For the following question the answer is given as option (a) . However both,option (a) and (c) has alpha H atoms and both will form 3° carbocations after rearrangement. So what makes (a) more favorable than (c)?
$\begingroup$
$\endgroup$
-
1$\begingroup$ Ring expansion will be favored in (c) over a hydride shift. $\endgroup$ – Avyansh Katiyar Sep 2 '18 at 6:19
-
1$\begingroup$ chemistry.stackexchange.com/questions/28939/… $\endgroup$ – Mithoron Sep 2 '18 at 23:40
$\begingroup$
$\endgroup$
Besides Avinsh's comment, there is another factor. When you have a carbocation center next to a cyclopropane ring or cyclobutane ring, the "bent bonds" of the ring partially overlap the "vacant" orbital on the cation center. To the extent that this occurs, the formally primary ion (c) is something like a more stable tertiary one. Ion (a) does not have this factor because the cation center is already inside the ring instead of next to it.
answered Sep 3 '18 at 16:20
-
$\begingroup$ Thank you..but could explain more precisely what do u exactly mean by" bent bonds"? $\endgroup$ – user67074 Sep 3 '18 at 16:25
-
-
$\begingroup$ The cyclobutyl cation itself (a) is probably non-classical, too, although I'm not sure how much the methyl group affects it $\endgroup$ – orthocresol♦ Sep 3 '18 at 16:35
