75 free MCQs on Organic Chemistry: Basic Principles and Techniques with worked answers and explanations. General Organic Chemistry covers the rules behind all organic reactions. Covers inductive effect, resonance, hyperconjugation, types of reactions (substitution, addition, elimination), and isomerism including optical and geometric forms.
Take the timed Organic Chemistry: Basic Principles and Techniques chapterwise test →Below are 75 practice questions on Organic Chemistry: Basic Principles and Techniques, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Organic Chemistry: Basic Principles and Techniques notes.
Geometric (cis-trans) isomers of 1,2-dichloroethene differ only in the relative arrangement of groups across the rigid C=C double bond.
Q1.
Which technique is most suitable to separate two miscible liquids whose boiling points differ by only a few degrees?
Answer: A. Fractional distillation
Why: When boiling points are close, a fractionating column is needed so repeated vaporisation-condensation cycles separate the components; simple distillation works only for wide boiling-point gaps.
Q2.
In Lassaigne's test, the organic compound is fused with a small piece of which metal?
Answer: B. Sodium
Why: Sodium fusion converts covalently bound N, S and halogens into ionic species (NaCN, Na<sub>2</sub>S, NaX) that can then be detected in aqueous extract.
Q3.
Steam distillation is applicable to compounds that are:
Answer: C. Steam volatile and immiscible with water
Why: The compound must be volatile in steam and insoluble in water so it distils over at a temperature below its normal boiling point, avoiding decomposition.
Q4.
Which of the following groups shows a +I (electron-releasing inductive) effect?
Answer: D. Alkyl (-CH<sub>3</sub>)
Why: Alkyl groups push electron density through sigma bonds (+I effect), whereas -NO<sub>2</sub>, -COOH and -CN are electron-withdrawing (-I) groups.
Q5.
Chromatography separates the components of a mixture mainly on the basis of their:
Answer: A. Different adsorption or partition tendencies
Why: In adsorption and partition chromatography components are separated by how strongly each is retained by the stationary phase relative to the mobile phase.
Q6.
What is the hybridisation of carbon in methane?
Answer: C. sp<sup>3</sup>
Why: Carbon in methane forms four sigma bonds, so it undergoes sp<sup>3</sup> hybridisation.
Q7.
How many sigma bonds are present in ethane?
Answer: C. 7
Why: Ethane contains six C-H sigma bonds and one C-C sigma bond, making a total of 7 sigma bonds.
Q8.
Which bond is formed by sidewise overlap of orbitals?
Answer: B. Pi bond
Why: Pi bonds are formed by sideways overlap of p orbitals.
Q9.
What is the hybridisation of carbon in ethene?
Answer: B. sp<sup>2</sup>
Why: Carbon atoms in ethene are sp<sup>2</sup> hybridised because each forms three sigma bonds.
Q10.
Which compound shows chain isomerism?
Answer: C. Butane
Why: Butane can exist as n-butane and isobutane, showing chain isomerism.
Q11.
Which functional group is present in alcohols?
Answer: B. -OH
Why: Alcohols contain the hydroxyl functional group represented by -OH.
Q12.
What is the IUPAC name of CH<sub>3</sub>CH<sub>2</sub>OH?
Answer: C. Ethanol
Why: CH<sub>3</sub>CH<sub>2</sub>OH contains two carbon atoms and an alcohol group, so it is ethanol.
Q13.
Compounds with the same molecular formula but different structures are called:
Answer: C. Isomers
Why: Isomers have the same molecular formula but different arrangements of atoms.
Q14.
What is the general formula of alkanes?
Answer: B. C<sub>n</sub>H<sub>2n</sub>+2
Why: Alkanes are saturated hydrocarbons with the formula C<sub>n</sub>H<sub>2n</sub>+2.
Q15.
What type of bond is present in all single covalent bonds?
Answer: C. Sigma bond
Why: Single covalent bonds are sigma bonds formed by direct overlap of orbitals.
Q16.
