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Organic Chemistry Multiple Choice Questions on “Physical Properties of Amines”.
1. Which of the following compound is expected to be most basic?
a) Aniline
b) Methylamine
c) Hydroxylamine
d) Ethylamine
Answer: d
Clarification: Alkyl groups are electron donating (EDG). So, the alkyl group stabilizes the positive charge on the attached nitrogen atom.
2. A solution of methyl amine shows which type of property with litmus paper?
a) Turns blue litmus red
b) Turns red litmus blue
c) Does not affect red or blue litmus
d) Bleaches litmus
Answer: b
Clarification: Methyl amine changes red litmus blue, this litmus paper test shows basic nature of methyl amine.
3. Which of the following statement is true about methyl amine?
a) Methyl amine is slightly acidic
b) Methyl amine is less basic than NH3
c) Methyl amine is stronger base than NH3
d) Methyl amine forms salts with alkali
Answer: c
Clarification: Presence of alkyl group increases electron density on nitrogen atom due to +I effect. Thus, basic nature increases.
4. By the presence of a halogen atom in the ring, what is the effect of this on basic properties of aniline?
a) Increased
b) Decreased
c) Unchanged
d) Doubled
Answer: a
Clarification: By the presence of a halogen atom in the ring, basic properties of aniline is increased because it is more electronegative so donation of electron will be easy, so basicity increases.
5. What is the characteristic smell of liquid amines?
a) Fish like smell
b) Rotten egg smell
c) Fruity smell
d) Bitter almond smell
Answer: a
Clarification: Amines possess a characteristic ammonia smell, liquid amines have a distinctive “fishy” smell.
6. Which of the following have less melting point than amine?
a) Alcohol
b) Phosphine
c) Carboxylic acid
d) Ether
Answer: b
Clarification: Hydrogen bonding significantly influences the properties of primary and secondary amines. Thus, the melting points and boiling points of amines are higher than those of the corresponding phosphines, but generally lower than those of the corresponding alcohols and carboxylic acids, ether.
7. Which of the following amine has highest boiling point?
a) Butyl amine
b) Diethylamine
c) Triethylamine
d) Dipropylamine
Answer: d
Clarification: Primary and secondary amines have higher boiling because they can engage in intermolecular hydrogen bonding. The boiling points of tertiary amines, which cannot engage in hydrogen bonding because they have no hydrogen atom on the nitrogen atom, and d is correct option because it has more molar mass than other primary, secondary and tertiary amine, more +I effect of propyl group.
8. What is the correct decreasing order of the basic character of the three amines and ammonia?
a) NH3 > CH3NH2 > C2H5NH2 > C6H5NH2
b) C2H5NH2 > CH3NH2 > NH3 > C6H5NH2
c) C6H5NH2 > C2H5NH2 > CH3NH2 > NH3
d) CH3NH2 > C2H5NH2 > C6H5NH2 > NH3
Answer: b
Clarification: All of the amines are more basic than ammonia, but primary and secondary amines are the most basic.
9. Unpleasant smelling carbylamines are formed by heating alkali and chloroform with which of the following?
a) Any amine
b) Any aliphatic amine
c) Any aromatic amine
d) Any primary amine
Answer: d
Clarification: The unpleasant smell is due to formation of isocyanide.
10. Which statement is an incorrect statement?
a) Amines form hydrogen bond
b) Ethyl amine has higher boiling point than propane
c) Methyl amine is more basic than ammonia
d) Dimethyl amine is less basic than methyl amine
Answer: d
Clarification: In methyl amine only one electron releasing group is present but in dimethyl amine two electron releasing groups are present which increase the basicity higher in dimethyl amine.
11. Which of the following is most basic?
a) C6H5NH2
b) (CH3)2NH
c) (CH3)3N
d) NH3
Answer: b
Clarification: In dimethyl amine two electron releasing groups are present which increase the basicity higher in dimethyl amine.
12. If methyl is alkyl group, then which order of basicity is correct?
a) R2NH > RNH2 > R3N > NH3
b) R2NH > R3N > RNH2 > NH3
c) RNH2 > NH3 > R2NH > R3N
d) NH3 > RNH2 > R2NH > R3N
Answer: a
Clarification: Alkyl groups are electron donating (EDG), so they stabilize the positively charge ammonium ions, i.e., the methyl ammonium ion is more stable than the parent ammonium ion because the alkyl group stabilizes the positive charge on the attached nitrogen atom. In dimethyl amine two electron releasing groups are present which increase the basicity higher in dimethyl amine.
