Basics of Organic Chemistry

The possible stereoisomers of the given compounds are as follows:

	Functional group (priority order)	suffix(s)	prefix
(1)		Oic acid(when carbon(C) of this functional group present in the main chain) or Carboxylic acid (when carbon(C) of this functional group is not present in the main chain)	Carboxy
(2)		sulphonic acid	sulpho
(3)		Oic anhydrid (when carbon(C) of this functional group present in the main chain) or Carboxylic anhydride (when carbon(C) of this functional group is not present in the main chain)	$ - $
(4)		Oate (when carbon(C) of this functional group present in the main chain) or Carboxylate (when carbon(C) of this functional group is not present in the main chain)	$ - $
(5)		Oyl halide (when carbon(C) of this functional group present in the main chain) or Carboxyl halide (when carbon(C) of this functional group is not present in the main chain)	haloflormyl
(6)		amide (when carbon(C) of this functional group present in the main chain) or carbamide (when carbon(C) of this functional group is not present in the main chain)	carbamoyl
(7)	cynide ($ - $ C $ \equiv $ N)	nitrile (when carbon(C) of this functional group present in the main chain) or carbonitrile (when carbon(C) of this functional group is not present in the main chain)	cyano
(8)		Carbylamine	isocyano
(9)		al (when carbon(C) of this functional group present in the main chain) or Carbaldehyde (when carbon(C) of this functional group is not present in the main chain)	formyl (oxo)
(10)		one	oxo
(11)	$ - $ OH	ol	hydroxy
(12)	$ - $ SH	thiol	mercapto
(13)	$ - $ NH$_2$	amine	amino

From this table you can see -COOH has highest priority and -NH₂ has lowest priority among all functional groups.

Stability of the following species depends upon the no. of $\alpha$ hydrogen which can undergoes hyperconjugation as well as resonance. Higher the no. of $\alpha$ hydrogen, higher will be the stability of the carbocation.

Stabilizes by resonance and have six $\alpha$-hydrogen atoms (hyperconjugation).

Stabilizes by resonance and have three $\alpha$-hydrogen atoms.

It have five $\alpha$-hydrogen atoms.

It have only two $\alpha$-hydrogen atoms.

$\therefore$ (I) > (III) > (II) > (IV)

Hyperconjugation effect is one of the type of permanent effect in which delocalisation of $\sigma$ electrons of C-H bond of an alkyl group directly attached to an atom of the unsaturated system or to an atom with an unshared p orbital. The main role of hyperconjugation is to stabilise the carbocation because it helps in the dispersal of positive charge. Higher the number of alkyl groups attached to a positively charged carbon atom, greater is the hyperconjugative effect and stability of carbonation. It generally involves the interaction of electrons in a C-H sigma orbital with nearby non-bonding p or anti-bonding $\sigma^*$ or $\pi^*$ orbitals to provide extended conjugation which further increases the stability of the carbocation. Sometimes, low lying anti-bonding $\sigma^*$ also interact with filled orbitals of lone pair character. This type of conjugation is termed as negative hyperconjugation.

The correct statements concerning the structures E, F and G are:

(B) E, F and G are tautomers.

E is in keto form (C=O) and F and G are in enol form (C=C$-$OH)

(C) F and G are geometrical isomers. F is Z isomer and E is E isomer.

(D) F and G are diastereomers. They are not mirror images of each other and non-superimposable.

If we add halogens on alkenes it is mostly anti addition. Anti addition of $\mathrm{Br_2}$ on $trans$ alkene gives meso compound.

So, here only 1 stereoisomer is formed.

$trans$-2-butene is

Bromination of $trans$-2-butene

Among the given resonance structures, the least stable resonance structure is given by option (A) as the two positive charges lie adjacent to each other and cause maximum repulsion.

(A)

Same charges are present at nearest position making the molecule less stable.

Structural Isomers

When 2-methyl butane is treated with $\mathrm{Cl}_2$, the chlorination of the compound occur (via free radical mechanism) and an isomeric mixture of products is obtained with chemical formula, $\mathrm{C}_5 \mathrm{H}_{11} \mathrm{Cl}$.

The products formed are shown below.

The chlorination yields total 6 different compounds. The compounds I, II, III and IV have chiral centres present in it. The compounds I and II, III and IV are a pair of enantiomers due to presence of chiral centres. Hence, value of N will be 6 .

The enantiomers have identical boiling points. The method of fractional distillation cannot be used for separating enantiomers.

Hence, only four compounds will be separated which are shown below. The compounds I and II, III and IV will remain as racemic mixture as fractional distillation will unable to separate each enantiomer.

Hence, value of $M$ will be 4 .