Q. 1
Do as directed:
- (i)Draw the contributing structure indicated by the curved arrow(s). Be certain to show all valence electrons and all formal charges. a) A structural drawing of formaldehyde (H₂C=O) — a carbon atom bonded to two hydrogens and double-bonded to an oxygen atom. The curved arrow shows electron movement from the C=O double bond toward oxygen, implying resonance where oxygen gains a negative formal charge and carbon becomes electron-deficient (positive formal charge). The resonance structure would show C⁺–O⁻ (single bond with charges). b) A drawing of cyclopentanone — a five-membered ring with a carbonyl group (C=O) at one position. The curved arrow again indicates electron movement from the C=O bond toward oxygen, producing a resonance contributor with a positively charged carbon and negatively charged oxygen (C⁺–O⁻).
- (ii)Explain the difference in reactivity between the two alcohols shown below: a) Benzyl Alcohol (a benzene ring with a –CH₂OH group attached) reacts with HCl to produce Benzyl Chloride (benzene ring with –CH₂Cl). The reaction proceeds readily because the benzyl carbocation intermediate formed is stabilized by resonance delocalization into the aromatic ring. b) CH₃CH₂OH (Ethanol) + HCl → No reaction (under normal conditions). Ethanol does not react because the primary carbocation that would form is not stabilized — there is no resonance or hyperconjugation sufficient to make the reaction favorable under mild conditions.
- (iii)Draw the structures of the following: Compound a) 1-Phenylethanone (Acetophenone — benzene ring bonded to a –C(=O)CH₃ group) b) 3-Methylbutan-2-one (CH₃–C(=O)–CH(CH₃)–CH₃) c) 2-Hydroxyphenylethanone (Acetophenone with –OH at ortho position on ring) d) Salicylaldehyde (Benzene ring with –CHO and –OH groups at ortho positions) e) 3-Oxopropanoic acid (OHC–CH₂–COOH — an aldehyde and carboxylic acid separated by one CH₂)
- (iv)Which compounds contain chiral centers? a)2-Chloropentane b)3-Chloropentane c)3-Chloro-1-pentene d)1,2-Dichloropropane