13. Among the following, which one is the strongest base?
a) C6H5NH2
b) p−NO2C6H4NH2
c) m−NO2−C6H4NH2
d) C6H5CH2NH2
Answer: d
Clarification: Alkyl groups are electron donating (EDG) and NO2 group is electron withdrawing group. So, the alkyl group stabilizes the positive charge on the attached nitrogen atom.
14. Which of the following is true characteristic feature of aniline?
a) Colourless solid
b) Brown coloured solid
c) Colourless liquid
d) Brown coloured liquid
Answer: c
Clarification: Pure aniline is a colourless liquid, but the colour of aniline changes to dark brown on storage due to atmospheric oxidation.
15. The correct order of basicity of amines in water is which of the following?
a) (CH3)2NH > (CH3)3N > CH3NH2
b) CH3NH2 > (CH3)2NH > (CH3)3N
c) (CH3)3N > (CH3)2NH > CH3NH2
d) (CH3)3N > CH3NH2 > (CH3)2NH
Answer: a
Clarification: Basicity of amines increase with increase in number of −CH3 groups (or any group which cause +I effect), due to increase in electron density on N atom. As a rule, the basicity of 3° amine should be more than that of 2° amine, but actually it is found to be lesser than 2° amines. This is due to more solvation by water in secondary amine. Hence the correct order of basicity is: (CH3)2NH > (CH3)3N > CH3NH2.
Organic Chemistry Multiple Choice Questions on “Five Membered Rings”.
1. Which of the following is not true about the five membered rings?
a) Five membered rings are more stable than 4 membered rings
b) Five membered rings are more stable than 6 membered rings
c) Five membered rings are more stable than 7 membered rings
d) Five membered rings are more stable than 8 membered rings
Answer: b
Clarification: 6-membered rings can have 3D conformations, such as the chair conformation (the more stable) and the boat conformation. These conformations relax the angles, getting them closer to the tetrahedral angle. Thus, cycle gains stability. That is why 5-membered rings are less stable than 6-membered.
2. Which of the following is a not a five membered ring?
a) Pyridine
b) Pyrrole
c) Furan
d) Thiophene
Answer: a
Clarification: Pyridine is a basic five membered heterocyclic organic compound with the chemical formula C5H5N. It is structurally related to benzene, with on (=CH–) group replaced by a nitrogen atom.
3. Which of the following five membered rings is most resonance stabilized?
a) Furan
b) Thiophene
c) Pyrrole
d) Pyridine
Answer: b
Clarification: Thiophene is most resonance stabilized five membered rings among above compounds. As thiophene has Sulphur and least electronegativity ring than nitrogen and oxygen in pyrrole and furan respectively.
4. Five membered rings come under which category of heterocycle classification on the basis of chemical behavior?
a) -excessive heterocycle
b) -deficient heterocycle
c) -equivalent heterocycle
d) Can’t say about the five membered rings
Answer: a
Clarification: In five membered ring, 6 π electrons are distributed between 5 atoms, each atom shares more than 1 e– (1.2e–).
5. What is the reactivity order in the following five membered heterocyclic compounds?
a) Pyrrole
b) Furan
c) Thiophene
d) Pyridine
Answer: a
Clarification: Pyrrole is more reactive than furan and thiophene in electrophilic reactions. Therefore; pyrrole is more prone to electrophilic substitution than furan. The nitrogen atom in pyrrole can conjugate with the π-electrons on the ring, so the density of the π-electrons on the ring will increase.
6. What is the product when thiophene reacts with Br2 in benzene?
a) 2-bromothiophene
b) 3-bromothiophene
c) 2,5-dibromothiophene
d) 3,4-dibromothiophene
Answer: c
Clarification: This is electrophilic substitution reaction, bromination of thiophene in presence of benzene leads to formation of 2,5-dibromothiophene.
