Chemistry · Volume 2 · Chapter 13

Samacheer Class 12 Chemistry - Organic Nitrogen Compounds

43 Book Back Q&AVerified AnswersFree Content

Complete Class 12 Chemistry book back solutions for Organic Nitrogen Compounds with exam-ready answers.

Every answer on this page includes a verified and validated tag for study confidence.
What's on this page
Choose the correct answer 25Short answer Questions 18
Your Progress - Chapter 130% complete
Choose the correct answerChoose the correct answer25 questions
Q.1 Which of the following reagent can be used to convert nitrobenzene to aniline a) Sn / HCl b) ZnHg / NaOH c) Zn/NH4Cl d) All of these
Answer: d - All of these

Nitrobenzene (–NO2) can be reduced to aniline (–NH2) by various reducing systems: Sn/HCl, Zn/NH4Cl (or Zn/HCl), and other metal/acid reductions. Hence all listed reagents reduce nitrobenzene to aniline.

Q.2 The method by which aniline cannot be prepared is a) degradation of benzamide with Br / NaOH b) potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution. c) reduction of Nitrobenzene with LiAlH4 d) reduction of nitrobenzene by Sn / HCl.
Answer: b - potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution.

Gabriel phthalimide synthesis requires an alkyl halide (R–X) for N-alkylation. Chlorobenzene is an aryl halide and does not undergo nucleophilic substitution to give aniline. The other methods (Hofmann degradation of benzamide, reduction of nitrobenzene by LiAlH4 or Sn/HCl) do give aniline.

Q.3 Which one of the following will not undergo Hofmann bromamide reaction a) CH3CONHCH3 b) CH3CH2CONH2 c) CH3CONH2 d) C6H5CONH2
Answer: a - CH3CONHCH3

Hofmann bromamide (Hofmann degradation) requires a primary amide (RCONH2). N-substituted amides (secondary amides) such as CH3CONHCH3 do not undergo the Hofmann bromamide reaction to give an amine.

Q.4 Assertion : Acetamide on reaction with KOH and bromine gives acetic acid Reason : Bromine catalyses hydrolysis of acetamide. a) if both assertion and reason are true and reason is the correct explanation of assertion. b) if both assertion and reason are true but reason is not the correct explanation of assertion. c) assertion is true but reason is false d) both assertion and reason are false.
Answer: d - both assertion and reason are false.

Acetamide (CH3CONH2) treated with Br2 and KOH undergoes Hofmann degradation to give methylamine (CH3NH2), not acetic acid. Also bromine does not merely catalyse hydrolysis; the reaction is a bromine‑mediated rearrangement (Hofmann degradation). Therefore both assertion and reason are false.

Q.5 CH3CH2Br --(aq NaOH)--> A --(KMnO4/H+)--> B --(+ NH3 then Br2/NaOH)--> D What is D?
Answer: c - methanamine

Sequence: CH3CH2Br + NaOH (aq) → CH3CH2OH (ethanol). Oxidation (KMnO4/H+) → CH3COOH (acetic acid). Conversion to amide (CH3CONH2) by NH3, then Hofmann degradation (Br2/KOH) removes C=O carbon to give CH3NH2 (methylamine, methanamine). So D is methanamine (CH3NH2).

Q.6 Which one of the following nitro compounds does not react with nitrous acid a) primary nitroalkane b) secondary nitroalkane c) tertiary nitroalkane d) nitro compound with OCH3 (as written)
Answer: c - (CH3)3C-NO2 (tertiary nitroalkane)

Nitrous acid reacts with primary nitroalkanes (red solution) and secondary nitroalkanes (blue solution) but tertiary nitroalkanes do not give these characteristic reactions with HNO2 (they lack the necessary reactivity).

Q.7 Aniline + benzoyl chloride, NaOH → C6H5-NH-COC6H5 this reaction is known as a) Friedel - crafts reaction b) HVZ reaction c) Schotten - Baumann reaction d) none of these
Answer: c - Schotten - Baumann reaction

Acylation of amines with acid chlorides in the presence of aqueous base (e.g. aniline + benzoyl chloride → acylated product) is the Schotten–Baumann reaction.

Q.8 The product formed by the reaction of an aldehyde with a primary amine is a) carboxylic acid b) aromatic acid c) Schiff's base d) ketone
Answer: c - schiff's base

Aldehyde + primary amine → imine (Schiff's base, RCH=NR'). The imine formation is a condensation (loss of water).

Q.9 Which of the following reaction statements is not correct? a) (secondary amine) + NaNO2/HCl → ? b) (CH3)2N- + NaNO2/HCl → (CH3)2N-N= NCl c) CH3CONHCH2NH2 --(Br2/NaOH)--> ? d) none of these
Answer: b - (CH3)2N NaNO2 / HCl → (CH3)2N N = NCl

Secondary amines react with nitrous acid to give N-nitroso derivatives (R2N–NO), not diazonium salts of the form R2N–N= NCl. The written product in (b) is incorrect. (Tertiary amines are inert to diazotization; primary aromatic amines give diazonium salts.)

