31 Year NEET Previous Questions: Some Basic Principles & Techniques - 2

4 ethyl - 3 methyl octane

Amino group is ring activating while nitro group is deactivating. Hence, correct order is  aniline > benzene > nitrobenzene.
I > II > III

–NO₂ is an electron attracting group hence decrease the electron density on ring, whereas –NH₂ group is electron releasing group hence increases electron density on ring. Benzene is also rich due to delocalization of electrons.

In the presence of UV rays or energy, by boiling chlorine, free radical is generated which attack the methyl carbon atom of the toluene.

Compound which are mirror image of each other and are not superimposable are termed as enantiomers.

are enantiomers

Carbon atoms from C₁ to C₅ are chiral

H2C = CHCl is capable of showing resonance which develops a partial double bond, character to C–Cl bond, thereby making it less reactive toward  nucleophilic substitution. ..

In diphenylmethane monochlorination at following positions will produce structured isomers

Clock wise rotation.

Hence configuration is R.
If the eye travel in a clockwise direction, the configuration is specified as the order of priority is Br > Cl > CH₃ > H

Optical and geometrical isomerism pair up to exhibit stereoisomerism. This is because the isomers differ only in their orientation in space.

SN₁ reaction is favour ed by heavy gr oups on the carbon atom attached to halogen i.e Benzyl > allyl > tertiary > primary > secondary > primary > alkyl halides

Obtained from SN₁ path.
This molecule is resonance stabilised.

More the number of alkyl groups, the greater the dispersal of positive charge and therefore more the stability of carbocation hence.

Phenols are more acidic than alcohol as they are resonance stabilised whereas alcohols are not .
Further nitro is an electron withdrawing which increases acidic character and facilitates release of proton, whereas –CH₃ is an electron donating group.Which decreases acidic character , thus removal of H⁺ becomes very difficult.

Correct IUPAC name of

  is 4-Ethyl-3–methyl heptane

Among the given compounds naph thelene is volatile but benzoic acid is non-volatile (it forms a dimer). So, the best method for their separation is sublimation, which is applicable to compounds which can be converted directly into the vapour phase from its solid state on heating and back to the solid state on cooling. Hence it is the most appropriate method.

General molecular formula of alkanols is

It is 2, 3 dimethyl pentanoyl chloride.

Among the three given hybrid orbitals, sp hybrid orbital is most electronegative.
Contribution of s in sp hybrid orbital is maximum (50%) so this orbital is closer to nucleus. Naturally it will have greater tendency to pull electron towards it. Hence it becomes more electronegative and sp3 becomes least electronegative as it have only 25% S character.

* marked are chiral carbons.

A is correct; the decreasing order of reactivity is NO₂ > CHO > H > CH₃.

Hydrolysis of an acyl chloride proceeds by nucleophilic attack at the carbonyl carbon. The rate increases when the carbonyl carbon is made more electrophilic.

Electron-withdrawing substituents increase electrophilicity by -I (inductive) and/or -M (mesomeric) effects; electron-donating groups decrease it.

NO₂ has a strong -I and strong -M effect, so it raises the positive character of the carbonyl carbon most and gives the fastest hydrolysis.

CHO (formyl) withdraws electron density by -I and -M, so it increases electrophilicity less than NO₂ but more than hydrogen.

Hydrogen is essentially neutral, while CH₃ is weakly electron donating by +I, which reduces electrophilicity and makes hydrolysis slowest.

Hence the verified order is NO₂ > CHO > H > CH₃, corresponding to option A.

Compounds which do not show optical activity inspite of the presence of chiral carbon atoms are called meso-compounds.
The absence of optical activity in these compound is due to the presence of a plane of symmetry in their molecules. e.g. mesotartaric acid is optically inactive.

(Plane of symmetry)

Nucleophilicity increases down the periodic table.   I^- > Br^- > Cl^- > F^-

Benzene having any activating group i.e, OH, R etc., undergoes electrophilic substitution very easily as compared to benzene itself. Thus toluene (C₆H₅CH₃), phenol (C₆H₅OH) undergo electrophilic substitution very readily than benzene.Chlorine with +E and +M effect deactivating the ring due to strong -I effect. So, it is difficult to carry out the substitution  in chlorobenzene than in benzene, so correct order is 

Phenol > Toluene > Benzene > Chlorobenzene

Out of the given compounds the most reactive towards nucleophilic attack

Phenoxide ion is stable due to resonance. i.e. the correct answer is option (c).

The amount of s-character in various hybrid orbitals is as follows. sp = 50%, sp² = 33% and sp³ = 25% Therefore character of the C – H bond in acetylene (sp) is greater than that of the C – H bond in alkene (sp² hybridized) which in turn has greater s character of the C – H bond in alkanes. Thus  owing to a highs character of the C – H bond in alkynes, the electrons constituting this bond are more strongly held by the carbon nucleus with the result the hydrogen present on such a carbon atom can be easily removed as proton. The acidic nature of three types of C – H bonds follows the following order ≡ C-H > = C - H > - C - H.
Further, as we know that conjugate base of a strong acid is a weak base, hence the correct order of basicity is

In the molecule

the number of stereoisomers is given by sum of geometrical isomers (because of presence of C = C) and optical isomers (because of presence of chiral carbon atom).
Number of geometrical isomers = 2 (one C = C is present).
Number of optical isomers = 2 (one chiral carbon atom).
Total number of stereoisomers = 2 + 2 = 4

A strong base can abstract an α-hydrogen from a ketone.

Correct option: D

The relative stability of carbanions is governed mainly by s-character of the orbital bearing the negative charge and by any possible resonance stabilization.

An orbital with greater s-character holds the negative charge closer to the nucleus and therefore stabilises the carbanion more; thus sp > sp2 > sp3.

The RC≡C^- carbanion is sp-hybridised, so it is strongly stabilised by high s-character and is the most stable of the four.

The Ph-CH^- (benzylic) carbanion is stabilised by resonance with the aromatic ring and is therefore more stable than a simple vinylic or alkyl carbanion, but less stable than the sp-acetylide.

The R2C=CH^- (vinylic) carbanion is sp2-hybridised and lacks effective resonance stabilisation in the same way as a benzylic centre, so it is less stable than the benzylic carbanion.

The R3C-CH2^- (alkyl) carbanion is sp3-hybridised and is destabilised by electron-releasing alkyl groups; it is the least stable.

Final order: RC≡C^- > Ph-CH^- > R2C=CH^- > R3C-CH2^- (option D).

C – 1 is sp hybridized (C ≡ C)
C – 3 is sp³ hybridized (C– C)
C – 5 is sp² hybridized (C = C) Thus the correct sequence is sp, sp³, sp².