Question

Draw the neutral organic products.

Answer 1

General guidance

Concepts and reason

S_N1 (Unimolecular reaction): This reaction proceeds into two steps. The first step is the formation of carbocation, and the second step is the nucleophile attack on the positive charge. During the formation of product, hybridization of carbocation will change the first sp³ (tetrahedral) to sp² (planar), then it will change again to sp³ hybridization.

Elimination reaction: Organic reactions involve the removal of a group of atoms from a molecule by breakage of bond between them, and this leads to the formation of unsaturation.

Elimination process may occur in two ways as follows:

1.${{\\rm{E}}_{\\rm{1}}}$elimination process

2.${{\\rm{E}}_{\\rm{2}}}$elimination process

Fundamentals

Reactivity order of the reaction is given as follows:

${\\rm{tert - alkyl}}\\,{\\rm{ > }}\\,{\\rm{sec - alkyl}}\\,{\\rm{ > }}\\,{\\rm{pri - alkyl}}$

Order of leaving group is given as follows:

${\\rm{R - X:}}\\,{\\rm{R - I}}\\,{\\rm{ > }}\\,{\\rm{R - Br}}\\,{\\rm{ > }}\\,{\\rm{R - Cl}}\\,{\\rm{ > }}\\,{\\rm{R - F}}$

Since it will form a stable carbocation, it has a good leaving group. Therefore, it will follow S_N1 reaction mechanism.

Br SNI carbocation intermediate Step-II tert-butyl alcohol

${{\\rm{E}}_{\\rm{1}}}$Elimination:${{\\rm{E}}_{\\rm{1}}}$ stands for unimolecular elimination. It is a two-step process. The two steps are as follows:

\u2022 Ionization

\u2022 Deprotonation.

During ionization, the carbon\u2013halogen bond breaks to give a carbocation intermediate. It is followed by deprotonation of the carbocation. It usually occurs in tertiary alkyl halides. It takes place in the complete absence of a base or in the presence of a weak base.

Order of reactivity for alkyl halides in E1 reaction:

For a ${{\\rm{E}}^{\\rm{1}}}$reaction, the order of reactivity is as follows:${3^o} > {2^o} > {1^o}$.

${{\\rm{E}}_{\\rm{1}}}$elimination could generally be represented as follows:

Stepl Step2 . Base

Step-by-step

Step 1 of 2

The possible mechanisms of the reaction can be written as follows:

SN1 pathway Step 1 : 0 -\u0397 OH Step 2 Sy1 product

E1 pathway Step 1 Step 2 \u043e\u043d. E1 product

Explanation | Common mistakes | Hint for next step

The given compound, cyclohexyl bromide, is a secondary alkyl halide. It can undergo both substitution and elimination reactions. The first step in the reaction is the formation of a carbocation. In the second step, either substitution or elimination occurs. Thus, two different products are expected to be produced.

Step 2 of 2

The products of the reaction are shown below:

OH

The neutral organic products are shown below:

Br_O.OH

The given product, cyclohexyl alkyl halide, on reaction with a base, hydroxide ion, produces two products. In the first step, the carbocation is formed. When water is added, cyclohexanol is formed as the substitution product and eliminates hydrogen to produce cyclohexene, which is the elimination product.

Answer

The neutral organic products are shown below:

Br_O.OH

Answer only

The neutral organic products are shown below:

Br_O.OH