300+ REAL TIME Organic Chemistry Objective Questions & Answers 2023

250+ TOP MCQs on Mass Spectroscopy and Answers

Organic Chemistry Multiple Choice Questions on “Mass Spectroscopy”.

1. Which of the following statement is false for mass spectroscopy?
a) Mass spectroscopy is used to identify unknown compounds within a sample, and to elucidate the structure and chemical properties of different molecules
b) Particle are characterized by their mass to charge ratios (m/z) and relative abundances
c) This technique basically studies the effect of ionizing energy on molecules
d) This technique can be used on all state of matter
Answer: d
Clarification: A mass spectrum measures the masses within a sample. Particles are characterized by their mass to charge ratios (m/z) and relative abundances. This technique basically studies the effect of ionizing energy on molecules and is applied to pure samples as well as complex mixtures. This technique is only possible to perform in the gaseous state.

2. Which of the following main component of mass spectroscopy deal with resolving the ions into their characteristics mass components according to their mass-to-charge ratio?
a) Ion Source
b) Analyzer
c) Detector System
d) Analyzer tube
Answer: b
Clarification: The complete process involves the conversion of the sample into gaseous ions, with or without fragmentation, which is then characterized by their mass to charge ratios (m/z) by an analyser. Eventually, the second analyzer will analyse fragment the selected ions and detect the ions emerging from the last analyzer and measure their and relative abundances with the detector that converts the ions into electrical signals.

3. Who discovered the mass spectrometer?
a) Francis Aston
b) J. J Thomson
c) Ernest O. Lawrence
d) Walter Kaufmann
Answer: b
Clarification: The mass spectrometer was invented by JJ THOMSON. He performed a series of experiments in 1897 designed to study the nature of electric discharge in a high-vacuum cathode-ray tube. Further modified by F. W. Aston shortly after World War I. He constructs the first velocity focusing mass spectrograph which has mass resolving power of 130.

4. In which state of matter mass spectroscopy is being performed?
a) solid
b) liquid
c) gaseous
d) plasma
Answer: c
Clarification: In the mass spectrometric analysis of compounds is firstly done by production of gas phase ions of the compound, basically by electron ionization. This molecular ion undergoes fragmentation.

5. A PMR spectrometer operates at 300 MHz. Find the value of magnetic field.
Given: g_N = 5.585 and B_N = 5.05 x 10^-27 JT^-I.
a) 7.05 T
b) 6.38 T
c) 7.58 T
d) 5.93 T
Answer: a
Clarification: According this formula of frequency:

6. What are the main criteria on which mass spectrometer used for?
a) Composition in sample
b) Relative mass of atoms
c) Concentration of elements in the sample
d) Properties of sample
Answer: b
Clarification: A mass spectrometer generates multiple ions from the sample under investigation, it then separates them according to their specific mass-to-charge ratio (m/z) or we can say on relative mass of atoms, and then records the relative abundance of each ion type.

7. A compound of molecular formula C₈H₇ClO shows a prominent band in its IR spectrum at 1690 cm^-1. ¹H NMR spectrum revealed only two major types of protons in the ratio of 5: 2. Which one of the following structures best fits the above data?
a)
b)
c)
d)
Answer: a
Clarification:
C₈H₇ClO
IR data: 1690 cm^–1
NMR data: two types of protons in 5: 2 ratio. Only two types of protons.
Ketone with conjugation shows peaks at 1690 cm^–1.

8. Which species of the following is used to bombard with the sample for which mass spectroscopy has been performed?
a) Alpha particles
b) Neutrons
c) Electrons
d) Protons
Answer: c
Clarification: In the mass spectrometer, the sample which is to be analysed is bombarded with electrons, which leads to the formation of the ions. The ions are sorted out by accelerating them through electric and magnetic field. A record of the number of different kinds of ions is called mass spectrum.

