Questions: Amines

Q1. How are amines classified, and how do you name a secondary amine?

Answer (English): The class depends on how many organic groups are attached to nitrogen. One group gives a primary amine, two a secondary, three a tertiary, and four a quaternary ammonium salt with a positive nitrogen and an anion. A secondary amine is named as an N-substituted primary amine, so CH3NHCH3 is N-methylmethanamine because the longer group supplies the parent name and the other group goes on N with a locant.

Hinglish: Class is baat par hai ki nitrogen se kitne organic groups jude hain. Ek group se primary, do se secondary, teen se tertiary aur chaar se quaternary ammonium salt banta hai. Secondary amine ko N-substituted primary amine ki tarah naam dete hain, isliye CH3NHCH3 N-methylmethanamine hai: bada group parent naam deta hai aur chhota group N par locant ke saath aata hai.

Q2. Why is aniline a weaker base than ammonia and than ethylamine?

Answer (English): In aniline the nitrogen lone pair is delocalised into the aromatic ring by resonance, so it is less available to accept a proton. That delocalisation also stabilises the free amine relative to the anilinium ion, which shifts the equilibrium away from the protonated form. Ethylamine has an electron releasing ethyl group and no resonance, so its lone pair is more available and it is a far stronger base than aniline; pKb values are about 3.29 for ethylamine and 9.38 for aniline.

Hinglish: Aniline mein nitrogen ka lone pair resonance se aromatic ring mein delocalise ho jata hai, isliye proton lene ke liye kam available hota hai. Ye delocalisation free amine ko anilinium ion ke muqable zyada stabil bhi karta hai. Ethylamine mein electron releasing ethyl group hai aur resonance nahi, isliye woh aniline se kaafi zyada strong base hai; pKb ethylamine 3.29 aur aniline 9.38 hai.

Q3. Why is the order of basic strength different in water and in the gas phase?

Answer (English): In the gas phase there is no solvent, so the ranking depends only on how the free base and the protonated ion are stabilised internally. Alkyl groups release electron density, so adding them raises basicity and the order is tertiary above secondary above primary above ammonia. In water the ammonium ion is solvated, and a primary or secondary ion has more N-H hydrogens available for hydrogen bonds with water than a tertiary ion, which has none. That solvation difference outweighs the inductive effect, so the aqueous order becomes secondary above primary above tertiary.

Hinglish: Gas phase mein solvent nahi hota, isliye ranking sirf molecules ki internal stability par hai; alkyl groups electron density dete hain, isliye tertiary > secondary > primary > ammonia. Paani mein ammonium ion solvate hota hai, aur primary ya secondary ion ke paas water se hydrogen bond banane ke liye zyada N-H hydrogens hote hain, jabki tertiary ke paas ek bhi nahi. Yahi solvation ka farq inductive effect ko beat kar deta hai, isliye aqueous order secondary > primary > tertiary ho jata hai.

Q4. How does the Hofmann bromamide reaction differ from the Gabriel synthesis?

Answer (English): Both give primary amines but they start from different compounds and the chain length changes differently. Hofmann bromamide starts with an amide and uses bromine with concentrated sodium hydroxide, and the product has one carbon fewer because the carbonyl carbon is lost as carbonate. Gabriel synthesis starts with phthalimide, converts it to the potassium salt, reacts that with an alkyl halide and then hydrolyses, and the product has the same number of carbons as the halide. Gabriel works only for aliphatic halides because aryl halides do not undergo the needed substitution, while Hofmann works from any amide.

Hinglish: Dono primary amine dete hain par shuruat alag hai. Hofmann bromamide amide se shuru hota hai, Br2 aur concentrated NaOH lagta hai, aur product mein ek carbon kam hota hai kyunki carbonyl carbon carbonate ban kar nikal jata hai. Gabriel phthalimide se shuru hota hai: pehle potassium salt, phir alkyl halide, phir hydrolysis, aur product mein utne hi carbons hote hain jitne halide mein. Gabriel sirf aliphatic halides ke liye chalta hai kyunki aryl halides substitution nahi dete.

Q5. How do you distinguish primary, secondary and tertiary amines by the Hinsberg test?

Answer (English): Benzenesulphonyl chloride reacts with each class differently. A primary amine forms a sulphonamide that still has one hydrogen on nitrogen, so the compound is acidic enough to dissolve in sodium hydroxide as a clear solution. A secondary amine forms a sulphonamide with no hydrogen on nitrogen, so it does not dissolve in alkali and separates as an insoluble solid or oil. A tertiary amine has no replaceable hydrogen on nitrogen, so no sulphonamide forms at all and the amine stays in the mixture. Shaking with alkali and watching what dissolves separates the three.

Hinglish: Benzenesulphonyl chloride teeno classes se alag react karta hai. Primary amine ka sulphonamide banata hai jismein nitrogen par ek hydrogen bacha rehta hai, isliye woh alkali mein ghul jata hai. Secondary ka sulphonamide N par hydrogen nahi rakhta, isliye alkali mein nahi ghulta aur alag ho jata hai. Tertiary amine ka sulphonamide hi nahi banta kyunki replace hone wala hydrogen nahi hota. Alkali ke saath shake karke dekho kaun ghulta hai, teeno alag ho jate hain.

Q6. What happens when each class of amine reacts with nitrous acid?

