300+ REAL TIME Organic Chemistry Objective Questions & Answers 2023

250+ TOP MCQs on Intramolecular Claisen Condensation and Answers

Organic Chemistry Multiple Choice Questions on “Intramolecular Claisen Condensation”.

1. Which type of precursor is used as reactant in intramolecular Claisen condensation?
a) One molecule with an ester end
b) Two molecules of ester
c) One molecule with two ester ends
d) One molecule of ester and enolate
Answer: c
Clarification: In intramolecular Claisen condensations, reactions occur for 1,6 and 1,7 diesters, as these substances result in the formation of compounds containing five and six membered rings, respectively.

2. Which is the main product of the following reaction?

a)
b)
c)
d)
Answer: a
Clarification: This is an example of an intramolecular Claisen condensation. These reactions occur for 1,6 and 1,7 diesters, as these substances result in the formation of compounds containing five and six membered rings, respectively.

3. What is the other name for the intramolecular Claisen condensation?
a) Perkin condensation
b) Stobbe condensation
c) Knoevenagel condensation
d) Dieckmann condensation
Answer: d
Clarification: Diester compounds can be used to give an intramolecular Claisen condensation which is known as the Dieckmann condensation.

4. Dieckmann Condensation is intramolecular condensation of ______________ to form cyclic product.
a) diamide
b) diol
c) diester
d) diketone
Answer: c
Clarification: These reactions occur for 1,6 and 1,7 diesters, as these substances result in the formation of compounds containing five and six membered rings.

5. What will be the product of the following intramolecular Claisen condensation?

a)
b)
c)
d)
Answer: c
Clarification: Favours the formation of the more stable 5- or 6-membered rings, as the ring stain in 5 and 6 membered rings are less.

6. Which is the main product of the following intramolecular Claisen condensation?

a)
b)
c)
d)
Answer: d
Clarification: This reaction is an intramolecular Claisen condensation, which is followed by attack of carbocation on carbon attached to ester group.

7. Which is the main product of the following intramolecular Claisen condensation?

a)
b)
c)
d)
Answer: b
Clarification: This reaction is an intramolecular Claisen condensation, in which the product is a β-keto ester which gets deprotonated by the strong base in the reaction mixture and get converted into simple keto group.

8. What will be the product(s) of the following intramolecular Claisen condensation?

a) I
b) II
c) I and II
d) II and III
Answer: d
Clarification: This reaction is an intramolecular Claisen condensation, favours the formation of the more stable 5- or 6-membered rings, as the ring stain in 5 and 6 membered rings are less. In both II and III five membered ring is formed.

9. Which is the main product of the following intramolecular Claisen condensation?

a)
b)
c)
d)
Answer: b
Clarification: This reaction is an intramolecular Claisen condensation, favours the formation of the more stable 5- or 6-membered rings, as the ring stain in 5 and 6 membered rings are less and 6-membered rings are more stable than 5 membered.

10. Which is the main product of the following intramolecular Claisen condensation?

a)
b)
c)
d)
Answer: a
Clarification: This reaction is an intramolecular Claisen condensation, favours the formation of the more stable 5- or 6-membered rings, as the ring stain in 5 and 6 membered rings are less and 6-membered rings are more stable than 5 membered. But in case bridge ring is present which is unstable as compared to 5 and 6 membered ring.

250+ TOP MCQs on Enantiomers and Answers

Organic Chemistry Multiple Choice Questions on “Enantiomers”.

1. Which of the following statements regarding enantiomers not true?
a) All (+) enantiomers are levorotatory
b) All (-) enantiomers rotate plane polarized light in a counterclockwise direction
c) (+) and (-) enantiomers rotate plane polarized light in opposite directions
d) All R enantiomers are dextrorotatory
Answer: d
Clarification: This statement is true. (+) and (-) enantiomers rotate plane polarized light in opposite directions.

2. Which of the following is/are the S-enantiomer of alanine?

a) Only 1
b) Only 2
c) 1 and 3
d) 2 and 3
Answer: d
Clarification: 1 and 3 have an S configuration, but 2 has an R configuration. The priority of the substituents on the chirality center of alanine is: -NH₂ > -COOH > -CH₃ > -H.