9. An organic compound Q exhibited the following spectral data obtained by mass spectroscopy.
IR: 1760 cm^–1
¹HNMR: chemical reference (ppm): 7.2 (IH, d, 16.0 Hz), 5.1 (IH, m), 2.1 (3H, s), 1.8 (3H, d, J = 7.0 Hz)
¹³CNMR chemical reference (ppm): 170 (carbonyl carbon). What is compound Q?
a)
b)
c)
d)
Answer: a
Clarification:
IR: 1760 cm^–1
Some options are not possible because of IR data.
1760 cm^–1 –> Represent either ester or anhydride
ester –>with cross conjugation or anhydride
chemical reference: 7.2 (1H), d, J = 16.0 Hz
This ‘J’ shows it is a trans alkene.

10. The spectrum of a compound with molecular formula C₅H₆0₂ is shown below. IR spectrum shows medium intensity band at 3270 and 2180 cm^-1. What will be the structure of compound? Chemical reference: 1.3 (³H, t); 2.8 (¹H, s), 4.3 (²H, q).
a)
b)
c)
d)
Answer: b
Clarification:
We have –>C₅H₆O₂
IR data: 3270 cm^–1, 2180 cm^–1 shows it can be an alkyne
NMR data: 1.3 (3H, t); 2.8 (1H, s); 4.3 (2H, q)
Now out of (a) & (b)
(a)

So, no such the electronegative difference between both cases C-C Bond.

(b)

So, electronegative difference between both cases C-C Bond is high. Hence 2H peak will at 4.3.

11. Separation of ions in mass spectrometer take place on the basis of which of the following?
a) Mass
b) Charge
c) Molecular weight
d) Mass to charge ratio
Answer: d
Clarification: Mass spectrometer separates ions on the basis of mass to charge ratio i.e. m/z. In the spectrum of a pure compound, the molecular ion, if present, appears at the highest value of m/z (followed by ions containing heavier isotopes) and gives the molecular mass of the compound. Most of the ions are singly charged. Hence, the mass to charge ratio is equal to the mass.

12. Which type of ionic species are allowed to pass through the slit and reach the collecting plate?
a) Negative ions of all masses
b) positive ions of the specific mass
c) Negative ions of the specific mass
d) Positive ions of all masses
Answer: b
Clarification: Positive ions of specific mass pass through the slit and reach the collecting plate. The ion currents are measured using sensitive electrometer tube. The ions reaching the collecting plate are measured.

250+ TOP MCQs on Alkyl Halides and Answers

Organic Chemistry Multiple Choice Questions on “Alkyl Halides”.

1. Which C-X bond has the highest bond energy per mole?
a) C-Br
b) C-Cl
c) C-F
d) C-I
Answer: c
Clarification: Bond energies depend on many factors: electron affinities, sizes of atoms involved in the bond, differences in their electronegativity, and the overall structure of the molecule. There is a general trend in that the shorter the bond length, the higher the bond energy.

2. Which alkyl halide has the highest reactivity for a particular alkyl group?
a) R-F
b) R-Cl
c) R-I
d) R-Br
Answer: c
Clarification: Reactivity order for the alkyl halides towards Sn2 reaction is R-I>R-Br>R-Cl>R-F. This can be explained by which halogen atom is a better leaving group compared to the other.

3. When ethyl chloride reacts with nascent hydrogen, what is the formed product?
a) Methane
b) Propane
c) Butane
d) Ethane
Answer: d
Clarification: When an alkyl halide is treated with zinc and hydrochloric acid, it is reduced to the respective alkane. The nascent hydrogen formed by the reaction between zinc and HCl acid reduces ethyl chloride to ethane.

4. Which alkyl halide out of the following may follow both S_N1 and S_N2 mechanism?
a) CH₃-X
b) (CH₃)₂CH-X
c) (CH₃)₃C-X
d) (CH₃)₃C-CH₂-X
Answer: b
Clarification: (CH₃)₂CH-X follows both S_N1 and S_N2 mechanism because the second CH₃ of (CH₃)₂CH-X further blocks a nucleophile (such as: OH) in backside S_N2 attack, but it increases the stability of the carbocation resulting from S_N1 ionization. As a result, S_N1 and S_N2 mechanisms are sometimes competitive for (CH₃)₂CH-X.

