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Organic Chemistry Multiple Choice Questions on “Nomenclature”.
1. Saturated hydrocarbons are otherwise referred as ________________
a) Alkanes
b) Alkenes
c) Alkynes
d) Alkaloids
Answer: a
Clarification: Saturated hydrocarbons contain large number of hydrogen atoms in them and hence they are known as alkanes.
2. Identify the correct alkane name for the molecular formula C30H62.
a) Propdecane
b) Eicosane
c) Triacontane
d) Dodecane
Answer: c
Clarification: Triacontane is the alkane with the molecular formula C30H62 as alkanes have the general formula CnH2n+2.
3. Identify the smallest alkane which can form a ring structure (cycloalkane)?
a) Cyclomethane
b) Methane
c) Cyclopropane
d) Propane
Answer: c
Clarification: Cyclopropane is the only smallest alkane that can form a successful ring structure with C-C bond on removal of a hydrogen atom.
4. In which among the following alkane, a carbon atom is displaced so as to form a compactly structure with the resemblance of a butterfly wing?
a) Cyclopropane
b) Cyclobutane
c) Cyclopentane
d) Cyclohexane
Answer: b
Clarification: If one of the carbon atoms of cyclobutane is not displaced, then the C-C bond would be exactly at right angles to each other and they do not form a cyclic structure.
5. The first step in IUPAC nomenclature is to identify the total number of carbon atoms present in the compound.
a) True
b) False
Answer: b
Clarification: The first step is to find the number of carbon atoms present only in the main/longest chain of the compound.
6. The substituent in the chain is named by replacing the “ane” in the alkanes by _________
a) ene
b) ic
c) one
d) yl
Answer: d
Clarification: ene, ic and one are used for representing alkenes, carboxylic acids, and ketones respectively, hence yl is the term which is to be substituted in place on ane in the alkanes.
7. The C=C bond in the chain of the compound considered is shown by _________
a) Specifying the number of carbon atoms associated with the bond
b) Specifying the number of carbon atoms at beginning of the C=C bond
c) Specifying the number of carbon atoms at end of the C=C bond
d) Specifying the number of carbon atoms in the entire chain
Answer: b
Clarification: According to the rules of IUPAC, The C=C bond in the chain of the compound considered is shown by specifying the number of carbon atoms at beginning of the C=C bond.
8. Dienes are the name given to compounds with _________
a) Exactly a double bond
b) Exactly a triple bond
c) Exactly two double bond
d) More than two double bond
Answer: c
Clarification: Alkenes are the name given to compound with one double bond and dienes are the one given to compounds with two double bond.
9. Triple bond with two carbon atoms on either side is called _________
a) Methnyl group
b) Ethynyl group
c) Propionyl group
d) Propargyl group
Answer: b
Clarification: Triple bond with two atoms on either side are called as ethynyl group and those with one carbon on one side and two carbon on another side are called as propargyl group.
10. The substituent groups that are commonly associated with benzene ring are _________
a) Phenyl and benzyl
b) Propyl and phenyl
c) Methyl and benzyl
d) Butyl and phenyl
Answer: a
Clarification: Phenyl and benzyl are commonly associated with benzene ring due to their closely associated structure with difference in hydrogen atom.
3. Which of the following is incorrect statement for the giver reaction?
a) Rate of formation of II and IV would be identical
b) Rate of formation of I would be slower than that of III
c) Formation of I would show primary isotope effect
d) Formation of III involves E1 reaction
Answer: d
Clarification: Rate of formation of II and IV similar because H & D are approximately of same size.
C–H & C–D bond breaking is the part of RDS hence rate in I is slower than III
It undergo E2 not via E1 because carbocation is not stable.
4. When the all-cis isomer of C6H6Cl6 (2, 3, 4, 5, 6-Hexachlorocyclohexane) is heated with alc.KOH, what will be the most probable product?
a) 
b) 
c) 
d) 
Answer: b
Clarification: Here three times Elimination(E2) occurs and three pie bond will be formed and three water molecule will be removed as shown below.
5. Which of the following statements is correct for alkyl halide?
a) Alkyl halide will always show SN1 mechanism
b) As branching at carbon increases, E1 mechanism is favoured as compared to SN1 mechanism
c) In unimolecular reaction, increasing the temperature donot favours E1 mechanism
d) In most unimolecular reactions of alkyl halide E1 reaction is favoured over SN1 reaction
Answer: b
Clarification: In most unimolecular reactions of alkyl halide SN1 reaction is favoured over E1 reaction. E1 mechanism is favoured as compared to SN1 mechanism by branching at carbon increases, as more steric hinderence at the attacking site will leads to E1 mechanism.
