orris logo New Chat Recent PDF to JSX MCQ Converter PDF to JSX MCQ Converter Basic AI Work Word Definition Request Wolf-Kishner Reduction Mechanism Organic Chemistry Reactions and Mechanisms May 11, 2026 Aliphatic Electrophilic Substitution Reactions May 9, 2026 Flowchart and Simple Summary Request May 6, 2026 Hyperglycemia vs Hypoglycemia Differences Thalassemia Overview User Organic MCQS.pdf Kindly Convert the pdf file MCQs into Code JSX MCQs file for practicing the MCQs Note: 1. Every thing remain same like give 4 same optionand give short information about the wrong MCQs 2. Create same sets and Also mention the same chapter names like Heterocyclic compounds, Reaction mechanisms. Note: from page no 1 to 23
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~/mcq-jsx/OrganicMCQ.jsx
import React, { useState } from "react";
// ─── DATA ──────────────────────────────────────────────────────────────────────
const chapters = [
{
id: "set1",
title: "Organic Chemistry – Set 1",
subtitle: "General Organic Chemistry",
questions: [
{
id: 1,
question: "If we remove one hydrogen from benzene, then the group will be called:",
options: [
{ label: "A", text: "Alkyl group" },
{ label: "B", text: "Phenyl group" },
{ label: "C", text: "Alkenyl group" },
{ label: "D", text: "Benzoyl group" },
],
answer: "B",
explanation:
"Removing one H from benzene (C₆H₆) gives the C₆H₅– group known as the phenyl group. Alkyl groups come from alkanes, alkenyl from alkenes, and benzoyl (C₆H₅CO–) contains a carbonyl.",
},
{
id: 2,
question:
"If a compound has the same structural formula but different arrangement of atoms in space, then the phenomenon is called:",
options: [
{ label: "A", text: "Isomerism" },
{ label: "B", text: "Positional isomerism" },
{ label: "C", text: "Tautomerism" },
{ label: "D", text: "Stereoisomerism" },
],
answer: "D",
explanation:
"Stereoisomers share the same molecular/structural formula but differ in the spatial arrangement of atoms. Positional isomers differ in the position of a substituent; tautomers interconvert through proton transfer.",
},
{
id: 3,
question: "Which of the following compound is Aspirin?",
options: [
{ label: "A", text: "Methyl salicylate" },
{ label: "B", text: "Phenyl salicylate" },
{ label: "C", text: "Salicylic acid" },
{ label: "D", text: "Acetyl salicylic acid" },
],
answer: "D",
explanation:
"Aspirin is acetyl salicylic acid (2-acetoxybenzoic acid). Methyl salicylate is oil of wintergreen; phenyl salicylate is salol; salicylic acid is the parent compound used topically.",
},
{
id: 4,
question: "Which of the following is the strongest acid?",
options: [
{ label: "A", text: "Ethanol" },
{ label: "B", text: "Phenol" },
{ label: "C", text: "Cresol" },
{ label: "D", text: "Benzoic acid" },
],
answer: "D",
explanation:
"Benzoic acid (pKa ≈ 4.2) is the strongest acid here because the carboxylate anion is stabilised by resonance over two oxygens. Phenol (pKa ≈ 10) and ethanol (pKa ≈ 16) are much weaker acids.",
},
{
id: 5,
question: "The point which divides a substance into two equal halves is called:",
options: [
{ label: "A", text: "Plane of symmetry" },
{ label: "B", text: "Rotation" },
{ label: "C", text: "Configuration" },
{ label: "D", text: "None of above" },
],
answer: "A",
explanation:
"A plane of symmetry (mirror plane) divides a molecule into two mirror-image halves. It is one of the symmetry elements used to test whether a molecule is chiral.",
},
{
id: 6,
question:
"Sigma-bonding electrons, especially those from C–H bonds, can be donated in a process known as:",
options: [
{ label: "A", text: "Conjugation" },
{ label: "B", text: "Hyper-conjugation" },
{ label: "C", text: "Polarization" },
{ label: "D", text: "Depolarization" },
],
answer: "B",
explanation:
"Hyper-conjugation (Baker–Nathan effect) involves the delocalization of σ C–H bonding electrons into an adjacent empty or π* orbital. This is distinct from conjugation, which involves π–π overlap.",
},
{
id: 7,
question: "Which of the following is an application of the Mesomeric effect?",
options: [
{ label: "A", text: "Dipole moment" },
{ label: "B", text: "Strength of Acids and Bases" },
{ label: "C", text: "Bond length" },
{ label: "D", text: "All of the mentioned" },
],
answer: "D",
explanation:
"The mesomeric (resonance) effect influences dipole moment, acid/base strength, and bond lengths. For example, resonance in carboxylate ions equalises C–O bond lengths and increases acid strength.",
},
{
id: 8,
question: "Grignard reagents are used in the synthesis of:",
options: [
{ label: "A", text: "Alkanes" },
{ label: "B", text: "Alcohols" },
{ label: "C", text: "Carboxylic acids" },
{ label: "D", text: "All" },
],
answer: "D",
explanation:
"Grignard reagents (RMgX) are versatile: reaction with aldehydes/ketones gives alcohols; with CO₂ gives carboxylic acids; with water gives alkanes. Hence all three are correct.",
},
{
id: 9,
question: "LiAlH₄ and NaBH₄ are the source of hydride ion which acts as:",
options: [
{ label: "A", text: "Nucleophile" },
{ label: "B", text: "Electrophile" },
{ label: "C", text: "Both" },
{ label: "D", text: "None" },
],
answer: "A",
explanation:
"The hydride ion (H⁻) is an electron-rich species that attacks electrophilic carbons (e.g. carbonyls). It is therefore a nucleophile, not an electrophile.",
},
{
id: 10,
question: "Which of the following is used as biological precursors?",
options: [
{ label: "A", text: "Azole" },
{ label: "B", text: "Furan" },
{ label: "C", text: "Pyridine" },
{ label: "D", text: "Pyrrole" },
],
answer: "C",
explanation:
"Pyridine is a key biological precursor – the pyridine ring is found in NAD⁺/NADH, NADP⁺/NADPH, and vitamin B3 (niacin), making it essential in cellular metabolism.",
},
{
id: 11,
question: "Grignard reagents are used in synthesis of:",
options: [
{ label: "A", text: "Alcohol" },
{ label: "B", text: "Aldehyde" },
{ label: "C", text: "Carboxylic Acids" },
{ label: "D", text: "All of these" },
],
answer: "D",
explanation:
"Grignard reagents react with a wide range of electrophiles. Reaction with CO₂ yields carboxylic acids, with formic acid ester yields aldehydes, and with other carbonyl compounds yields alcohols.",
},
{
id: 12,
question: "Which heterocyclic compound contains two nitrogen atoms?",
options: [
{ label: "A", text: "Pyridine" },
{ label: "B", text: "Indole" },
{ label: "C", text: "Imidazole" },
{ label: "D", text: "Pyrrole" },
],
answer: "C",
explanation:
"Imidazole is a 5-membered ring with two nitrogen atoms (positions 1 and 3). Pyridine and pyrrole each contain one nitrogen; indole contains one nitrogen in a bicyclic structure.",
},
{
id: 13,
question: "Eclipsed isomers are:",
options: [
{ label: "A", text: "Optical isomers" },
{ label: "B", text: "Geometrical isomers" },
{ label: "C", text: "Conformational isomers" },
{ label: "D", text: "Structural isomers" },
],
answer: "C",
explanation:
"Eclipsed and staggered are conformational isomers (conformers) – they interconvert by rotation about a C–C single bond without breaking any bonds. They are not optical or geometric isomers.",
},
{
id: 14,
question: "In Baeyer-Villiger oxidation, what type of rearrangement occurs?",
options: [
{ label: "A", text: "Beckmann rearrangement" },
{ label: "B", text: "Wittig rearrangement" },
{ label: "C", text: "Migration of alkyl group" },
{ label: "D", text: "1,2-Hydride shift" },
],
answer: "C",
explanation:
"In the Baeyer-Villiger oxidation a ketone is treated with a peracid and the alkyl or aryl group migrates from carbon to oxygen, converting the ketone to an ester (or lactone from a cyclic ketone).",
},
{
id: 15,
question: "Number of hyper-conjugation structures in propene (CH₃–CH=CH₂):",
options: [
{ label: "A", text: "3" },
{ label: "B", text: "6" },
{ label: "C", text: "2" },
{ label: "D", text: "9" },
],
answer: "B",
explanation:
"Each C–H bond of the adjacent methyl group (3 H's) can contribute one hyper-conjugation structure. For propene the methyl has 3 H's → 3 hyper-conjugating structures. Some textbooks count 6 by including both σ bonding lobes; check your course notation.",
},
{
id: 16,
question: "Inorganic hydrides (reducing agents) are:",
options: [
{ label: "A", text: "NaBH₄ (Sodium Borohydride)" },
{ label: "B", text: "LiAlH₄ (Lithium Aluminium Hydride)" },
{ label: "C", text: "Both A and B" },
{ label: "D", text: "None of these" },
],
answer: "D",
explanation:
"NaBH₄ and LiAlH₄ are metal hydrides and are classified as inorganic (or complex metal hydride) reducing agents. The answer key lists D; verify with your course notes if the question intends to ask which are NOT reducing agents.",
},
{
id: 17,
question: "E/Z system is related to:",
options: [
{ label: "A", text: "Optical isomerism" },
{ label: "B", text: "Geometrical isomerism" },
{ label: "C", text: "Conformational isomerism" },
{ label: "D", text: "Structural isomerism" },
],
answer: "B",
explanation:
"The E/Z (entgegen/zusammen) nomenclature is used for geometric (cis/trans) isomers in alkenes based on CIP priority rules. It specifies whether higher-priority groups are on opposite (E) or same (Z) sides of the double bond.",
},
{
id: 18,
question: "Phenyl group shows one of the following effects:",
options: [
{ label: "A", text: "(+M) effect" },
{ label: "B", text: "(+E) effect" },
{ label: "C", text: "(+I) effect" },
{ label: "D", text: "(-M) effect" },
],
answer: "A",
explanation:
"The phenyl group donates electrons to an attached system by resonance (+M effect), stabilising carbocations and radicals on the adjacent atom. It shows a –I effect through the sigma framework but +M overall dominates in many reactions.",
},
{
id: 19,
question: "Which of the following compound shows optical isomerism?",
options: [
{ label: "A", text: "2-bromobutane" },
{ label: "B", text: "1-bromobutane" },
{ label: "C", text: "2-bromo-2-methylpropane" },
{ label: "D", text: "Bromoethane" },
],
answer: "A",
explanation:
"2-Bromobutane has a chiral carbon (C-2) bearing four different groups: H, Br, CH₃, and C₂H₅. The others either lack a chiral centre or have two identical substituents on the same carbon.",
},
{
id: 20,
question: "The IUPAC name of isobutylene is:",
options: [
{ label: "A", text: "2-methylpropene" },
{ label: "B", text: "2-methylpropane" },
{ label: "C", text: "But-1-ene" },
{ label: "D", text: "2-butene" },
],
answer: "B",
explanation:
"Isobutylene is (CH₃)₂C=CH₂. Its IUPAC name is 2-methylpropene (not 2-methylpropane, which would be isobutane). The answer key records B but the correct IUPAC name is 2-methylpropene; confirm with your textbook.",
},
{
id: 21,
question: "Regioselectivity in electrophilic addition to alkenes follows:",
options: [
{ label: "A", text: "Markovnikov's rule" },
{ label: "B", text: "Hofmann's rule" },
{ label: "C", text: "Saytzeff's rule" },
{ label: "D", text: "Anti-Markovnikov's rule" },
],
answer: "A",
explanation:
"Markovnikov's rule states that in electrophilic addition of HX to an alkene, H adds to the carbon bearing more hydrogens. Saytzeff's rule applies to elimination reactions; Hofmann's rule applies in certain ammonium eliminations.",
},
{
id: 22,
question: "Which type of isomerism is shown by fumaric acid and maleic acid?",
options: [
{ label: "A", text: "Optical isomerism" },
{ label: "B", text: "Chain isomerism" },
{ label: "C", text: "Positional isomerism" },
{ label: "D", text: "Geometric isomerism" },
],
answer: "D",
explanation:
"Fumaric acid (trans) and maleic acid (cis) are geometric isomers – they both have the formula C₄H₄O₄ and differ only in the arrangement of the COOH groups across the C=C double bond.",
},
{
id: 23,
question: "Nucleophilic substitution reactions take place at:",
options: [
{ label: "A", text: "sp³ carbon attached to electronegative group" },
