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Draw a stepwise mechanism for the following reaction that illustrates how two substitution products are formed. Explain why 1-bromohex-2-ene reacts rapidly with a weak nucleophile (CH3OH) under reaction conditions, even though it is a 1°alkyl halide.

Short Answer

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Answer

The 1-bromo-hex-2-ene is a primary alkyl halide, but it reacts with the weak nucleophile under the SN1 reaction conditions because of the formation of the resonance stabilized carbocation.

The two substitution products are formed by the following reactions:

The nucleophilic attack on carbon 1
The nucleophilic attack on carbon 2

Step by step solution

01

Mechanism of the  SN1 reactions

The SN1 mechanism involves the following two steps:

  • The formation of a carbocation intermediate: The leaving group moves from the alkyl halide and leads to the formation of an intermediate carbocation.
  • The carbocation intermediate is planar in structure, due to which the nucleophile can attack either side of the carbocation. This results in the formation of a racemic mixture of the products.

The SN1 reactions are always favored by the polar protic solvents such as water, ethanol, etc.

02

The stepwise mechanism of the given reaction

The reaction of 1-bromo-hex-2-ene with ethanol takes place in the following way:

The above reaction proceeds in the following steps:

  • The removal of the leaving group: The bromide ion leaves the alkyl halide and results in the formation of a carbocation which is resonance stabilized.

Formation of resonance stabilized carbocation

The 1-bromo-hex-2-ene is a primary alkyl halide, but it reacts with the weak nucleophile under the SN1 reaction conditions because of the formation of the resonance stabilized carbocation.

  • The attack of the nucleophile: The attack of the nucleophile can result in the formation of two substitution products because the nucleophile is able to attack two different carbon atoms.

The nucleophilic attack on the carbon atoms is shown below:

The nucleophilic attack on carbon 1

The nucleophilic attack on carbon 2

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Most popular questions from this chapter

Question: What happens to the rate of an SN2 reaction under each of the following conditions?

a. [RX] is tripled, and:Nu- stays the same.

b. Both [RX] and role="math" localid="1648206216789" :Nu-are tripled.

c. [RX] is halved, androle="math" localid="1648206048742" :Nu-stays the same.

d. [RX] is halved, and role="math" localid="1648206067374" :Nu-is doubled.

Question: Consider the following SN2 reaction.

  1. Draw a mechanism using curved arrows.
  2. Draw an energy diagram. Label the axes, the reactants, products, Ea , and ∆H°. Assume that the reaction is exothermic.
  3. Draw the structure of the transition state.
  4. What is the rate equation?
  5. What happens to the reaction rate in each of the following instances? [1] The leaving group is changed from Br-to I-;[2] The solvent is changed from acetone to CH3CH2OH ; [3] The alkyl halide is changed from CH3(CH2)4Br to CH3CH2CH2CH(Br)CH3 ; [4] The concentration of CN- is increased by a factor of five; and [5] The concentrations of both the alkyl halide and CN-are increased by a factor of five.

Question: Classify each solvent as protic or aprotic

a.

b.

c.

Question: Draw the eight constitutional isomers having the molecular formula C5H11Cl .

  1. Give the IUPAC name for each compound (ignoring R and S designations).
  2. Classify each alkyl halide as 1°,2°or3°.
  3. Label any stereogenic centers.
  4. For each constitutional isomer that contains a stereogenic center, draw all possible stereoisomers, and label each stereogenic center as R or S.

Question: Draw the products of each nucleophilic substitution reaction.


b.

c.

d.

e.

f.

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