Given below are two statements : In the light of the above statements, choose the correct answer from the options given below :
- Both Statement I and Statement II are true
- Both Statement I and Statement II are false
- Statement I is true but Statement II is false
- Statement I is false but Statement II is true
Answer
(3) Statement I is true but Statement II is false
Solution
Statement I:
C6H5-CH2-CH=CH2 + HBr C6H5-CH2-CHBr-CH3 (Markovnikov addition via the 2° carbocation)
The C–Br carbon carries four different groups (H, Br, CH3, CH2C6H5) ⇒ a secondary alkyl bromide with a chiral carbon
(even if a hydride shift gives the benzylic cation, the product C6H5-CHBr-CH2-CH3 is again a chiral 2° bromide) ⇒ Statement I is true.
Statement II:
Sandmeyer: ArN2+X- CuCl/HCl ArCl and ArN2+X- CuCN/KCN ArCN
Gattermann: ArN2+X- Cu/HCl ArCl (or Cu/HBr ⇒ ArBr) — aryl cyanides are not prepared this way
⇒ Statement II is false.
Answer: (3)