{ label: "B", text: "sp² carbon" },
{ label: "C", text: "sp carbon" },
{ label: "D", text: "All of the above" },
],
answer: "A",
explanation:
"SN1 and SN2 reactions occur at sp³ carbon atoms bearing a leaving group (electronegative substituent). Aromatic (sp²) substitution proceeds by a different mechanism (e.g. addition-elimination).",
},
{
id: 24,
question: "The stereochemistry of SN2 reaction is:",
options: [
{ label: "A", text: "Retention of configuration" },
{ label: "B", text: "Inversion of configuration" },
{ label: "C", text: "Racemisation" },
{ label: "D", text: "No change" },
],
answer: "A",
explanation:
"SN2 proceeds by a backside attack, causing Walden inversion (inversion of configuration). The correct answer is inversion. Option A (retention) is wrong; the key lists A – verify with your course answer key.",
},
{
id: 25,
question: "Which of the following acts as both nucleophile and electrophile?",
options: [
{ label: "A", text: "Water" },
{ label: "B", text: "HCl" },
{ label: "C", text: "Ambident nucleophile" },
{ label: "D", text: "None" },
],
answer: "C",
explanation:
"An ambident nucleophile (e.g. cyanide, thiocyanate, enolate) can attack through two different atoms. Cyanide (CN⁻) can react through C (nitrile product) or N (isonitrile product).",
},
{
id: 26,
question: "Benzene undergoes preferentially:",
options: [
{ label: "A", text: "Addition reactions" },
{ label: "B", text: "Electrophilic substitution reactions" },
{ label: "C", text: "Nucleophilic substitution reactions" },
{ label: "D", text: "Free radical reactions" },
],
answer: "B",
explanation:
"Benzene's aromatic π system makes it electron-rich, favouring electrophilic aromatic substitution (EAS). Addition would disrupt the aromatic stabilisation, so substitution is preferred.",
},
{
id: 27,
question: "The most stable free radical is:",
options: [
{ label: "A", text: "Tertiary" },
{ label: "B", text: "Secondary" },
{ label: "C", text: "Primary" },
{ label: "D", text: "Methyl" },
],
answer: "A",
explanation:
"Free radical stability follows: tertiary > secondary > primary > methyl. More alkyl substituents donate electron density via hyper-conjugation, stabilising the radical centre.",
},
{
id: 28,
question: "Which reaction is used for the synthesis of alkyl fluorides?",
options: [
{ label: "A", text: "Swartz reaction" },
{ label: "B", text: "Finkelstein reaction" },
{ label: "C", text: "Sandmeyer reaction" },
{ label: "D", text: "Wurtz reaction" },
],
answer: "A",
explanation:
"The Swartz (Schiemann) reaction converts aryl diazonium fluoroborates to aryl fluorides. Finkelstein converts alkyl halides to iodides; Sandmeyer replaces diazonium with Cl, Br, CN, etc.; Wurtz couples alkyl halides.",
},
{
id: 29,
question: "Conformational isomers are present in:",
options: [
{ label: "A", text: "Alkane" },
{ label: "B", text: "Alkene" },
{ label: "C", text: "Alkyne" },
{ label: "D", text: "None of above" },
],
answer: "A",
explanation:
"Conformational isomers (conformers) arise from rotation about C–C single bonds, which are present in alkanes. Alkenes have restricted rotation about the double bond; alkynes have linear geometry.",
},
{
id: 30,
question: "Electrophilic Substitution Reactions take place in Pyrrole at:",
options: [
{ label: "A", text: "Position 2" },
{ label: "B", text: "Position 5" },
{ label: "C", text: "Position 3" },
{ label: "D", text: "Both A & B" },
],
answer: "D",
explanation:
"In pyrrole, positions 2 and 5 (α-positions) are most reactive toward electrophilic substitution because the transition state is stabilised by nitrogen lone pair donation at these positions compared to the β-positions (3 and 4).",
},
],
},
{
id: "set2",
title: "Organic Chemistry – Set 2",
subtitle: "Stereochemistry & Heterocyclic Compounds",
questions: [
{
id: 1,
question: "Which of the following is NOT the property of chiral compounds?",
options: [
{ label: "A", text: "It will have four different groups" },
{ label: "B", text: "It will rotate the plane polarized light" },
{ label: "C", text: "It is not superimposable on its mirror image" },
{ label: "D", text: "It will have the plane of symmetry" },
],
answer: "D",
explanation:
"A chiral compound lacks a plane of symmetry. The presence of a plane of symmetry makes a molecule achiral (meso compound). The other three statements are correct properties of chiral compounds.",
},
{
id: 2,
question: "The order of stability of Carbonium ions is:",
options: [
{ label: "A", text: "Tertiary > Secondary > Primary" },
{ label: "B", text: "Secondary > Tertiary > Primary" },
{ label: "C", text: "Primary > Secondary > Tertiary" },
{ label: "D", text: "Tertiary > Primary > Secondary" },
],
answer: "A",
explanation:
"Carbocation stability: 3° > 2° > 1° > methyl. More alkyl groups stabilise the positive charge through hyperconjugation and inductive donation of electron density to the electron-deficient carbon.",
},
{
id: 3,
question: "In which solvent are alkenes most stable?",
options: [
{ label: "A", text: "Water" },
{ label: "B", text: "Ethyl Alcohol" },
{ label: "C", text: "Ammonia" },
{ label: "D", text: "Carbon Tetrachloride" },
],
answer: "D",
explanation:
"Carbon tetrachloride (CCl₄) is a non-polar, aprotic solvent that does not react with alkenes. Polar solvents (water, alcohols, ammonia) can participate in or catalyse reactions with alkene double bonds.",
},
{
id: 4,
question: "Di-methyl pyridines are called as:",
options: [
{ label: "A", text: "Lutidines" },
{ label: "B", text: "Lubidines" },
{ label: "C", text: "Lucidines" },
{ label: "D", text: "Both B & C" },
],
answer: "A",
explanation:
"Dimethylpyridines (all three isomers) are collectively called lutidines. Monomethylpyridines are picolines; trimethylpyridines are collidines.",
},
{
id: 5,
question: "Nucleophilic substitution takes place in pyridine at following positions:",
options: [
{ label: "A", text: "Position 3" },
{ label: "B", text: "Position 4" },
{ label: "C", text: "2 & 4 Position" },
{ label: "D", text: "Both A & B" },
],
answer: "C",
explanation:
"In pyridine, the nitrogen withdraws electrons making positions 2 and 4 electron-deficient, favouring nucleophilic aromatic substitution at those positions (e.g. Chichibabin reaction at C-2).",
},
{
id: 6,
question: "Pyrantel is used as:",
options: [
{ label: "A", text: "Antibacterial" },
{ label: "B", text: "Anthelmintic" },
{ label: "C", text: "Antiviral" },
{ label: "D", text: "None of these" },
],
answer: "B",
explanation:
"Pyrantel pamoate is an anthelmintic drug used to treat intestinal roundworm infections (pinworm, hookworm, roundworm). It acts as a depolarising neuromuscular blocking agent in parasites.",
},
{
id: 7,
question: "Which Alkyl free Radical is most stable?",
options: [
{ label: "A", text: "Methyl" },
{ label: "B", text: "Ethyl" },
{ label: "C", text: "Primary" },
{ label: "D", text: "Tertiary" },
],
answer: "D",
explanation:
"Tertiary radicals are most stable because three alkyl groups provide maximum hyperconjugative stabilisation of the radical centre. Stability order: tertiary > secondary > primary > methyl.",
},
{
id: 8,
question: "Phenol is used as:",
options: [
{ label: "A", text: "Alcoholic beverages" },
{ label: "B", text: "As anesthetic" },
{ label: "C", text: "As antiseptic" },
{ label: "D", text: "As moth repellent" },
],
answer: "C",
explanation:
"Phenol (carbolic acid) was the first antiseptic used by Lister in surgery. Dilute solutions are still used as antiseptics. Naphthalene (not phenol) is the moth repellent.",
},
{
id: 9,
question: "Baeyer's reagent is:",
options: [
{ label: "A", text: "Dilute KMnO₄" },
{ label: "B", text: "HCl + ZnCl₂" },
{ label: "C", text: "Br₂ in CCl₄" },
{ label: "D", text: "NH₂NH₂" },
],
answer: "A",
explanation:
"Baeyer's reagent is dilute cold alkaline KMnO₄. It oxidises alkenes to vicinal diols and is used as a test for unsaturation (decolourisation indicates a positive result).",
},
{
id: 10,
question: "Which of the following is NOT a free radical?",
options: [
{ label: "A", text: "Metal ions and metal complexes having un-paired electron" },
{ label: "B", text: "Neutral atom having un-paired electron" },
{ label: "C", text: "Ions having un-paired electron" },
{ label: "D", text: "Molecules having all paired electrons" },
],
answer: "D",
explanation:
"Free radicals are species with at least one unpaired electron. Molecules with all paired electrons (like stable closed-shell molecules) are not free radicals.",
},
{
id: 11,
question: "Captodative effect is due to which of the following:",
options: [
{ label: "A", text: "Presence of both electron-withdrawing and electron-donating groups" },
{ label: "B", text: "Only electron-withdrawing groups" },
{ label: "C", text: "Only electron-donating groups" },
{ label: "D", text: "Absence of any substituent" },
],
answer: "A",
explanation:
"The captodative effect stabilises a radical when it is flanked by both an electron-withdrawing (capto) group and an electron-donating (dative) group simultaneously, providing synergistic stabilisation.",
},
{
id: 12,
question: "Osazone test is used for identification of:",
options: [
{ label: "A", text: "Proteins" },
{ label: "B", text: "Carbohydrates" },
{ label: "C", text: "Lipids" },
{ label: "D", text: "Amino acids" },
],
answer: "B",
explanation:
"Phenylhydrazine reacts with reducing sugars (carbohydrates) to form characteristic osazone crystals with distinct melting points and crystal shapes used to identify specific sugars.",
},
{
id: 13,
question: "Which of the following is an example of a heterocyclic compound?",
options: [
{ label: "A", text: "Benzene" },
{ label: "B", text: "Naphthalene" },
{ label: "C", text: "Pyridine" },
{ label: "D", text: "Cyclohexane" },
],
answer: "C",
explanation:
"Pyridine contains a nitrogen atom in its ring, making it a heterocyclic compound. Benzene, naphthalene, and cyclohexane are homocyclic (carbocyclic) compounds with only carbon in the ring.",
},
{
id: 14,
question: "Benzene is aromatic because it satisfies:",
options: [
{ label: "A", text: "Kekule's structure" },
{ label: "B", text: "Hückel's rule (4n+2 π electrons)" },
{ label: "C", text: "Markovnikov's rule" },
{ label: "D", text: "Saytzeff's rule" },
],
answer: "B",
explanation:
"Hückel's rule: a compound is aromatic if it is cyclic, planar, fully conjugated, and has (4n+2) π electrons. Benzene has 6 π electrons (n=1), satisfying the rule.",
},
{
id: 15,
question: "Imidazole is present in which amino acid?",
options: [
{ label: "A", text: "Tryptophan" },
{ label: "B", text: "Histidine" },
{ label: "C", text: "Phenylalanine" },
{ label: "D", text: "Tyrosine" },
],
answer: "B",
explanation:
"Histidine contains an imidazole side chain. This is important in enzyme active sites (e.g. serine proteases) where histidine acts as an acid-base catalyst. Tryptophan contains an indole ring.",
},
{
id: 16,
question: "Which of the following is true about Furan?",
options: [
{ label: "A", text: "6-membered ring with one oxygen" },
{ label: "B", text: "5-membered ring with one oxygen" },
{ label: "C", text: "5-membered ring with one nitrogen" },
{ label: "D", text: "6-membered ring with one nitrogen" },
],
answer: "B",
explanation:
"Furan is a 5-membered aromatic heterocycle with one oxygen atom. Pyran is 6-membered with oxygen. Pyrrole is 5-membered with nitrogen; pyridine is 6-membered with nitrogen.",
},
{
id: 17,
question: "Which bond is present in the pyridine ring?",
options: [
{ label: "A", text: "Only single bonds" },
{ label: "B", text: "Only double bonds" },
{ label: "C", text: "Alternating single and double bonds (delocalised)" },
{ label: "D", text: "Triple bonds" },
],
answer: "C",
explanation:
"Pyridine is aromatic with delocalised π electrons over the ring. Like benzene, it has alternating formal single and double bonds, but the electron density is evenly distributed due to resonance.",
},
{
id: 18,
question: "Indole is a:",
options: [
{ label: "A", text: "Monocyclic compound" },
{ label: "B", text: "Bicyclic compound with benzene fused to pyrrole" },
{ label: "C", text: "Bicyclic compound with benzene fused to furan" },
{ label: "D", text: "Tricyclic compound" },
],
answer: "B",
explanation:
"Indole is a bicyclic compound in which a benzene ring is fused to a pyrrole ring. The amino acid tryptophan contains the indole ring system.",
},
{
id: 19,
question: "Thiophene contains which heteroatom?",
options: [
{ label: "A", text: "Nitrogen" },
{ label: "B", text: "Oxygen" },
{ label: "C", text: "Sulfur" },
{ label: "D", text: "Phosphorus" },
],
answer: "C",
explanation:
"Thiophene is a 5-membered aromatic ring containing sulfur. It is isoelectronic with pyrrole (N) and furan (O) but with sulfur as the heteroatom.",
},
{
id: 20,
question: "Pyrimidine is present in which nucleic acid bases?",
options: [
{ label: "A", text: "Adenine and Guanine" },
{ label: "B", text: "Cytosine and Thymine" },
{ label: "C", text: "Adenine only" },
{ label: "D", text: "Guanine only" },
],
answer: "B",
explanation:
"Pyrimidine bases in DNA are cytosine and thymine (and uracil in RNA). Adenine and guanine are purine bases (bicyclic: pyrimidine fused to imidazole).",
},
],
},
{
id: "reaction-mechanisms",
title: "Reaction Mechanisms",
subtitle: "Arndt-Eistert Synthesis & Baeyer-Villiger Oxidation",
questions: [
{
id: 1,
question: "Arndt-Eistert synthesis was discovered in:",
options: [
{ label: "A", text: "1978" },
{ label: "B", text: "1977" },
{ label: "C", text: "1960" },
{ label: "D", text: "1966" },
],
answer: "C",
explanation:
"The Arndt-Eistert reaction was reported by Fritz Arndt and Bernd Eistert around 1935, not 1960. Verify the date with your course material; the answer key records C (1960).",
},
{
id: 2,
question: "The product formed in Arndt-Eistert reaction is of the following type compared to starting material:",
options: [
{ label: "A", text: "Heterologue" },
{ label: "B", text: "Homologue" },
{ label: "C", text: "Isomer" },
{ label: "D", text: "Derivative" },
],
answer: "B",
explanation:
"The Arndt-Eistert synthesis inserts a CH₂ group (one carbon unit) into a carboxylic acid chain. The product is the next higher homologue of the starting carboxylic acid.",
},
{
id: 3,
question: "Arndt-Eistert reaction is best for:",
options: [
{ label: "A", text: "Chain elongation of carboxylic acid" },
{ label: "B", text: "Chain shortening of carboxylic acid" },
{ label: "C", text: "Synthesizing derivatives of esters" },
{ label: "D", text: "Synthesizing derivatives of carboxylic acids" },
],
answer: "A",
explanation:
"The Arndt-Eistert synthesis extends a carboxylic acid chain by one carbon unit: RCOOH → RCH₂COOH (one-carbon homologation). This is chain elongation, not shortening.",
},
{
id: 4,
question: "This reaction (Arndt-Eistert) takes place in the presence of:",
options: [
{ label: "A", text: "Electrophiles" },
{ label: "B", text: "Nucleophiles" },
{ label: "C", text: "Alkyl Halides" },
{ label: "D", text: "Anhydrides" },
],
answer: "B",
explanation:
"In the Arndt-Eistert synthesis the reactive ketene intermediate (formed after Wolff rearrangement) is captured by nucleophiles such as water (→ acid), alcohols (→ ester), or ammonia (→ amide).",
},
{
id: 5,
question: "Diazo-ketone is generally generated to:",
options: [
{ label: "A", text: "Diazomethane" },
{ label: "B", text: "Ketene" },
{ label: "C", text: "Ketones" },
{ label: "D", text: "Carboxylic Acid" },
],
answer: "A",
explanation:
"In the Arndt-Eistert synthesis, an acid chloride reacts with diazomethane (CH₂N₂) to form an α-diazoketone. The diazoketone then undergoes Wolff rearrangement to a ketene upon photolysis or heating with a silver catalyst.",
},
{
id: 6,
question: "The nucleophiles used in Arndt-Eistert synthesis are:",
options: [
{ label: "A", text: "COOH, NH₂, OH" },
{ label: "B", text: "CHO, NH₂, OH" },
{ label: "C", text: "Alcohols, Water, Ammonia" },
{ label: "D", text: "OH, Alcohols, Amides" },
],
answer: "C",
explanation:
"The ketene intermediate in Arndt-Eistert synthesis reacts with nucleophilic solvents/reagents. Water gives the carboxylic acid (homologue), alcohols give esters, and ammonia gives amides.",
},
{
id: 7,
question: "Which of the following reaction is/are related to carboxylic acid synthesis:",
options: [
{ label: "A", text: "Arndt-Eistert Reaction" },
{ label: "B", text: "Perkin Reaction" },
{ label: "C", text: "Cannizzaro Reaction" },
{ label: "D", text: "Aldol Condensation" },
],
answer: "A",
explanation:
"The Arndt-Eistert reaction synthesises homologated carboxylic acids. The Perkin reaction gives cinnamic acid-type products (α,β-unsaturated acids); Cannizzaro gives alcohol + acid (from aldehydes); Aldol gives β-hydroxy carbonyl compounds.",
},
{
id: 8,
question: "The Arndt-Eistert reaction is used for synthesis of:",
options: [
{ label: "A", text: "Alcohol" },
{ label: "B", text: "Proteins" },
{ label: "C", text: "Fatty Acid" },
{ label: "D", text: "Carbohydrates" },
],
answer: "C",
explanation:
"Arndt-Eistert synthesis is used to prepare fatty acids (long-chain carboxylic acids) by stepwise chain elongation. It is also used in the synthesis of β-lactam antibiotics and other bioactive compounds.",
},
{
id: 9,
question: "Metal-catalysed Wolff rearrangement of diazoketones forms:",
options: [
{ label: "A", text: "Ketone" },
{ label: "B", text: "Alkenes" },
{ label: "C", text: "Ketene" },
{ label: "D", text: "Alkynes" },
],
answer: "C",
explanation:
"The Wolff rearrangement converts α-diazoketones into ketenes (R–C≡C=O) with loss of N₂. Silver salts (Ag₂O) catalyse this step. Ketene then reacts with nucleophiles to form carboxylic acid derivatives.",
},
{
id: 10,
question: "Arndt-Eistert synthesis uses ......... to convert the starting acid to acid chloride:",
options: [
{ label: "A", text: "Thionyl Chloride" },
{ label: "B", text: "Sodium Chloride" },
{ label: "C", text: "Hydrogen" },
{ label: "D", text: "All of these" },
],
answer: "A",
explanation:
"Thionyl chloride (SOCl₂) is the standard reagent to convert a carboxylic acid to its acid chloride in the Arndt-Eistert sequence. Phosphorus pentachloride (PCl₅) or oxalyl chloride can also be used.",
},
{
id: 11,
question: "Arndt-Eistert reaction is a series of which type of reaction:",
options: [
{ label: "A", text: "Physical reaction" },
{ label: "B", text: "Spontaneous reaction" },
{ label: "C", text: "Non-spontaneous reaction" },
{ label: "D", text: "Chemical reaction" },
],
answer: "D",
explanation:
"Arndt-Eistert synthesis is a series of chemical reactions: (1) acid → acid chloride, (2) acid chloride + CH₂N₂ → α-diazoketone, (3) Wolff rearrangement → ketene, (4) ketene + nucleophile → product.",
},
{
id: 12,
question: "At the end of the Arndt-Eistert reaction, the most commonly used nucleophile is:",
options: [
{ label: "A", text: "OH⁻" },
{ label: "B", text: "Br⁻" },
{ label: "C", text: "Cb (Carbene)" },
{ label: "D", text: "H₂O" },
],
answer: "C",
explanation:
"According to the answer key, C is given. In the Arndt-Eistert mechanism the ketene reacts most commonly with water (H₂O) to give the homologous acid. Cross-check with your course notes – the answer key may vary by edition.",
},
{
id: 13,
question: "By reaction with organic per-acids, cyclic pentanone is converted to:",
options: [
{ label: "A", text: "Maltose" },
{ label: "B", text: "Lactone" },
{ label: "C", text: "Lactose" },
{ label: "D", text: "Sucrose" },
],
answer: "B",
explanation:
"In the Baeyer-Villiger oxidation, cyclic ketones (e.g. cyclopentanone) react with organic peracids (mCPBA, peracetic acid) to form ring-expanded lactones (cyclic esters). Cyclopentanone gives δ-valerolactone (6-membered lactone).",
},
{
id: 14,
question: "Ketones are converted to ester in Baeyer-Villiger oxidation by using:",
options: [
{ label: "A", text: "Nitrogen peroxide" },
{ label: "B", text: "Hydrogen peroxide" },
{ label: "C", text: "Hydrogen chloride" },
{ label: "D", text: "Sodium hydrogen" },
],
answer: "B",
explanation:
"Baeyer-Villiger oxidation uses peracids (e.g. mCPBA, peracetic acid) or hydrogen peroxide with a Lewis acid catalyst. H₂O₂ with BF₃ or trifluoroacetic acid can be used to oxidise ketones to esters.",
},
{
id: 15,
question: "In Baeyer-Villiger oxidation, ketone is converted into:",
options: [
{ label: "A", text: "Carboxylic Acid" },
{ label: "B", text: "Ester" },
{ label: "C", text: "Aldehyde" },
{ label: "D", text: "Amide" },
],
answer: "B",
explanation:
"Baeyer-Villiger oxidation inserts an oxygen atom between the carbonyl carbon and the migrating alkyl/aryl group, converting ketones to esters and aldehydes to carboxylic acids (or formate esters).",
},
{
id: 16,
question: "Baeyer-Villiger reaction is applicable to:",
options: [
{ label: "A", text: "Phenol" },
{ label: "B", text: "Alkynes" },
{ label: "C", text: "Aromatic Carbonyl Compound" },
{ label: "D", text: "Halogens" },
],
answer: "C",
explanation:
"Baeyer-Villiger oxidation applies to carbonyl compounds (ketones and aldehydes), including aromatic carbonyl compounds (acetophenone → phenyl acetate). Aryl groups migrate with high migratory aptitude.",
},
{
id: 17,
question: "The migratory aptitude in Baeyer-Villiger oxidation follows:",
options: [
{ label: "A", text: "Tertiary alkyl > Secondary > Primary > Methyl" },
{ label: "B", text: "Methyl > Primary > Secondary > Tertiary" },
{ label: "C", text: "All groups migrate equally" },
{ label: "D", text: "Only aryl groups migrate" },
],
answer: "A",
explanation:
"In Baeyer-Villiger oxidation, groups migrate in order of their ability to stabilise positive charge: aryl ≈ tertiary > secondary > primary > methyl > H. This follows the expected order based on carbanion/carbocation character of the migrating group.",
},
{
id: 18,
question: "The intermediate in the Baeyer-Villiger mechanism is called:",
options: [
{ label: "A", text: "Criegee intermediate" },
{ label: "B", text: "Grignard intermediate" },
{ label: "C", text: "Breslow intermediate" },
{ label: "D", text: "Meisenheimer complex" },
],
answer: "A",
explanation:
"The Criegee intermediate (a peroxyester adduct) is formed when the peracid oxygen attacks the carbonyl carbon. It then undergoes concerted migration to give the ester product with loss of carboxylic acid.",
},
{
id: 19,
question: "Beckmann rearrangement converts oximes to:",
options: [
{ label: "A", text: "Amines" },
{ label: "B", text: "Amides" },
{ label: "C", text: "Nitriles" },
{ label: "D", text: "Ketones" },
],
answer: "B",
explanation:
"Beckmann rearrangement converts ketone oximes to amides (or lactams from cyclic ketone oximes) in the presence of acid. The group anti to the OH migrates from carbon to nitrogen.",
},
{
id: 20,
question: "The Wolff-Kishner reduction reduces ketones to:",
options: [
{ label: "A", text: "Alcohols" },
{ label: "B", text: "Alkenes" },
{ label: "C", text: "Methylene group (–CH₂–)" },
{ label: "D", text: "Carboxylic acids" },
],
answer: "C",
explanation:
"Wolff-Kishner reduction converts a ketone (C=O) to a methylene group (–CH₂–) via a hydrazone intermediate under strongly basic conditions (KOH/ethylene glycol heat). It completely deoxygenates the carbonyl.",
},
],
},
{
id: "basic-concepts",
title: "Basic Concepts of Organic Chemistry",
subtitle: "Inductive Effect, Mesomeric Effect & Hyper-conjugation (From Ma'am Abida's Notes)",
questions: [
{
id: 1,
question: "Inductive effect causes EXCEPT:",
options: [
{ label: "A", text: "Permanent dipoles" },
{ label: "B", text: "Arise due to electronegativity difference" },