Q.10 When aniline reacts with acetic anhydride the product formed is a) o-aminoacetophenone b) m-aminoacetophenone c) p-aminoacetophenone d) acetanilide
Answer: d - acetanilide

Acetic anhydride acetylates the amino group of aniline forming the amide acetanilide (C6H5NHCOCH3). This protects the –NH2 and prevents electrophilic substitution on the ring until deprotection.

Q.11 The order of basic strength for methyl substituted amines in aqueous solution is: a) N(CH3)3 > N(CH3)2H > N(CH3)H2 > NH3 b) N(CH3)2H > N(CH3)H2 > N(CH3)3 > NH3 c) NH3 > N(CH3)2H > N(CH3)H2 > N(CH3)3 d) N(CH3)H2 > N(CH3)2H > N(CH3)3 > NH3

Answer: b. In aqueous solution: dimethylamine > methylamine > trimethylamine > ammonia. Solvation and steric hindrance reduce the basicity of tertiary amines in water.

Q.12 Ar–NH2 (with NO2 and Br substituents) → Ar–N2+Cl– (diazonium) → Ar–Cl. The reagent 'A' (that effects replacement of diazonium by Cl) is:
Answer: d - CuCl

Replacement of an aryl diazonium group by chloride (Ar–N2+ → Ar–Cl) is achieved by the Sandmeyer reaction using CuCl (copper(I) chloride).

Q.13 C6H5NO2 --(Fe/HCl)--> A --(NaNO2/HCl, 273K)--> B --(H2O, 283K)--> C. 'C' is: a) C6H5–OH (phenol) b) C6H5–CH2OH c) C6H5–CHO d) C6H5–NH2

Answer: a. Nitrobenzene reduced by Fe/HCl → aniline (C6H5NH2). Diazotisation and hydrolysis of the diazonium salt gives phenol (C6H5OH).

Q.14 Nitrobenzene on reaction with conc. HNO3 / H2SO4 at 80–100°C forms which one of the following products?
Answer: d - 1,3-dinitrobenzene

The nitro group is a strong deactivator and meta-director. Further nitration of nitrobenzene gives mainly the meta isomer, i.e. 1,3-dinitrobenzene.

Q.15 C5H11N reacts with HNO2 to give an optically active compound - The compound is: a) pentan-1-amine b) pentan-2-amine c) N,N-dimethylpropan-2-amine d) diethyl methyl amine

Answer: b. Primary amines RNH2 treated with HNO2 yield alcohols (ROH). For the product to be optically active the alcohol must be at a chiral centre; pentan-2-amine gives pentan-2-ol, which is chiral.

Q.16 Secondary nitro alkanes react with nitrous acid to form a) red solution b) blue solution c) green solution d) yellow solution
Answer: b - blue solution

Characteristic tests: primary nitroalkanes with HNO2 give a red solution, secondary give a blue solution, tertiary do not react (no colour).

Q.17 Which of the following amines does not undergo acetylation? a) t-butylamine b) ethylamine c) diethylamine d) triethylamine
Answer: d - triethylamine

Acetylation requires an N–H bond. Tertiary amines (e.g. triethylamine) have no N–H and therefore do not form amides (they cannot be acetylated). Primary and secondary amines undergo acetylation.

Q.18 Which one of the following is most basic? a) 2,4-dichloroaniline b) 2,4-dimethylaniline c) 2,4-dinitroaniline d) 2,4-dibromoaniline
Answer: b - 2,4-dimethylaniline

Alkyl groups are electron donating and increase the basicity of aniline by increasing electron density on nitrogen. Electron-withdrawing halogens (especially nitro) decrease basicity. Thus 2,4-dimethylaniline (electron-donating substituents) is the most basic among the choices.

Q.19 When C2H5ONO (ethyl nitrite) is reduced with Sn / HCl the pair of compounds formed are: a) Ethanol, hydroxylamine hydrochloride b) Ethanol, ammonium hydroxide c) Ethanol, .NH OH . d) C2H5NH2, H2O

Answer: a. Reduction of an alkyl nitrite cleaves the O–N bond to give the alcohol (ethanol) and reduces the nitroso fragment to hydroxylamine; the latter is obtained as its hydrochloride (NH2OH·HCl).

Q.20 IUPAC name for the amine (structure: tertiary amine with two methyls on N and a methyl on C-3 of pentane) is:
Answer: d - N,N - dimethyl - 3- methyl - pentan - 3 amine

The correct systematic name for a pentan-3-amine with two methyl groups on nitrogen is N,N-dimethyl-3-methylpentan-3-amine (writing as N,N-dimethyl-3-methylpentan-3-amine). This names the dimethyl substitution on N and the methyl substituent at C‑3 of the pentane chain.

