1.In this reaction amide is treated with bromine in presence of sodium hydroxide to prepare primary amine.
2.This method is used to prepare only primary amine not secondary and tertiary amine.
3.In this reaction product has one carbon less than reactant that’s why it is called degradation reaction.
Answers are given in the end-
1.Amines are basic in nature . Why?
2.Arrange the following amines in decreasing order of their basic strength. Give reason also.
(CH3)3N > (CH3)2NH > CH3NH2 > NH3 In aqueous phase
3. Arrange the following amines in decreasing order of their basic strength. Give reason also.
C2H5NH2 > (C2H5)2NH > (C2H5)3N > NH3 In aqueous phase
4. Arrange the following amines in decreasing order of their basic strength. Give reason also.
(CH3)2NH > (CH3)3N > CH3NH2 > NH3 in gaseous phase
5.Arrange the following amines in decreasing order of their basic strength. Give reason also.
6. Methanamine is less basic than aniline. Why?
7.Anilene does not gives friedal craft,s reaction.Why?
8.Why does amine group is protected by acetylating it?
9. Explain briefly
(a) Carbylamine reaction( Isocyanide test)
10. pKb value of ethylamine is lower than benzylamine
Answer
1.Basic nature of amine is attributed to presence of lone pair of electron on nitrogen atom of amine group which can be donated.
2. (CH3)2NH > CH3NH2 > (CH3)3N > NH3 (This order is combine effect of three factors.In aqueous phase, inductive effect, solvation effect and steric hindrance also play important roles)
3.(C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3 (This order is combine effect of three factors.In aqueous phase, inductive effect, solvation effect and steric hindrance also play important roles)
4. (CH3)3N > (CH3)2NH > CH3NH2 > NH3 ( Due to Steric hinderance increases from primary to tertiary amine which hinder to donate lone pair of electron.)
5.
Methyl group, having +I effect increases the electron density on Nitrogen of amine whereas electron withdrawing(-I and –M effect) nitro group decreases electron density on nitrogen of benzene ring.
6.Due to +I effect of methyl group electron density increases in N of amine group whereas electron density decreases due to involvement of lone pair of electrons in resonance.
7. Because basic group amine (-NH2) reacts with friedal craft,s reagents which is acidic in nature and nitrogen bears +ive charge. This +vely charged nitrogen deactivate to the benzene ring due to its elecrtron withdrawing effect.
8. Amine group is protected by acetylating ( addition of -COCH3) it therefore it does not react with friedal craft,s reagens and able to perform friedal craft reaction.
9.When primary amine is treated with chloroform (CHCl3) in presence of base (KOH) yields isocyanide having offensive smell.
R-NH2 + CHCl3 +3 KOH –> R-NC + 3KCl + H2O
10.( pKb value lower mean basic strength is higher) Due to +I effect of ethyl group electron density increases in N of amine group, therefore ethamine is more basic whereas aniline is less basic as electron density decreases due to involvement of lone pair of electrons in resonance.
Aldehydes Ketones and Ethers:- Important and conceptual questions
Answers are given in the end-
1. Why does carbonyl compounds give Nucleophylic reaction?
2. Arrange the following carbonyl compounds in increasing order of their reactivities towards nucleophylic addition reaction.
CH3CHO, HCHO, CH3COCH3
3.Which of the following compounds give iodoform reaction and why?
CH3CHO, HCHO, CH3CH2COCH2CH3
4 Which of the following compounds give cannizaro,s reaction and why?
CH3CHO, HCHO, CH3CH2COCH2CH3
5. Which of the following compounds give aldol condensation reaction and why?
CH3CHO, HCHO, CH3CH2COCH2CH3
6. How many compounds are formed in the cross aldol condensation of
A) CH3CHO, CH3COCH3
B) Ph-CHO and CH3CHO
C) Ph-CO-Ph and CH3CHO
D) CH3COCH3 and CH3CH2COCH2CH3
E) Ph-CHO and Ph-CO-Ph
7.What are the disproportion reaction. Give example.
