Saved Bookmarks
| 1. |
Give reasons for the following: (i) Ethyl iodide undergoes S_(N)2reaction faster than ethyl bromide. (ii) (pm)2-Butanol is optically active. (iii) C - X bond length in halobenzene is smaller than C - X bond length in CH_(3) - X . |
|
Answer» Solution :(i) C- I bond is weaker than C - Br bond, therefore ethyl iodide undergoes `S_(N)2`reaction faster than ethyl bromide. (ii) `(pm)` 2-Butanol is a mixture of equal amounts of `(+)` 2-Butanol and `(-)` 2-Butanol. Both these compounds are separately active. Optical ROTATION of one cancels the opticals rotation of the other with the result that there is no NET rotation. (iii)  The carbon linked to the HALOGEN X in halobenzene is `sp^(2)`hybridised while that linked to X in `CH_(3) - X"is" sp^(3)`hybridised. An `sp^(2)`hybrid orbital is SMALLER in size compared to `sp^(3)`hybrid orbital. That is why C - X bond length in halobenzene is smaller than C-X bond length in `CH_(3) - X `. |
|