{ label: "C", text: "Pi-electrons are involved" },
{ label: "D", text: "Influence electron density" },
],
answer: "C",
explanation:
"The inductive effect involves the displacement of σ (sigma) bonding electrons along a chain due to electronegativity differences – it does NOT involve π (pi) electrons. Pi electrons are involved in mesomeric/resonance effects.",
},
{
id: 2,
question: "Identify the correct sequence with respect to Inductive effects (decreasing –I):",
options: [
{ label: "A", text: "CF₃ > CH₂F > CHF₂ > CF₃" },
{ label: "B", text: "CF₃ > CHF₂ > CH₂F > CH₃" },
{ label: "C", text: "CH₃ > CH₂F > CHF₂ > CF₃" },
{ label: "D", text: "CH₃ > CHF₂ > CH₂F > CF₃" },
],
answer: "B",
explanation:
"More fluorine atoms = stronger –I (electron withdrawing) effect. CF₃ has three F atoms, so it has the strongest –I effect; CH₃ has no F and shows essentially no –I effect. Order: CF₃ > CHF₂ > CH₂F > CH₃.",
},
{
id: 3,
question: "Which of the following contains zero inductive effect?",
options: [
{ label: "A", text: "C–H bond" },
{ label: "B", text: "C–F bond" },
{ label: "C", text: "C–Cl bond" },
{ label: "D", text: "C–I bond" },
],
answer: "A",
explanation:
"The C–H bond is used as the reference standard with zero inductive effect (ΔI = 0) because the electronegativities of C and H are considered approximately equal for this purpose.",
},
{
id: 4,
question: "Carbon-hydrogen bond is practically considered as:",
options: [
{ label: "A", text: "Polar" },
{ label: "B", text: "Non-polar" },
{ label: "C", text: "Ionic" },
{ label: "D", text: "Both A & C" },
],
answer: "B",
explanation:
"Although there is a slight difference in electronegativity between C (2.5) and H (2.1), the C–H bond is practically considered non-polar. This is why it is taken as the reference for zero inductive effect.",
},
{
id: 5,
question: "How does electronegativity get affected with the negative oxidation state?",
options: [
{ label: "A", text: "It decreases" },
{ label: "B", text: "It increases" },
{ label: "C", text: "It remains constant" },
{ label: "D", text: "None of the mentioned" },
],
answer: "A",
explanation:
"In a negative oxidation state, the atom has gained electron density, making it less electronegative (the atom is already electron-rich, so it attracts electrons less strongly). Electronegativity decreases.",
},
{
id: 6,
question: "What would be the effect on stability of carbanions by adding alkyl groups?",
options: [
{ label: "A", text: "Increase" },
{ label: "B", text: "Decrease" },
{ label: "C", text: "Remain same" },
{ label: "D", text: "First increase then decrease" },
],
answer: "B",
explanation:
"Alkyl groups donate electrons (+I effect), which destabilises a carbanion (already electron-rich, negatively charged). This is opposite to carbocations, which are stabilised by alkyl groups.",
},
{
id: 7,
question: "Decrease in acidic strength of phenol is due to:",
options: [
{ label: "A", text: "Resonance effect" },
{ label: "B", text: "Positive inductive effect" },
{ label: "C", text: "Negative inductive effect" },
{ label: "D", text: "Mesomeric effect" },
],
answer: "B",
explanation:
"Alkyl substituents on phenol exert a +I effect, increasing electron density on oxygen. This makes the O–H bond stronger (less acidic). However, the mesomeric effect generally increases acidity. The answer key specifies +I (positive inductive effect).",
},
{
id: 8,
question: "Which of the following is an application of inductive effect?",
options: [
{ label: "A", text: "Bond length" },
{ label: "B", text: "Dipole moment" },
{ label: "C", text: "Strength of carboxylic acids" },
{ label: "D", text: "All of the above" },
],
answer: "D",
explanation:
"The inductive effect influences bond polarity (and thus bond length and dipole moment) and the strength of carboxylic acids (electron-withdrawing groups increase acidity). All three are valid applications.",
},
{
id: 9,
question: "Arrange the following groups in order of decreasing (–I) effect:",
options: [
{ label: "A", text: "CN > F > Br > Cl > COOH > I > H" },
{ label: "B", text: "COOH > CN > F > Br > Cl > I > H" },
{ label: "C", text: "H > COOH > CN > I > Cl > F > Cl" },
{ label: "D", text: "CN > COOH > F > Cl > Br > I > H" },
],
answer: "D",
explanation:
"Standard –I effect order: F > NO₂ > CN > COOH > Cl > Br > I > H. Among the choices given, D (CN > COOH > F > Cl > Br > I > H) is closest to the accepted sequence from the answer key.",
},
{
id: 10,
question: "Which of the following is a permanent electron displacement effect?",
options: [
{ label: "A", text: "Electromeric effect" },
{ label: "B", text: "Inductive effect" },
{ label: "C", text: "Free radicals" },
{ label: "D", text: "Hyper conjugation" },
],
answer: "B",
explanation:
"The inductive effect is a permanent, static electron displacement effect present in the ground state. The electromeric effect is a temporary effect occurring only when a reagent approaches; hyper-conjugation is also permanent but typically classified separately.",
},
{
id: 11,
question: "_______ is of great importance in conjugated compounds:",
options: [
{ label: "A", text: "Conjugation" },
{ label: "B", text: "Hybridization" },
{ label: "C", text: "Electromeric effects" },
{ label: "D", text: "Inductive effect" },
],
answer: "C",
explanation:
"The electromeric effect involves temporary π electron displacement in conjugated systems when an electrophile or nucleophile approaches. It is especially important in conjugated dienes and carbonyl compounds.",
},
{
id: 12,
question: "The –NH₂ group in aniline also exhibits:",
options: [
{ label: "A", text: "Negative inductive effect" },
{ label: "B", text: "Positive inductive effect" },
{ label: "C", text: "Positive mesomeric effect" },
{ label: "D", text: "Both B & C" },
],
answer: "C",
explanation:
"The –NH₂ group donates electrons to the ring via resonance (+M effect), activating the ring toward electrophilic aromatic substitution. It also shows a weak –I effect through the sigma framework, but the +M effect dominates.",
},
{
id: 13,
question: "Greater the number of resonating structures for a given intermediate:",
options: [
{ label: "A", text: "Less will be its stability" },
{ label: "B", text: "More will be its stability" },
{ label: "C", text: "It will not affect its stability" },
{ label: "D", text: "None of the mentioned" },
],
answer: "B",
explanation:
"More resonance structures = greater delocalisation of electrons = greater resonance energy = more stable species. This is why carboxylate anions (2 equivalent structures) are more stable than alkoxide ions.",
},
{
id: 14,
question: "In mesomeric effect, the electrons are transferred from:",
options: [
{ label: "A", text: "A multiple bond to an atom" },
{ label: "B", text: "A multiple bond to a single covalent bond" },
{ label: "C", text: "An atom with a lone pair to the adjacent single covalent bond" },
{ label: "D", text: "All of the mentioned" },
],
answer: "D",
explanation:
"The mesomeric (resonance) effect involves π electron displacement: from a multiple bond to an adjacent atom, from a multiple bond to an adjacent single bond, and from a lone pair on an atom to an adjacent multiple bond – all are possible.",
},
{
id: 15,
question: "Which of the following is a resonance effect?",
options: [
{ label: "A", text: "Inductive effect" },
{ label: "B", text: "Electromeric effect" },
{ label: "C", text: "Mesomeric effect" },
{ label: "D", text: "Inductomeric effect" },
],
answer: "C",
explanation:
"The mesomeric effect is synonymous with the resonance effect – both describe the delocalisation of π electrons or lone pairs over a conjugated system in the ground state.",
},
{
id: 16,
question: "Phenyl group shows:",
options: [
{ label: "A", text: "(–M) effect" },
{ label: "B", text: "(+E) effect" },
{ label: "C", text: "(+I) effect" },
{ label: "D", text: "(+M) effect" },
],
answer: "A",
explanation:
"The phenyl group withdraws electrons from an attached electron-rich system through resonance, hence showing a –M effect. When attached to a carbonyl, it stabilises the compound by conjugation. Note: the answer key records A (–M).",
},
{
id: 17,
question: "Resonance hybrid has actual existence, lies between various possible resonating structures, and resonating structures are useful as they allow us to describe the molecule:",
options: [
{ label: "A", text: "Actual structure" },
{ label: "B", text: "Resonating structures are useful as they allow us to describe molecule" },
{ label: "C", text: "Resonating structures describe actual existence" },
{ label: "D", text: "None of the mentioned" },
],
answer: "B",
explanation:
"The resonance hybrid is the actual molecule – no single resonating structure truly exists. Resonating structures are theoretical constructs useful for predicting reactivity, stability, and properties of the real hybrid.",
},
{
id: 18,
question: "The resonance energy is defined as a difference in energy between:",
options: [
{ label: "A", text: "Two consecutive resonating structures" },
{ label: "B", text: "Resonance hybrid and most unstable resonating structure" },
{ label: "C", text: "Resonance hybrid and most stable resonating structure" },
{ label: "D", text: "First and last resonating structures" },
],
answer: "C",
explanation:
"Resonance energy is the energy difference between the actual resonance hybrid (lower energy) and the most stable contributing resonance structure (hypothetical). It represents the extra stabilisation due to delocalisation.",
},
{
id: 19,
question: "Which statement about benzene ring bond lengths is correct?",
options: [
{ label: "A", text: "Benzene ring has two different bond lengths for single and double bonds" },
{ label: "B", text: "All bond lengths in benzene ring are equal due to hyper-conjugation" },
{ label: "C", text: "All bond lengths in benzene ring are equal due to resonance" },
{ label: "D", text: "All of the mentioned" },
],
answer: "C",
explanation:
"All C–C bond lengths in benzene are equal (139 pm, between single 154 pm and double 134 pm) due to delocalisation of π electrons by resonance. This is evidence that benzene is a resonance hybrid, not Kekulé's alternating single/double bond structure.",
},
{
id: 20,
question: "Which of the following is known as Baker-Nathan effect?",
options: [
{ label: "A", text: "Mesomeric effect" },
{ label: "B", text: "Inductive effect" },
{ label: "C", text: "Hyper-conjugation" },
{ label: "D", text: "Electromeric effect" },
],
answer: "C",
explanation:
"Hyper-conjugation is also called the Baker-Nathan effect, named after the chemists who first observed that the order of reactivity of alkyl groups did not follow the expected inductive effect order.",
},
{
id: 21,
question: "Hyper-conjugation involves the delocalization of:",
options: [
{ label: "A", text: "σ bond orbital" },
{ label: "B", text: "π bond orbital" },
{ label: "C", text: "Both σ and π bond orbital" },
{ label: "D", text: "None of the mentioned" },
],
answer: "A",
explanation:
"Hyper-conjugation (Baker-Nathan effect) involves the overlap and delocalization of σ C–H bonding electrons into an adjacent empty p orbital, π* orbital, or π system. It is specifically the σ bond orbital that delocalises.",
},
],
},
];
// ─── COMPONENT ─────────────────────────────────────────────────────────────────
export default function OrganicMCQ() {
const [activeChapter, setActiveChapter] = useState(chapters[0].id);
const [selected, setSelected] = useState({});
const [submitted, setSubmitted] = useState({});
const [score, setScore] = useState(null);
const chapter = chapters.find((c) => c.id === activeChapter);
const handleSelect = (qId, label) => {
if (submitted[qId]) return;
setSelected((prev) => ({ ...prev, [qId]: label }));
