Chapter 17: Problem 375
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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