7. What is the product when pyrrole reacts with Br2 in ethanol?
a) 2,3-dibromopyrrole
b) 2,3,4,5-tetrabromopyrrole
c) 2,5-dibromopyrrole
d) 3,4-dibromopyrrole
Answer: b
Clarification: This is electrophilic substitution reaction, bromination of pyrrole in presence of ethanol leads to formation of 2,3,4,5-tetrabromopyrrole.
8. What is the name of the following reaction?
a) Gattermann reaction
b) Riemer tiemann reaction
c) Friedal craft reaction
d) Blanc’s chloromethylation
Answer: b
Clarification: This is a Riemer tiemann reaction, in which boiling pyrrole will react with aqueous or alcoholic potash and chloroform.
9. What is the name of the following reaction?
a) Gattermann reaction
b) Riemer tiemann reaction
c) Friedal craft reaction
d) Blanc’s chloromethylation
Answer: a
Clarification: This is a Gattermann reaction, in which boiling pyrrole will react with HCN and HCl followed by addition of water and leads to formation of pyrole-2-carbaldehyde.
10. What will be the reagent used for the completion of the following reaction?
a) Concentrated acid
b) Dilute acid
c) Concentrated base
d) Dilute base
Answer: b
Clarification: In dilute acid for a brief time, pyrrole form trimer.
Organic Chemistry Multiple Choice Questions on “Classification of Organic Compounds”.
1. Organic compounds are broadly classified as _________
a) Open chain compounds and acyclic compounds
b) Open chain compounds and linear chain compounds
c) Cyclic compounds and alicyclic compounds
d) alicyclic compounds and acyclic compounds
Answer: d
Clarification: Organic compounds are broadly classified into open chain and closed chain compounds.
2. Aliphatic compound is the other name for _________
a) Acyclic compounds
b) Alicyclic compounds
c) Ring compounds
d) Closed chain compounds
Answer: a
Clarification: open chain compounds or acyclic compounds are otherwise called as aliphatic compounds.
3. Which among the following is not an example of Acyclic compound?
a) Acetaldehyde
b) Ethane
c) Cyclopropane
d) Isobutane
Answer: c
Clarification: Cyclopropane is a ring (cyclic) compound and hence it does not come with the examples of open chain compounds.
4. Which among the following is not an example of alicyclic compound?
a) Cyclohexane
b) Cyclohexene
c) Tetrahydrofuran
d) Acetic acid
Answer: d
Clarification: Acetic acid is a linear chain compound (acyclic) and hence it is not an example of ring compound (alicyclic).
5. Which among the following is not an aromatic compound (in specific)?
a) Naphthalene
b) Aniline
c) Pyridine
d) Tropolone
Answer: c
Clarification: Pyridine is heterocyclic aromatic compound. Whereas naphthalene and aniline are benzenoid aromatic compounds and tropolone is a non-benzenoid aromatic compound.
6. Find the odd one among the following.
a) Alicyclic compounds
b) Heterogeneous compounds
c) Branched chain compounds
d) Aromatic compounds
Answer: c
Clarification: Branched chain compound is a classification of open-chain compounds. Whereas, alicyclic, aromatic and heterogeneous compounds are sub-classifications of cyclic compounds.
7. Identify the odd one among the following.
a) Indene
b) Anthracene
c) o,m,p-xylene
d) Azulene
Answer: d
Clarification: Azulene is a non- benzenoid compound. Whereas, Indene, anthracene, and o,m,p-Xylene are examples of benzenoid aromatic compounds.
8. Organic compounds can be classified even based upon the function groups. Identify the one which is not a functional group?
a) Isocyanide
b) Isocyano
c) Carboxyl
d) Carbonyl
Answer: a
Clarification: Isocyanide is a compound and it is not a functional group.
9. Which among the following is not a class of organic compound?
a) Carbonyl compound
b) Nitro compound
c) Amides
d) Electro compounds
Answer: d
Clarification: Classes of organic compounds are those which involves organic compounds such as carbon, hydrogen and oxygen. Hence, electro compounds is not a class of organic compounds.
10. Which among these is not associated with aliphatic compounds?
a) They contain (4n+2)pi electrons
b) Contain straight chain compounds
c) Contain branched chain compounds
d) Has appropriate number of H-atoms and functional groups
Answer: a
Clarification: The aromatic compounds (4n+2)pi electrons, which comes under the classification of cyclic compounds and hence they are not associated with aliphatic compounds.
