GENERAL ORGANIC CHEMISTRY
LEVEL – I
- Amides are less basic than amines because lone pair on nitrogen are involved in resonance with C = O, π electrons resulting in less availability of electrons on nitrogen.
- 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
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
- 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.
- C4H10O can represent alcohol and ether. Ether shows metamerism, ether and alcohol shows functional isomerism and alcohols show position isomerism.
- Total number of structural isomers = 7
(i) CH3 (ii)
- Since S is larger than O, the charge density of negative charge is less hence CH3S– is weaker base hence its conjugate acid CH3SH is stronger than CH3OH.
- Vinyl < Methyl <1° < 2°< 3° < Allyl
- Order of basic strength is as follows:–
LEVEL – II
- 2-butene has two dissimilar groups attached to each unsaturated carbon,
2-butyne is linear so it can not have geometrical isomer
- On heating first will give anhydride at comparatively lower temperature
- Benzyl carbonium ion is more stable due to resonance
- 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.
- (i) S (II) R
- Due to more covalent bonds also octet of every atom in R—C≡O+ is complete
- Mechanism of addition of HCl on 1,3-butadiene is as follows
In step 1 proton adds to one of the terminal cabons of 1,3 butadiene to form, as usual, the more stable carbonium ion, in this case a resonance stabilized allylic cation. Addition of one of the inner carbon atoms would have produced a much less stable primary cation that could not be stabilized by resonance;
In step (2), a chloride ion forms a bond to one of the carbon atoms of the allylic cation that bears a partial positive charge. Reaction at one carbon atom results in the 1,2 – addition product, reaction at the other gives that 1,4 addition product.
- (i) E (II) Z
LEVEL – III
- In aliphatic amines lone pair lies in sp3 hybridized nitrogen whereas in pyridine lone pair lies in sp2hybridized nitrogen. So in pyridine nitrogen atom has more hold over this lone pair of electrons, decreasing its availability.
- 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.
- In 2,6-diterbutyl phenol bulkyo-tert butyl groups prevent solvation of corresponding phenoxide anion thus making it less stable than phenoxide ion.
- 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.
- 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).
- 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.
- In enol form of Ar2CH – CHO two bulky Ar groups are attached to sp2 hybridized carbon atom whereas in keto form they are attached to sp3 hybridized carbon atom.
- 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.
R-2-butanol when attacked by H+ forms a carbocation which is symmetrical and therefore can be attacked by water molecule with equal probability giving a racemic mixture of R-2-butanol and S-2-butanol.
- i) Number of optical isomers = 22 = 4 (2dl pairs)
Number of geometrical isomers = 0
(Since, compound does not contain a ring or a double bond )
∴ total number of isomers = 4 + 0 = 4
- ii) The compound has two chiral carbons and it can be divided into two equal halves. The restricted rotation associated with the ring structure also makes geometrical isomers a possibility
∴ Number of optically active forms = 22–1 = 2
Number of optically active forms = 2(2–2)/2 = 2= 1 (meso)
∴ total number of stereoisomers = 2+1= 3
Since, the two chiral carbons coincide with two ring carbons responsible for geometrical isomerism, the total number of stereoisomers is the same as the number of optical isomers
iii) The compound contains three chiral carbons (n=3) but the molecule can be divided into two equal halves.
∴ Number of optically active forms = 23–1–2(3–1)/2 = 22—21 = 4–2 = 2
Number of optically inactive forms = 2(3–1)/2 = 2 (meso)
Number of geometrical isomers =- 0
∴ Total number of stereoisomers = 2+2+0 = 4