What is the hybridisation of carbon in ethyne?
Answer: A. sp
Why: Carbon atoms in ethyne are sp hybridised due to triple bonding.
Q17.
Which effect involves displacement of sigma electrons?
Answer: B. Inductive effect
Why: Inductive effect is the permanent displacement of sigma electrons due to electronegativity differences.
Q18.
Which element is the most electronegative?
Answer: C. Fluorine
Why: Fluorine is the most electronegative element in the periodic table.
Q19.
What type of isomerism is shown by ethanol and dimethyl ether?
Answer: C. Functional isomerism
Why: They have the same molecular formula but different functional groups.
Q20.
A homologous series differs by which group?
Answer: A. CH<sub>2</sub>
Why: Successive members of a homologous series differ by a CH<sub>2</sub> group.
Q21.
How many pi bonds are present in ethyne?
Answer: C. 2
Why: Ethyne contains one sigma bond and two pi bonds between carbon atoms.
Q22.
Which functional group is present in aldehydes?
Answer: B. -CHO
Why: Aldehydes contain the aldehyde functional group represented by -CHO.
Q23.
What is the IUPAC name of CH<sub>3</sub>COOH?
Answer: B. Ethanoic acid
Why: CH<sub>3</sub>COOH contains two carbon atoms and a carboxylic acid group, so it is ethanoic acid.
Q24.
Which hybrid orbital arrangement gives linear geometry?
Answer: A. sp
Why: sp hybridisation results in linear geometry with a bond angle of 180 degrees.
Q25.
What is the bond angle in sp<sup>2</sup> hybridisation?
Answer: C. 120 degrees
Why: sp<sup>2</sup> hybridisation produces trigonal planar geometry with 120 degree bond angles.
Q26.
In paper chromatography, the stationary phase is:
Answer: B. Water held on the paper
Why: In paper chromatography the stationary phase is the water trapped in the cellulose of the paper; the solvent is the mobile phase.
Q27.
What is the degree of unsaturation (rings plus pi bonds) for a compound of molecular formula C<sub>6</sub>H<sub>6</sub>?
Answer: C. 4
Why: Degree of unsaturation = (2x6 + 2 - 6)/2 = 4, matching benzene with one ring and three pi bonds.
Q28.
Which of the following carboxylic acids is the strongest?
Answer: D. Cl<sub>2</sub>CHCOOH
Why: Two chlorine atoms exert a combined strong -I effect that stabilises the carboxylate anion most, making dichloroacetic acid the strongest (lowest pKa).
Q29.
Which of the following carbocations is the most stable?
Answer: A. Benzyl cation
Why: The benzyl cation is stabilised by extensive resonance delocalisation into the aromatic ring, making it more stable than allyl and the alkyl cations.
Q30.
The acetate ion (CH<sub>3</sub>COO<sup>-</sup>) owes its stability to how many equivalent resonance structures?
Answer: B. 2
Why: The negative charge is delocalised equally over the two oxygen atoms, giving two equivalent resonance contributors and equal C-O bond lengths.
Q31.
Which effect explains the stability of benzene through electron delocalisation?
Answer: C. Resonance
Why: Benzene is stabilised by resonance because its pi electrons are delocalised over the ring.
Q32.
Which carbocation is most stable?
Answer: D. Tertiary carbocation
Why: Tertiary carbocations are most stable due to hyperconjugation and electron donating alkyl groups.
Q33.
Which species acts as an electron pair donor?
Answer: B. Nucleophile
Why: Nucleophiles donate electron pairs to electron deficient species.
Q34.
Which intermediate contains a positively charged carbon atom?
Answer: C. Carbocation
Why: A carbocation is an intermediate where carbon carries a positive charge.
Q35.
What is the geometry of a carbocation?
Answer: B. Trigonal planar
Why: Carbocations are sp<sup>2</sup> hybridised and have trigonal planar geometry.
Q36.
Which effect involves complete transfer of pi electrons due to an attacking reagent?
