Question

Given below are two statements:

Statement I: 3-phenylpropene reacts with $HBr$ and gives secondary alkyl bromide having a chiral carbon atom as the major product.

Statement II: Aryl chlorides and aryl cyanides can be prepared by Sandmeyer reaction as well as Gattermann reaction.

In the light of the above statements, choose the correct answer from the options given below

• A. Both Statement I and Statement II are true
• B. Both Statement I and Statement II are false
• C. Statement I is true but Statement II is false
• D. Statement I is false but Statement II is true

Hint:

Think about carbocation rearrangement to the more stable benzylic position for the first statement. For the second, distinguish the reagents used in Sandmeyer and Gattermann reactions, specifically regarding cyanide.

Answer 1

The Correct Option is C

In this problem, we evaluate two chemical statements regarding organic reactions and mechanisms.

Statement I: 3-phenylpropene has the chemical structure $C_6H_5-CH_2-CH=CH_2$. When it reacts with $HBr$ (hydrobromic acid), the reaction proceeds via an ionic electrophilic addition mechanism. In the first step, the $\pi$ electrons of the alkene attack the proton ($H^+$) from $HBr$ to form a carbocation intermediate. To minimize energy, the more stable carbocation is formed, which in this case is the secondary carbocation at the second carbon:
$$C_6H_5-CH_2-CH=CH_2 + H^+ \rightarrow C_6H_5-CH_2-CH^+-CH_3$$
However, this secondary carbocation can undergo a 1,2-hydride shift from the adjacent carbon (the one attached to the phenyl ring) to form an even more stable carbocation. This new carbocation is benzylic, meaning the positive charge is stabilized by resonance with the phenyl ring:
$$C_6H_5-CH_2-CH^+-CH_3 \xrightarrow{1,2-H~shift} C_6H_5-CH^+-CH_2-CH_3$$
In the final step, the nucleophilic bromide ion ($Br^-$) attacks this stable benzylic carbocation to form the major product, 1-bromo-1-phenylpropane:
$$C_6H_5-CH^+-CH_2-CH_3 + Br^- \rightarrow C_6H_5-CH(Br)-CH_2-CH_3$$
This product is a secondary alkyl bromide because the carbon bearing the bromine is attached to two other carbons. Furthermore, this central carbon is chiral because it is bonded to four different groups: a hydrogen atom, a bromine atom, a phenyl group, and an ethyl group. Thus, Statement I is true.

Statement II: The Sandmeyer reaction is a well-known method to replace the diazonium group ($-N_2^+$) of an aryl diazonium salt with substituents like chlorine, bromine, or cyanide using copper(I) salts (e.g., $CuCl$ for chloride, $CuCN$ for cyanide). The Gattermann reaction is a variation that uses copper powder and the corresponding halogen acid ($HCl$ or $HBr$) to introduce chloride or bromide. While Gattermann works for aryl halides, it is not the standard or practical method for preparing aryl cyanides; the Sandmeyer reaction ($CuCN/KCN$) is the standard choice for cyanide introduction. Therefore, the assertion that both reactions can prepare both types of compounds is technically incorrect regarding aryl cyanides. Thus, Statement II is false.

Based on this analysis, Statement I is true and Statement II is false.