General Organic Chemistry Questions & solutions

Solved Objectives

 

Problem 1:       Nitrophenol, C₆H₄(OH) (NO₂) can have:

                        (A) No isomer ( only a single compound is possible)

                        (B) Two isomers

                        (C) Three isomers

                        (D) Four isomers

 

Solution:        o-,m- and p-isomers, i.e., position isomers

                        \(C)

 

Problem 2:       Geometrical isomerism is possible in:

                        (A) Butene-2                                   (B) Ethene

                        (C) Propane                                    (D) Propene

 

Solution:

If either of the two doubly bonded carbon atoms has some group or atoms attached on it, it will not show geometrical isomerism.

                        \ (A)

 

Problem 3:       Which of the following compounds will exhibit cis-trans isomerism.

                        (A) 2-butene                                    (B) 2-butyne

                        (B 2-butanol                                    (D) Butanol

Solution:        Due to the presence of asymmetric carbon atom.

                        \(B)

 

Problem 4:       Dichloro ethylene shows.

                        (A) Geometrical isomerism             (B) Position isomerism

                        (C) Both                                          (D) None

 

Solution:        CH₂ = CCl₂ and CHCl = CHCl are position isomers; CHCl = CHCl also show geometrical isomerism.

                        \(C)

 

Problem 5:       The compound having molecular formula C₄H₁₀O can show

                        (A) Metamerism                              (B) Functional isomerism

                        (C) Positional isomerism                (D) All

 

Solution:        Alcohols show position isomerism; Ethers show metamerism; Alcohol and ethers shows functional isomerism.

                        \(D)

 

Problem 6:       A compound contains 2 dissimilar asymmetric carbon atoms. The number of optical isomers is:

                        (A) 2                                               (B) 3

                        (C) 4                                               (D) 5

 

Solution:        a = 2ⁿ; where n is no. of dissimilar asymmetric  carbon atoms.

                        \(C)

 

Problem 7:       The greater the s-character in an orbital the ———— is its energy

                        (A) Greater                                      (B) Lower

                        (C) Both                                          (D) None

 

Solution:        Bond energy order

sp–sp > sp²-sp² > sp³-sp³

                        \(A)

 

Problem 8:     The type of isomerism observed in urea molecule is

                        (A)  Chain                                       (B) Position

                        (C) Geometrical                              (D) Functional

 

Solution:        NH₄CNO is functional isomer of urea.

                        \(D)

 

Problem 9:       Number of possible isomers of glucose  are

                        (A) 10                                             (B) 14

                        (B) 16                                             (D) 20

 

Solution:        Glucose has four dissimilar asymmetric carbon atoms; a = 2⁴.

                        \ (C)

Problem 10:     Which of the following statement is correct:

                        (A) Allyl carbonium ion (CH₂=CH–) is more stable than propyl carbonium ion

                        (B) Propyl carbonium ion is more stable than allyl carbonium ion

                        (C) Both are equally stable

                        (D) None

 

Solution:        Allyl carbonium undergoes resonance stabilization.

                        \(A)

 

 

Solved Subjectives

Problem 1:       Out of the following pairs, specify which of them are chiral-enantiomers or which of them are achiral-identical mirror images?

 

 

 

 

 

 

 

 

 

Solution:        (a) achiral – identical mirror images

(b) chiral – enantiomers

(c) chiral – enantiomers

(d) no chiral carbon – same structure

(e) chiral – enantiomers

(f)  chiral –  enantiomers

 

Problem 2:       What do you think about the following compounds – they are having chirality or not

 

 

 

 

 

 

 

Solution:        (a) It is optically active due to presence of chiral carbon.

(b) It has two chiral carbon atoms but due to plane of symmetry it is optically in-active.

(c) It has two chiral carbon atoms ()

 

hence it is chiral compound.

(d) There are no chiral carbon, but the molecule is chiral (an allene)

(e) There are no chiral carbon; this has a plane of symmetry – it is not a chiral molecule.

(f) Planar molecule – achiral

 

Problem 3:       Identify the end products in the following

 

 

 

 

 

Solution:

 

 

 

 

 

The intermediate can combine with another molecule to form higher alkane

 

Can you think about one other hydrocarbon?

