Grade 11 : Chemistry - Organic Chemistry Basics (Set 3)

Explanation: The central carbon forms two $\sigma$ bonds and two $\pi$ bonds with adjacent carbons. A carbon forming two $\sigma$ bonds requires sp hybridization and linear geometry.

Explanation: The ketone group ($-CO-$) has higher priority than the alcohol group ($-OH$). Numbering begins from the end closer to the principal functional group: C1 is on the right, making C2 the ketone and C4 the hydroxy substituent.

Explanation: When both double and triple bonds are present at equivalent distances from the ends, the double bond gets the lower locant. Numbering starts from the double bond end giving pent-1-en-4-yne.

Explanation: Metamerism arises due to different alkyl chain distributions around a polyvalent functional group (ether oxygen). Ethoxyethane has two ethyl groups while methoxypropane has a methyl and a propyl group.

Explanation: Geometrical isomerism requires each carbon of the double bond to bear two distinct groups. In propene, terminal carbon C1 carries two identical hydrogen atoms ($CH_2=$), preventing cis-trans isomerism.

Explanation: Electron-withdrawing group inductive effect ($-I$) stabilizes the carboxylate anion. Electronegativity order $F > Cl > Br > H$ results in the order $FCH_2COOH > ClCH_2COOH > BrCH_2COOH > CH_3COOH$.

Explanation: The tert-butyl carbocation $(CH_3)_3C^+$ is $3^\circ$ and possesses 9 hyperconjugative $\alpha$-hydrogens along with $+I$ effects from three methyl groups, making it the most stable.

Explanation: Hyperconjugation (no-bond resonance) involves delocalization of $\sigma$ electrons of a $C-H$ bond directly attached to an unsaturated system or electron-deficient carbon.

Explanation: Homolytic cleavage involves symmetrical splitting of a shared electron pair, where each departing atom retains one electron, producing neutral free radicals with unpaired electrons.

Explanation: Boron trifluoride ($BF_3$) has an incomplete octet on boron (6 valence electrons) and acts as an electron pair acceptor (electrophile). $NH_3, H_2O, CN^-$ act as nucleophiles.

Explanation: Hydroxyl group ($-OH$) takes precedence over nitro group ($-NO_2$). Phenol is the parent compound, numbering C1 at $-OH$, placing $-NO_2$ at C4: 4-nitrophenol.

Explanation: Sodium cyanide formed in sodium fusion extract reacts with ferrous sulfate and ferric ions to form iron(III) hexacyanoferrate(II), $Fe_4[Fe(CN)_6]_3$, giving Prussian blue.

Explanation: Mass of $N_2 = \frac{50}{22400} \times 28 g = 0.0625 g$. Percentage of $N = \frac{0.0625}{0.30} \times 100 = 20.83%$.

Explanation: Normality of 1 M $H_2SO_4 = 2 \text{ N}$. Using formula: $%N = \frac{1.4 \times N \times V}{m} = \frac{1.4 \times 2 \times 10}{0.50} = 56%$.

Explanation: Benzene contains 6 $C-C$ $\sigma$ bonds, 6 $C-H$ $\sigma$ bonds (total 12 $\sigma$ bonds), and 3 delocalized $\pi$ bonds within the aromatic ring.

Explanation: The Electromeric effect ($E$ effect) is a temporary effect involving complete transfer of a shared pair of $\pi$-electrons to one of the atoms in a multiple bond upon approach of an attacking reagent.

Explanation: Baeyer's reagent (cold dilute alkaline $KMnO_4$) oxidizes alkenes and alkynes to vicinal diols, decoloring the purple solution, indicating unsaturation.

Explanation: Cyanide ion ($CN^-$) possesses electron lone pairs and a negative charge, making it electron-rich and capable of donating an electron pair (nucleophile).

Explanation: Triple bond ($C \equiv C \approx 1.20 \text{ \AA}$) is shorter than double bond ($C = C \approx 1.34 \text{ \AA}$), which is shorter than single bond ($C - C \approx 1.54 \text{ \AA}$). Thus: Ethyne < Ethene < Ethane.

Explanation: Atomic ratios: $C = 40/12 = 3.33$, $H = 6.67/1 = 6.67$, $O = 53.33/16 = 3.33$. Dividing by 3.33 gives $C:H:O = 1:2:1$, yielding empirical formula $CH_2O$.