5. When two moles of ethyl chloride react with two moles of sodium in the presence of ether what will be formed?
a) 2 moles of ethane
b) 1 moles of ethane
c) 2 moles of butane
d) 1 moles of butane
Answer: d
Clarification: Wurtz reaction is method of preparation of higher alkanes from lower alkyl halides. This is coupling reaction. In this reaction alkyl halides are reacted with sodium metal in presence of dry ether and higher alkanes with even number of carbon atoms only are formed, by this method.

6. Which of the following halide can give best S_N2 reaction?
a) Primary alkyl halide
b) Tertiary alkyl halide
c) Secondary alkyl halide
d) All can give S_N2 reaction at same rate
Answer: a
Clarification: A 1° alkyl halide has only one alkyl group, so it is relatively unstable. It is unlikely to form a 1° carbocation in an S_N1 reaction. Instead, it will take the lower-energy S_N2 path as 1° alkyl is sterically unhindered. More the steric hinderance the substrate becomes less susceptible to S_N2 attack.

7. Why alkyl halides are considered to be very reactive compounds towards nucleophile?
a) they have an electrophilic carbon & a bad leaving group
b) they have a nucleophilic carbon & a good leaving group
c) they have an electrophilic carbon
d) they have an electrophilic carbon & a good leaving group
Answer: d
Clarification: Alkyl halides are considered to be very reactive compounds towards nucleophile because they have an electrophilic carbon & a good leaving group as we go down the periodic table, halides that are larger in size will also be able to distribute their charge over a larger volume, making them less reactive (less basic). This is why fluoride is a much poorer leaving group than any of the other halides. So, alkyl halides are good for nucleophilic substitution reactions.

8. In primary alkyl halides, carbon attached to the halogen atom is further attached to how many carbon atoms?
a) 1
b) 2
c) 3
d) 4
Answer: a
Clarification: As we can see below, carbon attached to the halogen atom is further attached to one carbon atom.
Example: CH₃-X → Methyl halide
CH₃-CH₂-X → Ethyl halide
CH₃-CH₂-CH₂-X → n-Propyl halide.

9. Which of the following is not the method of preparation of alkyl halide?
a) Darzen’s method
b) Halogenation of alkene
c) Addition of HX on alkenes
d) Hydration of alkene
Answer: d
Clarification: Hydration of alkene is electrophilic addition of H₂O to alkenes which forms alcohol not alkyl halides.

10. Which of the following reactant gives the best method of preparation of alkyl halides when reacts with alcohol?
a) Zn/HCl
b) PCl₅
c) SOCl₂/ Pyridine
d) PCl₃
Answer: c
Clarification: The best method of preparation of alkyl halides is a reaction of alcohol with SOCl₂/ Pyridine because by-products formed in the reaction are SO₂ and HCl which are in gaseous form and escape into the atmosphere leaving behind pure alkyl chlorides.
CH₃CH₂−OH + SOCl₂ → CH₃CH₂−Cl + SO₂↑ + HCl↑

250+ TOP MCQs on Prepartions of Glycols and Answers

Organic Chemistry Multiple Choice Questions on “Prepartions of Glycols”.

1. Which of the following reagent can be used to carry out this synthesis of glycol?

a) Benedict reagent
b) Baeyer’s reagent
c) Barfoed reagent
d) Seliwanoff reagent
Answer: b
Clarification: Cold dilute alkaline solution of Bayer’s reagent can be used to carry out this synthesis of glycol.

2. Which of the following is the industrial method of formation of glycol?
a) Hydroxylation of alkene by Bayer’s reagent
b) From 1,2-dibromoethane
c) Oxidation of ethylene and using Ag as catalyst
d) Ethylene treatment with HOCl
Answer: c
Clarification: In the industrial preparation of ethylene glycol, ethylene (IUPAC name: ethene) is oxidized to ethylene oxide (IUPAC name: oxirane) using oxygen and a silver catalyst. Ethylene oxide is then reacted with water at high temperature or in the presence of an acid catalyst to produce ethylene glycol. Diethylene glycol is a useful by-product of this process.

