Question

Determine whether each of these reactions occur through an S N 1 , S N 2 , E1, or E2 mechanism. A. A tertiary alkyl halide is treated with ethanol and heating to give an alkene. A. The mechanism of reaction A is: S N 1 E1 S N 2 E2 B.A tertiary alkyl halide is treated with potassium tert butoxide in tert butanol to give an alkene. B. The mechanism of reaction B is: E1 S N 1 S N 2 E2 C. A tertiary alkyl halide is treated with ethanol and heating to give an alkene. C. The mechanism of reaction C is: E2 E1 S N 1 S N 2

Answer 1

Answer:

See explanation below

Explanation:

First to all, we need to remember the difference between Sn and E reactions.

Reactions where a Sn1 or Sn2 occur, it's where you have a reactant with a very good leaving group and a nucleophyle, and the product or products of the reaction always results that the nucleophyle is now taking the place where the leaving group of the innitial reactant was. This can be done in acid medium (SN1 carbocation intermediate 2 steps) or basic medium (SN2 fast reaction 1 step)

Reactions where a E1 or E2 occur, we also have a reactant with a very good leaving group and a nucleophyle, and the product of reaction always result in the fact that not only the leaving group is out of the innitial reactant but also an electrophyle from the reactant is no longer there, and the product form a new pi bond (a double bond or triple bond) and the nucleophile gets the electrophyle that leaves the molecule. Depending of the conditions it can be either E1 or E2 (same conditions as Sn1 and Sn2).

According to this very brief explanation, let's see what we have in the options:

A. Tertiary alkyl halide with ethanol, heating and the product is an alkene.

According to the explanation, we have a tertiary alkyl halide (a carbon with all of it's bond different than hydrogen), reacting with ethanol and heating. The product formed was an alkene. In this case, the halide leaves the molecule forming a carbocation intermediate and the ethanol just react with an electrophyle of the bromide alkyl, in this case, an atom of hydrogen, and then the lone pairs form a double bond giving the alkene. Because of the carbocation intermediate this is a E1 reaction.

B. Tertiary alkyl halide treated with potassium terbutoxide in tert butanol and give an alkene.

This is similar as case A, the difference is that we are doing this in basic medium, therefore, the mechanism involved is either E2 or Sn2. However, the final product is also an alkene, which means that the tert-butoxide substract the electrophyle of the alkyl halide, the lone pairs goes down to the molecule promoving the leaving of the halide to form the alkene. This then, would be an E2 mechanism reaction.

C. is the same as case A so it's a E1 reaction.