Test: Reaction Intermediates, Attacking reagents - JEE MCQ

Order of Acidic strength ∝ stability of conjugate base.

so bond dissociation energy order is

- Igroup destablises carbocation and since inductive effect decreases with increasing length of carbon chain. Therefore (b) is the correct option.

The stability of free radicals follows the order:-

Thus, stability order will be :-

In I the unshared pair of electrons on N always available for protonation in III due to presence of electronegative O atom the electron density on N is decreased . In II and IV resonance suppresses the basic character.

The order of stability of carbocations is :

The correct option is (2): Triphenylmethyl cation.
Stability of carbonium ions follows the order:
tertiary alkyl > secondary alkyl > primary alkyl > methyl
More the number of alkyl groups, the greater the dispersal of positive charge and therefore, more the stability of carbonium ion is observed.
In triphenylmethyl cation, dispersal of positive charge increases with the increase in the number of benzene ring. Hence, it is the most stable carbonium ion.

CH₃ group give +M and +I but +M work only at ortho and para position. Reactivity ∝ positive charge on carbonyl carbon so that's why reactivity II > I > III.

Stability of Alkene ∝ No. of H-atoms or more hyperconjugation.

very stable carbocation.

Friedel-crafts reaction is an electrophilic substitution reaction and presence of an electron releasing substituent like Me, makes the benzene nucleus more reactive towards such reactions) Thus toluene is most reactive among the given.

NO₂ group is benzene deactivator, it is meta directing.

Among all given reagents -CCl₃ is strongest electron withdrawing group so it will reduce electron density at ortho and para position and it will direct reacting group at meta position.
So, -CCl₃ will yield maximum meta product.

The carbanion (CH₃)₃C⁻is expected to be least stable due to electron-releasing nature of methyl groups. The carbanion Cl₃C⁻is expected to be most stable due to electron-attracting nature of chlorine atoms Keeping only these two facts in mind, the choice (a) seems to be correct.

More the ∝ H, more will be its stability of free radicals. So, the correct order is [B] < [C] < [D] < [A].

The reaction proceeds as follows.

There is a formation of intermediate carbocation.

C₆H₅CH₂NH₂ is an aliphatic amine. It is most basic amongst the given compounds.

In substituted phenols, the presence of electron-withdrawing groups such as nitro group enhances the acidic strength of phenol. This effect is more pronounced when such a group is present at ortho and para positions. It is due to the effective delocalisation of negative charge in phenoxide ion.

Also, the acidity will increase with the number of electron-withdrawing groups. Thus, the order of acidity of the given compounds is