3. Which type of product is formed when Cold dilute alkaline solution of Bayer’s reagent reacts with alkene?
a) Syn-glyol
b) Syn- and anti-geometry will not be there in glycol
c) Anti-glyol
d) Trans glycol
Answer: a
Clarification: On hydroxylation product is formed when Cold dilute alkaline solution of Bayer’s reagent reacts with alkene is syn-glycol.

4. Which of the following statement is not true about preparation of alcohol?
a) French chemist Charles-Adolphe Wurtz (1817–1884) first prepared ethylene glycol in 1856
b) First synthesis of glycol was from “ethylene iodide” (C₂H₄I₂) with silver acetate and then hydrolyzed the resultant “ethylene diacetate” with potassium hydroxide
c) It was synthesized from ethylene dichloride in Germany and used as a substitute for glycerol in the explosives industry
d) There is no biological process for the synthesis of glycol
Answer: d
Clarification: Glycol can be synthesised biologically, example: The caterpillar of the Greater wax moth, Galleria mellonella, has gut bacteria with the ability to degrade polyethylene (PE) into ethylene glycol.

5. What will be the compound A which can be used to carry out this synthesis of glycol?

a) RCOR’
b) RCO₂OH
c) RCHO
d) RCOOH
Answer: b
Clarification: A peroxy acid can be used to carry out this synthesis of glycol as shown in below reaction.

6. What will be the catalyst A which can be used to carry out this synthesis of glycol?

a) Catalyst Cu, 200 – 400℃
b) Catalyst Pt, 100 – 200℃
c) Catalyst Ni, 200 – 400℃
d) Catalyst Ag, 200 – 400℃
Answer: d
Clarification: Catalyst Ag at 200 – 400℃ can be used to carry out this synthesis of glycol as shown in below reaction.

7. What will be the product A and B for the given reaction?

a) ethane, glycol
b) glycol, ethylene chlorohydrine
c) Ethylene chlorohydrine, glycol
d) ethan-1-ol, Glycol
Answer: c
Clarification: The product A and B for the given reaction are Ethylene chlorohydrine and glycol respectively.

8. Which compound is used with 1,2- dibromoethane for the formation of glycol?
a) Na₂CO₃
b) NaHCO₃
c) NaOH
d) CH₃COONa
Answer: a
Clarification: Na₂CO₃ compound is used with 1,2- dibromoethane for the formation of glycol, as shown in given reaction.

9. Which of the following is the most convenient and inexpensive method of formation of glycol?
a) Hydroxylation of alkene by Bayer’s reagent
b) From 1,2-dibromoethane
c) Oxidation of ethylene and using Ag as catalyst
d) Ethylene treatment with HOCl
Answer: a
Clarification: The most convenient and inexpensive method of preparing a glycol in the laboratory is to react an alkene with cold dilute potassium permanganate, KMnO₄. Yields from this reaction are often poor and better yields are obtained using osmium tetroxide, OsO₄. However, this reagent has the disadvantages of being expensive and toxic.

10. Which of the following is not a method of preparation of glycol?
a) Shell’s omega method
b) From carbon monoxide
c) From Dimethyl oxalate
d) From nitrogen
Answer: d
Clarification: In the OMEGA process, the ethylene oxide is first converted with carbon dioxide (CO₂) to ethylene carbonate. Ethylene glycol is produced from carbon monoxide in countries with large coal reserves and less stringent environmental regulations. Dimethyl oxalate can be converted into ethylene glycol in high yields (94.7%) by hydrogenation with a copper catalyst.

250+ TOP MCQs on Preparation of Carboxylic Acids and Answers

Organic Chemistry Multiple Choice Questions on “Preparation of Carboxylic Acids”.