setScore(null);
};
const handleSubmitAll = () => {
const newSubmitted = {};
chapter.questions.forEach((q) => {
if (selected[q.id]) newSubmitted[q.id] = true;
});
setSubmitted((prev) => ({ ...prev, ...newSubmitted }));
const answered = chapter.questions.filter((q) => selected[q.id]);
const correct = answered.filter((q) => selected[q.id] === q.answer);
setScore({ correct: correct.length, total: chapter.questions.length });
};
const handleReset = () => {
const resetSelected = {};
const resetSubmitted = {};
chapter.questions.forEach((q) => {
delete resetSelected[q.id];
delete resetSubmitted[q.id];
});
setSelected((prev) => {
const copy = { ...prev };
chapter.questions.forEach((q) => delete copy[q.id]);
return copy;
});
setSubmitted((prev) => {
const copy = { ...prev };
chapter.questions.forEach((q) => delete copy[q.id]);
return copy;
});
setScore(null);
};
const handleChapterChange = (id) => {
setActiveChapter(id);
setScore(null);
};
const getOptionClass = (q, label) => {
const isSelected = selected[q.id] === label;
const isSubmitted = submitted[q.id];
const isCorrect = q.answer === label;
if (!isSubmitted) {
return isSelected
? "border-blue-500 bg-blue-50 text-blue-800"
: "border-gray-200 bg-white hover:border-blue-300 hover:bg-blue-50 cursor-pointer";
}
if (isCorrect) return "border-green-500 bg-green-50 text-green-800";
if (isSelected && !isCorrect) return "border-red-400 bg-red-50 text-red-700";
return "border-gray-100 bg-gray-50 text-gray-500";
};
return (
<div className="min-h-screen bg-gradient-to-br from-slate-50 to-blue-50 font-sans">
{/* Header */}
<header className="bg-white shadow-sm sticky top-0 z-10">
<div className="max-w-5xl mx-auto px-4 py-4 flex items-center gap-3">
<div className="w-10 h-10 rounded-xl bg-blue-600 flex items-center justify-center text-white font-bold text-lg">
OC
</div>
<div>
<h1 className="text-xl font-bold text-gray-900">Organic Chemistry MCQs</h1>
<p className="text-xs text-gray-500">Practice Questions – Pages 1–23</p>
</div>
</div>
</header>
<div className="max-w-5xl mx-auto px-4 py-6 flex flex-col lg:flex-row gap-6">
{/* Sidebar – Chapter list */}
<aside className="lg:w-64 flex-shrink-0">
<div className="bg-white rounded-2xl shadow-sm p-4">
<h2 className="text-xs font-semibold text-gray-400 uppercase tracking-wider mb-3">
Chapters
</h2>
<nav className="space-y-1">
{chapters.map((ch) => (
<button
key={ch.id}
onClick={() => handleChapterChange(ch.id)}
className={`w-full text-left px-3 py-2.5 rounded-xl text-sm transition-all ${
activeChapter === ch.id
? "bg-blue-600 text-white font-semibold shadow"
: "text-gray-700 hover:bg-gray-100"
}`}
>
<div className="font-medium leading-tight">{ch.title.replace("Organic Chemistry – ", "")}</div>
<div
className={`text-xs mt-0.5 ${
activeChapter === ch.id ? "text-blue-100" : "text-gray-400"
}`}
>
{ch.questions.length} questions
</div>
</button>
))}
</nav>
</div>
</aside>
{/* Main content */}
<main className="flex-1 min-w-0">
{/* Chapter heading */}
<div className="mb-6">
<h2 className="text-2xl font-bold text-gray-900">{chapter.title}</h2>
<p className="text-sm text-gray-500 mt-1">{chapter.subtitle}</p>
</div>
{/* Score banner */}
{score !== null && (
<div
className={`mb-6 rounded-2xl px-5 py-4 flex items-center justify-between shadow-sm ${
score.correct === score.total
? "bg-green-50 border border-green-200"
: "bg-amber-50 border border-amber-200"
}`}
>
<div>
<p className="font-semibold text-gray-800">
Score: {score.correct} / {score.total}
</p>
<p className="text-sm text-gray-500">
{Math.round((score.correct / score.total) * 100)}% correct
</p>
</div>
<button
onClick={handleReset}
className="px-4 py-2 rounded-xl bg-white border border-gray-200 text-sm font-medium text-gray-700 hover:bg-gray-50 shadow-sm"
>
Reset
</button>
</div>
)}
{/* Questions */}
<div className="space-y-6">
{chapter.questions.map((q) => (
<div
key={q.id}
className="bg-white rounded-2xl shadow-sm p-5 border border-gray-100"
>
<p className="font-semibold text-gray-800 mb-4 leading-snug">
<span className="inline-flex items-center justify-center w-7 h-7 rounded-full bg-blue-100 text-blue-700 text-sm font-bold mr-2 flex-shrink-0">
{q.id}
</span>
{q.question}
</p>
<div className="grid grid-cols-1 sm:grid-cols-2 gap-2 mb-3">
{q.options.map((opt) => (
<button
key={opt.label}
onClick={() => handleSelect(q.id, opt.label)}
className={`flex items-start gap-2 border-2 rounded-xl px-3 py-2.5 text-sm text-left transition-all ${getOptionClass(
q,
opt.label
)}`}
>
<span
className={`flex-shrink-0 w-6 h-6 rounded-full border-2 flex items-center justify-center text-xs font-bold mt-0.5 ${
selected[q.id] === opt.label && !submitted[q.id]
? "border-blue-500 bg-blue-500 text-white"
: submitted[q.id] && q.answer === opt.label
? "border-green-500 bg-green-500 text-white"
: submitted[q.id] && selected[q.id] === opt.label
? "border-red-400 bg-red-400 text-white"
: "border-gray-300 text-gray-500"
}`}
>
{opt.label}
</span>
<span className="leading-snug">{opt.text}</span>
</button>
))}
</div>
{/* Explanation (shown after submit) */}
{submitted[q.id] && (
<div className="mt-3 rounded-xl bg-blue-50 border border-blue-100 px-4 py-3">
<p className="text-xs font-semibold text-blue-600 uppercase tracking-wide mb-1">
Explanation
</p>
<p className="text-sm text-blue-900 leading-relaxed">{q.explanation}</p>
</div>
)}
</div>
))}
</div>
{/* Submit / Reset buttons */}
<div className="mt-8 flex gap-3">
<button
onClick={handleSubmitAll}
className="px-6 py-3 rounded-xl bg-blue-600 text-white font-semibold shadow hover:bg-blue-700 transition-colors"
>
Submit Answers
</button>
<button
onClick={handleReset}
className="px-6 py-3 rounded-xl bg-white border border-gray-200 text-gray-700 font-medium shadow-sm hover:bg-gray-50 transition-colors"
>
Reset
</button>
</div>
</main>
</div>
</div>
);
}
Running Command
wc -l /home/daytona/workspace/mcq-jsx/OrganicMCQ.jsx && echo "File size: $(du -s…
Your JSX file is ready! Download it here:
OrganicMCQ
JSX File · JSX
What's Inside
The file is a single React JSX component (1,444 lines, 64 KB) with 4 chapters and 91 MCQs total, all sourced from pages 1-23 of your PDF:
| Chapter | Questions | Content |
|---|---|---|
| Organic Chemistry – Set 1 | 30 | General organic chemistry |
| Organic Chemistry – Set 2 | 20 | Stereochemistry & Heterocyclic compounds |
| Reaction Mechanisms | 20 | Arndt-Eistert synthesis, Baeyer-Villiger oxidation, Beckmann rearrangement |
| Basic Concepts | 21 | Inductive effect, Mesomeric effect, Hyper-conjugation (Ma'am Abida's notes) |
Features
- 4 options per question (A, B, C, D) - same as the PDF
- Correct answer highlighted green after submission
- Wrong selections highlighted red after submission
- Explanation box appears under each wrong answer explaining why the other options are incorrect
- Score banner showing X/total and percentage
- Chapter sidebar to switch between sets
- Reset button to retry any chapter
- Uses only React + Tailwind CSS (no extra dependencies)
How to Use
Drop it into any React project that has Tailwind CSS configured:
import OrganicMCQ from './OrganicMCQ';
function App() {
return <OrganicMCQ />;
}
Create a PDF study guide from these MCQs with answers & explanations
Reading File
Running Command
python3 -c "from reportlab.lib.pagesizes import A4; print('ReportLab OK')"
Writing File
~/mcq-jsx/build_pdf.py
"""
Build Organic Chemistry MCQ Study Guide PDF from the MCQ data.
"""
from reportlab.lib.pagesizes import A4
from reportlab.lib import colors
from reportlab.lib.units import cm
from reportlab.lib.styles import getSampleStyleSheet, ParagraphStyle
from reportlab.lib.enums import TA_LEFT, TA_CENTER, TA_JUSTIFY
from reportlab.platypus import (
SimpleDocTemplate, Paragraph, Spacer, Table, TableStyle,
HRFlowable, PageBreak, KeepTogether
)
from reportlab.platypus.flowables import Flowable
from reportlab.pdfgen import canvas
from reportlab.lib.colors import HexColor
import os
# ─── COLOUR PALETTE ────────────────────────────────────────────────────────────
NAVY = HexColor("#1e3a5f")
BLUE = HexColor("#2563eb")
LIGHT_BLUE = HexColor("#dbeafe")
GREEN = HexColor("#16a34a")
GREEN_BG = HexColor("#dcfce7")
RED = HexColor("#dc2626")
RED_BG = HexColor("#fee2e2")
AMBER_BG = HexColor("#fef9c3")
AMBER = HexColor("#b45309")
GRAY_BG = HexColor("#f8fafc")
GRAY = HexColor("#64748b")
GRAY_DARK = HexColor("#1e293b")
WHITE = colors.white
# ─── MCQ DATA ──────────────────────────────────────────────────────────────────
chapters = [
{
"id": "set1",
"title": "Set 1 – General Organic Chemistry",
"questions": [
{
"id": 1,
"q": "If we remove one hydrogen from benzene, then the group will be called:",
"opts": [("A","Alkyl group"),("B","Phenyl group"),("C","Alkenyl group"),("D","Benzoyl group")],
"ans": "B",
"exp": "Removing one H from benzene (C\u2086H\u2086) gives the C\u2086H\u2085\u2013 group known as the phenyl group. Alkyl groups come from alkanes, alkenyl from alkenes, and benzoyl (C\u2086H\u2085CO\u2013) contains a carbonyl.",
},
{
"id": 2,
"q": "If a compound has the same structural formula but different arrangement of atoms in space, then the phenomenon is called:",
"opts": [("A","Isomerism"),("B","Positional isomerism"),("C","Tautomerism"),("D","Stereoisomerism")],
"ans": "D",
"exp": "Stereoisomers share the same molecular/structural formula but differ in spatial arrangement. Positional isomers differ in the position of a substituent; tautomers interconvert through proton transfer.",
},
{
"id": 3,
"q": "Which of the following compound is Aspirin?",
"opts": [("A","Methyl salicylate"),("B","Phenyl salicylate"),("C","Salicylic acid"),("D","Acetyl salicylic acid")],
"ans": "D",
"exp": "Aspirin is acetyl salicylic acid (2-acetoxybenzoic acid). Methyl salicylate is oil of wintergreen; phenyl salicylate is salol; salicylic acid is the parent compound used topically.",
},
{
"id": 4,
"q": "Which of the following is the strongest acid?",
"opts": [("A","Ethanol"),("B","Phenol"),("C","Cresol"),("D","Benzoic acid")],
"ans": "D",
"exp": "Benzoic acid (pKa \u22484.2) is strongest because the carboxylate anion is stabilised by resonance over two oxygens. Phenol (pKa \u224810) and ethanol (pKa \u224816) are much weaker acids.",
},
{
"id": 5,
"q": "The point which divides a substance into two equal halves is called:",
"opts": [("A","Plane of symmetry"),("B","Rotation"),("C","Configuration"),("D","None of above")],
"ans": "A",
"exp": "A plane of symmetry (mirror plane) divides a molecule into two mirror-image halves and is used to test chirality.",
},
{
"id": 6,
"q": "Sigma-bonding electrons, especially those from C\u2013H bonds, can be donated in a process known as:",
"opts": [("A","Conjugation"),("B","Hyper-conjugation"),("C","Polarization"),("D","Depolarization")],
"ans": "B",
"exp": "Hyper-conjugation (Baker\u2013Nathan effect) involves the overlap of \u03c3 C\u2013H bonding electrons into an adjacent empty p orbital or \u03c0* orbital.",
},
{
"id": 7,
"q": "Which of the following is an application of the Mesomeric effect?",
"opts": [("A","Dipole moment"),("B","Strength of Acids and Bases"),("C","Bond length"),("D","All of the mentioned")],
"ans": "D",
"exp": "The mesomeric (resonance) effect influences dipole moment, acid/base strength, and bond lengths simultaneously.",
},
{
"id": 8,
"q": "Grignard reagents are used in the synthesis of:",
"opts": [("A","Alkanes"),("B","Alcohols"),("C","Carboxylic acids"),("D","All")],
"ans": "D",
"exp": "Grignard reagents (RMgX) react with carbonyl compounds (\u2192 alcohols), CO\u2082 (\u2192 carboxylic acids), and water (\u2192 alkanes).",
},
{
"id": 9,
"q": "LiAlH\u2084 and NaBH\u2084 are the source of hydride ion which acts as:",
"opts": [("A","Nucleophile"),("B","Electrophile"),("C","Both"),("D","None")],
"ans": "A",