Answer (English): A primary aliphatic amine gives a diazonium salt that is unstable and decomposes at once, releasing nitrogen gas and leaving an alcohol; the bubbles identify it. A primary aromatic amine gives a diazonium salt that survives if the mixture is kept between 273 and 278 K, which is why the diazonium chemistry in this chapter is done in ice. A secondary amine gives an N-nitroso amine, a yellow oily liquid, with no gas. A tertiary aliphatic amine simply dissolves as a nitrite salt, and a tertiary aromatic amine gives a green colour on the ring.

Hinglish: Aliphatic primary amine se bana diazonium salt turant toot jata hai, nitrogen gas nikalta hai aur alcohol bachta hai; bubbles se pahchan hoti hai. Aromatic primary amine ka diazonium salt 273-278 K par bacha rehta hai, isliye is chapter ki diazonium chemistry ice mein hoti hai. Secondary amine se N-nitroso amine banta hai jo yellow oily liquid hai aur gas nahi nikalti. Tertiary aliphatic amine sirf nitrite salt ban kar ghul jata hai, aur tertiary aromatic amine ring par green colour deta hai.

Q7. Write three replacement reactions of benzene diazonium chloride.

Answer (English): First, with copper(I) chloride or bromide the diazonium group is replaced by chlorine or bromine, which is the Sandmeyer reaction; copper powder can replace the copper salt in the Gattermann variant. Second, with potassium iodide the group is replaced by iodine with no copper catalyst at all. Third, on warming with water the group is replaced by OH and phenol forms. Two more are worth knowing: hypophosphorous acid removes the group and gives benzene, and fluoroboric acid followed by heating gives fluorobenzene in the Balz Schiemann reaction.

Hinglish: Pehla, CuCl ya CuBr ke saath diazonium group ki jagah chlorine ya bromine aa jata hai; yahi Sandmeyer reaction hai, aur copper powder wala version Gattermann hai. Doosra, KI ke saath bina copper catalyst iodine aa jata hai. Teesra, paani ke saath garam karne par OH aa jata hai aur phenol banta hai. Hypophosphorous acid se group hat kar benzene milta hai, aur fluoroboric acid se garam karne par fluorobenzene banta hai (Balz Schiemann).

Q8. Why does aniline not give a clean Friedel Crafts reaction, and how is the problem solved?

Answer (English): The NH2 group has a lone pair that grabs the Lewis acid catalyst, aluminium chloride, and forms a complex. That turns the strongly activating free amine into a deactivated complexed group, so the ring stops reacting the way it should. The solution is to protect the amine by acetylation. Acetyl chloride converts aniline to acetanilide, whose amide nitrogen is far less nucleophilic and does not complex the catalyst, so the Friedel Crafts or bromination reaction now runs on the ring and gives the para product. Hydrolysis of the amide then restores the amine.

Hinglish: NH2 group ka lone pair Lewis acid catalyst AlCl3 ko pakad leta hai aur complex bana deta hai. Isse activating NH2 group deactivating complex ban jata hai aur ring theek se react nahi karta. Solution hai acetylation se protection: acetyl chloride se aniline acetanilide ban jata hai, jiska amide nitrogen catalyst ko complex nahi karta. Ab ring par Friedel Crafts ya bromination ho sakta hai, aur baad mein hydrolysis se amine wapas mil jati hai.

Q9. Why do amines have lower boiling points than alcohols of similar mass?

Answer (English): Both classes hydrogen bond, but the hydrogen bond in an amine is weaker because nitrogen is less electronegative than oxygen. Nitrogen holds the shared pair less tightly, so the N-H hydrogen is less positive and the attraction to a neighbouring lone pair is weaker. The result is that an amine boils below the alcohol with the same carbon count. Ethanamine at 290 K and ethanol at 351 K show the gap, and the alkane ethane at 184 K shows how much the hydrogen bond is worth in the first place.

Hinglish: Dono hydrogen bond banate hain, par amine ka bond kamzor hota hai kyunki nitrogen oxygen se kam electronegative hai. Nitrogen shared pair ko dheela pakadta hai, isliye N-H hydrogen kam positive hota hai aur attraction kam hoti hai. Isliye amine ka boiling point same carbons wale alcohol se kam hota hai. Ethanamine 290 K par aur ethanol 351 K par ubalta hai, aur ethane 184 K par, jo dikhata hai hydrogen bond ka kitna asar hai.

Q10. How is aniline converted into p-bromoaniline?

Answer (English): Direct bromination of aniline gives 2,4,6-tribromoaniline because the NH2 group activates all three positions strongly. To stop at one bromine at the para position, protect the amine first. Acetyl chloride converts aniline to acetanilide, which is much less activating, and bromination then gives p-bromoacetanilide. Hydrolysis with dilute acid or alkali removes the acetyl group and gives p-bromoaniline. The protection step is the key idea: it changes the reactivity so the substitution can be controlled, and it is reversed easily afterwards.

Hinglish: Aniline ka direct bromination 2,4,6-tribromoaniline deta hai kyunki NH2 group teeno positions ko strongly activate karta hai. Ek hi bromine para par chahiye to pehle protection karo: acetyl chloride se acetanilide banao, jo kam activating hai, aur bromination se p-bromoacetanilide milta hai. Phir dilute acid ya alkali se hydrolysis karke acetyl group hatao aur p-bromoaniline mil jati hai.

Quiz below: five questions on basicity, preparation, identification and diazonium salts. Check the pKb table before you start.