3.11 Amines
暂无描述。系统推荐的高质量记忆内容，适合每天坚持背诵学习。
卡片预览 (22 张)
Why can primary aliphatic amines act as Broasted-Lowry bases?
Because the lone pair of electrons on the nitrogen is readily available for forming a dative covalent bond with a H+ so is accepting a proton.
Why are primary aliphatic amines weak bases?
as only a low concentration of hydroxide ions is produced
Why are primary aliphatic amines stronger bases than ammonia?
Because the alkyl groups are electron releasing and push electrons towards the nitrogen atom and so making the lone pair of electrons on the nitrogen more readily available.
why do primary aromatic amines, such as phenylamine, not form basic solutions?
Because the lone pair of electrons on the nitrogen delocalise with the ring of electrons in the benzene ring. This means the N is less able to accept protons.
Whats the overall order of base strength.
Weak base Aromatic amines < Ammonia < Primary amines < Tertiary amines < Secondary amines strong base
What do all amines react with to become ___?
All amines react with acids to become ammonium salts
What will the addition of NaOH to an ammonium salt do?
Convert it back to the amine
How do you form a basic buffer with an amine?
A weak base with a salt of that weak base.
Why is reactions of halogenoalkanes with ammonia not a good method of making a primary amine?
Because of the further reactions. It would mean the desired product would have to be separated from the other products
What happens in the second step of the mechanism with ammonia and a halogenoalkane?
A second ammonia removes a proton from the intermediate (acts as a base) to form the amine.
Why do we use excess ammonia in the reaction of ammonia and a halogenoalkane?
Using an excess of ammonia can limit the further subsequent reactions and will maximise the amount of primary amine formed
Why will the primary amine formed from the reaction of ammonia with a halogenoalkane react further with the halogenoalkane?
It has a lone pair of electrons on the nitrogen.
What will using an excess of halgoneoalkane do?
Promote the formation of the quaternary salt.
Describe the 2 step reaction scheme to form an amine from a nitrile.
Step 1: Convert halogenoalkane to nitrile by using KCN in aqueous ethanol (heat under reflux) Step 2: Reduce nitrile to amine by using LiAlH4 in ether or by reducing with H2 using a Ni catalyst
What is a disadvantage of using the 2 step reaction method to form amines?
It is a two step reaction that may therefore have a low yield. Also KCN is toxic.
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
Further reaction - impure product. With nitrile no further reactions - single product.
1,6-Diaminohexane can also be formed in a two-stage synthesis starting from 1,4-dibromobutane. Suggest the reagent and a condition for each stage in this alternative synthesis. Stage 1 reagent and condition Stage 2 reagent and condition
Stage 1 reagent KCN or NaCN condition aqueous alcohol Stage 2 reagent & condition H2 and Ni
Justify the statement that there are no chiral centres in 3-aminopentane.
No carbon (atom is) attached to 4 different groups
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.
(Strength depends on availability of) lone pair on N (atom) E N (next to ring): (lp) delocalised into ring (lp) less available (to donate to or to accept a H+) 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+) order of increasing base strength E<G<F
There are three secondary amines that contain four carbon atoms per molecule. Draw the skeletal formulas of these three secondary amines.
Amine F can be prepared in a three-step synthesis starting from methylbenzene. Suggest the structures of the two intermediate compounds. For each step, give reagents and conditions only. Equations and mechanisms are not required.
Intermediate compounds Product of step 1 C6H5CH1Cl Allow C6H5CH2Br Product of step 2 C6H5CH2CN Step 1 Cl2 & UV Step 2 KCN alcoholic & aq Step 3 H2 /N