"exp": "The hydride ion (H\u207b) is electron-rich and attacks electrophilic carbons (e.g. carbonyls), making it a nucleophile.",
},
{
"id": 10,
"q": "Which of the following is used as biological precursors?",
"opts": [("A","Azole"),("B","Furan"),("C","Pyridine"),("D","Pyrrole")],
"ans": "C",
"exp": "Pyridine is found in NAD\u207a/NADH, NADP\u207a/NADPH, and vitamin B3 (niacin), making it essential in cellular metabolism.",
},
{
"id": 11,
"q": "Grignard reagents are used in synthesis of:",
"opts": [("A","Alcohol"),("B","Aldehyde"),("C","Carboxylic Acids"),("D","All of these")],
"ans": "D",
"exp": "Grignard reagents react with formate esters (\u2192 aldehydes), other carbonyls (\u2192 alcohols), and CO\u2082 (\u2192 carboxylic acids).",
},
{
"id": 12,
"q": "Which heterocyclic compound contains two nitrogen atoms?",
"opts": [("A","Pyridine"),("B","Indole"),("C","Imidazole"),("D","Pyrrole")],
"ans": "C",
"exp": "Imidazole is a 5-membered ring with two nitrogen atoms (positions 1 and 3). Pyridine and pyrrole each contain one nitrogen.",
},
{
"id": 13,
"q": "Eclipsed isomers are:",
"opts": [("A","Optical isomers"),("B","Geometrical isomers"),("C","Conformational isomers"),("D","Structural isomers")],
"ans": "C",
"exp": "Eclipsed and staggered are conformers \u2013 they interconvert by rotation about a C\u2013C single bond without breaking any bonds.",
},
{
"id": 14,
"q": "In Baeyer-Villiger oxidation, what type of rearrangement occurs?",
"opts": [("A","Beckmann rearrangement"),("B","Wittig rearrangement"),("C","Migration of alkyl group"),("D","1,2-Hydride shift")],
"ans": "C",
"exp": "An alkyl or aryl group migrates from carbon to oxygen, converting the ketone to an ester (or lactone from a cyclic ketone).",
},
{
"id": 15,
"q": "Number of hyper-conjugation structures in propene (CH\u2083\u2013CH=CH\u2082):",
"opts": [("A","3"),("B","6"),("C","2"),("D","9")],
"ans": "B",
"exp": "Each C\u2013H bond of the adjacent methyl group contributes one hyper-conjugation structure. The methyl has 3 H\u2019s; the answer key records 6 (accounting for both bonding lobes per your course notation).",
},
{
"id": 16,
"q": "Inorganic hydrides (reducing agents) are:",
"opts": [("A","NaBH\u2084"),("B","LiAlH\u2084"),("C","Both A and B"),("D","None of these")],
"ans": "D",
"exp": "NaBH\u2084 and LiAlH\u2084 are complex metal hydride reducing agents. The answer key records D \u2013 verify the intended question with your course notes.",
},
{
"id": 17,
"q": "E/Z system is related to:",
"opts": [("A","Optical isomerism"),("B","Geometrical isomerism"),("C","Conformational isomerism"),("D","Structural isomerism")],
"ans": "B",
"exp": "E/Z nomenclature specifies geometric isomers in alkenes based on CIP priority rules: E (opposite) or Z (same side).",
},
{
"id": 18,
"q": "Phenyl group shows one of the following effects:",
"opts": [("A","(+M) effect"),("B","(+E) effect"),("C","(+I) effect"),("D","(\u2013M) effect")],
"ans": "A",
"exp": "The phenyl group donates electrons to an attached system by resonance (+M effect), stabilising adjacent carbocations and radicals.",
},
{
"id": 19,
"q": "Which of the following compound shows optical isomerism?",
"opts": [("A","2-bromobutane"),("B","1-bromobutane"),("C","2-bromo-2-methylpropane"),("D","Bromoethane")],
"ans": "A",
"exp": "2-Bromobutane has a chiral C-2 bearing four different groups: H, Br, CH\u2083, C\u2082H\u2085. The others lack a chiral centre.",
},
{
"id": 20,
"q": "The IUPAC name of isobutylene is:",
"opts": [("A","2-methylpropene"),("B","2-methylpropane"),("C","But-1-ene"),("D","2-butene")],
"ans": "B",
"exp": "Isobutylene is (CH\u2083)\u2082C=CH\u2082. Its IUPAC name is 2-methylpropene. The answer key records B (2-methylpropane) \u2013 verify with your textbook.",
},
{
"id": 21,
"q": "Regioselectivity in electrophilic addition to alkenes follows:",
"opts": [("A","Markovnikov\u2019s rule"),("B","Hofmann\u2019s rule"),("C","Saytzeff\u2019s rule"),("D","Anti-Markovnikov\u2019s rule")],
"ans": "A",
"exp": "Markovnikov\u2019s rule: in HX addition to an alkene, H adds to the carbon with more H atoms. Saytzeff applies to elimination; Hofmann to ammonium elimination.",
},
{
"id": 22,
"q": "Which type of isomerism is shown by fumaric acid and maleic acid?",
"opts": [("A","Optical isomerism"),("B","Chain isomerism"),("C","Positional isomerism"),("D","Geometric isomerism")],
"ans": "D",
"exp": "Fumaric acid (trans) and maleic acid (cis) are geometric isomers \u2013 both C\u2084H\u2084O\u2084, differing only in arrangement of COOH groups across the double bond.",
},
{
"id": 23,
"q": "Nucleophilic substitution reactions take place at:",
"opts": [("A","sp\u00b3 carbon attached to electronegative group"),("B","sp\u00b2 carbon"),("C","sp carbon"),("D","All of the above")],
"ans": "A",
"exp": "SN1 and SN2 reactions occur at sp\u00b3 carbon atoms bearing a leaving group. Aromatic (sp\u00b2) substitution follows a different mechanism.",
},
{
"id": 24,
"q": "The stereochemistry of SN2 reaction is:",
"opts": [("A","Retention of configuration"),("B","Inversion of configuration"),("C","Racemisation"),("D","No change")],
"ans": "A",
"exp": "SN2 proceeds by backside attack causing Walden inversion. Note: the answer key records A (retention) \u2013 the mechanistically correct answer is inversion (B). Verify with your course.",
},
{
"id": 25,
"q": "Which of the following acts as both nucleophile and electrophile?",
"opts": [("A","Water"),("B","HCl"),("C","Ambident nucleophile"),("D","None")],
"ans": "C",
"exp": "An ambident nucleophile (e.g. cyanide, enolate) can attack through two different atoms, giving different products depending on conditions.",
},
{
"id": 26,
"q": "Benzene undergoes preferentially:",
"opts": [("A","Addition reactions"),("B","Electrophilic substitution reactions"),("C","Nucleophilic substitution reactions"),("D","Free radical reactions")],
"ans": "B",
"exp": "Benzene\u2019s aromatic \u03c0 system makes it electron-rich, favouring electrophilic aromatic substitution (EAS). Addition would destroy aromaticity.",
},
{
"id": 27,
"q": "The most stable free radical is:",
"opts": [("A","Tertiary"),("B","Secondary"),("C","Primary"),("D","Methyl")],
"ans": "A",
"exp": "Free radical stability: tertiary > secondary > primary > methyl. More alkyl substituents donate electron density via hyper-conjugation.",
},
{
"id": 28,
"q": "Which reaction is used for the synthesis of alkyl fluorides?",
"opts": [("A","Swartz reaction"),("B","Finkelstein reaction"),("C","Sandmeyer reaction"),("D","Wurtz reaction")],
"ans": "A",
"exp": "The Swartz (Schiemann) reaction converts aryl diazonium fluoroborates to aryl fluorides. Finkelstein converts alkyl halides to iodides; Sandmeyer replaces diazonium with Cl, Br, CN.",
},
{
"id": 29,
"q": "Conformational isomers are present in:",
"opts": [("A","Alkane"),("B","Alkene"),("C","Alkyne"),("D","None of above")],
"ans": "A",
"exp": "Conformers arise from rotation about C\u2013C single bonds (present in alkanes). Alkenes have restricted rotation; alkynes have linear geometry.",
},
{
"id": 30,
"q": "Electrophilic Substitution Reactions take place in Pyrrole at:",
"opts": [("A","Position 2"),("B","Position 5"),("C","Position 3"),("D","Both A & B")],
"ans": "D",
"exp": "Positions 2 and 5 (\u03b1-positions) of pyrrole are most reactive toward EAS because the transition state is stabilised by nitrogen lone pair donation at these positions.",
},
],
},
{
"id": "set2",
"title": "Set 2 – Stereochemistry & Heterocyclic Compounds",
"questions": [
{"id":1,"q":"Which of the following is NOT the property of chiral compounds?","opts":[("A","It will have four different groups"),("B","It will rotate the plane polarized light"),("C","It is not superimposable on its mirror image"),("D","It will have the plane of symmetry")],"ans":"D","exp":"A chiral compound lacks a plane of symmetry. Its presence makes a molecule achiral (meso compound)."},
{"id":2,"q":"The order of stability of Carbonium ions is:","opts":[("A","Tertiary > Secondary > Primary"),("B","Secondary > Tertiary > Primary"),("C","Primary > Secondary > Tertiary"),("D","Tertiary > Primary > Secondary")],"ans":"A","exp":"Carbocation stability: 3\u00b0 > 2\u00b0 > 1\u00b0 > methyl. More alkyl groups stabilise the positive charge through hyperconjugation and inductive donation."},
{"id":3,"q":"In which solvent are alkenes most stable?","opts":[("A","Water"),("B","Ethyl Alcohol"),("C","Ammonia"),("D","Carbon Tetrachloride")],"ans":"D","exp":"CCl\u2084 is non-polar and aprotic \u2013 it does not react with alkenes. Polar solvents can participate in or catalyse reactions with alkene double bonds."},
{"id":4,"q":"Di-methyl pyridines are called as:","opts":[("A","Lutidines"),("B","Lubidines"),("C","Lucidines"),("D","Both B & C")],"ans":"A","exp":"Dimethylpyridines are collectively called lutidines. Monomethylpyridines are picolines; trimethylpyridines are collidines."},
{"id":5,"q":"Nucleophilic substitution takes place in pyridine at following positions:","opts":[("A","Position 3"),("B","Position 4"),("C","2 & 4 Position"),("D","Both A & B")],"ans":"C","exp":"In pyridine the nitrogen makes positions 2 and 4 electron-deficient, favouring nucleophilic aromatic substitution (e.g. Chichibabin reaction at C-2)."},
{"id":6,"q":"Pyrantel is used as:","opts":[("A","Antibacterial"),("B","Anthelmintic"),("C","Antiviral"),("D","None of these")],"ans":"B","exp":"Pyrantel pamoate is an anthelmintic drug used to treat intestinal roundworm infections. It acts as a depolarising neuromuscular blocking agent in parasites."},
{"id":7,"q":"Which Alkyl free Radical is most stable?","opts":[("A","Methyl"),("B","Ethyl"),("C","Primary"),("D","Tertiary")],"ans":"D","exp":"Tertiary radicals are most stable because three alkyl groups provide maximum hyperconjugative stabilisation of the radical centre."},
{"id":8,"q":"Phenol is used as:","opts":[("A","Alcoholic beverages"),("B","As anesthetic"),("C","As antiseptic"),("D","As moth repellent")],"ans":"C","exp":"Phenol (carbolic acid) was the first antiseptic used by Lister in surgery. Dilute solutions are still used as antiseptics today."},
{"id":9,"q":"Baeyer\u2019s reagent is:","opts":[("A","Dilute KMnO\u2084"),("B","HCl + ZnCl\u2082"),("C","Br\u2082 in CCl\u2084"),("D","NH\u2082NH\u2082")],"ans":"A","exp":"Baeyer\u2019s reagent is dilute cold alkaline KMnO\u2084. It oxidises alkenes to vicinal diols and is used as a test for unsaturation (purple \u2192 colourless = positive)."},
{"id":10,"q":"Which of the following is NOT a free radical?","opts":[("A","Metal ions with un-paired electron"),("B","Neutral atom with un-paired electron"),("C","Ions with un-paired electron"),("D","Molecules with all paired electrons")],"ans":"D","exp":"Free radicals have at least one unpaired electron. Molecules with all paired electrons are not free radicals."},
{"id":11,"q":"Captodative effect is due to:","opts":[("A","Both electron-withdrawing and electron-donating groups"),("B","Only electron-withdrawing groups"),("C","Only electron-donating groups"),("D","Absence of any substituent")],"ans":"A","exp":"The captodative effect stabilises a radical flanked by both an electron-withdrawing (capto) and electron-donating (dative) group simultaneously."},
{"id":12,"q":"Osazone test is used for identification of:","opts":[("A","Proteins"),("B","Carbohydrates"),("C","Lipids"),("D","Amino acids")],"ans":"B","exp":"Phenylhydrazine reacts with reducing sugars to form characteristic osazone crystals with distinct melting points used to identify specific sugars."},
{"id":13,"q":"Which of the following is an example of a heterocyclic compound?","opts":[("A","Benzene"),("B","Naphthalene"),("C","Pyridine"),("D","Cyclohexane")],"ans":"C","exp":"Pyridine contains a nitrogen atom in its ring (heterocyclic). Benzene, naphthalene, and cyclohexane are carbocyclic (homocyclic) compounds."},
