Test: Reaction Intermediate Level - 1 - Chemistry MCQ

The most nucleophile is dependent on the compound which has a tendency to give an electron. In the compound there is oxygen and one has sulphur. The sulphur has big in size that why it has more tendency to donate an electron. The correct answer is PhS-

Correct Answer :- a

Explanation : In a polar protic solvent (CH3OH), nucleophilicity increases down a column of the periodic table. So S- is more nucleophilic than O-.

For two species with the same attacking atom, the more basic is the more nucleophilic, so CH3CH2O- is more nucleophilic than CH3CO2-

CO2- give identical str resonance hybrids more solvated, very less nucleophilic.

I is the shortest because of double bond.

III is next because of resonance.

II is last as there is no kind of bond strengthening.

Hence, B is correct.

Correct Answer :- d

Explanation : Resonance energy per ring decreases as ring size increases

So, III > I > II (Resonance energy/ring)

Anthracene>Benzene>Naphthalene.

Correct Answer :- d

Explanation : The dispersal of the charge stabilizes the carbocation. More the number of alkyl groups, the greater the dispersal of positive charge and therefore, more the stability of carbocation,is an electron donating group, thus it will increases the stability of carbocation, hence the expected order is, (iii)>(ii)>(i).

Since, a free radical does the homolysis of pi bond.
The adjacent free radicals later form a bend.

Since after deprotonation cyclopentadiene becomes stable by aromaticity, So, it will definitely have a higher acidic strength than that of cyclopentane and consequently lower pKa value.

Further, no such stability factor has a role in the case of cyclopentane so it has lower acidic strength comparatively.

A is the correct option as 9- alpha Hydrogens are present. More alpha hydrogen, more hyperconjugation, more stability.

In option B and C, after carbocation rearrangement, it will remain 2°, therefore, rearrangement will not take place.
In option D, after carbocation rearrangement 3° will become 2° or 1°, therefore, rearrangement will not take place.
In option A, after rearrangement, carbocation will become 3° from 2°, therefore, this rearrangment will take place.

Correct Answer :- c

Explanation : The carbocation here is stabilized when there is an electron-donating group present.

More the activating effect of the group, more stable is the cation. Out of given options, I is most stable due to ability of oxygen to activate the ring with its lone pair of electrons.

Out of III and I,III is more stable as the lone pair of nitrogen in II is less available due to Ac group. IV is least stable due to electron- withdrawing effect of Cl atom.

The correct answer is : IV < II < III < I

Ans: d

Explanation: As we can see, the first step in the presence of light carbene is formed.

Correct Answer :- b

Explanation : I have an sp2 hybridized carbocation which is highly unstable. Hence, I will be the least stable of all. III will be the most stable as it is tertiary carbocation.

Among II and IV, IV has 4 alpha− hydrogens whereas II has only, α− hydrogens. Also IV is a secondary carbocation while II is a primary one. Hence IV will be more stable than II.

Hence, the decreasing order of stability or stabilization energy is III>IV>II>I