene to phenylamine
what is the reducing agent
Sn / HCl
Give a use for phenylamines
making dyes OR making quaternary ammonium salts OR making (cationic) surfactants OR making hair conditioner OR making fabric softener OR making detergents
Compound K (C6H5CH2NH2) is a structural isomer of J (CH3C6H4NH2) Explain why J is a weaker base than K.
lone or electron pair on N. M1
If no mention of lone pair CE = 0
in J spread / delocalised into ring (or not delocalised in K)
Ignore negative inductive effect of benzene. M2
less available (for protonation or donation in J)
M3
OR
in K there is a positive inductive effect / electron releasing). M2
more available (for protonation or donation in K) M3