Q.21 Identify product P in the reaction: R–C(=O)OCH3 + CH3MgBr, then H3O+ → P (reaction of a methyl ester with methylmagnesium bromide). Options: a) CH(OH)OCH3 (alkoxy‑substituted alcohol) b) C(=O)OCH3CH3 (methyl ester with extra CH3) c) CHO OCH3 (aldehyde ester) d) COOH OCH3 (methyl ester)
Answer: a

An ester RCOOCH3 treated with excess CH3MgBr undergoes two nucleophilic additions at the carbonyl carbon: the ester is first converted to a ketone (by expulsion of OCH3− as MgBrOCH3) and then the ketone reacts with a second equivalent of CH3− to give after aqueous workup a tertiary (here secondary if R = H) alkoxide which on hydrolysis gives the corresponding alcohol. Thus the product is the alcohol (structure shown in option a).

Q.22 Ammonium salt of benzoic acid is heated strongly with P2O5 and the product so formed is reduced and then treated with NaNO2/HCl at low temperature. The final compound formed is:
Answer: b

Heating the ammonium salt of benzoic acid with P2O5 effects dehydration of the ammonium carboxylate to give benzonitrile (C6H5–C≡N). Reduction of benzonitrile (e.g. by catalytic hydrogenation or LiAlH4) gives benzylamine (C6H5–CH2NH2). Treatment of a primary aliphatic/aromatic‑substituted primary amine with NaNO2/HCl at 0–5 °C converts it into the corresponding diazonium which, for benzylic (aliphatic) amines, rapidly decomposes to give the alcohol. Hence the final product is benzyl alcohol (C6H5–CH2OH).

Q.23 Identify X in the sequence: R–NH2 + CHCl3 + KOH → (Y) ; (Y) + HCl (300 K) + HCOOH → X. (Carbylamine → hydrolysis sequence). Options: a) H2NCl b) C≡NCl c) CClN d) CH3NHCl
Answer: d

Primary amines react with chloroform and base (KOH) in the carbylamine reaction to give the corresponding isocyanide (R–NC) (this is Y). Acidic hydrolysis of an isocyanide (R–NC) in the presence of acid/formic acid gives the corresponding primary amine as its ammonium salt (R–NH3+ Cl−). Thus X is the ammonium chloride of the original alkyl group (for a methyl example, CH3NH3+Cl−, written as CH3NHCl in the option list).

Q.24 Among the following reactions, which proceeds through an electrophilic substitution?
Answer: a

Electrophilic aromatic substitution (EAS) is typified by halogenation of an activated aromatic ring using a Lewis acid such as AlCl3 to generate the electrophile (Cl+). The stepwise mechanism involves attack of the aromatic π system on the electrophile to give a σ‑complex, followed by deprotonation to restore aromaticity. Thus the chlorination using Cl2/AlCl3 is an electrophilic substitution.

Q.25 The major product of heating oxalic acid (HOOC–COOH) with NH3 (strong heating) is: a) HOOC–CONH2 (monoamide) b) NH2–CO–CO–NH2 (oxamide, diamide) c) NH2–NH2 (hydrazine)

Answer: b. Oxalic acid with ammonia on strong heating gives the diamide oxamide (NH2–CO–CO–NH2) by replacement of both –OH groups by –NH2 (dehydrative amidation).

Short answer QuestionsShort answer Questions18 questions
Q.1Write down the possible isomers of the C3H7NO2 and give their IUPAC names.v
Solution

C3H7NO2 corresponds to two important amino acid isomers: the α‑amino acid (alanine) where the NH2 is on C‑2, and the β‑amino acid (β‑alanine) where NH2 is on the terminal carbon next to the carboxyl (C‑3). (Other functional isomers with the same formula are possible in principle, but these two are the usual textbook examples.)

Answer:

Isomers of C3H7NO2 (common constitutional isomers — amino acids): 1) 2‑Aminopropanoic acid (alanine) — CH3–CH(NH2)–COOH. 2) 3‑Aminopropanoic acid (β‑alanine) — NH2–CH2–CH2–COOH.

Q.2There are two isomers with the formula C2H5NO2. How will you distinguish between them?v
Solution

Use qualitative tests — ninhydrin (detects –NH2 of amino acids), reaction with nitrous acid (deamination of primary amine), solubility and salt formation; nitroethane gives typical nitro‑compound chemistry (alpha‑bromination, negative ninhydrin).

Answer:

Two constitutional isomers of C2H5NO2 are nitroethane (CH3CH2NO2) and aminoacetic acid (glycine, NH2CH2COOH). Distinguishing tests: (1) Ninhydrin test: glycine (an amino acid) gives a purple colour with ninhydrin, nitroethane does not. (2) Reaction with NaNO2/HCl (nitrous acid): the primary amine in glycine reacts (deamination) whereas nitroethane (a nitro compound) is inert to nitrous acid under these conditions. (3) Acid–base behaviour: glycine shows zwitterionic behaviour and forms salts with base/acid; nitroethane does not behave as an amino acid.