Organic Chemistry Multiple Choice Questions on “Nuclear Magnetic Resonance – 1”.
1. Which of the following organic compound with molecular formula C3H C12 exhibits only one signal in the IH NMR spectrum?
a) 2, 2-dichloropropane
b) 1, 2-dichloropropane
c) 1, 3-dichloropropane
d) 1, 1-dichloropropane
Answer: a
Clarification: Only (a) compound has all chemically equivalent hydrogen. So, give only one peak.
2. The 1H NMR spectrum of CH3OCHCICH2Cl will exhibit ________
a) A three proton doublet. One proton singlet and a two proton doublet
b) A three proton singlet. One proton singlet and a two proton doublet
c) A three proton singlet. One proton triplet and a two proton doublet
d) A three proton triplet. One proton triplet and a two proton triplet
Answer: c
Clarification: The 1H NMR spectrum of CH3OCHCICH2Cl will exhibit three proton singlet, one proton triplet and two proton doublet.
No. of peaks = n + 1
(n = no. of neighbouring chemically equivalent Hydrogen nuclii)
3. Which of the following 1H-NMR spectrum of compound with molecular formula C4H9NO2 shows delta 5.30 (broad, IH), 4.10(q, 2H), 2.80 (d, 3H), 1.20 (t, 3H) ppm?
a) CH3NHCOOCH2CH3
b) CH3CH2NHCOOCH3
c) CH3OCH2CONHCH3
d) CH3CH2OCH2CONH2
Answer: a
Clarification: C4H9NO2;
chemical shift = 5.30 (broad, 1H)
chemical shift = 4.10 (q, 2H)
chemical shift = 2.80 (d, 3H)
chemical shift = 1.20 (t, 3H)
As shown in the above compound, doublet, because of neighbouring 1H attached to N and one will be broad as because of hydrogen attached to N, quatret becauseof neighbouring 3H and finally triplet, because
of neighbouring 2H.
4. Which of the following has three types of hydrogens in the following compounds?
a) Br-CH = CH2
b) CH3 – CH2 – CH3
c) C6H5CH2
d) CH3– CH2 – CH(CH3) – N02
Answer: a
Clarification:
We will replace them by D and check the structure if they are different then they are different.
both are different
so, total 3 type of hydrogen
So, three different type of hydrogen is present in compound a.
5. Which of the compound show only one signal is present in the PMR spectra?
a) C3H4, C3H6
b) C4H6, C5H12
c) C8H18, C2H6O
d) All of the mentioned
Answer: d
Clarification: All of the below compounds have only type of proton, so only one signal.
6. How many Hertz does 1 ppm correspond to for an PMR spectrometer operating at a radio frequency of 60 MHz and 100 MHz?
a) 6 Hz, 10 Hz
b) 60 Hz, 100 Hz
c) 100 Hz, 60 Hz
d) 10Hz, 100Hz
Answer: b
Clarification:
7. Compound C4H10O gave PMR spectrum consisting of two groups of lines (multiplets) with relative intensities in the ratio 3 : 2. Other compound of the same formula exhibited two lines with relative area of 9 : 1. What are these compounds.
a) Diethyl ether and n-butanol
b) t-Butyl alcohol and 1-Methoxypropane
c) diethyl ether and t-Butyl alcohol
d) Diethyl ether and 2-Methoxypropane
Answer: c
Clarification: NMR data: Two groups of lines in 3 : 2 and two groups of lines in 9 : 1 for same molecular formula
a) CH3–CH2–O–CH2–CH3
6Ha -> 1 group of line;
4Hb -> 1 group of line
Therefore, Intensity => 6 : 4 => 3 : 2.
b) 
9H -> 1 group of line and 1H -> 1 group of line / single peak.
Therefore, Intensity => 9 : 1.
8. The distance between the centers of the peaks of doublet is called as?
a) Coupling constant
b) Spin constant
c) Spin-spin coupling
d) Chemical shift
Answer: a
Clarification: When the proton under observation is split into several peaks by neighboring protons, the distance between these peaks is a constant called the coupling constant (J). Any two sets of protons that exhibit the same coupling constant are most likely splitting eachother signals and are said to be coupled. 