Answer: C. Electromeric effect
Why: Electromeric effect involves temporary complete transfer of pi electrons.
Q37.
Which of the following is a nucleophile?
Answer: C. OH-
Why: OH- donates an electron pair, so it behaves as a nucleophile.
Q38.
Optical isomerism is caused by the presence of:
Answer: C. Chiral carbon
Why: A chiral carbon bonded to four different groups causes optical isomerism.
Q39.
Mirror image stereoisomers are called:
Answer: C. Enantiomers
Why: Enantiomers are non-superimposable mirror image isomers.
Q40.
Which type of isomerism is shown by cis and trans forms?
Answer: C. Geometrical isomerism
Why: Cis and trans forms arise due to restricted rotation around double bonds.
Q41.
Which carbocation is stabilised by resonance?
Answer: C. Benzyl carbocation
Why: Benzyl carbocation is resonance stabilised by the benzene ring.
Q42.
Which species contains an unpaired electron?
Answer: C. Free radical
Why: Free radicals contain one unpaired electron.
Q43.
Which effect stabilises alkenes through overlap with adjacent sigma bonds?
Answer: B. Hyperconjugation
Why: Hyperconjugation stabilises alkenes through sigma electron delocalisation.
Q44.
Which compound can show geometrical isomerism?
Answer: C. But-2-ene
Why: But-2-ene has restricted rotation around the double bond and different groups attached.
Q45.
Which of the following is an electrophile?
Answer: C. BF<sub>3</sub>
Why: BF<sub>3</sub> is electron deficient and accepts electron pairs, so it acts as an electrophile.
Q46.
Which carbanion is most stable?
Answer: A. Methyl carbanion
Why: Methyl carbanion is most stable because alkyl groups destabilise negative charge.
Q47.
Which type of isomers are not mirror images of each other?
Answer: B. Diastereomers
Why: Diastereomers are stereoisomers that are not mirror images.
Q48.
The delocalisation of electrons over adjacent atoms is called:
Answer: B. Resonance
Why: Resonance spreads electron density across multiple atoms, increasing stability.
Q49.
Which free radical is most stable?
Answer: D. Tertiary radical
Why: Tertiary radicals are stabilised by hyperconjugation from surrounding alkyl groups.
Q50.
Which effect is permanent in nature?
Answer: B. Inductive effect
Why: Inductive effect is a permanent displacement of sigma electrons in a molecule.
Q51.
In Lassaigne's test for nitrogen, the Prussian blue colour is due to the formation of:
Answer: C. Fe<sub>4</sub>[Fe(CN)<sub>6</sub>]<sub>3</sub>
Why: Cyanide from sodium fusion forms ferrocyanide, which reacts with Fe(III) to give ferric ferrocyanide, Fe<sub>4</sub>[Fe(CN)<sub>6</sub>]<sub>3</sub>, the Prussian blue precipitate.
Q52.
In Liebig's method for estimating carbon and hydrogen, the carbon content is calculated from the mass of:
Answer: D. CO<sub>2</sub> absorbed
Why: Carbon is oxidised to CO<sub>2</sub>, absorbed in KOH; its mass gives carbon, while the H<sub>2</sub>O absorbed in anhydrous CaCl<sub>2</sub> gives hydrogen.
Q53.
In the Carius method, the halogen in an organic compound is estimated by precipitating it as:
Answer: A. Silver halide (AgX)
Why: The compound is heated with fuming HNO<sub>3</sub> and AgNO<sub>3</sub> so the halogen is converted to insoluble silver halide, which is weighed.
Q54.
Which order correctly represents the decreasing -I (electron-withdrawing) inductive effect?
Answer: B. -NO<sub>2</sub> > -CN > -F
Why: The accepted -I order places -NO<sub>2</sub> above -CN, which is above the halogens, so -NO<sub>2</sub> > -CN > -F.
Q55.
