Draw structures showing stereochemistry where necessary for the products you would expect from the following reactions.

Short Answer

Expert verified
In summary, the products for the given cases are: 1. Addition Reaction: 2-bromopropane (CH3CH(Br)CH3) 2. Substitution Reaction: (S)-2-cyanobutane (CH3CH(CN)CH2CH3) 3. Elimination Reaction: trans-2-butene (CH3CH=CHCH3)

Step by step solution

01

Identify the reactants

We will consider the addition of HBr to propene CH3CH=CH2 as an example. The reactants are propene and HBr.
02

Understand the mechanism

This is an electrophilic addition reaction: HBr dissocates into H+ and Br-, and propene forms a bond with H+, generating an intermediate carbocation.
03

Analyze the potential products

There will be two possible carbocations: a secondary carbocation CH3CH(Br)CH3 if the H+ reacts with the outer double bond carbon and a primary carbocation CH3(CH2)CH2+ if the H+ reacts with the inner carbon.
04

Determine the most stable carbocation

A secondary carbocation is more stable than a primary carbocation, so the major product will consist of the following structure: CH3CH(Br)CH3.
05

Draw the product and consider stereochemistry

In this case, stereochemistry doesn't matter since there is no chiral carbon. The product will be 2-bromopropane: CH3CH(Br)CH3. #Case 2: Substitution Reaction - SN2 reaction#
06

Identify the reactants

Let us consider the reaction of sodium cyanide, NaCN, with (R)-2-bromobutane (CH3CH(Br)CH2CH3) as an example.
07

Understand the mechanism

The mechanism of this reaction is SN2, where CN- attacks the chiral carbon from the backside and Br- leaves.
08

Determine the product with stereochemistry

The product will be (S)-2-cyanobutane since the stereochemistry will be inverted at the chiral center. The structure is: CH3CH(CN)CH2CH3. #Case 3: Elimination Reaction - E2 reaction#
09

Identify the reactants

Let's consider the reaction of (S)-2-bromobutane (CH3CH(Br)CH2CH3) with potassium tert-butoxide (KOtBu) as an example.
10

Understand the mechanism

This is an E2 reaction, a concerted process where the base abstracts the beta-hydrogen, and the halogen leaves, forming a double bond in the process.
11

Identify the beta-hydrogens and potential products

There are two possible beta-hydrogens, one on the CH2 between CH(Br)CH2CH3 and another on CH3CH(Br)CH2CH3.
12

Determine the more substituted and stable alkene

The more substituted alkene formed when KOtBu abstracts the beta-hydrogen from CH3CH(Br)CH2CH3 will be the major product.
13

Draw the major product and consider stereochemistry

The product is trans-2-butene where the double bond is formed between CH3 and CH2-CH3. No stereochemistry is needed as there is no chiral center.

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