1. Hydrolysis of CH₃CH₂NO₂ with 85% H₂SO₄ gives which of the following compound?
a) CH₃CH₂OH
b) C₂H₆
c) CH₃CH=NOH
d) CH₃COOH
Answer: d
Clarification: Hydrolysis of CH₃CH₂NO₂ with 85% H₂SO₄ gives carboxylic acid as shown in below chemical equation:

2. Acetic acid is obtained when which of the given reaction takes place?
a) Methyl alcohol is oxidised with potassium permanganate
b) Calcium acetate is distilled in the presence of calcium formate
c) Acetaldehyde is oxidised with potassium dichromate and sulphuric acid
d) Glycerol is heated with sulphuric acid
Answer: c
Clarification: Acetic acid is obtained when acetaldehyde is oxidised with potassium dichromate and sulphuric acid.

3. Acetic acid is manufactured by the fermentation of which of the following reaction?
a) Ethanol
b) Methanol
c) Ethanal
d) Methanal
Answer: a
Clarification: Ethanol fermentation leads to the formation of acetic acid.

4. When benzyl alcohol is oxidised with KMnO₄, the product obtained is which of the following compound?
a) Benzaldehyde
b) Benzoic acid
c) CO₂ and H₂O
d) Benzophenone
Answer: a
Clarification: When benzyl alcohol is oxidised with KMnO₄, the product obtained is benzoic acid.

5. Which of the following gives benzoic acid on oxidation?
a) Chlorophenol
b) Chlorotoluene
c) Chlorobenzene
d) Benzyl chloride
Answer: d
Clarification: Benzyl chloride gives benzoic acid on oxidation.
C₆H₅CH₂CI + 2 KOH + 2 [O] → C₆H₅COOK + KCl + H₂O.

6. In the below sequence of reactions, A and B are which of the following?

a) (CH₃)₂C(OH)CN, (CH₃)₂C(OH)COOH
b) (CH₃)₂C(OH)CN, (CH₃)₂C(OH)₂
c) (CH₃)₂C(OH)CN, (CH₃)₂CHCOOH
d) (CH₃)₂C(OH)CN, (CH₃)₂C=O
Answer: a
Clarification: In the below sequence of reactions, A and B are a shown below (a):

7. Formic acid is obtained when which of the given reaction occurs?
a) Calcium acetate is heated with conc. H₂SO₄
b) Calcium formate is heated with calcium acetate
c) Glycerol is heated with oxalic acid at 110℃
d) Acetaldehyde is oxidised with K₂Cr₂O₇ and H₂SO₄
Answer: c
Clarification: Glycerol is heated with oxalic acid at 110℃ gives formic acid.

8. o-xylene when oxidised in presence of V₂O₅ the product is which carboxylic acid?
a) Benzoic acid
b) Phenyl acetic acid
c) Phthalic acid
d) Acetic acid
Answer: c
Clarification: o-xylene when oxidised in presence of V₂O₅ the product is phthalic acid.

9. The below reaction is called as which of the following name reaction?

a) Wurtz reactions
b) Koch reaction
c) Clemenson’s reduction
d) Kolbe’s reaction
Answer: b
Clarification: The Koch reaction is an organic reaction for the synthesis of tertiary carboxylic acids from alcohols or alkenes.

10. By aerial oxidation, which one of the following gives phthalic acid?
a) Naphthalene
b) Benzene
c) Mesitylene
d) Toluene
Answer: a
Clarification: By aerial oxidation naphthalene, we can get phthalic acid.

250+ TOP MCQs on Preparation of Aromatic Amines and Answers

Organic Chemistry Multiple Choice Questions on “Preparation of Aromatic Amines”.

1. Aniline is usually purified by which of the following method?
a) Steam distillation
b) Simple distillation
c) Vacuum distillation
d) Extraction with a solvent
Answer: a
Clarification: Aniline is usually purified by Steam distillation. The separation is long, tedious and potentially dangerous – involving steam distillation, solvent extraction and a final distillation.

2. Which of the following method cannot be used for preparation of aromatic amine?
a) Gabriel phthalimide synthesis
b) Reduction of nitro compound
c) Reduction of nitrile with LiAl4
d) Decarboxylation of amino acids
Answer: a
Clarification: Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.