6. Which of the following order is incorrect for the rate of E2 reaction?
a) 5-Bromocycloheptene > 4-Bromocycloheptene
b) 2-Bromo- I -phenylbutane > 3-Bromo- I-phenylbutane
c) 3-Bromocyclohexene > Bromocyclohexane
d) 3-Bromo-2-methylpentane > 2-Bromo-4-methylpentane
Answer: a
Clarification: For 5-Bromocycloheptene > 4-Bromocycloheptene, more stable product leads to faster rate of reaction. Product formed by E2 of 4-Bromocycloheptene is more stable than 5-Bromocycloheptene. So, rate of reaction 4-Bromocycloheptene is fater than 5-Bromocycloheptene. So option 5-Bromocycloheptene > 4-Bromocycloheptene is incorrect.
For 2-Bromo- I -phenylbutane > 3-Bromo- I-phenylbutane, it is true rate of reaction for 2-Bromo- I -phenylbutane is more than 3-Bromo- I-phenylbutane as product formed in former is more stable due to resonance.
For 3-Bromocyclohexene > Bromocyclohexane, it is true rate of reaction for 3-Bromocyclohexeneis more than Bromocyclohexane as product formed in former is more stable.
For 3-Bromo-2-methylpentane > 2-Bromo-4-methylpentane, it is true rate of reaction for 3-Bromo-2-methylpentane more stable than 2-Bromo-4-methylpentane as product formed in former is more stable.
7. Which of the following statement is correct?
a) E2 is a concerted reaction in which bonds break and new bonds form at the same time in a single step
b) Order of reactivity of alkyl halides towards E2 dehydrohalogenation is found to be 3o > 2o > 1o
c) In E2 elimination different stereoisomer (diastereomer) converts into different stereo product
d) All of the mentioned
Answer: d
Clarification: E2 is a concerted reaction in which bonds break and new bonds form at the same time in a single step. Order of reactivity of alkyl halides towards E2 dehydrohalogenation is found to be 3o > 2o > 1o. In E2 reaction both hydrogen and leaving group should be antiperiplanar and that’s why In E2 elimination different stereoisomer (diastereomer) converts into different stereo product.
8. Which of the following statement is correct regarding the following reaction?
a) Major product is endocyclic alkene formed according to Saytzeff
b) Major product is exocyclic alkene formed according to Saytzeff
c) Major product is exocyclic alkene formed according to Hoffmann
d) Major product is endocyclic alkene formed according to Hoffmann
Answer: c
Clarification: Exocyclic alkene are more stable than endo beacuse of but due to “steric hindrance” the less substituted alkene is formed.
9. Which of the following statement (s) is/are true about the following eliminations?
(1) Hoffmann product is major product in I (2) Saytzeff product is major product in I (3) Hoffmann product is major product in II (4) Saytzeffproduct is major product in II
a) 1 and 2
b) 1 and 4
c) 2 and 3
d) 3 and 4
Answer: b
Clarification: Here correct is statement 1 and 4 are true. Due to steric hinderence tert-Butyl alcohol anion will attack less substituted C-H bond.
Ethanol anion will attack more substituted C-H bond to form more stable alkene.
10. Which incorrect about alkyl bromide having molecular formula C5H11Br?
a) One isomeric alkyl bromide undergoes E1 elimination at the fastest rate
b) Only one is incapable of reacting by the E2 mechanism
c) Only one isomer gives a single alkene on E2 elimination
d) 2-Bromopentane gives the most complex mixture of alkenes on E2 elimination
Answer: c
Clarification:
From the following isomers of C5H11Br, except IX all are capable of E2 elimination.
11. In which reaction product formation takes place by hoffmann rule?
a) I and II
b) II and III
c) I, II and III
d) I, II, III and IV
Answer: d
Clarification: All of the above molecules will form product by hoffmann rule that is less substituted C-H bond will be attacked because more stable product will be formed via this in all cases.
I.
II.
When leaving group are F & NH2 type then they give hoffmann product.
III.
IV.
12. Which of the following compounds cannot give E1CB reaction?
a) CF3 – CHCl2
b) C6h5 – CH2 – CH2F
c) CH3 – CH2 – CH2Br
d) 
Answer: d
Clarification:
Compound d will give E2 reaction, Br is a better leaving group so formation of carboanion and removal of leaving group occurs simultaneously. But E1CB reaction is shown by poor leaving group like in compounds a, b and c.
