Question

uploaded imageI have the SN2! I need the SN1!

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Answer #1

General guidance

Concepts and reason

The concept used to solve this problem is SN1{{\\rm{S}}_{\\rm{N}}}{\\rm{1}}and SN2{{\\rm{S}}_{\\rm{N}}}{\\rm{2}}. The substrate is a secondary alkyl halide and the nucleophile is a strong base.

In SN1{{\\rm{S}}_{\\rm{N}}}{\\rm{1}} reactions, rearrangement of carbocation takes place. Here, racemic mixture of products will be formed.

In SN2{{\\rm{S}}_{\\rm{N}}}{\\rm{2}}, the attack of nucleophile takes place from the back side, so inversion of configuration takes place in the product.

Fundamentals

SN1{{\\rm{S}}_{\\rm{N}}}{\\rm{1}}reaction is a two-step reaction. The rate of the reaction depends only on the concentration of substrate. It does not depend on the concentration of incoming nucleophile.

SN2{{\\rm{S}}_{\\rm{N}}}{\\rm{2}} reaction is a single step reaction. The rate of the reaction depends on both concentration of substrate and concentration of nucleophile.

Step-by-step

Step 1 of 2

(a)

The reaction is as follows:

: \u043e\u043dBrSNI2-methylpentan-2-ol3-bromo-2-methylpentane

Here, the formed product is 2-methylpentan-2-ol.

Part a

The major product of the reaction proceeds by [{{ m{S}}_{ m{N}}}1]mechanism is as follows:

HO:


Explanation | Hint for next step

Treatment of 3-bromo-2-methylpentane with hydroxide ion gives 2-methyl-2-pentanol. Here, the secondary carbocation is formed by the heterolytic cleavage of C-Br bond. Next, by the 1,2-hydride shift, the secondary carbocation is converted to more stable tertiary carbocation. Now, the hydroxide ion attacks the tertiary carbocation and forms the corresponding product.

Step 2 of 2

(b)

The reaction is as follows:

Br:: \u043e\u043dSN23-bromo-2-methylpentane2-methylpentan-3-ol

Here, the formed product is 2-methylpentan-3-ol.

Part b

The major product of the reaction proceeds by [{{ m{S}}_{ m{N}}}{ m{2}}]mechanism is as follows:

OH


Explanation

Treatment of 3-bromo-2-methylpentane with the hydroxide ion gives 2-methylpentan-3-ol. This reaction proceeds through SN2{{\\rm{S}}_{\\rm{N}}}{\\rm{2}} mechanism. Here, the nucleophile attacks from the back the side of the carbon bearing the leaving group, so, the bromide ion expels out and forms the required product.

Answer

Part a

The major product of the reaction proceeds by [{{ m{S}}_{ m{N}}}1]mechanism is as follows:

HO:

Part b

The major product of the reaction proceeds by [{{ m{S}}_{ m{N}}}{ m{2}}]mechanism is as follows:

OH

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