The number of alpha-hydrogens responsible for the hyperconjugative stabilisation of the tert-butyl cation, (CH<sub>3</sub>)<sub>3</sub>C<sup>+</sup>, is:
Answer: C. 9
Why: Each of the three methyl groups attached to the positive carbon has three C-H bonds, giving 9 alpha C-H bonds available for hyperconjugation.
Q56.
How many stereoisomers are possible for a compound with two chiral carbons?
Answer: C. 4
Why: Maximum stereoisomers = 2n where n is the number of chiral carbons. Here 2 squared = 4.
Q57.
Which rule is used to assign R and S configuration?
Answer: C. Cahn-Ingold-Prelog rule
Why: R and S configurations are assigned using Cahn-Ingold-Prelog priority rules.
Q58.
Which compound follows Huckel's rule for aromaticity?
Answer: B. Cyclopropenyl cation
Why: Cyclopropenyl cation contains 2 pi electrons and satisfies the 4n+2 rule with n=0.
Q59.
How many pi electrons are required for aromaticity according to Huckel's rule?
Answer: C. 4n+2
Why: Aromatic compounds must contain 4n+2 pi electrons in a cyclic conjugated system.
Q60.
Which conformation of ethane is most stable?
Answer: B. Staggered
Why: Staggered conformation has minimum torsional strain and is most stable.
Q61.
Which conformation of cyclohexane is most stable?
Answer: B. Chair
Why: Chair conformation has minimum angle and torsional strain.
Q62.
The interconversion between keto and enol forms is called:
Answer: C. Tautomerism
Why: Tautomerism is the equilibrium between keto and enol forms of a compound.
Q63.
Which aromatic compound has 6 pi electrons?
Answer: B. Benzene
Why: Benzene contains six delocalised pi electrons and satisfies Huckel's rule.
Q64.
Which intermediate is most stable?
Answer: D. Benzyl carbocation
Why: Benzyl carbocation is highly stabilised by resonance with the aromatic ring.
Q65.
What is the IUPAC name of CH3CH(CH<sub>3</sub>)CH<sub>2</sub>CH<sub>3</sub>?
Answer: A. 2-methylbutane
Why: The longest chain contains four carbons with a methyl substituent at carbon 2.
Q66.
Which species is antiaromatic?
Answer: B. Cyclobutadiene
Why: Cyclobutadiene contains 4 pi electrons and is antiaromatic.
Q67.
Which notation is used for geometrical isomers based on CIP priority?
Answer: C. E-Z
Why: E-Z notation is based on priority rules assigned to groups around a double bond.
Q68.
What does Z configuration indicate?
Answer: C. Higher priority groups on same side
Why: Z comes from the German word zusammen meaning together, indicating same-side arrangement.
Q69.
Which cyclic compound is non-aromatic?
Answer: C. Cyclooctatetraene
Why: Cyclooctatetraene is non-planar and does not undergo continuous conjugation.
Q70.
Ring-chain isomerism is shown by:
Answer: A. Propane and cyclopropane
Why: Propane and cyclopropane have the same molecular formula but differ as open chain and cyclic structures.
Q71.
Which Newman projection represents eclipsed conformation?
Answer: C. Dihedral angle 0 degrees
Why: In eclipsed conformation, bonds align directly behind each other with 0 degree dihedral angle.
Q72.
Which carbocation undergoes maximum hyperconjugation?
Answer: D. Tertiary
Why: Tertiary carbocations have the maximum number of adjacent C-H bonds for hyperconjugation.
Q73.
Which compound can exhibit tautomerism?
Answer: B. Acetaldehyde
Why: Acetaldehyde contains alpha hydrogen and can show keto-enol tautomerism.
Q74.
What is the total number of stereoisomers possible for tartaric acid?
Answer: B. 3
Why: Tartaric acid has two chiral carbons but one meso form, giving a total of 3 stereoisomers.
Q75.
Which compound is aromatic?
Answer: A. Cyclopentadienyl anion
Why: Cyclopentadienyl anion contains 6 pi electrons and satisfies Huckel's rule for aromaticity.