3. What are A and B in the given sequence, respectively?

a) Aldehyde, nitro compound
b) Nitro compound, phenyl amine
c) Phenyl amine, nitro compound
d) Phenthalene, phenyl amine
Answer: b
Clarification: Nitrobenzene is reduced to phenyl ammonium(A) ions using a mixture of tin and concentrated hydrochloric acid. The phenylamine(B) is formed together with a complicated mixture of tin compounds from reactions between the sodium hydroxide solution and the complex tin ions formed.

4. Which reducing agent is used for the reduction of nitro compound to phenyl amine?
a) LiAlH₄
b) Sn/HCl
c) Na/alcohol
d) H₂/Ni
Answer: b
Clarification: Aromatic amines were prepared in good yields by a novel reduction of aromatic nitro compounds with tin metal in hydrochloric acid.

5. What is the known name of the reaction given below?

(where, X=Cl, Br, I, OTf; R₂=Alkyl, aryl, H; R₃=alkyl, aryl)
a) Gabriel phthalimide synthesis
b) Buchwald-Hartwig Reaction
c) Chan-Lam coupling
d) Ullmann reaction
Answer: b
Clarification: The Buchwald–Hartwig amination is a chemical reaction used in organic chemistry for the synthesis of carbon–nitrogen bonds via the palladium-catalyzed cross-coupling of amines with aryl halides.

6. What is the known name of the reaction given below?

a) Gabriel phthalimide synthesis
b) Buchwald-Hartwig Reaction
c) Chan-Lam coupling
d) Ullmann reaction
Answer: d
Clarification: The Ullmann reaction or Ullmann coupling is a coupling reaction between aryl halides and copper and substituted ammonia to form aromatic amines. The traditional version of the Ullmann reaction requires harsh reaction conditions, and the reaction has a reputation for erratic yields.

7. What is the known name of the reaction given below?

a) Gabriel phthalimide synthesis
b) Buchwald-Hartwig Reaction
c) Chan-Lam coupling
d) Ullmann reaction
Answer: c
Clarification: This reaction allows aryl carbon-heteroatom bond formation via an oxidative coupling of boronic acids, stannanes or siloxanes with N-H containing compounds in air. The reaction is induced by a stoichiometric amount of copper(II) or a catalytic amount of copper catalyst which is reoxidized by atmospheric oxygen or another primary oxidant.

8. Which type medium is required for the formation of aniline by reaction of aryl boric acid and HAS?
a) Acidic
b) Basic aqueous
c) Neutral dry
d) aqueous
Answer: b
Clarification: Various anilines are prepared by treatment of functionalized aryl boronic acids with H₂N-OSO₃H (HSA) as a common, inexpensive source of electrophilic nitrogen, under basic aqueous conditions.

9. What will be the product for the following reaction?

a) secondary aliphatic amine
b) primary aromatic amine
c) tertiary aromatic amine
d) secondary aromatic amine
Answer: c
Clarification: this reaction is N- arylation of acyclic secondary amine, in which CuI and DMPAO are catalyst.

10. In the direct amination of alkyl and aryl pinacol boronates to form aromatic amine, which of the following is used as reagent with pinacol?
a) Aryl boric acid
b) Copper oxide
c) Pd(OAc)₂
d) Lithiated methoxyamine
Answer: d
Clarification: The direct amination of alkyl and aryl pinacol boronates with lithiated methoxyamine provides aliphatic and aromatic amines, stereospecifically, and without preactivation of the boronate substrate.

250+ TOP MCQs on Electrophilic Substitution in Pyridine and Answers

Organic Chemistry Assessment Questions and Answers on “Electrophilic Substitution in Pyridine”.

1. Electrophilic substitution reaction of pyridine is most effective in which of the conditions?
a) In slightly acidic conditions
b) In slightly basic conditions
c) In neutral medium
d) In vigorous conditions
Answer: d
Clarification: When electrophilic substitution reaction of pyridine is carried out in acidic conditions then pyridine ring will further get deactivated forming pyridinium ion. However, under more vigorous conditions the usual electrophilic substitution reaction can be possible but yield will be less.

