Test: Reaction Intermediate Level - 2 - Chemistry MCQ

Radical is stabilized by α hydrogen. More the α-hydrogen more will be hyperconjugation more will be radical stabilization resonance. 

Resonance is also a good stabilizing factor and is given preference than hyperconjugation.

Higher the stability of cation more will be its half-life. As, in the carbocation, p orbital overlap with the compound, will be more stabilizing factor than aromaticity.

a. EWG ( e−− withdrawing groups) increases the acidic strength, whereas EDG ( e−− donating groups) decrease the acidic strength.

b. Nearer is the EWG to the source [(−COOH)group], stronger is the acid, i.e., α− substituted halo acid stronger than β−orγ− substituted halo acid.
Increasing order of acidic strength:
(III)<(II)<(I).

Both (IV) and (III) member ring are aromatic due to charge separation
in case of IV member ring total 6 electron and 3 member ring 2 electron are present
according to 4n+2 rule they are aromatic
and all the carbon are sp² hybridize

Concept:

• An organic compound having a positive charge on the carbon atom and six valence electrons is known as carbocation.
• Carbocations may be formed by the attack of a positively charged species on an unsaturated molecule.
• Carbocations may be classified as primary, secondary, and tertiary.
• Methyl carbocation having only one carbon is an exception.
• The positively charged carbon in the carbocation is sp2 hybridized.
• The three hybrid orbitals are used to form bonds with three other atoms.
• The carbocation has a planar structure. 

Explanation:

• Carbocations are highly reactive and are used up as soon as they are generated in the reaction process.
• The relative order of their stabilities are:
  
  

As from IMAGE 01 it is clear that here IMAGE 02 is not a resonance structure of others.

The given reaction is : (+)CH₃CH(I)CH₂CH₃ + Nal* → CH₃CH(I*)CH₂CH₃ + Nal

This is an example of an SN₂ type reaction.

• SN₂ reaction involves the product formation with inversion of configuration. When one molecule of the original substances undergoes inversion, one molecule of product is formed with inversion of configuration.
• This trier that when one original optically active molecule undergoes inversion, the inverted product and another origins, optically active molecules lose their optical activity due to the occurrence of racemisation i.e. when one molecule is inverted, actually two molecules are racemised.
• Hence, it can be concluded that the rate of racemisation is double to the rate of inversion. Hence, mathematically, 2k_i = k_r

Correct Answer :- C

Explanation : The correct option is A. Since −OE+ which increases the acidity of the compound and is distance dependent. In I −OE+ group is nearer to the substituents. So, its more acidic than the other two. SO the correct order will be i>ii>iii.