9. H2, CH4, C2H6 and C6H6 exhibit which PMR spectra?
a) Singlet
b) Doublet
c) Triplet
d) Quintet
Answer: a
Clarification: All structures have only one type of hydrogen. So, all give single peak.
10. A proton Hb is coupled to four equivalent protons Ha. The multiplicity and the relative intensity of lines in the signal Hb is?
a) Doublet, I : 4
b) Triplet, I : 4 : 6
c) Quintet, 1 : 4 : 6 : 4 : 1
d) Quartet, 1 : 4 : 6 : 4
Answer: c
Clarification: Each signal in a proton NMR spectrum may or may not be split into one or more peaks, this is called as multiplicity. The most common concept associated with signal multiplicity is the n+1 rule. According to this rule, the signal for the proton under observation will be split into n+1 peaks by protons attached to adjacent carbons, where n is the number of such protons.
No. of peaks = n + 1
(n = no. of neighbouring chemically equivalent Hydrogen nuclii)
So, Hb —-> 4Ha
Multiplicity —->n + 1 = 5 (Quintet)
And, according to Pascal triangle(shown below): Intensity–> 1 : 4 : 6 : 4 : 1
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
Organic Chemistry Multiple Choice Questions on “Bond Cleavage”.
1. When a bond is dissociated, what is the change in energy in surrounding?
a) Energy is absorbed
b) Energy is released
c) No change in energy
d) Firstly, energy will be absorbed and then released
Answer: a
Clarification: When a chemical bond is broken energy will be absorbed from surrounding and this is called as bond dissociation energy.
2. Energy released or absorbed in a chemical bond formation or dissociation is measured in which of the following units?
a) Kelvin
b) Joule
c) Pascal
d) mol
Answer: b
Clarification: Energy is measured in Joules (J). However, we often express the quantity as kilojoules (kJ) – that is, the value in Joules divided by 1000.
3. What will be the change in energy for the following reaction?
H2(g)+F2(g)→ 2HF
Given: To break one mole of H2, energy absorbed is 436 kJ. To break one mole of F2, energy absorbed is 158 kJ. To form one moles of HF, energy released is 568 kJ.
a) 542KJ
b) -542KJ
c) 26 KJ
d) -26 KJ
Answer: b
Clarification: So, applying the equation,
∆H = ∑∆H (bonds broken in reactants) − ∑∆H (bonds made in products)
ΔH reaction = (436+158)–(2X568) = −542kJ
The overall enthalpy of the reaction is negative, i.e., it’s an exothermic reaction where energy is released in the form of heat.
4. Compound undergoing homolytic bond cleavage will lead to formation of which chemical species?
a) Anion
b) Cation
c) Free radical
d) Atoms
Answer: c
Clarification: During homolytic fission of a neutral molecule with an even number of electrons, two free radicals will be generated.
5. In which form energy is stored in chemical bonds?
a) Kinetic energy
b) Chemical energy
c) Potential energy
d) Thermal energy
Answer: c
Clarification: The first law of thermodynamics essentially states that the total energy in a chemical universe is constant but it can be converted from one form to another. The energy stored in chemical bonds is also a type of potential (stored) energy, not a type of kinetic energy.
6. Which one of the following statements best describes the enthalpy change of a reaction?
a) The energy released when chemical bonds are formed during a chemical reaction
b) The energy consumed when chemical bonds are broken during a chemical reaction
c) The difference between the energy released by bond formation and the energy consumed by bond cleavage during a chemical reaction
d) The increase in disorder of the system as a reaction proceeds
Answer: c
Clarification: The enthalpy change of a reaction is the overall change in enthalpy as a reaction proceeds, taking into account the energy consumed by bond cleavage and the energy liberated by bond formation. The increase in disorder of the system is represented by its entropy, not its enthalpy.
7. Which of the following terms describes a reaction in where more energy is released by bond formation than is consumed by bond cleavage?
a) Exothermic
b) Endothermic
c) Isothermal
d) Adiabatic
Answer: a
Clarification: Exothermic is the flow of energy out of the system to its surroundings. By contrast endothermic’ describes a reaction in which there is a flow of energy into the system from its surrounding.
8. Which energy is required for homolytic cleavage?
a) Singlet
b) Doublet
c) Triplet
d) Quadruplet
Answer: c
Clarification: The triplet excitation energy of a sigma bond is the energy required for homolytic dissociation, but the actual excitation energy may be higher than the bond dissociation energy due to the repulsion between electrons in the triplet state.