2. Which of the following electrophilic substitution reaction is not possible in pyridine?
a) Nitration
b) Sulphonation
c) Bromination
d) Friedel craft reaction
Answer: d
Clarification: Common alkylations and acylations, such as Friedel–Crafts alkylation or acylation, usually fail for pyridine because they lead only to the addition at the nitrogen atom. Substitutions usually occur at the 3-position, which is the most electron-rich carbon atom in the ring and is, therefore, more susceptible to an electrophilic addition.

3. At which position of pyridine electrophilic substitution reaction is most preferred?
a) First
b) Second
c) Third
d) Forth
Answer: c
Clarification: Electrophilic substitution reaction at position the 3-position is preferred over attack at 2 and 4-position because the intermediate found by electrophilic addition at 3-position is more stable as it will have 3-resonating structure and none of will have positive charge on N.

4. Which of the following is the product for the below reaction?

a) 3- Nitropyridine
b) 2- Nitropyridine
c) 4- Nitropyridine
d) 2,4- dinitropyridine
Answer: a
Clarification: Electrophilic substitution reaction at position the 3-position is preferred over attack at 2 and 4-position because the intermediate found by electrophilic addition at 3-position is more stable as it will have 3-resonating structure and none of will have positive charge on N.

5. Which of the following is the product for the below reaction?

a) 3- bromopyridine
b) 2- bromopyridine
c) 2- bromopyridine and 3- bromopyridine
d) 3- bromopyridine and 3,5- dibromo pyridine
Answer: d
Clarification: Electrophilic substitution reaction at position the 3-position 5 position is preferred over attack at 2 and 4-position because the intermediate found by electrophilic addition at 3-position is more stable as it will have 3-resonating structure and none of will have positive charge on N.

6. Which of the following is the product for the below reaction?

a) 3- pyridinesulphonic acid
b) 2- pyridinesulphonic acid
c) 4- pyridinesulphonic acid
d) 2,4- pyridinedisulphonic acid
Answer: a
Clarification: Electrophilic substitution reaction at position the 3-position is preferred over attack at 2 and 4-position because the intermediate found by electrophilic addition at 3-position is more stable as it will have 3-resonating structure and none of will have positive charge on N.

7. Which of the following is the product for the below reaction?

a) 3- methyl-pyridine-5-sulphonic acid
b) 3- methyl-pyridine-4-sulphonic acid
c) 3- methyl-pyridine-6-sulphonic acid
d) 3- methyl-pyridine-1-sulphonic acid
Answer: a
Clarification: Alkyl group activate the pyridine ring towards electrophilic substitution, however, directive influence of the ring predominates regardless of the position of alkylation i.e. attack at 3-position.

8. Which of the following is the product for the below reaction?

a) 5- bromo-2-aminopyridine
b) 3- bromo-2-aminopyridine
c) 4- bromo-2-aminopyridine
d) 6- bromo-2-aminopyridine
Answer: a
Clarification: Amino group dominates the ring orientation effect and direct the incoming electrophile to o- and p- position. 2-amino group directs the incoming electrophile predominantly to the 5-position.

9. Which of the following is the product for the below reaction?

a) 2- Nitro-3-aminopyridine
b) 4- Nitro-3-aminopyridine
c) 5- Nitro-3-aminopyridine
d) 6- Nitro-3-aminopyridine
Answer: a
Clarification: Amino group dominates the ring orientation effect and direct the incoming electrophile to o- and p- position. 3-amino group directs the incoming electrophile predominantly to the 2-position.

10. Which of the following is the product for the below reaction?

a) 4-amino -pyridine-5-sulphonic acid
b) 4- amino -pyridine-3-sulphonic acid
c) 4- amino -pyridine-6-sulphonic acid
d) 4- amino -pyridine-2-sulphonic acid
Answer: a
Clarification: Amino group dominates the ring orientation effect and direct the incoming electrophile to o^– and p^– position. 4-amino group directs the incoming electrophile predominantly to the 3-position.

Organic Chemistry Assessment Questions,