3. Would the following compound have an enantiomer.

a) True
b) False
Answer: a
Clarification: Yes, this compound has enantiomer because of the steric bulk of the phenyl groups, the molecule would be held in place.

4. Which of the following is true of any (S)-enantiomer?
a) It rotates plane-polarized light to the right
b) It rotates plane-polarized light to the left
c) It is a racemic form
d) It is the mirror image of the corresponding (R)-enantiomer
Answer: d
Clarification: Priorities are assigned to each of the four different groups attached to a given chiral center (one through four, one being the group of highest priority). (It should be understood that each chiral center has to be treated separately.) Orient the molecule so that the group of priority four (lowest priority) points away from the observer. Draw a circular arrow from the group of first priority to the group of second priority. If this circular motion is counterclockwise, it is the S enantiomer.

5. Which of the following is the pair of enantiomer from the following substance?

a) I and II
b) II and III
c) III and IV
d) I and V
Answer: c
Clarification: Stereocenters, (sometimes called chiral centers) are carbons that have four non-identical substituents on them and are designated as either of R stereochemistry or S stereochemistry. If a molecule has one stereocenter of R configuration, then in the mirror image of that molecule, the stereocenter would be of S configuration, and vice-versa.

6. Is the given pair of compound enantiomers?

a) True
b) False
Answer: a
Clarification: Stereocenters, (sometimes called chiral centers) are carbons that have four non-identical substituents on them and are designated as either of R stereochemistry or S stereochemistry. If a molecule has one stereocenter of R configuration, then in the mirror image of that molecule, the stereocenter would be of S configuration, and vice-versa.

7. Which of the following is the enantiomer for the following substance?

a)
b)
c)
d)
Answer: a
Clarification: Stereocenters, (sometimes called chiral centers) are carbons that have four non-identical substituents on them and are designated as either of R stereochemistry or S stereochemistry. If a molecule has one stereocenter of R configuration, then in the mirror image of that molecule, the stereocenter would be of S configuration, and vice-versa.

8. Which of the following molecules exists as a pair of enantiomers?
a) 2-Bromopropane
b) 1-Bromo-3-methylbutane
c) 2-Cyclohexen-1-ol
d) cis-1,2-Dichlorocyclobutane
Answer: c
Clarification: 2-Cyclohexen-1-ol can exist as enantiomers as shown below tis compound is called as (1R)-2-Cyclohexen-1-ol.

9. Which of the following diols exists as a pair of enantiomers?
a) cis-1,3-Cyclohexanediol
b) trans-1,3-Cyclohexanediol
c) cis-1,4-Cyclohexanediol
d) trans-1,4-Cyclohexanediol
Answer: b
Clarification: Trans-1,3-Cyclohexanediol exist as enantiomer. (S,S)-cyclohexane-1,3-diol is shown below.

10. Which of the following is capable of existing as a pair of enantiomers?
a) 2-methylpropane
b) 2-methylpentane
c) 3-methylpentane
d) 3-methylhexane
Answer: d
Clarification: 3-methylhexane exists as a pair of enantiomer. This carbon atom has four non-identical substituents around it, making this carbon a chiral center, and as proof of its chirality the molecule has a non-superimposable mirror image. A fancy term used in textbooks and in the literature to describe molecules that are mirror images of each other is enantiomers, as in “the enantiomer of the left molecule above is the molecule on the right, its mirror image.”

250+ TOP MCQs on Chlorination and Answers

Organic Chemistry Multiple Choice Questions on “Chlorination”.

1. When chlorine gas reacts with methane, Which of the product is formed?
a) CHCl
b) CH₂Cl
c) CH₃Cl
d) CH₄Cl
Answer: c
Clarification: Under the influence of UV light or with a temperature of 250-400(degree) chlorine and methane react vigorously to give hydrogen chloride and chloromethane (CH₃Cl).

2. The methyl chloride undergoes substitution to form __________
a) CHCl
b) CH₂Cl
c) CHCl₂
d) CH₂Cl₂
Answer: d
Clarification: Chloromethane undergoes further substitution to form hydrogen chloride and dichloromethane (CH₂Cl₂).