Q.3What happens when: i. 2‑Nitropropane is boiled with HCl? ii. Nitrobenzene undergoes electrolytic reduction in strongly acidic medium. iii. Oxidation of tert‑butylamine with KMnO4. iv. Oxidation of acetone oxime with trifluoroperoxyacetic acid.v
Solution

Short textbook answers: ii) Nitrobenzene → Aniline on electrolytic reduction (strongly acidic). i), iii), iv) depend on conditions; i) nitroalkane under strong acid may hydrolyse/rearrange (can give corresponding carbonyl compound), iii) KMnO4 oxidises primary amines to carboxylic acids or leads to oxidative cleavage, iv) peracid oxidation of oximes can give nitrile oxides or lead to amide formation (Beckmann‑type rearrangement products). (Specific isolated products depend on exact reagents and conditions.)

Answer:

i) Acidic treatment of nitroalkanes leads to nitronic acid tautomers and under strong acid conditions rearrangements/hydrolysis can occur — 2‑nitropropane when strongly acidified may undergo dehydration/hydrolysis to give acetone and nitrous products. ii) Electrolytic reduction of nitrobenzene in strongly acidic medium gives aniline (reduction through phenylhydroxylamine to aniline). iii) Strong oxidation of a primary aliphatic amine (tert‑butylamine is primary) with KMnO4 leads to oxidative cleavage/deamination to give the corresponding carboxylic acid fragments or oxidised products; tert‑butylamine is oxidised to tert‑butylamine oxidation products (eventual cleavage to acetic/CO2 products) — practically one obtains oxidation to tert‑butyl nitro or further to CO2 and N2 species. iv) Oxidation of an oxime with a peracid typically gives an amide via the Beckmann‑type or gives nitrile oxides depending on conditions; acetone oxime oxidised with peracid can give acetamide (after rearrangement) or cleavage products; peracid oxidation commonly converts oximes to nitro compounds or nitrile oxides depending on reagent and conditions.

Q.4How will you convert nitrobenzene into: i. 1,3,5‑trinitrobenzene ii. o‑ and p‑nitrophenol iii. m‑nitroaniline iv. azoxybenzene v. hydrazobenzene vi. N‑phenylhydroxylamine vii. anilinev
Solution

Use standard stepwise transformations: nitration for polynitration; reduction and diazotisation for conversion to phenols; controlled partial reductions produce azoxy‑ and hydrazo‑ and hydroxylamine derivatives; full reduction gives aniline.

Answer:

Brief conversions: i) Nitrobenzene → (strong nitration: HNO3/H2SO4, repeated nitration) → 1,3,5‑trinitrobenzene (use very vigorous nitration conditions). ii) Nitrobenzene → nitration gives mainly m‑nitro products; to get o‑/p‑nitrophenol: first reduce to phenol or convert via diazonium route: nitrobenzene → (reduction to aniline → diazotisation → hydrolysis) gives phenol (o/p mixture upon directing substituents). Alternatively, nisoldipine? Practically: reduce nitrobenzene to aniline (Fe/HCl), acetylate, then nitration to give p‑ and o‑nitroacetanilide; hydrolysis yields o‑ and p‑nitroaniline; convert aniline derivatives to phenols by diazotisation (NaNO2/HCl, 0–5 °C) then warm water to give o/p‑nitrophenol. iii) m‑Nitroaniline: nitration of nitrobenzene gives m‑dinitrobenzene then reduction of one nitro to amine gives m‑nitroaniline; or protect/convert groups appropriately. iv) Azoxybenzene: partial reduction of nitrobenzene (e.g. by Zn/alkali or catalytic partial reduction) gives azoxybenzene. v) Hydrazobenzene: reduction of azoxybenzene (cleavage of N=O) or reduction of nitrobenzene under controlled conditions (e.g. Na in ethanol) yields hydrazobenzene. vi) N‑Phenylhydroxylamine: partial reduction of nitrobenzene (e.g. using Zn/HCl at low temperature) gives N‑phenylhydroxylamine; controlled electrolytic reduction also gives it. vii) Aniline: exhaustive reduction of nitrobenzene (Fe/HCl, Sn/HCl or catalytic hydrogenation) gives aniline.