13. Which of the following are incorrect statement?
a) X being electronegative, makes the H (on C-2) more acidic
b) Due to electron withdrawal nature of X, it stabilises the carbanion
c) (X) destabilises the carbanion due to the presence of lone pairs
d) The reaction proceeds by an E1CB pathway
Answer: c
Clarification: Due to inductive effect of halogen it stabilize the negative charge on adjacent to carbon atom, as X will withdraw the electron towards itself.
14. Which of the following is not the examples of E1CB reaction?
a) 
b)
c) 
d) 
Answer: d
Clarification: Compound d will give E2 reaction, CL is a better leaving group so formation of carboanion and removal of leaving group occurs simultaneously. But E1CB reaction shown by poor leaving group like in compounds a, b and c.
15. Which of the following reaction will not show addition elimination mechanism?
a) 
b) 
c) 
d) 
Answer: d
Clarification: Compounds a, b and c will follow Addition- Elimination mechanism.
Compound will follow elimination -addition mechanism, as shown in below reaction. NH– anion first attack on the ring which lead to formation of benzyne ring and further addition of -NH2 will take place.
.
Tricky Organic Chemistry Questions and Answers on “Electrophilic Aromatic Substitution”.
1. What will be the product formed when phenol reacts with Br2 in CCl4 medium?
a) 3-Bromophenol
b) 4- Bromophenol
c) 3,5-Dibromophenol
d) 2,4,6-Tribromophenol
Answer: b
Clarification: Because the non-polar solvent (i.e. CCl4) induces a weaker dipole in Br-Br as compared to the polar solvent (i.e. water), a weaker electrophile (Br+) is produced that undergoes electrophilic aromatic substitution with the benzene ring. Consequently, we get mono-bromination with Br (in CCl4) with compared with tri-bromination Br (in aq.).
2. What is the electrophile in the electrophilic substitution reaction of benzene using oleum and conc. H2SO4?
a) SO3
b) NO3
c) NO2+
d) NO+
Answer: a
Clarification: This reaction is sulphonation of benzene ring in which SO3 is the real attacking electrophile and oleum will increase the concentration of SO3.
3. What will be the attacking electrophile in this reaction?
a) Carbanion
b) Halonium
c) Radical
d) Transition state
Answer: b
Clarification: In halogenation of benzene ring halonium ion (Cl+) will attack at benzene ring.
4. Which of the following aromatic compounds undergo Friedel–Crafts alkylation with methyl chloride and aluminum chloride?
a) Benzoic acid
b) Nitrobenzene
c) Toluene
d) Aniline
Answer: c
Clarification: Toluene will undergo Friedel–Crafts alkylation with methyl chloride and aluminum chloride as because as weak electron withdrawing functional group — will not make the ring less electron dense, hence more reactive in an aromatic electrophilic substitution as compared to electron withdrawing group (benzoic acid, nitrobenzene, aniline).
5. What will be the product in the given reaction?
a) m-chlorophenol
b) o-chlorophenol and p-chlorophenol
c) o-hydroxytoluene and p-hydroxytoluene
d) m-hydroxytoluene
Answer: c
Clarification: -OH is a ortho and para directing group so attack of electrophile (carbocation- CH3) will attack there, as shown below:
6. What is the electrophile in the acylation of benzene?
a) AlCl3
b) CO+
c) Cl+
d) R-CO+
Answer: d
Clarification: The electrophile in the electrophilic substitution reaction of acetyl chloride (CH3COCl) and AlCl3 reacting with benzene is R-CO+.
7. How would you synthesis the following reaction?
a) conc. HNO3 + conc. H2SO4
b) conc. HNO3
c) anhydrous AlCl3 + Ph-NO2
d) conc. H2SO4 + Oleum
Answer: a
Clarification: This reaction is nitration of benzene and conc. HNO3 and conc. H2SO4 is used to carry out this synthesis although HNO2 + HNO3 can also be used.
8. Which of the following reactants can be used to carry out following reaction?
a) conc. HNO3 + conc. H2SO4
b) HCl + CO + AlCl3
c) anhydrous AlCl3 + Ph-NO2
d) conc. H2SO4 + Oleum
Answer: b
Clarification: This is Gattermann Koch Reaction in which HCl + CO + AlCl3 are used as reactant. H-CO+ is the real attacking electrophile.
9. What will be the product in the given reaction?
a) m-chlorotoluene
b) o-chlorotoluene and p-chlorotoluene
c) 1-chloro-3-methylbenzene
d) no reaction
Answer: d
Clarification: Here no reaction will take place because CCl4 + Cl2 show addition reaction and benzene don’t give addition reaction.
