Question

Refer Image

Which statement is true?

(i) The product is (ref. image- i)

(ii) The Product is (ref. image- ii)

(iii) The reaction is called alkoxymercuration-demercuration (a type of solvomercuration-demercuration).

(iv) It proceeds via Markovnikov addition, anti-addition of $(F_{3}C-\stackrel{O}{\stackrel{||}{C}}-O-{)}_2$ and ROH (a nucleophilic solvent), and no rearrangement.

• A. $\left(i\right)$
• B. $\left(ii\right)$
• C. $\left(i\right),\left(iii\right),$ and $\left(iv\right)$
• D. $\left(ii\right),\left(iii\right),$ and $\left(iv\right)$

Answer 1

The correct option is C $\left(i\right),\left(iii\right),$ and $\left(iv\right)$

The statements $\left(i\right),\left(iii\right),$ and $\left(iv\right)$ are true.

The product is $2-methoxypropane$ $C{H}_{3}CH(-OC{H}_3)C{H}_3$
The reaction is alkoxy mercuration-demercuration. Net result is addition of a molecule of methanol to $C=C$ double bond. $-OC{H}_3$ group is added to less substituted internal $C$ atom. Thus addition follows Markovnikov rule. It is an example of anti addition and the reaction proceeds without rearrangement.