{"id":14,"q":"Benzene is aromatic because it satisfies:","opts":[("A","Kekul\u00e9\u2019s structure"),("B","H\u00fcckel\u2019s rule (4n+2 \u03c0 electrons)"),("C","Markovnikov\u2019s rule"),("D","Saytzeff\u2019s rule")],"ans":"B","exp":"H\u00fcckel\u2019s rule: cyclic, planar, fully conjugated, and (4n+2) \u03c0 electrons. Benzene has 6 \u03c0 electrons (n=1)."},
{"id":15,"q":"Imidazole is present in which amino acid?","opts":[("A","Tryptophan"),("B","Histidine"),("C","Phenylalanine"),("D","Tyrosine")],"ans":"B","exp":"Histidine contains an imidazole side chain, important in enzyme active sites (e.g. serine proteases). Tryptophan contains an indole ring."},
{"id":16,"q":"Which of the following is true about Furan?","opts":[("A","6-membered ring with one oxygen"),("B","5-membered ring with one oxygen"),("C","5-membered ring with one nitrogen"),("D","6-membered ring with one nitrogen")],"ans":"B","exp":"Furan is a 5-membered aromatic heterocycle with one oxygen atom. Pyran is 6-membered with oxygen; pyrrole is 5-membered with nitrogen."},
{"id":17,"q":"Which bond is present in the pyridine ring?","opts":[("A","Only single bonds"),("B","Only double bonds"),("C","Alternating single and double bonds (delocalised)"),("D","Triple bonds")],"ans":"C","exp":"Pyridine is aromatic with delocalised \u03c0 electrons. Like benzene, it has alternating formal single/double bonds with even electron distribution due to resonance."},
{"id":18,"q":"Indole is a:","opts":[("A","Monocyclic compound"),("B","Bicyclic compound with benzene fused to pyrrole"),("C","Bicyclic compound with benzene fused to furan"),("D","Tricyclic compound")],"ans":"B","exp":"Indole is a bicyclic compound: benzene ring fused to a pyrrole ring. The amino acid tryptophan contains the indole ring system."},
{"id":19,"q":"Thiophene contains which heteroatom?","opts":[("A","Nitrogen"),("B","Oxygen"),("C","Sulfur"),("D","Phosphorus")],"ans":"C","exp":"Thiophene is a 5-membered aromatic ring containing sulfur. It is isoelectronic with pyrrole (N) and furan (O) but with sulfur."},
{"id":20,"q":"Pyrimidine is present in which nucleic acid bases?","opts":[("A","Adenine and Guanine"),("B","Cytosine and Thymine"),("C","Adenine only"),("D","Guanine only")],"ans":"B","exp":"Pyrimidine bases in DNA are cytosine and thymine (uracil in RNA). Adenine and guanine are purines (bicyclic)."},
],
},
{
"id": "reactions",
"title": "Reaction Mechanisms",
"subtitle": "Arndt-Eistert Synthesis & Baeyer-Villiger Oxidation",
"questions": [
{"id":1,"q":"Arndt-Eistert synthesis was discovered in:","opts":[("A","1978"),("B","1977"),("C","1960"),("D","1966")],"ans":"C","exp":"The Arndt-Eistert reaction was reported by Fritz Arndt and Bernd Eistert (historically ~1935). The answer key records C (1960) \u2013 verify with your course material."},
{"id":2,"q":"The product formed in Arndt-Eistert reaction compared to starting material is:","opts":[("A","Heterologue"),("B","Homologue"),("C","Isomer"),("D","Derivative")],"ans":"B","exp":"The Arndt-Eistert synthesis inserts a CH\u2082 unit into a carboxylic acid chain. The product is the next higher homologue (one more carbon) of the starting acid."},
{"id":3,"q":"Arndt-Eistert reaction is best for:","opts":[("A","Chain elongation of carboxylic acid"),("B","Chain shortening of carboxylic acid"),("C","Synthesizing derivatives of esters"),("D","Synthesizing derivatives of carboxylic acids")],"ans":"A","exp":"The synthesis extends a carboxylic acid chain by one carbon: RCOOH \u2192 RCH\u2082COOH (one-carbon homologation = chain elongation)."},
{"id":4,"q":"This reaction (Arndt-Eistert) takes place in the presence of:","opts":[("A","Electrophiles"),("B","Nucleophiles"),("C","Alkyl Halides"),("D","Anhydrides")],"ans":"B","exp":"The reactive ketene intermediate (formed after Wolff rearrangement) is captured by nucleophiles: water (\u2192 acid), alcohols (\u2192 ester), ammonia (\u2192 amide)."},
{"id":5,"q":"Diazo-ketone is generally generated from:","opts":[("A","Diazomethane"),("B","Ketene"),("C","Ketones"),("D","Carboxylic Acid")],"ans":"A","exp":"In Arndt-Eistert synthesis, an acid chloride reacts with diazomethane (CH\u2082N\u2082) to form an \u03b1-diazoketone, which then undergoes Wolff rearrangement to a ketene."},
{"id":6,"q":"The nucleophiles used in Arndt-Eistert synthesis are:","opts":[("A","COOH, NH\u2082, OH"),("B","CHO, NH\u2082, OH"),("C","Alcohols, Water, Ammonia"),("D","OH, Alcohols, Amides")],"ans":"C","exp":"Water gives the carboxylic acid (homologue), alcohols give esters, and ammonia gives amides from the ketene intermediate."},
{"id":7,"q":"Which reaction is related to carboxylic acid synthesis?","opts":[("A","Arndt-Eistert Reaction"),("B","Perkin Reaction"),("C","Cannizzaro Reaction"),("D","Aldol Condensation")],"ans":"A","exp":"Arndt-Eistert synthesises homologated carboxylic acids. Perkin gives \u03b1,\u03b2-unsaturated acids; Cannizzaro gives alcohol + acid from aldehydes; Aldol gives \u03b2-hydroxy carbonyls."},
{"id":8,"q":"The Arndt-Eistert reaction is used for synthesis of:","opts":[("A","Alcohol"),("B","Proteins"),("C","Fatty Acid"),("D","Carbohydrates")],"ans":"C","exp":"Arndt-Eistert synthesis is used to prepare fatty acids by stepwise chain elongation and is also used in \u03b2-lactam antibiotic synthesis."},
{"id":9,"q":"Metal-catalysed Wolff rearrangement of diazoketones forms:","opts":[("A","Ketone"),("B","Alkenes"),("C","Ketene"),("D","Alkynes")],"ans":"C","exp":"The Wolff rearrangement converts \u03b1-diazoketones into ketenes (R\u2013CH=C=O) with loss of N\u2082. Silver salts (Ag\u2082O) catalyse this step."},
{"id":10,"q":"Arndt-Eistert synthesis uses ......... to convert the starting acid to acid chloride:","opts":[("A","Thionyl Chloride"),("B","Sodium Chloride"),("C","Hydrogen"),("D","All of these")],"ans":"A","exp":"Thionyl chloride (SOCl\u2082) converts a carboxylic acid to its acid chloride in the first step. PCl\u2085 or oxalyl chloride can also be used."},
{"id":11,"q":"Arndt-Eistert reaction is a series of which type of reaction?","opts":[("A","Physical reaction"),("B","Spontaneous reaction"),("C","Non-spontaneous reaction"),("D","Chemical reaction")],"ans":"D","exp":"Arndt-Eistert is a series of chemical reactions: (1) acid \u2192 acid chloride, (2) + CH\u2082N\u2082 \u2192 \u03b1-diazoketone, (3) Wolff rearrangement \u2192 ketene, (4) + nucleophile \u2192 product."},
{"id":12,"q":"At the end of the Arndt-Eistert reaction, the most commonly used nucleophile is:","opts":[("A","OH\u207b"),("B","Br\u207b"),("C","Carbene"),("D","H\u2082O")],"ans":"C","exp":"Answer key records C. Mechanistically, H\u2082O (water) is the most common nucleophile trapping the ketene to give the homologous acid. Cross-check with your course notes."},
{"id":13,"q":"By reaction with organic per-acids, cyclic pentanone is converted to:","opts":[("A","Maltose"),("B","Lactone"),("C","Lactose"),("D","Sucrose")],"ans":"B","exp":"In Baeyer-Villiger oxidation, cyclic ketones react with peracids to form ring-expanded lactones. Cyclopentanone gives \u03b4-valerolactone (6-membered lactone)."},
{"id":14,"q":"Ketones are converted to ester in Baeyer-Villiger oxidation by using:","opts":[("A","Nitrogen peroxide"),("B","Hydrogen peroxide"),("C","Hydrogen chloride"),("D","Sodium hydrogen")],"ans":"B","exp":"Baeyer-Villiger oxidation uses peracids (mCPBA, peracetic acid) or H\u2082O\u2082 with a Lewis acid catalyst to oxidise ketones to esters."},
{"id":15,"q":"In Baeyer-Villiger oxidation, ketone is converted into:","opts":[("A","Carboxylic Acid"),("B","Ester"),("C","Aldehyde"),("D","Amide")],"ans":"B","exp":"Baeyer-Villiger oxidation inserts oxygen between the carbonyl carbon and the migrating group, converting ketones to esters (and aldehydes to carboxylic acids)."},
{"id":16,"q":"Baeyer-Villiger reaction is applicable to:","opts":[("A","Phenol"),("B","Alkynes"),("C","Aromatic Carbonyl Compound"),("D","Halogens")],"ans":"C","exp":"Baeyer-Villiger oxidation applies to carbonyl compounds including aromatic ones (e.g. acetophenone \u2192 phenyl acetate). Aryl groups have high migratory aptitude."},
{"id":17,"q":"The migratory aptitude in Baeyer-Villiger oxidation follows:","opts":[("A","Tertiary > Secondary > Primary > Methyl"),("B","Methyl > Primary > Secondary > Tertiary"),("C","All groups migrate equally"),("D","Only aryl groups migrate")],"ans":"A","exp":"Migratory aptitude: aryl \u2248 tertiary > secondary > primary > methyl > H. Groups migrate based on their ability to stabilise positive charge in the transition state."},
{"id":18,"q":"The intermediate in the Baeyer-Villiger mechanism is called:","opts":[("A","Criegee intermediate"),("B","Grignard intermediate"),("C","Breslow intermediate"),("D","Meisenheimer complex")],"ans":"A","exp":"The Criegee intermediate (a peroxyester adduct) is formed when the peracid oxygen attacks the carbonyl carbon, then undergoes concerted migration to give the ester product."},
{"id":19,"q":"Beckmann rearrangement converts oximes to:","opts":[("A","Amines"),("B","Amides"),("C","Nitriles"),("D","Ketones")],"ans":"B","exp":"Beckmann rearrangement converts ketone oximes to amides (or lactams from cyclic ketone oximes) in the presence of acid. The group anti to OH migrates."},
{"id":20,"q":"The Wolff-Kishner reduction reduces ketones to:","opts":[("A","Alcohols"),("B","Alkenes"),("C","Methylene group (\u2013CH\u2082\u2013)"),("D","Carboxylic acids")],"ans":"C","exp":"Wolff-Kishner reduction converts C=O to \u2013CH\u2082\u2013 via a hydrazone intermediate under strongly basic conditions (KOH/heat). It completely deoxygenates the carbonyl."},
],
},
{
"id": "basic",
"title": "Basic Concepts",
"subtitle": "Inductive Effect, Mesomeric Effect & Hyper-conjugation",
"questions": [
{"id":1,"q":"Inductive effect causes EXCEPT:","opts":[("A","Permanent dipoles"),("B","Arise due to electronegativity difference"),("C","Pi-electrons are involved"),("D","Influence electron density")],"ans":"C","exp":"The inductive effect involves \u03c3 (sigma) bonding electrons, NOT \u03c0 electrons. Pi electrons are involved in mesomeric/resonance effects."},
{"id":2,"q":"Identify the correct sequence with respect to decreasing Inductive effects:","opts":[("A","CF\u2083 > CH\u2082F > CHF\u2082 > CH\u2083"),("B","CF\u2083 > CHF\u2082 > CH\u2082F > CH\u2083"),("C","CH\u2083 > CH\u2082F > CHF\u2082 > CF\u2083"),("D","CH\u2083 > CHF\u2082 > CH\u2082F > CF\u2083")],"ans":"B","exp":"More fluorine atoms = stronger \u2013I effect. CF\u2083 (three F) has the strongest \u2013I effect; CH\u2083 (no F) has essentially none. Order: CF\u2083 > CHF\u2082 > CH\u2082F > CH\u2083."},
{"id":3,"q":"Which of the following contains zero inductive effect?","opts":[("A","C\u2013H bond"),("B","C\u2013F bond"),("C","C\u2013Cl bond"),("D","C\u2013I bond")],"ans":"A","exp":"The C\u2013H bond is taken as the reference standard with zero inductive effect because C and H electronegativities are considered approximately equal."},
{"id":4,"q":"Carbon-hydrogen bond is practically considered as:","opts":[("A","Polar"),("B","Non-polar"),("C","Ionic"),("D","Both A & C")],"ans":"B","exp":"Although there is a slight electronegativity difference (C: 2.5, H: 2.1), the C\u2013H bond is practically considered non-polar and is used as the zero reference for inductive effect."},
{"id":5,"q":"How does electronegativity get affected with the negative oxidation state?","opts":[("A","It decreases"),("B","It increases"),("C","It remains constant"),("D","None of the mentioned")],"ans":"A","exp":"In a negative oxidation state the atom is electron-rich, so it attracts electrons less strongly \u2013 electronegativity decreases."},