10. Which of the following is the most activating in electrophilic aromatic substitution?
a) -NO2
b) -NHCOCH3
c) -CN
d) -NH2
Answer: d
Clarification: Groups with unshared pairs of electrons, such as the amino group of aniline, are strongly activating and ortho/para-directing by resonance. Such activating groups donate those unshared electrons to the pi system, creating a negative charge on the ortho and para positions. These positions are thus the most reactive towards an electron-poor electrophile. The highest electron density is located on both ortho and para positions, although this increased reactivity might be offset by steric hindrance between substituent and electrophile.
To practice tricky questions and answers on all areas of Organic Chemistry,
Organic Chemistry Multiple Choice Questions on “Physical Properties of Ethers”.
1. Which of the following is distinguish test for ether?
a) Lucas test
b) FeCl3 test
c) Molisch test
d) Zeisel test
Answer: d
Clarification: The Zeisel determination or Zeisel test is a chemical test for the presence of esters or ethers in a chemical substance. In a qualitative test a sample is first reacted with a mixture of acetic acid and hydrogen iodide in a test tube.
2. Which of the following statement is not true about ethers?
a) The lower ethers are highly volatile and flammable
b) Lower ethers also act as anaesthetics
c) Ethers are not organic solvents
d) Simple ethers (such as diethyl ether) are tasteless
Answer: c
Clarification: Because like dissolve like, Diethyl ether is very non-polar compared to water so it will dissolve the non-polar substances and precipitate the ionic compounds. Since like dissolves like, your barely-polar organic ether is often happy to dissolve your non-polar organic molecules.
3. Why ether is more volatile than an alcohol having the same molecular formula?
a) dipolar character of ethers
b) alcohols having resonance structures
c) inter-molecular hydrogen bonding in ethers
d) inter-molecular hydrogen bonding in alcohols
Answer: d
Clarification: The ether are less volatile than alcohol of same molecular mass because in alcohols there is intermolecular H-bonding which makes it less volatile and High B.P. But in case of ether there is no intermolecular H-Bonding which makes it more volatile and less B.P.
4. Why the boiling point of ethyl alcohol is much higher than that of dimethyl ether though both have the same molecular weight?
a) Ether is insoluble in water
b) Methyl groups are attached to oxygen in ether
c) Dipole moment of ethyl alcohol is less
d) Ethyl alcohol shows hydrogen bonding
Answer: d
Clarification: Alcohols have much higher boiling points than the comparable alkanes or the related ethers because molecules of alcohols can interact through hydrogen bonding while those of alkanes and ethers cannot. Thus, Ethanol has a higher boiling point (78℃) than dimethyl ether (−25℃) though they have the same molecular weight.
5. Which of the following has lower boiling point than dimethyl ether?
a) Pentane
b) Propane
c) Tetrahydrofuran
d) n-butanol
Answer: b
Clarification: The difference in molecular weight made difference between boiling point here mainly. It would be best to set up a better comparison and then ask the question. diethyl ether is a chain of 5 atoms in a row. The appropriate alkane to compare it to is pentane, also 5 atoms in a row. Ethane only has two carbons (CH3-CH3).
6. Which of the following statements about ethers is incorrect?
a) Ethers are flammable
b) Ethers form peroxides by free radical oxidation
c) Ethers are widely used as extraction solvents
d) An ether oxygen can only be found within an acyclic carbon chain
Answer: d
Clarification: Although widely used as extraction solvents ethers are highly flammable and readily form peroxides by free radical oxidation. An ether oxygen can be present in an open chain or as part of a cyclic system.
7. Which of the following product is formed, when ether is exposed to air?
a) Oxide
b) Alkanes
c) Alkenes
d) Peroxide of diethyl ether
Answer: d
Clarification: On exposure to air ether form peroxide of diethyl ether, as shown in below reaction.
8. An organic compound A reacts with sodium metal and forms B. On heating with conc. H2SO4, A gives diethyl ether. What are A and B?
a) C2H5OH and C2H5ONa
b) C3H7OH and CH3ONa
c) CH3OH and CH3ONa
d) C4H9OH and C4H9ONa
Answer: d
Clarification: When C2H5OH reacts with sodium metal and forms C2H5ONa and on heating with H2SO4 giver diethyl ether.
9. What is the smell of the diethyl ether?
a) Vinegar
b) Sweet smell
c) Rotten egg
d) Fish like smell
Answer: b
Clarification: Diethyl ether has a rather sweet, ethereal smell. It might have a hint of an alcohol-like odor, very penetrating.
