Question

An organic chemist combines 2-methyl-2-butanol with sodium dichromate in sulfuric acid and water. Which of the following best describes the sequence of mechanistic steps for the potential reaction that would/might occur? Please note the following abbreviations: p.t. = proton transfer; Nu attack = nucleophilic attack; loss of LG = loss of leaving group; SN2 = a concerted nucleophilic attack and loss of a leaving group; E2 = a concerted bimolecular elimination reaction; SN1 = a stepwise loss of a leaving group to form a carbocation followed by nucleophilic attack; E1 = a stepwise loss of a leaving group to form a carbocation followed by abstraction of a βH to form an alkene; E1cB = a stepwise loss of an αH (through the action of a base) followed by the resonance-stabilized carbanion of the conjugate base "kicking out" a β leaving group; and N.R. = no reaction.

Answer 1

Answer:

NR = No reaction

Explanation:

To solve this, and understand it, we need to explain what happen with the sodium dicromate in acid.

First, the 2-methyl-2-butanol is an alcohol. By the position of the OH in the structure, it's on a tertiary carbon which makes this a tertiary alcohol.

Now, alcohols react with sodium dicromate in acid medium to promove an oxidation of the alcohol might occur only when this alcohol is secondary and primary. However in a tertiary alcohol this, do not occur.

Now why this do not occur? basically because the only way to promote the oxidation of an alcohol, comes with the movement of alpha or beta hydrogens. In the case of a tertiary alcohol, it doesn't have an alpha hydrogen to move, therefore, oxidation of tertiary alcohol do not occur.

The only way to "oxidize" a tertiary alcohol is promoving first the dehidratation of this alcohol in acid medium. This will turn the alcohol into an alkene, and then, this alkene can be oxidized to a ketone or aldehyde.

Answer 2

Answer:

SN1 = a stepwise loss of a leaving group to form a carbocation followed by nucleophilic attack

Explanation:

Since 2-methyl-2-butanol is tertiary alcohol, the first step will be loss of leaving group to form a 3° carbocation which is very stable and favours SN1, followed by attack of nucleophile