 

Problem 4:       Explain mechanism of 1 : 4 addition of HBr on 1,3-butadiene.

 

Solution:

 

Due to Mesomeric effect, there is polarity on C₁ and C₄ resulting in the 1,4 addition at these sites. This type of behaviour is generally obtained in conjugated systems.

 

Problem 5:       What are the products when HCl adds to 2,4- hexadiene?

Solution: 

 

First we decide which is more stable. Naturally (I) which is 2° allylic and in which p-electrons can be further delocalised. Hence addition of Cl^– takes place on (I) in following way:

 

 

 

 

 

 

 

 

 

 

   Assignment Level – III

 

1. Toluene on treatment with methyl chloride and AlCl₃ at 0°C gives O – xylene and p – xylene, where as at 80°C the main product obtained is m – xylene, explain for the above deviation with a suitable mechanism ?

 

2. What shape will you expect for the following molecules

(i) Amide ion                (b) (CH₃)₃N

(iii)                       (iv)

(v)

 

3. Give reasons for the following

(a) Lithium acetyl acetonate has very higher melting point than beryllium acetyl acetonate.

(b) n – butyl alcohol has much higher boiling point than diethyl ether but both of them have same solubility in water.

(c) Why does every organic compound containing oxygen dissolves in Con H₂SO₄ and from the resulting solution the compound can be recovered by dilution with water.

 

4. Explain the following observations.
5. i) CH₃ – I   CH₃OH + I^–
6. ii) CF₃ – I   CF₃H + IO^–

 

5. Identify the configuration of products and also specify whether the products are optically active or inactive.

(a)  1- chloropentane  + Cl₂ (300°C) ¾®  C₅H₁₀Cl₂

(b) (S) –3- chloro –1- butene + HCl ¾® 2,3 dichloro 2-methyl butane

(c) (R)       -2- chloro –2,3 – dimethyl pentane + Cl₂ (300°C) ¾® C₇H₁₄Cl₂

 

6. Account for the following observations

(a) In benzene there are six equivalent carbon – carbon bonds, on the other hand in naphthalene (C₁ – C₂ bond length is 1.365Å, C₂ – C₃ is 1.404 Å) there are two bond lengths

(b) Acetylene is more acidic than benzene

(c) Aromatic bromination catalysed by Lewis acid thallium acetate gives only para isomer

 

7. What is relation between following pairs of structure indicate whether they are identical or enatiomers.

 

 

8. Which of the following compound is stronger acid and why?

 

9.

Why in (i) enol form is more stable whereas (ii) mainly exists in keto form.

10. Why is N, N-dimethyl –o-toluidine is stronger bas than N, N-dimethyl aniline?

 

 

Answers

 

1. At 0°C rate controlled product is obtained while at 80°C equilibrium controlled product i.e. m – xylene is obtained.

Although methyl group activates strongly at ortho and para positions, but it also favours dealkylation at these positions, but as meta isomer is formed more slowly and tends to persist with undergoing dealkylation HCl provides H^Å for reversal of alkylation.

 

In the case of smaller group like metyl, mechanism is slightly different

 

 

2. (i) Flat with tetrahedral angle

(ii)  Pyramidal

(iii) Trigonal

(iv) Pyramidal

(v)  Trigonal

 

4. In the first case I is more electronegative than C and hence the positive charge is an the C atom. So OH^– attacks C and displaces I^–.

In the second case due to the presence of three F atoms, the electron deficiency of C increases. Hence it gives a positive charge on I. Hence OH^– attacks I and displaces CF₃.

 

5.(a)

 

 

(b)

 

6.   (a)

By viewing the above structures the C₁ – C₂ bond is having double bond character in two  structures and single bond in the remaining one the C – 2 bond is single bond in two structures and double in only one \ C₁ – C₂ has more double bond character than C₂ – C₃ and hence the bond is short.

(b) s–character is more in Acetylene  than in benzene

(c)        An acid base complex is formed, which transfers halogen without its electrons because of its bulkiness reagent attacks least hindered position.