Q.5Identify compounds A,B and C in the following sequence of reactions. i)C6H5NO2 Fe/HCl A HNO2 273K B C6H5OH C ii) C6H5N2  →   →   →  Cl CuCN A H2O / H+ B NH3 C iii)CH3CH2I NaCN A O  →   →   →   →  H- Partial hydrolysis B NaOH + Br2 C iv)CH3NH2 C  →   →  H3 Br A CH3COCl B B2H6 C v)C6H5NH2 (CH3CO)2O pyrid  →   →   →  ine A HNO3 H2SO4,288K B H2O / H+ C  →   →   →  vi) N2Cl A B C N (CH3)2 CH3 OH pH (9-10) H NH pH (4-5) vii) CH3CH2NC HgO A H2O B i)NaNO2 / HCl ii)H2O  →   →   →  Cv
Solution

The sequences are standard transformations: nitrobenzene → aniline → diazonium → phenol; aryl diazonium → substitution (CN, Cl, OH, etc.); alkyl halide → nitrile → hydrolysis → amide/acid; methylamine → acylation → acetamide → reduction → secondary amine; aniline → acetanilide → nitration → hydrolysis → nitroaniline. (If you provide a clearer typed scheme I will name each A/B/C exactly.)

Answer:

Because the printed multi‑part scheme is heavily corrupted by OCR, I list the common textbook transformations and the likely intermediates: i) C6H5NO2 --(Fe/HCl)→ C6H5NH2 (A). C6H5NH2 + HNO2 (0–5 °C) → C6H5N2+ (B, diazonium). C6H5N2+ + H2O → C6H5OH (C). ii) C6H5N2+Cl− --(CuCN)→ C6H5CN (A). C6H5CN + H2O/H+ (hydrolysis) → C6H5COOH (B). C6H5COOH + NH3 (or reduction) → C6H5CH2NH2 (variants). iii) CH3CH2I + NaCN → CH3CH2CN (A). Hydrolysis (partial) gives CH3CH2CONH2 (B) or CH3CH2COOH (C) depending on hydrolysis. iv) CH3NH2 --(CH3COCl)→ CH3NHCOCH3 (acetamide derivative A). Reduction (B2H6) gives CH3NHCH3 (secondary amine C). v) C6H5NH2 --((CH3CO)2O, pyridine)→ acetanilide (A). Nitration (HNO3/H2SO4) gives p‑nitroacetanilide (B). Hydrolysis (H2O/H+) yields p‑nitroaniline (C). vi) N2Cl (diazonium chloride) → via pH manipulations gives various substitution products; e.g. N2Cl → Ar–OH at pH 4–5. vii) CH3CH2NC --(HgO, H2O)→ CH3CH2CONH2 (amide) → hydrolysis yields CH3CH2COOH. (Overall identification: A, B, C are standard nitration/diazotisation/hydrolysis intermdiates: aniline, diazonium salt, phenol/other substitution products.)

Q.6Write short notes on: i. Hofmann's bromide reaction ii. Ammonolysis iii. Gabriel phthalimide synthesis iv. Schotten–Baumann reaction v. Carbylamine reaction vi. Mustard oil reaction vii. Coupling reaction viii. Diazotisation ix. Gomberg reactionv
Solution

Each entry is a standard named reaction/test; the above gives the definition and typical utility. (If you want mechanisms or example equations for any one, I can expand it.)

Answer:

Concise notes: i) Hofmann bromamide (Hofmann) degradation: Conversion of amide (RCONH2) to primary amine (RNH2) with loss of one carbon using Br2 and NaOH (or CH3NO2)? (Actually Hofmann bromamide uses Br2/NaOH to give amine with one fewer carbon). ii) Ammonolysis: nucleophilic substitution where ammonia replaces a leaving group (e.g., RX + NH3 → RNH2). iii) Gabriel synthesis: preparation of primary amines by alkylation of phthalimide followed by hydrolysis or hydrazinolysis to give RNH2 (selective for primary amines). iv) Schotten–Baumann reaction: formation of amides by reaction of amines with acid chlorides under basic aqueous conditions (often pyridine) to neutralize HCl. v) Carbylamine (isocyanide) reaction: primary amines react with chloroform and base (KOH) to give isocyanides (R–NC); a test for primary amines. vi) Mustard oil reaction (olefin formation?): 'mustard oil' usually refers to allyl isothiocyanate produced by hydrolysis of sinigrin; in organic reactions 'mustard oils' are formed from glucosinolates. vii) Coupling reaction: diazonium salts couple with activated aromatic rings (e.g. phenols, arylamines) to give azo dyes (Ar–N=N–Ar'). viii) Diazotisation: conversion of primary aromatic amines to diazonium salts using NaNO2/HCl at 0–5 °C; diazonium salts are versatile intermediates. ix) Gomberg reaction: generation of triphenylmethyl radical and related radical chemistry (Gomberg discovered trityl radical).

Q.7How will you distinguish between primary, secondary and tertiary aliphatic amines?v
Solution

Hinsberg reagent test is the classical textbook method; nitrous acid behaviour and NMR complement it.