Answer-
1. Nucleophylic reaction reaction occure due to presence of partial positive charge on carbonyl carbon and addition reaction occure due to presence of double bond.
2. CH3COCH3 < CH3CHO < HCHO
3. CH3CHO, The compounds have three alpha hydrogen atoms to carbonyl groups gives iodoform reaction.
4. HCHO, The compounds have no any alpha hydrogen atoms to carbonyl groups gives cannizaro,s reaction.
5. CH3CHO and CH3CH2COCH2CH3 The compounds have atleast one alpha hydrogen atoms to carbonyl groups gives aldol condensation reaction.
6. A) Four aldol products
B) Two aldol products
C)Two aldol products
D) Four aldol products
E) No aldol products
7. The reaction in which same species is oxidized and reduced simultaneously.
Example is cannizaro,s reaction in which carbonyl compounds having no alpha hydrogen is oxidized in carboxylate ion and reduced to alcohol.
Ph-CHO + KOH à Ph-COO– + Ph-CH2-OH
Answers are given in the end-
1. How will you distinguish butanol, butan-2-ol and 2-Methylbutan-2-ol
2. How will you distinguish aniline and phenol?
3.How will you distinguish phenol and ethanol?
4.How will you distinguish ethanamine and N-Methyl methanine.
5.How will you distinguish Propanamine and N,N-Dimethylethanamine.
6.How will you distinguish Propanamine, N-Methyethanamine and N,N-Dimethylmethanamine.
7. How will you distinguish Ethanol and Ethanoic acid.
8.How will you distinguish Methanal and Ethanal?
9.How will you distinguish Propanal and Propanol.
10.How will you distinguish Propanol and Ethanol?
11.How will you distinguish Choloromethane and Chlorobenzene.
Answer-
1.When Licas reagent (ZnCl2 + HCl ) is added than turbidity appears immediately in Methylbutan-2-ol ( tertiary alcohol), turbidity appears after 5 minutes in butan-2-ol ( Secondary alcohol) and no turbidity appears in butanol ( Primary alcohol).
2.Phenol gives violet colour with neutral FeCl3 while aniline does not.
3.Ethanol gives red colour with ceric ammonium nitrate while phenol does not.
4.Ethanamine reacts with benzenesulphonyl chloride( Hinesburg’s Reagent) which is soluble in dilute solution of NaOH, whereas N-Methyl methanine reacts but insoluble in dilute solution of NaOH.
5.Propanamine reacts with benzenesulphonyl chloride ( Hinesburg’s Reagent) which is soluble in dilute solution of NaOH, whereas N,N-Dimethylethanamine does not react.
6.Propanamine reacts with benzenesulphonyl chloride ( Hinesburg’s Reagent) which is soluble in dilute solution of NaOH whereas N-Methyl ethanine reacts but insoluble in dilute solution of NaOH whereas N,N-Dimethylethanamine does not react.
7.Ethanol gives red colour with ceric ammonium nitrate while Ethanoic acid does not.
Or Ethanol gives frutty smell of esters with ethanoic acid and few drops of conc. H2SO4 whereas ethanoic acid does not. (Note- This is ester test)
8.Ethanal gives yellow ppt with a mixture of NaOH and I2 while methanol does not ( Iodoform,s Test)
9.Propanal gives silver mirror or black ppt with Tollen,s reagent( AgNO3 + NH4OH) while Propanol does not.
Or Propanal gives red ppt with mixture of Fehling,s solution A ( solution of CuSO4) and B ( solution of sodium potassium tartarate and NaOH) while Propanol does not.
( Note- Aldehydes gives Tollens test and Fehling,s solution test while ketones does not.)
10.Ethanol gives yellow ppt with a mixture of NaOH and I2 while propanol does not (Note-This is Iodoform,s Test given by compounds having three alpha hydrogen to carbonyl group or three beta hydrogen to alcoholic group)
11.Choloromethane gives white precipitate with AgNO3 but Chlorobenzene does not.
CHEMIGOD 