General Organic Chemistry


Subjective Problems 


  1. Pentane’s three isomers
i) v)
ii) vi)
iii) vii)
iv) viii)

Total no. of isomers = 8


Phenyl acetic acid has greater acidic character due to –I effect of phenyl ring.

  1. In aniline the lone pair of electrons on nitrogen involves in the resonance of benzene ring while in alkyl amine the nitrogen has increased electron density due to +I effect of R group.
  2. In Guanidine lone pair on any one of the nitrogen atoms is always available for donation as a base.

If any of the nitrogen gets protonated, even then also we have a lone pair available for donation as a base.

Hence Guanidine is a stronger base

6. i)

There are two (even) asymmetric carbon atoms (*), but the molecule cannot be divided into symmetrical halves. So the number of d- and l- isomers, a = 2n = 22 = 4, and the no. of meso forms m = 0

So, total number of stereoisomers = a + m = 4 + 0 = 4


There are two (even asymmetric carbon (*) atoms and the molecule can be divided into symmetrical halves, so the number of d- and l-isomers a = 2n–1 = 22–1 = 2, and the number of meso forms, m = 2(n/2)-1 = 20 = 1

Total number of stereoisomers = 2 + 1 = 3

  1. a) The allene has non-planar character. The two double bond systems in 2,3-pentadiene are at right angle to each other. Due to this it has no plane of symmetry and its mirror images are not super-imposable and the allene is capable of existence in (+) and (–) enantiomers.

In allene, the central carbon atom is sp hybridised and linear while the two terminal carbon atoms are sp2 hybridized and trigonal. The two unhybridized p-orbitals on a sp hybrid carbon atom are perpendicular. So the two π-bonds must also be perpendicular. Allene itself is achiral, however, if some substitutents are added to allene the situation is different and it becomes chiral as in the case of 2,3-pentadiene.

b) i) ii)
iii) iv)


  1. C4H10O can represent alcohol and ether. Ether shows metamerism, ether and alcohol shows functional isomerism and alcohols show position isomerism.
  2. Total number of structural isomers = 7



(i)         CH3 (ii)




    |     |

    OH     CH3

(iii) (Iv)



(v) (vi)









  1. Specific rotation [α]D = given C (conc) =

[α]D  = = +16°

Where l = length of the solution in decimetres

  1. [I, II] and [II, III] are diastereomers [I, III] are enantiomers.
  2. 2-butene has two dissimilar groups attached to each unsaturated carbon,
    2-butyne is linear so it can not have geometrical isomer 
  3. Benzyl carbonium ion is more stable due to resonance 
  4. A gives iodoform test presence of

Thus only structure of A is and B is 

Since d-and l-are formed due to formation of chiral carbon on reaction with HCN and also by subsequent hydrolysis, mixture is optically inactive (reacemic). This process is called racemisation.

  1. Due to more covalent bonds  also octet of every atom in R—CO+ is complete 



  1. The main driving force for enolization is relief of the electrostatic repulsion that occurs where the two electrophilic carbonyl groups are adjacent to each other.

Enol form will also stabilize due to intramolecular H-bonding’.

  1. < < <

In all singlet carbene, the stability increase as it is attached with halogen compound because halogen can have pπ – pπ conjugation with carbon. pπ-pπ conjugation for as it is 2pπ-2pπ, where as in it is 2pπ – 3pπ.


As the fluorine is more electron withdrawing than chlorine p-flourophenol should be more acidic but it is actually less acidic. The reason is

The conjugate base of p-chlorophenol is more stable as the negative charge on oxygen is dispersed by resonance and chlorine has vacant d-orbital to take up the lone pair. Where as in p-flouro phenoxite ion the negative charge is dispersed by –I effect of fluorine only.


Enol form is much more stable than keto form because of great stability associated with aromatic ring which is absent in keto form.

  1. First requirement for a compound to show optical isomerism is that it should have at least one asymmetric carbon atom (the carbon marked as * is asymmetric). Therefore the above compound having are asymmetric). Therefore the above compound having one asymmetric carbon atom will show optical isomerism. First requirement for a compound to show geometrical isomerism is it should have hindered rotation (the above compound has one double bond). Second requirement is no two same groups should be attached to double bonded carbon atoms (The above compound has two – CH3 groups attached to double bond). Therefore, the compound does not fulfil the second requirement, hence does not exist as cis and trans isomers.
  2. 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.
  3. Anion derived from 3,5-dimethyl-4-nitrophenol is less stable because due to steric hindrance nitrogroup is forced out of the plane of ring and thus is unable to stabilize negative charge on anion through resonance.
  4. In p-fluorobenzoic acid +R and –I effect cancel each other whereas in p-chlorobenzoic acid +R effect is much weaker than –I effect (because of greater bond length of C – Cl bond +R effect is weaker).
  5. 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.
  6. 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.