Q.66. Given below are two statements : Statement-I : 3-phenylpropene reacts with HBr and gives secon

Q: What is the correct answer for JEE Main Chemistry Q66?

The correct answer is C. Statement-I: 3-phenylpropene (PhCH$_2$-CH=CH$_2$) reacts with HBr via electrophilic addition. The reaction proceeds through a carbocation intermediate. The primary carbocation (PhCH$_2$-CH$^+$-CH$_3$) is formed first, which can rearrange to a more stable secondary carbocation (PhCH(Br)-CH$_2$-CH$_3$) or a benzylic carbocation (PhCH$^+$-CH$_2$-CH$_3$). The benzylic carbocation (PhCH$^+$-CH$_2$-CH$_

Q: Q.66. 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…

The correct answer is C. Statement-I: 3-phenylpropene (PhCH$_2$-CH=CH$_2$) reacts with HBr via electrophilic addition. The reaction proceeds through a carbocation intermediate. The primary carbocation (PhCH$_2$-CH$^+$-CH$_3$) is formed first, which can rearrange to a more stable secondary carbocation (PhCH(Br)-CH$_2$-CH$_3$) or a benzylic carbocation (PhCH$^+$-CH$_2$-CH$_3$). The benzylic carbocation (PhCH$^+$-CH$_2$-CH$_

Q: What are the options for this JEE Main Chemistry question?

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.

Question

Q.66. 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.

FirstInTest · Accepted Answer

Statement-I: 3-phenylpropene (PhCH$_2$-CH=CH$_2$) reacts with HBr via electrophilic addition. The reaction proceeds through a carbocation intermediate. The primary carbocation (PhCH$_2$-CH$^+$-CH$_3$) is formed first, which can rearrange to a more stable secondary carbocation (PhCH(Br)-CH$_2$-CH$_3$) or a benzylic carbocation (PhCH$^+$-CH$_2$-CH$_3$). The benzylic carbocation (PhCH$^+$-CH$_2$-CH$_3$) is stabilized by resonance. The addition of Br$^-$ to this carbocation would give 1-bromo-1-phenylpropane. The solution shows a hydride shift leading to PhCH(Br)-CH$_2$-CH$_3$, which is 1-bromo-1-phenylpropane. This product has a chiral carbon at position 1. So, the major product is 1-bromo-1-phenylpropane, which is a secondary alkyl bromide and has a chiral carbon. Statement-I is true.

Statement-II: Aryl chlorides and aryl cyanides can be prepared by the Sandmeyer reaction (from diazonium salts). However, the Gattermann reaction (using Cu powder and HCl/HBr) is used to prepare aryl halides (chlorides, bromides) but not aryl cyanides. For aryl cyanides, the Sandmeyer reaction uses CuCN/KCN. Therefore, Statement-II is false.

Thus, Statement-I is true but Statement-II is false.

Answer

Both Statement-I and Statement-II are true.

Answer

Both Statement-I and Statement-II are false.

Answer

Statement-I is true but Statement-II is false.

Answer

Statement-I is false but Statement-II is true.

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