Answer any two questions out of four questions:
(A) (i) Which one between aniline (C6H5NH2) and acetanilide (C6H5NHCOCH3) is less reactive towards aromatic electrophilic substitution reaction? Justify your answer. (ii) Arrange the following compounds in increasing order of their

Question

Answer any two questions out of four questions:
(A) (i) Which one between aniline (C6H5NH2) and acetanilide (C6H5NHCOCH3) is less reactive towards aromatic electrophilic substitution reaction? Justify your answer. (ii) Arrange the following compounds in increasing order of their basic nature: [four structures drawn - aniline, o-toluidine (2-methylaniline), m-toluidine (3-methylaniline), p-toluidine (4-methylaniline)]
(B) (i) Write the IUPAC name of CH3-(CH2)4-CN. (ii) Which is more polar between R-CN and R-NC? Justify your answer.
(C) Mention the chemical reaction with equation which happens in the following cases: (i) [benzenediazonium chloride: benzene ring with N2Cl group] + SnCl2 / conc. HCl -> ? (ii) CH3CN + H2O2 / NaOH -> ?
(D) Identify organic compounds A, B, C & D used in following sequence of reaction steps: C7H7NO (A) --P2O5--> B --LiAlH4--> C --HNO2--> D

Accepted Answer

> > [!TLDR] > Amide resonance makes acetanilide less activating than aniline; ring +I of -CH3 raises toluidine basicity (with an ortho steric anomaly); nitrile naming counts the CN carbon; SnCl2/HCl reduces diazonium to phenylhydrazine, alkaline H2O2 hydrates nitrile to amide, and benzamide steps down to benzyl alcohol. > > **(A)(i) Aniline vs acetanilide toward aromatic electrophilic substitution** > > Acetanilide (C6H5-NH-CO-CH3) is **less reactive**. In aniline the nitrogen lone pair is fully available to the ring, making it strongly activating. On acetylation the lone pair is delocalised into the acetyl C=O group (amide resonance), so less electron density is donated to the ring; the ring is therefore less activated and acetanilide undergoes electrophilic substitution more slowly (this is why acetylation is used to moderate aniline before nitration/bromination). > > **(A)(ii) Increasing order of basic nature** > > The -CH3 group is electron-releasing (+I) and raises basicity when meta or para. The ortho isomer, though it carries +I, is anomalously weak because of the ortho effect (steric + inductive crowding next to -NH2). Using standard pKb data (o-9.6, aniline 9.4, m-9.3, p-8.9; lower pKb = more basic): > > **o-toluidine < aniline < m-toluidine < p-toluidine** > > **(B)(i)** CH3-(CH2)4-CN: counting the nitrile carbon as C-1 gives a 6-carbon chain -> **hexanenitrile**. > > **(B)(ii)** **R-CN is more polar** than R-NC. In the cyanide the highly electronegative terminal N of the C=N triple bond produces a strong permanent dipole (acetonitrile mu ~ 3.9 D), slightly larger than that of the isocyanide (~3.8 D), so alkyl cyanides are more polar than alkyl isocyanides. > > **(C)(i)** Benzenediazonium chloride reduced by SnCl2 / conc. HCl gives phenylhydrazine: …

Amine	Gas-phase order (inductive only)	Aqueous order (all effects)	Reason
$NH_3$	Weakest	Weakest	No alkyl donation; only 3 H-bonds
$1^\circ$ ( $RNH_2$ )	Second	Second	One alkyl group donates; 3 H-bonds after protonation
$2^\circ$ ( $R_2NH$ )	Third	Strongest	Two alkyl groups donate; 2 H-bonds; steric hindrance still low
$3^\circ$ ( $R_3N$ )	Strongest	Third	Three alkyl groups donate most, but only 1 H-bond and high steric hindrance

Q.Answer any two questions out of four questions:
(A)

Concept understanding — Amine Basicity Order

Amine Basicity Order – From Intuition to Precision

The Intuition – What Would You Expect?

The Complication – Solvation and Steric Hindrance

The Precise Statement – Aqueous Phase Order

The Full Picture – A Table

Q.Answer any two questions out of four questions: (A)

🎓Unlock everything free for 14 days

Q.Answer any two questions out of four questions:
(A)

Unlock everything free for 14 days