Solved Examples for JEE: General Organic Chemistry | Mock Tests for JEE Main and Advanced 2025 PDF Download

Q.1. Considering the basic strength of amines in aqueous solution, which one has

the smallest pK_b value?

(1) (CH₃)₃N

(2) C₆H₅NH₂

(3) (CH₃)₂NH

(4) CH₃NH₂

Correct answer is option (3).

• Because of resonance, aryalamines are less basic than alkyl amines. 
• In aryalamines, the lone pair of electrons on N is partly shared with the ring and is less available for sharing with a proton. 
• In alkylamines, the electron releasing alkyl group increases the electron density on nitrogen atom. Hence increases the protonation ability of amines. So, more the number of alkyl groups, higher will be basicity of amine. 
• Because of steric effect, a slight discrepancy occurs in case of trimethyl amine. So (CH₃)₂NH has the smallest pK_b value.
• Hence option (3) is the answer.

Q.2 The compound formed in the positive test for nitrogen with the Lassaigne solution of an organic compound is

(1) Fe₄[Fe(CN)₆]₃

(2) Na₄[Fe(CN)₅NOS]

(3) Fe(CN)₃

(4) Na₃[Fe(CN)₆]

Correct option is Option (1)
The compound formed in the positive test for nitrogen with the Lassaigne solution of an organic compound is prussian blue coloured complex compound ferric ferrocyanide.

Hence option (1) is the answer.

Q.3. Arrange the following compounds in order of decreasing acidity:  (IIT- JEE 2013)
(1) I  > II > III > IV
(2) III > I > II > IV
(3) IV > III > I > II
(4) II > IV > I > III

Correct Answer is Option (2)

• Electron donating groups increase the acidity of phenol when present as a substituent on the benzene ring while electron-withdrawing groups decrease the acidity of phenol.
• Cl and NO₂ are electron-withdrawing groups while -CH₃ and -OCH₃ are electron donating groups.
• However, nitro undergoes resonance, which allows it to better stabilize the negative charge of the oxygen compared to chlorine. Thus, acidic character of nitrophenol would be higher than that of chlorophenol.
• So, Nitrophenol is the most acidic compound here followed by chlorophenol.
  Hence, the correct option is 2.

Q.4. Which of the given statement(s) about N, O, P and Q with respect to M is (are) correct?    (IIT-JEE 2012)
(1) M and N are non–mirror image stereoisomers.
(2) M and O are identical.
(3) M and P are enantiomers.
(4) M and Q are identical.

Correct Answer is Option (1), (2) & (3)
Fischer Projections of the molecules:

It is clear that:

• M and N are non-mirror image stereoisomers.
• M and O are the same molecules i.e. Identical.
• M and P are the non-superimposable mirror images of hence they are enantiomers.
• N and Q are identical.
• M and Q are also are non–mirror image stereoisomers i.e diastereomers.
  Hence 1, 2 and 3 are the correct statements.

Q.5. Among the following oxoacids, the correct decreasing order of acid strength is :

(1) HClO₄ > HClO₃ > HClO₂ > HOCl

(2) HClO₂ > HClO₄ > HClO₃ > HOCl

(3) HOCl > HClO₂ > HClO₃ > HClO₄

(4) HClO₄ > HOCl > HClO₂ > HClO₃

Correct option is Option (1)

Acidic strength is directly proportional to the oxidation number. Increasing acid strength is because of an increase in the oxidation state of the central atom.

Hence option (1) is the answer.

Q.6. The correct order of increasing basicity of the given conjugate base (R =CH₃) is:

(1) RCOO^– < HC ≡ C^– < NH₂^– < R^–

(2) RCOO^– < HC ≡ C^– < R^– < NH₂

(3) R^– < HC ≡ C^– < RCOO^– < NH₂^–

(4) RCOO < NH₂ < HC ≡ C^– < R^–

Correct option is Option (1)

• The basic strength is inversely proportional to the stability of conjugate base.
•  In basicity, if the availability of the electrons is more, then more readily they can be donated to form a new bond to the proton and, and hence the stronger base. 
• If bronsted acid is a strong acid then its conjugate base is a weak base .
• The correct order of increasing basic strength is RCOO^– < HC ≡ C^– < NH₂^– < R^–
• Hence option (1) is the answer.

Q.7. In gaseous triethyl amine the "−C−N−C−" bond angle is _________ degree. (JEE 2021)

Solution: In gaseous triethyl amine the "−C−N−C−" bond angle is 108 degree.  

Q.8. Consider the below chemical reaction. The total number of stereoisomers possible for Product 'P' is _____________.                                      (JEE 2021)

Solution: The total number of products possible = 2 

Q.9. Which one the following does not have sp² hybridized carbon?

(1) Acetone

(2) Acetamide

(3) Acetonitrile

(4) Acetic acid

Correct option is Option (3)

Acetonitrile is having only sp³ and sp hybridized carbon atoms.

Hence option (3) is the answer.

Q.10. The total number of cyclic isomers possible for a hydrocarbon with the molecular formula C₄H₆ is ____   (IIT-JEE 2010)

5 cyclic isomers are possible for C₄H₆, possible.