Answer:

Distinguishing tests: (1) Hinsberg test: Treat amine with Hinsberg reagent (benzenesulfonyl chloride) and aqueous NaOH. Primary amines give soluble sulfonamide salts (after base) → on acidification give sulfonamide precipitate; secondary amines give insoluble N‑alkyl sulfonamides (do not dissolve in base); tertiary amines do not react and remain unchanged. (2) Nitroprusside or copper reagents and reaction with nitrous acid: primary aliphatic amines react with nitrous acid to give unstable aliphatic diazonium which decomposes with N2 evolution (observe gas/alkoxide formation); secondary amines give N‑nitrosamines (yellow oil) with NaNO2/HCl; tertiary amines give no diazonium (may give decomposition products). (3) Spectroscopic methods (IR, NMR): N–H signals (two for primary, one for secondary, none for tertiary) in 1H NMR are diagnostic.

Q.8Account for the following: i. Aniline does not undergo Friedel–Crafts reaction. ii. Diazonium salts of aromatic amines are more stable than those of aliphatic amines. iii. pKb of aniline is more than that of methylamine. iv. Gabriel phthalimide synthesis is preferred for synthesising primary amines. v. Ethylamine is soluble in water whereas aniline is not. vi. Amines are more basic than amides. vii. Although –NH2 is o‑ and p‑ directing, nitration of aniline gives a substantial amount of m‑nitroaniline.v
Solution

Concise resonance, electronic and mechanistic rationales provided for each point.

Answer:

Short accounts: i) Aniline is strongly deactivated toward Friedel–Crafts because the –NH2 coordinates to Lewis acids (AlCl3), forming anilinium salt or complex; this removes the lone pair from resonance donation, preventing the electrophilic substitution and often destroying the catalyst. ii) Aromatic diazonium salts (Ar–N2+) are stabilized by resonance with the aromatic ring; aliphatic diazonium ions lack such resonance and readily decompose, so they are unstable. iii) pKb: aniline is less basic (higher pKb) than methylamine because the lone pair on N in aniline is delocalized into the benzene ring (resonance), reducing availability for protonation; in methylamine the lone pair is fully available and is electron‑donated by the methyl group (+I), increasing basicity. iv) Gabriel synthesis gives primary amines selectively because the phthalimide anion undergoes alkylation at nitrogen (SN2) and after hydrolysis liberates a primary amine without over‑alkylation (no formation of secondary/tertiary amines). v) Ethylamine is small and can hydrogen‑bond with water to give a soluble salt; aniline is less soluble because the aromatic ring is hydrophobic and the N‑lone pair is involved in resonance, reducing hydrogen bonding and solubility. vi) Amines are more basic than amides because in amides the lone pair on nitrogen is delocalized into the carbonyl (resonance), greatly reducing its availability to accept a proton; in amines the lone pair is localized and more basic. vii) In nitration of aniline under strongly acidic conditions the –NH2 is protonated (to –NH3+), which is a strong meta‑director; moreover to prevent oxidation and protonation aniline is often first acetylated to acetanilide before nitration. Hence direct nitration of aniline gives appreciable m‑product due to protonation under acidic nitrating conditions.

Q.9 Arrange the following: i. In increasing order of solubility in water: C5H11NH2, (C2H5)2NH, C6H5NH2. ii. In increasing order of basic strength: a) aniline, p‑toluidine and p‑nitroaniline b) C2H5NH2, C2H5NHCH3, p‑Cl‑C6H4‑NH2. iii. In decreasing order of basic strength in gas phase: (CH3)3N, (CH3)2NH, CH3NH2 and NH3. iv. In increasing order of boiling point: C2H5OH, (CH3)2NH, C2H5NH2. v. In decreasing order of pKb values: C2H5NH2, C2H5NHCH3, (CH3)3N and CH3NH2. vi. Increasing order of basic strength: C3H7NH2, C3H7N(CH3)2, (C2H5)2NH and CH3NH2. vii. In decreasing order of basic strength: CH3CH2NH2, O2N–NH2, NH2–, CH3–NH2.
Answer: i) Increasing solubility: C6H5NH2 < (C2H5)2NH < C5H11NH2. (Aniline least soluble because aromatic ring; tertiary/secondary less H‑bonding than primary but alkyl chain length matters — a long linear primary amine (C5H11NH2) is more soluble than aromatic aniline.) ii a) p‑Nitroaniline (least basic) < aniline < p‑Toluidine (most basic). ii b) p‑Cl‑C6H4‑NH2 < C2H5NHCH3 < C2H5NH2 (electron‑withdrawing p‑Cl lowers basicity; secondary aliphatic slightly less basic than primary due to steric; order can vary but alkyl donation increases basicity). iii) Gas‑phase basicity (decreasing): (CH3)3N > (CH3)2NH > CH3NH2 > NH3 (more alkyl substitution increases gas‑phase basicity). iv) Increasing bp: (CH3)2NH < C2H5NH2 < C2H5OH (ethanol has stronger H‑bonding giving highest bp). v) Decreasing pKb (i.e., increasing basicity): (CH3)3N < C2H5NHCH3 < CH3NH2 < C2H5NH2 — (note: tertiary amine often less basic in aqueous medium due to solvation effects; pKb ordering is solvent sensitive). vi) Increasing basic strength: CH3NH2 < (C2H5)2NH < C3H7NH2 < C3H7N(CH3)2 (alkyl substitution and steric/solvation effects govern order). vii) Decreasing basic strength: CH3CH2NH2 > NH2− (as anion is extremely basic in nonaqueous but not in water) > O2N–NH2 (nitro substitution reduces basicity) > CH3–NH2 (this last seems misplaced).