10. Which of the following Is the correct reason for cyclic ether are miscible in water?
a) Planer structure
b) Larger size
c) Molecular weight
d) Hydrogen bonding
Answer: d
Clarification: Cyclic ethers such as tetrahydrofuran and 1,4-dioxane are miscible in water because of the more exposed oxygen atom for hydrogen bonding as compared to linear aliphatic ethers.
Organic Chemistry Multiple Choice Questions on “Enolates”.
1. Which of the following is the structure of the most stable enol form of 2-methylpentan-3-one?
a) 
b) 
c) 
d) 
Answer: a
Clarification: A more substituted double bond is generally more stable.
2. Which of the following is the least abundant aldol adduct formed from an equimolar mixture of ethanal and propanone in aqueous NaOH solution?
a) 
b) 
c) 
d) 
Answer: d
Clarification: Adduct (d) is the propanone dimer which is destabilized by the steric strain caused by the geminal methyl groups; the aldol reaction is reversible and a less stable product will be less abundant.
3. An acetoacetic ester synthesis of a ketone proceeds by alkylation of the enolate of the acetoacetic ester followed by ester hydrolysis and decarboxylation of the β-ketoacid. Which of the following methyl ketones is difficult to prepare by this method?
a) 
b) 
c)
d)
Answer: a
Clarification: The alkylation of the enolate is an SN2 reaction, and sterically hindered alkyl groups are difficult to introduce. However, even dialkylation is possible with reactive alkylating agents.
4. A malonic ester synthesis of a carboxylic acid proceeds by alkylation of the enolate of the malonic ester followed by ester hydrolysis and decarboxylation of the β-dicarboxylic acid. Which of the following carboxylic acids is difficult to prepare by this method?
a) PhCH2CH2CO2H
b) (CH3)2CHO2H
c) PhCH2CO2H
d)
Answer: c
Clarification: The alkylation of an enolate is an SN2 reaction, and halobenzene do not react by this mechanism; consequently, phenyl cannot be introduced in this way. However, even dialkylation is sometimes possible with primary alkyl halides.
5. Which of the following is the most stable enolate derived from 4-methylhexan-3-one?
a) 
b) 
c) 
d) 
Answer: c
Clarification: A more substituted double bond is generally more stable. The difference in stability between the stereoisomeric enolates.
6. Which enol form of ethyl 3-oxobutanoate is predominant at equilibrium?
a) 
b)
c) 
d) 
Answer: b
Clarification: Conjugation of the C=C double bond with the remaining carbonyl provides a more stable enol, which rules out. Enols may all be obtained by protonation of the same enolate or the readily interconverting stereoisomeric enolates. Of these, It allow stabilizing intramolecular hydrogen bonding; it is the more stable of the two owing to the conjugative stability of the remaining ester group.
7. Which is a wrong structure as a resonance contributor to the most stable enolate form of pentan-2,4-dione?
a) 
b) 
c) 
d) 
Answer: d
Clarification: Structure is of a different enolate ion (the position of a hydrogen atom is changed).
8. Which enol form of the following dicarbonyl compound is predominant at equilibrium?
a) 
b) 
c) 
d) 
Answer: b
Clarification: For enolate ions of 1,3-dicarbonyl compounds. Conjugation of the C=C double bond with the remaining carbonyl provides a more stable enol, which rules out. Enols may all be obtained by protonation of the same enolate or the readily interconverting stereoisomeric enolates. Of these, allow stabilizing intramolecular hydrogen bonding. NMR evidence indicates that is the more stable of the two presumably owing to the conjugative stability of the remaining amide group.
9. Which enolate derived from 1-phenylbutan-2-one is most stable?
a) 
b) 
c)
d) 
Answer: c
Clarification: Formation of the double bond in conjugation with the phenyl group leads to a more stable enolate, and it is probably more stable (less sterically strained) than the stereoisomeric.
10. Which of the following statements is wrong?
a) The base-catalysed α-halogenation of propanone is first order in the concentration of the base
b) The rate constant for the base-catalysed α-halogenation of propanone decreases in the order Cl2 > Br2 > I2
c) The base-catalysed α-halogenation of propanone proceeds easily to give 1,1,1-trihalopropanone
d) Polyhalogenation of propanone is difficult under acidic conditions, but the products are the same as those obtained under basic conditions
Answer: b
Clarification: The rate-determining step of the base-catalyzed α-halogenation is usually the initial proton abstraction to give the enolate, so the rate constant is independent of the identity (or concentration) of the halogen.
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.