3. Methane reacts more readily with chlorine than with Fluorine.
a) True
b) False
Answer: b
Clarification: Methane reacts with chlorine but not as vigorously as with fluorine. The reactivity order is as follows: F₂ > Cl₂ > Br₂ > I₂.

4. Identify the one which is not a type of chlorination?
a) Free radical chlorination
b) Electrophobic chlorination
c) Ketone chlorination
d) Chlorine addition reaction
Answer: b
Clarification: Electrophobic is not a kind of halogenation. It is electrophilic chlorination, in which aromatic substitutions takes place.

5. Which among the following on chlorination undergoes substitution at the alkyl group?
a) Ethers
b) Hydroxyl groups
c) Carbonyl compounds
d) Carboxylic acids
Answer: a
Clarification: In the absence of sunlight, the alpha hydrogen of ethers undergoes substitution when treated with chlorine.

6. In the presence of red phosphorous, chlorine converts the fatty acids having alpha hydrogen atoms into __________
a) Halo acids
b) Alpha-halo acids
c) Alpha,alpha-dihalo acids
d) Trihalo acids
Answer: b
Clarification: The propionic acid, in the presence of Cl₂/P gets converted into alpha chloro propionic acid due to the action of the halogen chlorine.

7. Chlorine in the presence of which among the following generates positively charged species?
a) Ferric chloride
b) Anhydrous ferric chloride
c) Ferrous chloride
d) Anhydrous ferrous chloride
Answer: b
Clarification: Ferric chloride is a lewis acid catalyst and in the presence of chlorine, it converts nitrobenzene into 3- chloro nitro benzene.

8. p-nitrotoluene on reaction with chlorine forms hydrogen chloride and __________
a) p-Nitrobenzyl chloride
b) o-Nitrobenzyl chloride
c) m-Nitrobenzyl chloride
d) Nitrobenzyl chloride
Answer: a
Clarification: p-Nitrotoluene reacts with chlorine in the presence of CCl₄ at 80 (degree) C to form p-nitrobenzyl chloride under the conditions of photochemical initiation.

9. Chlorination of cyclobutane gives which among the following in addition to hydrogen chloride?
a) Cyclobutyl chloride
b) Cyclobutyl chlorite
c) 1-chlorobutene
d) 1,1-chlorobutene
Answer: b
Clarification: All the hydrogens of cyclobutane are equivalent and substitution of any gives the same product as that of others.

10. The step in which Cl-Cl bond homolysis occurs is called __________
a) Initiation step
b) Propagation step
c) Intermediate step
d) Termination step
Answer: a
Clarification: Each Cl atom in the initiation step has several valence electrons and is very reactive and hence it abstracts a hydrogen atom from methane and homolysis occurs.

250+ TOP MCQs on Nucleophilic Substitution Reaction and Answers

Organic Chemistry Multiple Choice Questions on “Nucleophilic Substitution Reaction”.

1. Which of the following compound shows the correct decreasing order of solvolysis with aqueous ethanol?

The correct choice is:
a) III > II > I > IV
b) III > II > IV > I
c) II > III > IV > I
d) III > I > IV > II
Answer: b
Clarification: The solvation of above compound can be compared by this trend, i.e. Solvation trend: 3^O halide > 2^O halide > 1^O halide > vinylic halide. So, (III) molecule is an example of 3^O halide, (II) molecule is an example of 2^O halide, (IV) molecule is an example of 1^O halide and (I) molecule is an example of vinylic halide.

2. Which of the following reaction will go faster if the concentration of the nucleophile is raised?

a) I and III
b) II and IV
c) I and II
d) I, II and IV
Answer: c
Clarification: I & II reaction undergo via S_N2 mechanism. III & IV reaction undergo via S_N1 mechanism. As we know the rate of reaction in S_N2 reaction is directly proportional to the concentration of nucleophile.