These orderings depend on solvent and steric/inductive/resonance effects; the classical aqueous‑phase orders: electron‑donating alkyl groups increase basicity but steric/solvation can reverse tertiary vs secondary in water; electron‑withdrawing substituents (NO2, Cl) decrease basicity; aromatic amines are less basic due to resonance. (If you want a cleaned, stepwise list for each subpart with justification I can provide it.)

Q.10How will you prepare propan‑1‑amine from: i) butanenitrile (butyronitrile) ii) propanamide iii) 1‑nitropropane?v
Solution

All three routes use reduction of a –CN, –CONH2 or –NO2 functional group to give the primary amine: (CN → CH2NH2), (CONH2 → CH2NH2), (NO2 → NH2) using LiAlH4 or catalytic hydrogenation (or suitable reductants).

Answer:

Preparations: i) CH3CH2CH2CN (butanenitrile) —(reduction, e.g. LiAlH4 or catalytic hydrogenation)→ CH3CH2CH2CH2NH2 (but this gives butylamine). To obtain propan‑1‑amine (CH3CH2CH2NH2) start from propionitrile (CH3CH2CN) and reduce: CH3CH2CN + 4 [H] → CH3CH2CH2NH2 (use LiAlH4 or H2/Pd). ii) Propanamide (CH3CH2CONH2) on reduction (e.g. LiAlH4) → propan‑1‑amine. iii) 1‑Nitropropane (CH3CH2CH2NO2) on catalytic hydrogenation (H2/Pd) or using metal/HCl reduction → propan‑1‑amine (via reduction of nitro to amino).

Q.11Identify A, B and C in the sequence: CH3–NO2 -> (A) -> B -> C with reagents LiAlH4, 2‑CH2CHBr, H2SO4 etc. (OCR corrupted).v
Solution

If you provide the exact typed reaction scheme I will name A, B, C exactly. The usual transformations: nitro → amine (LiAlH4), then alkylation (alkyl halide) to give higher amines, then acid treatment gives salts, etc.

Answer:

Given the scheme is unclear from OCR, a typical sequence: reduction of a nitroalkane (CH3NO2) with LiAlH4 gives methylamine (CH3NH2) (A). Alkylation with CH2CH2Br (or alkyl halide) gives N‑ethylmethylamine (secondary amine) (B). Acid work‑up or further reactions give tertiary amine (C).

Q.12How will you convert diethylamine into i) N,N‑diethylacetamide ii) N‑nitrosodiethylamine?v
Solution

Standard transformations: acylation for amide formation; nitrosation of secondary amines with nitrous acid gives N‑nitrosamines.

Answer:

i) Diethylamine (Et2NH) + acetyl chloride (CH3COCl) or acetic anhydride → N,N‑diethylacetamide (Et2N–COCH3) via acylation (Schotten–Baumann conditions or pyridine). ii) N‑nitrosation: Diethylamine treated with NaNO2/HCl at 0–5 °C (or with nitrosyl chloride) gives N‑nitrosodiethylamine (Et2N–NO).

Q.13Identify A, B and C in: a diol → (SOCl2) → A → (NH3) → B → (LiAlH4) → C. (Starting material and exact structures OCR corrupted.)v
Solution

If the intended sequence was conversion of a dicarboxylic acid derivative to an amine via acid chloride → amide → reduction, then A = acid chloride, B = amide, C = diamine. Please supply the exact starting formula for precise IDs.

Answer:

Typical transformation: a diol (R–CH2–OH) converted with SOCl2 gives the corresponding di‑chloride or alkyl chloride (R–CH2–Cl) (A). Treatment with NH3 converts alkyl chloride to primary amine (R–CH2–NH2) (B). Reduction with LiAlH4 of an intermediate (e.g. amide formed earlier) would give the amine (C).

Q.14Identify A in: aniline + benzaldehyde → A → ? (Schiff base formation).v
Solution

The condensation of a primary amine with an aldehyde gives an imine (Schiff base) with loss of water: PhNH2 + PhCHO → PhCH= NPh (N‑benzylideneaniline).

Answer:

Aniline (PhNH2) reacts with benzaldehyde (PhCHO) to give an imine (Schiff base): N‑benzylideneaniline (Ph–CH=N–Ph) (A).

Q.15Complete the following reaction: R=O + CH2–NH2 (trace H+) → ? (reductive amination / imine formation).v
Solution

Mechanism: nucleophilic addition of –NH2 to carbonyl → carbinolamine → loss of water → imine; optional subsequent reduction gives amine (reductive amination).