9. Which energy is required for heterolytic cleavage?
a) Singlet
b) Doublet
c) Triplet
d) Quadruplet
Answer: a
Clarification: The singlet excitation energy of a sigma bond is the energy required for heterolytic dissociation, but the actual singlet excitation energy may be lower than the bond dissociation energy of heterolysis as a result of the Coulombic attraction between the two ion fragments.
10. How the bond dissociation enthalpy changes as one goes along the series of diatomics Li2, B2, C2, N2, O2 and F2?
a) Increases
b) Decreases
c) Increases then decreases
d) Decreases then increases
Answer: c
Clarification: On going Li2, B2, C2, N2, O2, F2 the complexity of the molecular orbitals changes as the p-orbitals can interact with each other end on to give σ-bonding and σ-antibonding molecular orbitals OR they can interact side-ways on to give π-bonding and π-antibonding molecular orbitals that come in degenerate (equal energy) pairs. Electrons continue to be added according to aufbau, Pauli & Hund.
Organic Chemistry Multiple Choice Questions on “Dihydric Alcohols”.
1. How many hydroxyl groups are present in diols?
a) One
b) Two
c) Three
d) Four
Answer: b
Clarification: Any of a class of alcohols that have two hydroxyl groups in each molecule are diols or dihydric alcohol.
2. Which of the following is not the example of dihydric alcohol?
a) Glycerin
b) Parahydroxyphenol
c) Resorcinol
d) Catechol
Answer: a
Clarification: Glycerin is a trihydroxy alcohol i.e. Glycerol, is a simple polyol compound. It is a colorless, odorless, viscous liquid that is sweet-tasting and non-toxic. Structure of glycerin is
3. Which of the following is the correct structure of catechol?
a) 
b) 
c) 
d) 
Answer: b
Clarification: Catechol is dihydric alcohol, in which hydroxyl group is attached adjacent to each other in benzene ring.
4. Which of the following is the correct structure of parahydroxyphenol?
a)
b)
c)
d)
Answer: c
Clarification: Parahydroxyphenol is dihydric alcohol, in which hydroxyl group is attached at 1, 4 positions to each other in benzene ring.
5. Which of the following is the correct structure of resorcinol?
a)
b)
c)
d)
Answer: a
Clarification: Resorcinol is dihydric alcohol, in which hydroxyl group is attached at 1, 3 positions to each other in benzene ring.
6. What is the general formula for dihydric alcohol?
a) (CH2)n(OH)2 where n = 2,3…etc
b) (CH2)n(OH)2 where n = 1,2,4… etc
c) CnH2n+1OH where n = 1, 2 …etc
d) (CH2)n(OH)3 where n = 3, 4, 5 …etc
Answer: b
Clarification: Dihydric alcohol have general formula (CH2)n(OH)2, where n = 2,3,4 etc.
7. Which of the following reagent can be used to carry out this synthesis?
a) Seliwanoff reagent
b) Baeyer’s reagent
c) Barfoed reagent
d) Benedict reagent
Answer: b
Clarification: Cold dilute alkaline solution of Bayer’s reagent can be used to carry out this synthesis.
8. Which type of product is formed when Cold dilute alkaline solution of Bayer’s reagent reacts with alkene?
a) Syn-diol
b) Anti-diol
c) syn- and anti-geometry will not be there in diol
d) Trans diol
Answer: a
Clarification: On hydroxylation product is formed when Cold dilute alkaline solution of Bayer’s reagent reacts with alkene is syn.
9. What will be the compound A which can be used to carryout this synthesis of diols?
a) RCO2OH
b) RCHO
c) RCOR’
d) RCOOH
Answer: a
Clarification: A peroxy acid can be used to carryout this synthesis as shown in below reaction.
10. How can we detect the presence of resorcinol in the solution?
a) Ceric ammonium nitrate test
b) Lucas test
c) Phthalic acid test
d) Remini’s test
Answer: c
Clarification: Phthalic acid test:The anhydrides of aromatic 1,2-dicarboxylic acids on heating with resorcinol gives a dye fluorescein. This dye in NaOH solution gives a yellowish red solution with green fluorescence.