 

7. a,b,d enantiomers, C identical

 

8. Carbanion which results from loss of H⁺ from (ii) is less stable because rings lose coplanarity due to steric repulsion and therefore are not involved in stabilization of negative charge through resonance whereas this is not possible in (i) where all rings are tied to each other.
9. In rotation about the single bond is possible which helps in minimizing repulsion between two – C = O dipoles whereas in  this rotation is not possible and so in order to reduce two – C = O dipole repulsion molecule prefers enol form which is also stabilized by hydrogen bonding.

 

10. In N, N-dimethyl-o-toluidine due to steric repulsion from –N(CH₃) group is pushed upward and does not remain in the plane of ring so that electrons on nitrogen are not involved in resonance with the ring, increasing availability of electrons on nitrogen and hence basic nature is enhanced.

 

 

 

Objectives

 

1. Which is the strongest carboxylic acid among the following ?

(A) Cl₃CCO₂H                                                      (B) Br₃CCO₂H

(C) F₃CCO₂H                                                  (D) Cl₂CH-CO₂H

 

2. The number of optically active isomers observed in 2,3, – dichlorobutane is :

(A) 0                                                                (B) 2

(C) 3                                                                (D) 4

 

3. Allyl isocyanide has:

(A) 9 sigma bonds and 4 pi bonds

(B) 9 sigma bonds, 3 pi bonds and 2 non-bonding electrons

(C) 8 sigma bonds and 5 pi bonds

(D) 8 sigma bonds, 3 pi bonds and 4 non-bonding electrons

 

4. Keto – enol tautomerism is observed in

(A) H₅C₆-CHO                                                      (B) H₅C₆-CO-CH=CH₂

(C) H₅C₆-CO-C₆H₅                                               (D) H₅C₆-CO-CH₂-CO-CH₃

 

5. The compound which gives the most stable carbonium ion on dehydration is :

(A) CH₃-CH-CH₂OH                                      (C) CH₃-CH₂-CH₂-CH₂OH

|

CH₃

CH₃

|

(B) CH₃-C-OH                                                      (D) CH₃-CH-CH₂-CH₂OH

|                                                                     |

CH₃                                                               OH

 

6. Examine the followings two structures for the anilinium ion and choose the correct statement from the ones given below

 

 

(A) II is not an acceptable canonical structure because carbonium ions are less stable than ammonium ion.

(B) II is not an acceptable canonical structure because it is non-aromatic

(C) II is not an acceptable canonical structure because the nitrogen has 10 valence electrons.

(D) II is an acceptable canonical structures

 

7. meso-Tartaric acid is optically inactive due to the presence of

(A) two chiral carbon  atoms                    (B) molecular unsymmetry

(C) molecular symmetry                          (D) external compensation

8. Arrange following compounds in decreasing order of basicity.

 

(A)                                                             4 > 1 > 3  > 2   (B) 3 > 1> 4  > 2

(C) 2 > 1 > 3 > 4                                        (D) 1 > 3 > 2 > 4

 

9. Which of the following has the maximum resonance energy?

	(D) None

 

10. Which of the following ions in aromatic?

(A)		(B)
(C)		(D)

11. The maximum number of carbon atom arranged linearly in the molecule, CH₃CºC—CH=CH₂ is

(A) 2                                                          (B) 3

(C) 4                                                          (D) 5

 

12. Which of the following is aromatic in nature

(A)		(B)
(C)		(D)

 

13. A compound has the formula C₂HCl₂Br. The number of non identical structures that are possible is

(A) 1                                                          (B) 2

(C) 3                                                          (D) 4

 

14. Which one of the following can exhibit cis – trans isomerism

(A) CH₃CHCl—COOH                             (B) H—CºC—Cl

(C) ClCH=CHCl                                        (D) ClCH₂—CH₂Cl

 

15. How many total isomers are possible by replacing one hydrogen atom of propane with chlorine

(A) 2                                                          (B) 3

(C) 4                                                          (D) 5

Answers

 

 

1. C                                                         2.         B
2. B                                                         4.         D
3. B                                                         6.         c
4. c                                                         8.         D
5. D                                                         10.       B
6. C                                                         12.       D
7. C                                                         14.       c
8. A