3. S_N1 reaction undergoes through a carbocation intermediate as follows:

[R = t-Bu, iso-Pr, Et, Me] (X = Cl, Br, I)
The correct statements are:
I. The decreasing order of rate of S_N1reaction is t-BuX > iso-PrX > EtX > MeX
II. The decreasing order of ionization energy is MeX > EtX > iso-PrX > t-BuX
III. The decreasing order of energy of activation is t-BuX > iso-PrX > EtX > MeX
a) I & II are correct
b) I & III are correct
c) II and III are correct
d) I, II & III are correct
Answer: a
Clarification:
(1) Formation of carbocation is RDS, more stable the cation more easily it will form. Hence rate order is t-BuX > iso-PrX > EtX > MeX
(2) Formation of carbocation CH₃ is most difficult hence ionization energy is highest. Similarly, EtX has lesser ionization energy than MeX and so on.
(3) Order of energy of activation is MeX > EtX > iso-PrX > t-BuX.

4. Which statement is incorrect about the following reaction?

a) The rate of these reaction depends on both [R—I] and [131_I-]
b) Loss of optical activity was twice as fast as the gain of radioactivity
c) Each molecule undergoing substitution suffers Inversion of configuration
d) The Final solution has radioactive iodine only
Answer: d
Clarification:

Here the rate of reaction is directly proportional to the concentration of RI and (131_I-), so statement is correct.
If any molecule undergoes substitution, then it forms inversion product due to SN₂ reaction. Hence statement is correct.
For statement, below reaction one (+) molecule converts into (–) molecular and it also converts one optical reaction of (+) molecule.

Final solution contains both Iodine (I¹³¹) and normal (I). So, we can say this is the only incorrect statement about the above reaction.

5. Which one is an excellent substrate for SN₂ reaction?
a)
b)
c) CH₃ – CH₂ – Cl
d)
Answer: a
Clarification: Alpha-halo carbonyl undergoes fastest S_N2 reaction due to the coupling of π* of CO bond and π* of C–X bond.

6. Which of the following curve correctly represents S_N1 vs S_N2?
a)
b)
c)
d)
Answer: b
Clarification: As we know rate order in S_N2 is CH₃X > 1^o > 2^o > 3^o halide and rate order in S_N1 is CH₃X ^o^o^o halide. So according to the following representation.

7. Which configuration will be adopted by X and Y respectively?

a) R, R
b) R, S
c) S, S
d) S, R
Answer: b
Clarification: When ether reacts with the reactant nucleophilic substitution conjugate base reaction (i.e. S_NCB) will take place and there will be retention of configuration and it will be R product. When C₅H₅NH reacts with reactant S_N2 reaction will take place and configuration inversion will take place and it will be S product.

8. What will be the correct order of S_N2/E2 ratio for the %yield of the product of the following halide?

a) III > IV > II > I
b) III > II > IV > I
c) I > III > IV > II
d) II > I > III > IV
Answer: a
Clarification: Least hindered halide give fastest S_N2 reaction as the hindrance increases. As the hindrance increases, the occurrence of S_N2 reaction decreases.

9. Which of the following is an incorrect statement?

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 undergoes E2 not via E1 because carbocation is not stable.

10. Among the following reactions, which one is incorrect?
a)
b)
c)
d)
Answer: b
Clarification:

α-halocarbonyl are the best S_N2 substrate.
Here nucleophilic substitution reaction will happen and the above product will be formed. Hence this reaction is correct.

Here (PrO) and (CH₃) will be at same carbon atom and (OH) will be at adjacent carbon atom. Hence this reaction is wrong.

n-PrO^– will attack the carbon atom attached to oxygen above the plane and the above product will be formed. Hence the reaction is correct.

Attack of Bromine ion (Br^–) will be at that carbon atom where less substitution is present, the above product will form. Hence this reaction is correct.

250+ TOP MCQs on Nucleophiles and Answers

Organic Chemistry Multiple Choice Questions on “Nucleophiles”.

1. Which of the following is not true about nucleophile?
a) donates an electron pair to an electrophile to form a chemical bond
b) all molecules or ions with a free pair of electrons or at least one pi bond can act as nucleophiles
c) nucleophile are Lewis acids by definition
d) a nucleophile becomes attracted to a full or partial positive charge
Answer: c
Clarification: Because nucleophiles donate electrons, they are by definition Lewis bases.

