FBISE Class 9th (SSC-l)
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FBISE Class 9th (SSC-l)
Answer: HCN adds across the C=O bond of propanal: the CN- attacks the carbonyl carbon and H+ protonates the oxygen, forming a cyanohydrin.
Answer: In a tetrahedral complex, the metal is surrounded by 4 ligands at the corners of a tetrahedron, giving a coordination number of 4.
Answer: SOCl2 converts alcohols to alkyl chlorides with inversion of configuration (if chiral). The byproducts SO2 and HCl are gases, driving the reaction forward.
Answer: Chromium is an exception: one 4s electron promotes to 3d to achieve the extra stability of half-filled d-orbitals, giving [Ar] 3d5 4s1.
Answer: The Wurtz reaction uses sodium metal to couple two alkyl halide molecules via a free radical mechanism: 2R-X + 2Na -> R-R + 2NaX.
Answer: Low-spin d6: t2g6 eg0. CFSE = 6(-0.4 Delta_o) + 0(+0.6 Delta_o) = -2.4 Delta_o, with 3 pairing energies (P) required for pairing all six electrons in t2g.
Answer: Ozonolysis cleaves the C=C double bond. Reductive workup (Zn/H2O or DMS) of cyclopentene gives pentanedial (glutaraldehyde), a dialdehyde with 5 carbons.
Answer: Oxygen in ether has two bonding pairs (two C-O bonds) and two lone pairs, giving 4 electron domains and sp3 hybridization with a bent geometry.
Answer: AlCl3 is a Lewis acid that coordinates with the acyl chloride, generating the electrophilic acylium ion (R-C≡O+) that attacks the aromatic ring.
Answer: The methyl group is an ortho/para director (activating group) due to hyperconjugation and inductive effects, so nitration gives a mixture of ortho and para products.
Answer: In cell notation: anode (oxidation) | anode solution || cathode solution | cathode (reduction). Zn is oxidized (anode) and Cu2+ is reduced (cathode).
Answer: DU = (2C + 2 - H)/2 = (12 + 2 - 6)/2 = 4. This corresponds to benzene\'s three double bonds and one ring.
Answer: In the Cannizzaro reaction, an aldehyde lacking alpha-hydrogens (like benzaldehyde) undergoes base-induced disproportionation to form an alcohol and a carboxylate salt.
Answer: Mg+ has configuration [Ne]3s1. Removing the second electron from the stable 3s1 is very difficult. Na+ already has a noble gas config [Ne], making its second IE very high too, but Mg2+ removing from [Ne]3s1 to [Ne] is harder than Na+ removing from [Ne] to [He]2s2 2p5.
Answer: The haloform reaction converts methyl ketones (CH3-CO-) to iodoform (yellow precipitate CHI3) and a carboxylate salt (sodium acetate) using I2 and NaOH.
Answer: Bond dissociation energy increases with bond order: C-C single (~347 kJ/mol) < C=C double (~614 kJ/mol) < C≡C triple (~839 kJ/mol).
Answer: E2 requires anti-periplanar geometry. In the meso compound, the two bromines are anti to each other, and anti-elimination gives (E)-stilbene (trans product).
Answer: The parent chain has 3 carbons with COOH at C1 and OH at C2, giving the IUPAC name 2-hydroxypropanoic acid (lactic acid is the common name).
Answer: The Sandmeyer reaction converts aryl diazonium salts to aryl halides (Ar-Cl, Ar-Br, Ar-CN) using copper(I) salts via a free radical mechanism.
Answer: Ni2+ is d8. With the strong field CN- ligand, all electrons pair up (low spin), and dsp2 hybridization gives a square planar, diamagnetic complex.