List of top Chemistry Questions asked in CBSE Class XI

Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?
Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene?
An alkene ‘A’ contains three C – C, eight C – H σ bonds and one C – C \(\pi\) bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of ‘A’.
An alkene 'A' on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of 'A'.
Write IUPAC names of the products obtained by the ozonolysis of the following compounds:
  1. Pent-2-ene 
  2. 3,4-Dimethyl-hept-3-ene 
  3. 2-Ethylbut-1-ene
  4. 1-Phenylbut-1-ene
For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated: 
  1. C4H8 (one double bond) 
  2. C5H8 (one triple bond)
The ionization constant of dimethylamine is 5.4 × 10–4. Calculate its degree of ionization in its 0.02M solution. What percentage of dimethylamine is ionized if the solution is also 0.1M in NaOH?
Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74. How is the degree of dissociation affected when its solution also contains (a) 0.01M (b) 0.1M in HCl ?
What is the pH of 0.001M aniline solution? The ionization constant of aniline can be taken from Table 6.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
How will you convert benzene into 
  1. p-nitrobromobenzene
  2. m-nitrochlorobenzene 
  3. p -nitrotoluene
  4. acetophenone
The pH of 0.005M codeine (C18H21NO3) solution is 9.95. Calculate its ionization constant and pKb.
The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKa of bromoacetic acid.
Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.
Out of benzene, m-dinitrobenzene and toluene which will undergo nitration most easily and why?
Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E+
  1. Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene 
  2. Toluene, p-H3C-C6H4-NO2, p-O2N-C6H4-NO2.
Write structures of all the alkenes which on hydrogenation give 2-methylbutane.
How would you convert the following compounds into benzene? 
  1. Ethyne
  2. Ethene
  3. Hexane
Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?
Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support Kekulé structure for benzene?
Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour.
Assuming complete dissociation, calculate the pH of the following solutions: (a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH
It has been found that the pH of a 0.01M solution of an organic acid is 4.15
Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.
Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:
\(CH_3COOH (l) + C_2H_5OH (l) ⇋ CH_3COOC_2H_5 (l) + H_2O (l) \)
  1. Write the concentration ratio (reaction quotient), Qc , for this reaction (note: water is not in excess and is not a solvent in this reaction)
  2. At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant. 
  3. Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?
Predict which of the following reaction will have appreciable concentration of reactants and products:
  1. \(Cl_2 (g) ⇋ 2Cl (g): \  K_c = 5×10_{–39}\)
  2. \(Cl_2 (g) + 2NO (g) ⇋ 2NOCl (g); \ K_c = 3.7×10^8\)
  3. \(Cl_2 (g) + 2NO_2 (g) ⇋ 2NO_2Cl (g); \  K_c = 1.8\)