2. Which halogen nucleophile is weakest in polar, aprotic solvents?
a) I^–
b) F^–
c) Cl^–
d) Br^–
Answer: a
Clarification: In polar, aprotic solvents, I^– the weakest and this is reversed in polar, protic solvents.

3. Which of the following nucleophile can be used for partial enolization 1,3-Dicarbonyl compound?
a) Lithium di-isopropyl amide (LDA)
b) Et-O^–
c) Weak base like NaOH
d) Sodium acetate
Answer: d
Clarification: Sodium acetate will do partial enolization of 1,3-Dicarbonyl compound. Weak base like NaOH and Et-O^– will form an alcohol and LDA will do complete enolization.

4. What will be the number of products (excluding stereoisomers) in the given reaction?
C₆H₅CHO + CH₃-CHO + NaOH → Product
a) One
b) Three
c) Two
d) Four
Answer: c
Clarification: First product will be formed by cross aldol where nucleophile (^–OH) will attack at different type or reactant molecule.
Second type of product will be formed aldol condensation where reaction will occur between same type of molecule.

5. Where will nucleophile (-OH) will attack to form which of the following the product X?

a)
b)
c)
d)
Answer: d
Clarification: Nucleophile (-OH) will attack at carbon adjacent to carbonyl bond and form the below product.

6. Which reagent is a good nucleophile?
a) NH₃
b) BH₃
c) Br₂
d) HBr
Answer: a
Clarification: Less electronegative atoms make better nucleophiles as they are more willing to share electrons.

7. Which of the following statements is true about ammonia and water?
a) ammonia is more basic and more nucleophilic than water
b) ammonia is less basic and less nucleophilic than water
c) ammonia is more basic but less nucleophilic than water
d) ammonia is less basic but more nucleophilic than water
Answer: a
Clarification: Because oxygen is more electronegative than nitrogen, it holds onto its lone pairs more tightly than nitrogen, and hence is less likely to donate its lone pairs to form a covalent bond with a carbon atom during a nucleophilic attack.

8. Which of the following statements is true about the following two anionic molecules?

a) I is more basic and more nucleophilic than II
b) I is less basic and less nucleophilic than II
c) I is more basic but less nucleophilic than II
d) I is less basic but more nucleophilic than II
Answer: c
Clarification: Larger atoms make better nucleophiles due to polarizability so II containing S will be more nucleophilic than I but less basic.

9. Which halide ion is the best nucleophile in dimethyl sulfoxide solution?
a) I^–
b) F^–
c) Cl^–
d) Br^–
Answer: b
Clarification: In polar, aprotic solvents, F^– strongest is the nucleophile, and I^– the weakest.

10. Which of the following cannot react as a nucleophile?
a) CH₃NH₂
b) (CH₃)₂NH
c) (CH₃)₃N
d) (CH₃)₄N⁺
Answer: d
Clarification: Nucleophile donates electron to the electrophiles and there is no lone pair present on nitrogen (CH₃)₄N⁺, so donation of electron is not possible.

250+ TOP MCQs on Preparation of Ethers and Answers

Organic Chemistry Multiple Choice Questions on “Preparation of Ethers”.

1. Chloroethane reacts with X to form diethyl ether. What is X?
a) NaOH
b) H₂SO₄
c) C₂H₅ONa
d) Na₂S₂O₃
Answer: c
Clarification: The reaction of sodium ethoxide with chloroethane to form diethyl ether and sodium chloride:
[Na]⁺[C₂H₅O]^– + C₂H₅Cl → C₂H₅OC₂H₅ + [Na]⁺[Cl]^–

2. The reaction given below is known as by which o the following reaction?
C₂H₅ONa+IC₂H₅→C₂H₅OC₂H₅+NaI
a) Kolbe’s synthesis
b) Wurtz’s synthesis
c) Williamson’s synthesis
d) Grignard’s synthesis
Answer: c
Clarification: Williamson Ether Reactions involve an alkoxide that reacts with a primary haloalkane or a sulfonate ester. Alkoxides consist of the conjugate base of an alcohol and are comprised of an R group bonded to an oxygen atom. They are often written as RO^–, where R is the organic substituent.

