Question

For $3 - {{methyl}} - 2 - {{butanol}}$ on treatment with $\xrightarrow {{dryHCl}}$gives predominantly:
A.$2 -$chloro$2 -$methylbutane
B.$2 -$chloro$2 -$methylbutane
C.$2 -$ $2 -$dimethylbutane
D.None of the above.

Answer 1

To attempt this question first draw the structure of $3 -$methyl\ [- 2 - ]butanol and treat it with hydrochloric acid or$HCl$. The product will be a carbocation, after getting the product we have to look for the stability of carbocation and rearrange it accordingly.

Complete step by step answer:

The mechanism of this reaction involves $4$ steps
Step: 1
Protonation: is also known as hydro nation is addition of proton or a hydrogen cation to an atom, molecule. This forms a conjugate acid of that particular compound. In this question the $3 -$Methyl\ [- 2 - ]butanol accepts a pair of protons which gets attached to the $'OH'$ ion.
Step: 2
Now the formation of carbocation takes place. Carbocation is a molecule in which a carbon atom has $3$ positive charges. This is also known as carbonium ion. It is divided into $3$ groups, primary, secondary and tertiary. After the attachment of the proton to the $'OH'$ ion a secondary carbocation is formed by the elimination of water ${H_2} O$.
Step: 4
After the elimination of water, a hydride shift takes place. Hydride shift is a type of rearrangement in a carbocation. In this type of shift the hydrogen atom in a carbocation migrates to the carbon atom which has the formal charge of $+ 1$(carbon$2$) from an adjacent carbon (carbon$1$).
Step: 5
After the hydride shift the tertiary carbocation is formed which is more stable. Now the $'Cl'$ molecule attacks on the tertiary carbocation which is also known as the nucleophilic attack.
The answer is A. $2 -$ $2 -$chloro$2 -$methylbutane

Note:
Unimolecular substitution reactions are favoured when tertiary substrates such as 3-methyl2-butanol are reacted with a good nucleophile such as chlorine ion under protic aprotic conditions that is in acidic medium.