Question

 

Given the following mechanism:

Step 1:           2-iodo-2-methylpropane                     à        2-methylpropane cation + iodide

Step 2:             2-methylpropyl cation + water            à        protonated 2-methylpropanol

Step 3:             protonated 2-methylpropanol + water à       2-methylpropanol + hydronium

19. Which compound is an intermediate?

                        a. iodide

                        b. protonated 2-methylpropanol

                        c. water

                        d. hydronium

20. Which compound is a reactant?

                        a. iodide

                        b. protonated 2-methylpropanol

                        c. water

                        d. hydronium

Answer 1

(i) Firstly 2-Iodo-2-methylpropane solvolyses to a very stable Tertiary Carbocation

                         (ii) This stable carbocation then attacked by solvent Water molecule H₂O. Then the species formed i.e. Protonated 2-Methylpropan-2-ol. Which has low stability and readily deprotonates to give neutral 2-Methylpropan-2-ol.

                         (iii) Beccause of it’s transitory existence it’s an Intermediate.

Ans 20 ) Question has answer here as Water as one have to choose answer from given option.

Let’s rule out other options

Iodide – Not the reactant as it departed in solvation step I. Rather it is a product of step I

Protonated 2-Methylpropanol is intermediate and not actual reactant

Hydronium ion is Product of step III

This way only Water find to be Reactant.

SO reactant is water. Is the right answer.