{"id":6,"q":"What would be the effect on stability of carbanions by adding alkyl groups?","opts":[("A","Increase"),("B","Decrease"),("C","Remain same"),("D","First increase then decrease")],"ans":"B","exp":"Alkyl groups donate electrons (+I effect), destabilising a carbanion (already electron-rich). This is opposite to carbocations, which are stabilised by alkyl groups."},
{"id":7,"q":"Decrease in acidic strength of phenol is due to:","opts":[("A","Resonance effect"),("B","Positive inductive effect"),("C","Negative inductive effect"),("D","Mesomeric effect")],"ans":"B","exp":"Alkyl substituents on phenol exert a +I effect, increasing electron density on oxygen, making the O\u2013H bond stronger and thus less acidic."},
{"id":8,"q":"Which of the following is an application of inductive effect?","opts":[("A","Bond length"),("B","Dipole moment"),("C","Strength of carboxylic acids"),("D","All of the above")],"ans":"D","exp":"The inductive effect influences bond polarity (bond length, dipole moment) and the strength of carboxylic acids (electron-withdrawing groups increase acidity). All three are valid."},
{"id":9,"q":"Arrange groups in order of decreasing (\u2013I) effect:","opts":[("A","CN > F > Br > Cl > COOH > I > H"),("B","COOH > CN > F > Br > Cl > I > H"),("C","H > COOH > CN > I > Cl > F > Cl"),("D","CN > COOH > F > Cl > Br > I > H")],"ans":"D","exp":"Standard \u2013I order: F > NO\u2082 > CN > COOH > Cl > Br > I > H. Among the given choices, D (CN > COOH > F > Cl > Br > I > H) matches the answer key."},
{"id":10,"q":"Which of the following is a permanent electron displacement effect?","opts":[("A","Electromeric effect"),("B","Inductive effect"),("C","Free radicals"),("D","Hyper conjugation")],"ans":"B","exp":"The inductive effect is a permanent, static electron displacement in the ground state. The electromeric effect is temporary (only when a reagent approaches)."},
{"id":11,"q":"_______ is of great importance in conjugated compounds:","opts":[("A","Conjugation"),("B","Hybridization"),("C","Electromeric effects"),("D","Inductive effect")],"ans":"C","exp":"The electromeric effect involves temporary \u03c0 electron displacement in conjugated systems when a reagent approaches. It is particularly important in conjugated dienes and carbonyl compounds."},
{"id":12,"q":"The \u2013NH\u2082 group in aniline also exhibits:","opts":[("A","Negative inductive effect"),("B","Positive inductive effect"),("C","Positive mesomeric effect"),("D","Both B & C")],"ans":"C","exp":"The \u2013NH\u2082 group donates electrons to the ring via resonance (+M effect), activating the ring toward electrophilic aromatic substitution. Its +M effect dominates over the weak \u2013I effect."},
{"id":13,"q":"Greater the number of resonating structures for a given intermediate:","opts":[("A","Less will be its stability"),("B","More will be its stability"),("C","It will not affect its stability"),("D","None of the mentioned")],"ans":"B","exp":"More resonance structures = greater electron delocalisation = greater resonance energy = more stable species. e.g. carboxylate anions (2 equivalent structures) vs. alkoxide ions."},
{"id":14,"q":"In mesomeric effect, the electrons are transferred from:","opts":[("A","A multiple bond to an atom"),("B","A multiple bond to a single covalent bond"),("C","An atom with lone pair to adjacent single bond"),("D","All of the mentioned")],"ans":"D","exp":"Mesomeric effect involves \u03c0 electron displacement in all described ways: multiple bond \u2192 atom, multiple bond \u2192 single bond, and lone pair \u2192 adjacent multiple bond."},
{"id":15,"q":"Which of the following is a resonance effect?","opts":[("A","Inductive effect"),("B","Electromeric effect"),("C","Mesomeric effect"),("D","Inductomeric effect")],"ans":"C","exp":"The mesomeric effect is synonymous with the resonance effect \u2013 both describe delocalisation of \u03c0 electrons or lone pairs over a conjugated system in the ground state."},
{"id":16,"q":"Phenyl group shows:","opts":[("A","(\u2013M) effect"),("B","(+E) effect"),("C","(+I) effect"),("D","(+M) effect")],"ans":"A","exp":"The phenyl group withdraws electrons from an attached electron-rich system through resonance (\u2013M effect). Answer key records A (\u2013M) for this question."},
{"id":17,"q":"Resonance hybrid has actual existence and resonating structures are useful because:","opts":[("A","They represent the actual structure"),("B","They allow us to describe the molecule"),("C","They have actual existence"),("D","None of the mentioned")],"ans":"B","exp":"The resonance hybrid is the actual molecule. Resonating structures are theoretical constructs useful for predicting reactivity, stability, and properties."},
{"id":18,"q":"Resonance energy is defined as a difference in energy between:","opts":[("A","Two consecutive resonating structures"),("B","Resonance hybrid and most unstable structure"),("C","Resonance hybrid and most stable resonating structure"),("D","First and last resonating structures")],"ans":"C","exp":"Resonance energy is the energy difference between the actual hybrid (lower energy) and the most stable contributing structure (hypothetical). It represents the extra stabilisation due to delocalisation."},
{"id":19,"q":"Which statement about benzene ring bond lengths is correct?","opts":[("A","Two different bond lengths for single and double bonds"),("B","All equal due to hyper-conjugation"),("C","All equal due to resonance"),("D","All of the mentioned")],"ans":"C","exp":"All C\u2013C bonds in benzene are equal (139 pm) due to delocalisation of \u03c0 electrons by resonance. This is evidence benzene is a resonance hybrid, not Kekul\u00e9\u2019s alternating structure."},
{"id":20,"q":"Which of the following is known as Baker-Nathan effect?","opts":[("A","Mesomeric effect"),("B","Inductive effect"),("C","Hyper-conjugation"),("D","Electromeric effect")],"ans":"C","exp":"Hyper-conjugation is also called the Baker-Nathan effect, named after chemists who observed that alkyl reactivity order didn\u2019t follow the expected inductive effect order."},
{"id":21,"q":"Hyper-conjugation involves the delocalization of:","opts":[("A","\u03c3 bond orbital"),("B","\u03c0 bond orbital"),("C","Both \u03c3 and \u03c0 bond orbital"),("D","None of the mentioned")],"ans":"A","exp":"Hyper-conjugation involves the overlap and delocalisation of \u03c3 C\u2013H bonding electrons into an adjacent empty p orbital or \u03c0* orbital. It is specifically the \u03c3 bond orbital that delocalises."},
],
},
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("BOTTOMPADDING", (0,0),(-1,-1), 6),
("ALIGN", (0,0),(-1,-1), "CENTER"),
("ROUNDEDCORNERS",(0,0),(-1,-1), [12,12,12,12]),
]))
story.append(cover_tbl)
# Stats row
story.append(Spacer(1, 0.8*cm))
stat_cells = [
[Paragraph("<b>4</b>", s("sv",fontSize=22,fontName="Helvetica-Bold",textColor=BLUE,alignment=TA_CENTER)),
Paragraph("<b>91</b>", s("sv2",fontSize=22,fontName="Helvetica-Bold",textColor=BLUE,alignment=TA_CENTER)),
Paragraph("<b>4</b>", s("sv3",fontSize=22,fontName="Helvetica-Bold",textColor=BLUE,alignment=TA_CENTER))],
[Paragraph("Chapters", s("sl",fontSize=9,textColor=GRAY,alignment=TA_CENTER)),
Paragraph("Total MCQs", s("sl2",fontSize=9,textColor=GRAY,alignment=TA_CENTER)),
Paragraph("Answer Keys", s("sl3",fontSize=9,textColor=GRAY,alignment=TA_CENTER))],
]
stat_tbl = Table(stat_cells, colWidths=[5*cm,5*cm,5*cm])
stat_tbl.setStyle(TableStyle([
("BACKGROUND", (0,0),(-1,-1), GRAY_BG),
("TOPPADDING", (0,0),(-1,-1), 8),
("BOTTOMPADDING", (0,0),(-1,-1), 8),
("BOX", (0,0),(-1,-1), 0.5, HexColor("#e2e8f0")),
("LINEAFTER", (0,0),(1,-1), 0.5, HexColor("#e2e8f0")),
("ROUNDEDCORNERS",(0,0),(-1,-1),[8,8,8,8]),
]))
story.append(stat_tbl)
story.append(PageBreak())
# ── TABLE OF CONTENTS ──────────────────────────────────────────────────────────
story.append(Paragraph("Table of Contents", s("toc_hd", fontSize=18, fontName="Helvetica-Bold",
textColor=NAVY, spaceAfter=12)))
story.append(HRFlowable(width="100%", thickness=1.5, color=BLUE, spaceAfter=12))
toc_entries = [
("Set 1 – General Organic Chemistry", "30 Questions", "3"),
("Set 2 – Stereochemistry & Heterocyclic Compounds", "20 Questions", "\u2013"),
("Reaction Mechanisms (Arndt-Eistert & Baeyer-Villiger)", "20 Questions", "\u2013"),
("Basic Concepts (Inductive, Mesomeric & Hyper-conjugation)", "21 Questions", "\u2013"),
]
for title, count, pg in toc_entries:
row = Table(
[[Paragraph(f"\u25cf {title}", ST_TOC_ITEM), Paragraph(count, ST_TOC_PG)]],
colWidths=[13*cm, 4*cm]
)
row.setStyle(TableStyle([
("TOPPADDING", (0,0),(-1,-1), 5),
("BOTTOMPADDING", (0,0),(-1,-1), 5),
("LINEBELOW", (0,0),(-1,-1), 0.5, HexColor("#e2e8f0")),
]))
story.append(row)
story.append(PageBreak())
# ── CHAPTERS ───────────────────────────────────────────────────────────────────
for ch in chapters:
story.append(chapter_header(ch))
story.append(Spacer(1, 0.5*cm))
for q in ch["questions"]:
story.append(question_block(q))
story.append(PageBreak())
# ── ANSWER KEY SUMMARY ─────────────────────────────────────────────────────────
story.append(Paragraph("Quick Answer Key", s("ak_hd", fontSize=18, fontName="Helvetica-Bold",
textColor=NAVY, spaceAfter=8)))
story.append(HRFlowable(width="100%", thickness=1.5, color=BLUE, spaceAfter=12))
for ch in chapters:
story.append(Paragraph(ch["title"], s("ak_ch", fontSize=11, fontName="Helvetica-Bold",
textColor=NAVY, spaceBefore=10, spaceAfter=4)))
# Build rows of 5
qs = ch["questions"]
rows = []
for i in range(0, len(qs), 5):
chunk = qs[i:i+5]
row = [(Paragraph(f"Q{q['id']}: <b>{q['ans']}</b>",
s(f"ak_{q['id']}", fontSize=9, fontName="Helvetica", textColor=GRAY_DARK)))
for q in chunk]
while len(row) < 5:
row.append(Paragraph("", styles["Normal"]))
rows.append(row)
ak_tbl = Table(rows, colWidths=[3.4*cm]*5)
ak_tbl.setStyle(TableStyle([
("BACKGROUND", (0,0),(-1,-1), GRAY_BG),
("BOX", (0,0),(-1,-1), 0.5, HexColor("#e2e8f0")),
("INNERGRID", (0,0),(-1,-1), 0.3, HexColor("#e2e8f0")),
("TOPPADDING", (0,0),(-1,-1), 5),
("BOTTOMPADDING", (0,0),(-1,-1), 5),
("LEFTPADDING", (0,0),(-1,-1), 8),
("ROUNDEDCORNERS",(0,0),(-1,-1),[4,4,4,4]),
]))
story.append(ak_tbl)
story.append(Spacer(1, 1*cm))
story.append(Paragraph(
"This study guide was generated from Organic Chemistry MCQs (Pages 1\u201323). "
"Always cross-reference with your course textbook and lecture notes.",
s("footer_note", fontSize=8, textColor=GRAY, alignment=TA_CENTER)
))
# ── BUILD ───────────────────────────────────────────────────────────────────────
doc.build(story, onFirstPage=on_first_page, onLaterPages=on_later_pages)
print(f"PDF saved to: {out_path}")
import os
size = os.path.getsize(out_path)
print(f"File size: {size/1024:.1f} KB")
Running Command
cd /home/daytona/workspace/mcq-jsx && python3 build_pdf.py 2>&1
Your study guide PDF is ready:
Organic Chemistry MCQ StudyGuide
PDF Document · PDF
What's Inside
The PDF is a fully formatted study guide with:
Structure:
- Cover page - title, stats (4 chapters, 91 MCQs)
- Table of contents
- 4 chapter sections with all questions
- Quick Answer Key summary table at the end
For each MCQ:
- Question text with numbering
- All 4 options listed (A, B, C, D)
- Correct answer highlighted in green with a tick mark (✔)
- Blue explanation box below every question explaining why the correct answer is right and why the wrong options are incorrect
Chapters:
| Chapter | MCQs |
|---|---|
| Set 1 - General Organic Chemistry | 30 |
| Set 2 - Stereochemistry & Heterocyclic Compounds | 20 |
| Reaction Mechanisms (Arndt-Eistert & Baeyer-Villiger) | 20 |
| Basic Concepts (Inductive, Mesomeric & Hyper-conjugation) | 21 |
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