Question

Rank the transition states that occur during the following reaction steps in order of increasing stability (least $\to$ most stable)
1.$H_{3}C-\stackrel{+}{O}{H}_2\to C{H}_3^{+}+H_{2}O$
2. $(C{H}_3{)}_{3}C-\stackrel{+}{O}{H}_2\to (C{H}_3{)}_3{C}^{+}+H_{2}O$
3. $(C{H}_3{)}_{2}C-\stackrel{+}{O}{H}_2\to (C{H}_3{)}_{2}C{H}^{+}+H_{2}O$

• A. $1<2<3$
• B. $2<3<1$
• C. $1<3<2$
• D. $2<1<3$

Answer 1

Answer: (C)

$1<3<2$

The equilibrium constant of reactions is high if the carbocation formed is stable.

$3^{o}cation>2^{o}cation>1^{o}cation$ [stability]

(1) $C{H}_3-\stackrel{\oplus }{O{H}_2}⟶\underset{1^o}{\stackrel{\oplus }{C{H}_3}}+H_{2}O$

(2) $(C{H}_3{)}_{3}C-\stackrel{\oplus }{O{H}_2}⟶\underset{3^o}{(C{H}_3{)}_3}\stackrel{\oplus }{C}+H_{2}O$

(3) $(C{H}_3{)}_{2}C-\stackrel{oplus}{O}{H}_2⟶\underset{2^o}{(C{H}_3{)}_2{C}^{\oplus }}-H_{2}O$

So, $2>3>1$ .