Organic Chemistry Class 11 & 12 - Most Important Questions for NEET | NEET 2023 Chemistry Prep

NEET 2023 is approaching, and it is time to do thorough revisions and give your best for the entrance exam. We often overlook the most important questions in Organic Chemistry while reading the theory. This comprehensive session has provided you with the most important questions of Complete Organic Chemistry for NEET 2023.

Question 1: The IUPAC name of the following compound is ________. (CH₃)₂CH–CH₂CH=CH–CH=CH–CHCH₃–C₂H₅

1. 1, 1, 7, 7-tetramethyl-2, 5-octadiene
2. 2, 8-dimethyl-3, 6-decadiene
3. 1, 5-diisopropyl-1, 4-hexadiene
4. 2, 8-dimethyl-4, 6-decadiene

Answer: (d) 2, 8-dimethyl-4, 6-decadiene

Explanation: Identification of parent chain: First, determine which carbon chain in the molecule is the longest. The longest chain in the given compound has ten carbons and is referred to as the parent or root chain.

Assign numbers to the parent chain: The functional group is connected to the carbon atom with the lowest possible number in the longest chain of carbon atoms carrying the functional group.

The C=C functional groups are found at carbon atoms 4 and 6, while the substituent (–CH3) is present at carbon atoms 2 and 8. The C=C functional groups are indicated by the suffixes ‘ene’.

Hence, the name of the compound is 2, 8-dimethyl-4, 6-decadiene.

Question 2: The most stable carbanion among the following is __________.

Answer: (d)

Explanation: Dispersion of –ve charge by –M and –I group stabilises the carbanions. NO₂ group on benzene stabilises the carbanion.

Question 3: Which of the following phenols has the largest pK_a value?

Answer: (c)

Explanation: Acidic strength ∝ K_a ∝ 1/pK_a

The presence of an electron-withdrawing group makes phenol more acidic because it stabilises the phenoxide ion generated when the proton is removed. Electron-donating groups, on the other hand, reduce the acidic intensity of phenol.

As a result, p–methyl phenol will have the least acidity and the highest pKa value.

Question 4: Which of the following has inductive, mesomeric, and hyperconjugation effects?

1. CH₃Cl
2. CH₃–CH=CH₂
3. CH₃–CH=CH–CO–CH₃
4. CH₂=CH–CH=CH₂

Answer: (c) CH₃–CH=CH–CO–CH₃

Explanation: Because of the carbonyl group, double bond resonance alpha hydrogens are also present, due to which hyperconjugation occurs.

Question 5: The correct stability order for the following species is _________.

1. II > IV > I > III
2. I > II > III > IV
3. II > I > IV > III
4. I > III > II > IV

Answer: (d) I > III > II > IV

Explanation:

The lone pair on oxygen stabilises the carbocation in the Ist option, making it more stable than the tertiary carbocation.

The stability of carbocations should be in the order: tertiary alkyl > secondary alkyl > primary alkyl > methyl carbonium ion.

The stability of a carbocation increases when an electron donating group (+I effect) is present.

Question 6: Which among the given molecules can exhibit tautomerism?

1. Both I and II
2. Both II and III
3. III only
4. Both I and III

Answer: (c) III only

Explanation:

α-Hydrogen at bridge carbon does not ever participate in tautomerism. Thus, only (III) exhibits tautomerism.

Question 7: What is the total number of structural isomers C₄H₁₀O?

1. 7
2. 5
3. 4
4. 6

Answer: (a) 7

Explanation: C₄H₁₀O will have seven structural isomers, four of which will be alcohols and three of which will be ethers.

• n-butanol
• butan-2-ol
• 2-methyl-propanol
• 2-methyl propane-2-ol
• diethyl ether
• 1-methoxy propane
• 2-methoxy propane

As a result, 7 is the right answer.

Question 8: For the given reaction, what is ‘A’?

Answer: (c)

Question 9: An unsaturated hydrocarbon ‘X’ on ozonolysis gives ‘A’. Compound A when warmed with ammoniacal silver nitrate, forms a bright silver mirror along the sides of the test tube. The unsaturated hydrocarbon X is _________.

Answer: (c)

Question 10: The compound which reacts faster in the S_N2 mechanism is _________.

1. CH₃C–Cl
2. H₂C=CH–Cl
3. H₂C=CH–CH₂–Cl
4. C₆H₅–CH₂–Br

Anwer: (c) H₂C=CH–CH₂–Cl

Explanation: H₂C=CH–CH₂–Cl will easily participate in the Substitution reaction. Hence, it reacts faster in S_N2 mechanism as it is primary and has less stearic hindrance.

Question 11: The product ‘A’ in the given reaction is __________.

Answer: (b)

Question 12: 3-Methylpent-2-ene on reaction with HBr in the presence of peroxide forms an additional product. The number of possible stereoisomers for the product is _________.

1. Six
2. Three
3. Two
4. Four

Answer: (d) Four

Since the product (X) includes 2 chiral centres, it is asymmetrical.

As a result, the total stereoisomers are 2² = 4.

Question 13: The most suitable reagent for the given conversion is __________.

R–CH₂OH → R–CHO

1. PCC
2. KMnO₄
3. K₂Cr₂O₇
4. Anhydrous CrO₃

Answer: (a) PCC

Explanation: R–CH₂OH → R–CHO is an oxidation reaction.

Pyridinium chlorochromate (PCC) is a mild oxidising agent that converts primary and secondary alcohols to aldehydes and ketones, respectively. It has a great selectivity for oxidation of alcohols because it has no effect on any other functional group. Acidic permanganate, acidic dichromate, and chromic anhydride in glacial acetic acid are strong oxidising agents that can further oxidise and produce carboxylic acid.

The most suitable reagent for R–CH₂OH → R–CHO is pyridinium chlorochromate (PCC)

Question 14: Match the chemical conversions in List I with appropriate events in List II and select the correct answer using the code given.

Answer: (a)

Question 15: In the given transformation, which of the following is the most appropriate reagent?

1. Zn(Hg)-HCl
2. Na, Liq.NH₃
3. NaBH₄
4. NH₂NH₂,OH^–

Answer: (d) NH₂NH₂,OH^–

Explanation: The OH group attached to the compound is unaffected by the Wolff Kishner reaction.

Question 16: The number of aldol reaction(s) that occur in the given transformation is _______.

1. 1
2. 2
3. 3
4. 4

Answer: (c) 3

Explanation:

Due to the presence of α–H in steps 1, 2, and 3, an aldol addition reaction occurs. Because α–H isn’t present, step 4 is a Cannizzaro reaction.

Question 17: The total number of amines among the following that can be synthesised by Gabriel synthesis is __________.

1. 1
2. 2
3. 3
4. 4

Answer: (c) 3

Explanation: A total of 3 amines can be synthesised by Gabriel synthesis. The Gabriel phthalimide synthesis is commonly used to produce 1° aliphatic/alicyclic amine. As a result, the amine that can be synthesised using the Gabriel Phthalimide synthesis technique is (A) Me₂CH-CH₂-NH₂ (B) CH₃CH₂NH₂ (C) ph-CH₂-NH₂

Question 18: Match List I with List II.

1. 3,2,1
2. 3,1,2
3. 1,3,2
4. 2,1,3

Answer: (d) 2,1,3

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