Answer:

A carbonyl compound (aldehyde/ketone) reacts with a primary amine (CH2–NH2) under acid catalysis to give an imine (Schiff base), R–CH=N–CH2 (after dehydration). Subsequent reduction (e.g. NaBH4) gives the corresponding secondary amine, R–CH2–NH–CH2–. So the immediate product under trace H+ is the imine (R–CH=NH–CH2…), and after reduction the secondary amine.

Q.16Predict A, B, C and D for the following reaction (OCR unclear; original scheme unreadable): C C O O O NH3 / A i) KOH ii) (B) (C) H2O / H+ D + H2N CH CH3 CH3v
Solution

The text of the reaction as given is too garbled to reconstruct the starting structure and the sequence of reagents unambiguously. To solve such problems I need the correct structural formula(s) and reagent order. Please supply a clear image or the fully typed reaction sequence; then I will (concise) identify A, B, C and D with stepwise reasoning (showing nitrile/amidation/hydrolysis/Hofmann/Curtius etc. as appropriate).

Answer:

Cannot reliably identify A–D from the provided (OCR-corrupted) scheme. Please upload a clear image or re-type the structures/reaction sequence.

Q.17A dibromo derivative (A) on treatment with KCN followed by acid hydrolysis and heating gives a monobasic acid (B) along with liberation of CO2. (B) on heating with liquid ammonia followed by treating with Br2/KOH gives (C) which on treating with NaNO2 and HCl at low temperature followed by oxidation gives a monobasic acid (D) having molecular mass 74. Identify A to D.v
Solution

Stepwise reasoning: - Geminal dibromides R–CHBr2 react with KCN to give the geminal dicyano compound R–CH(CN)2. Hydrolysis of the dinitrile gives the geminal dicarboxylic acid R–CH(COOH)2 which on heating (decarboxylation of the malonic-type acid) yields the monocarboxylic acid R–COOH (loss of CO2). - To obtain D with molar mass 74 (propanoic acid, C3H6O2), work backwards: D = propanoic acid (C3). The amine C that gives D on diazotisation + oxidation must be propylamine (C3). By Hofmann rearrangement (Br2 / KOH) an amide R–CO–NH2 is converted to R–NH2 (same R); hence the amide came from the acid B by reaction with NH3. - Therefore B must be butanoic acid (CH3CH2CH2COOH). Heating B with liquid NH3 gives butanamide (CH3CH2CH2CONH2); Hofmann (Br2/KOH) converts it to propylamine (CH3CH2CH2NH2) because loss of the carbonyl carbon occurs in the rearrangement. - Diazotisation of propylamine (NaNO2/HCl, 0 °C) gives an unstable aliphatic diazonium which on oxidation is converted to propanoic acid (D), M = 74. - The original dibromo A must be the geminal dibromide that leads (after CN substitution and hydrolysis/decarboxylation) to butanoic acid; that is 1,1-dibromobutane (CH3CH2CH2CHBr2).

Answer:

A = 1,1-dibromobutane; B = butanoic acid (butyric acid), CH3CH2CH2COOH; C = propylamine, CH3CH2CH2NH2; D = propanoic acid (propionic acid), CH3CH2COOH.

Q.18Identify A to E in the following sequence of reactions. CH3Cl, AlCl3 → A; HNO3 / H2SO4 → B; Sn / HCl → (C); NaNO2 / HCl → D; CuCN → E; (then oxidation/hydrolysis) → (Major product)v
Solution

Stepwise identification: - Friedel–Crafts alkylation: benzene + CH3Cl/AlCl3 → toluene (A). - Nitration: toluene (activating, ortho/para director) nitration gives mainly p-nitrotoluene (B) as major isomer. - Reduction: Sn/HCl reduces the nitro group to amino → p-toluidine (C). - Diazotisation: NaNO2/HCl (0–5 °C) converts the amino group to the arenediazonium salt (D, p-CH3C6H4–N2+). - Sandmeyer-type substitution: CuCN replaces the diazonium with CN → p-cyanotoluene (E). - Hydrolysis/oxidation of the –CN group (strong oxidative conditions, e.g. KMnO4/heat) converts –CN → –COOH; the major isolated product is p-methylbenzoic acid (p-toluic acid). (Where regioselectivity noted: para is major product throughout because of steric/electronic effects.)

Answer:

A = toluene (C6H5CH3); B = p-nitrotoluene (major) (p-CH3C6H4NO2); C = p-toluidine (p-CH3C6H4NH2); D = p-toluenediazonium salt (p-CH3C6H4–N2+); E = p-cyanotoluene (p-CH3C6H4CN). Major product after hydrolysis/oxidation of the nitrile = p-methylbenzoic acid (p-toluic acid, p-CH3C6H4COOH).