3. The Williamson ether synthesis produces ethers by reacting which of the following?
a) alcohol with a metal
b) alkoxide with a metal
c) alkoxide with an alkyl halide
d) alkyl halide with an aldehyde
Answer: c
Clarification: It’s a type of reaction we’ve already seen many times before – an S_N2 reaction between a deprotonated alcohol [“alkoxide”] and an alkyl halide that forms an ether.

4. In which case would a Williamson ether synthesis fail?
a) sodium ethoxide + iodomethane
b) sodium ethoxide + iodoethane
c) sodium ethoxide + 2-iodopropane
d) sodium ethoxide + 2-iodo-2-methylpropane
Answer: d
Clarification: Because alkoxides are strong bases, competition with elimination [E2] pathways becomes a concern once the alkyl halide becomes more sterically hindered. For this reason, trying to perform a Williamson on a secondary alkyl halide is a bit more problematic than it is for a primary alkyl halide.

5. Which of the following reagents should be used to prepare tert-butyl ethyl ether?
a) tert-butyl bromide and sodium ethoxide
b) tert-butyl alcohol and ethyl bromide
c) tert-butyl alcohol and ethanol
d) potassium tert-butoxide and ethyl bromide
Answer: d
Clarification: When ethyl bromide(CH₃CH₂Br) is added to potassium tert-butoxide ((CH₃)_3-O^– + K), the product is ethyl tert-butyl ether(CH₃)₃-O-CH₂CH₃).

CH₃CH₂Br + (CH₃)₃-O^– + K → (CH₃)₃-O-CH₂CH₃.

6. Which of the following reagents should be used to prepare tert-butyl propyl ether?
a) 2-methylpropene and 1-propanol
b) 2-methylpropene, acetic acid and 1-propanol
c) 2-methylpropene, H₂SO₄ and 1-propanol
d) 1-propyl bromide and tert-butyl alcohol
Answer: c
Clarification: The dehydration reaction of alcohols to generate alkene proceeds by heating the alcohols in the presence of a strong acid, such as sulfuric or phosphoric acid, at high temperatures. If the reaction is not sufficiently heated, the alcohols do not dehydrate to form alkenes, but react with one another to form ethers.

7. What will be the reactant and reaction condition required for the given reaction?
CH₃CH₂OH +? → C₂H₅OC₂H₅
a) H₂SO₄, 413K
b) H₂SO₄, 443K
c) HNO₃, 413K
d) HNO₃, 443K
Answer: a
Clarification: In the presence of protic acids (sulphuric acid), alcohols undergo dehydration to produce alkenes and ethers under different conditions. For example: in the presence of sulphuric acid, dehydration of ethanol at 443 K yields ethene whereas it yields ethoxyethane at 413 K. This is an ideal method of preparation through primary alcohols.

8. Which of the following is the method of preparation of ether?
a) Dows process
b) Williamson synthesis
c) By decarboxylation of salicylic acid
d) Hock method
Answer: b
Clarification: In this method, an alkyl halide is reacted with sodium alkoxide which leads to the formation of ether. The reaction generally follows S_N2 mechanism for primary alcohol.

9. Williamson synthesis proceed visa which of the following mechanism?
a) S_N1
b) S_N2
c) E1
d) E2
Answer: b
Clarification: Ethers can be synthesized in standard S_N2 conditions by coupling an alkoxide with a haloalkane/sulfonate ester. The alcohol that supplies the electron rich alkoxide can be used as the solvent, as well as dimethyl sulfoxide (DMSO) or hexamethylphosphoric triamide (HMPA).

10. Which of the following is the commercial method of formation of ether?
a) Dehydration of alcohol
b) Williamson ether synthesis
c) Ullmann condensation
d) Preparation of epoxides
Answer: b
Clarification: The Williamson reaction is of broad scope, is widely used in both laboratory and industrial synthesis, and remains the simplest and most popular method of preparing ethers. Both symmetrical and asymmetrical ethers are easily prepared.