Q.1 - Organic Chemistry

Question

Label the $\mathbf{\text{α}}$ and $\mathbf{\text{β}}$ carbons in each alkyl halide. Draw all possible elimination products formed when each alkyl halide is treated with ${\mathbf{\text{K}}}^{\mathbf{+}\mathbf{-}}{\mathbf{\text{OC(CH}}}_{\mathbf{3}}{\mathbf{\text{)}}}_3$.

a.

b.

c.

Step-by-Step Solution

Answer

The $\text{α}$ and  $\text{β}$ carbons are marked in the following:

a.

b.   

c.

The possible products obtained after elimination are as follows:

a.

b.

c.

Choosing the α and β carbons

The halide which is directly attached to the carbon is considered as the $\mathbf{\text{α}}$ carbon. While the carbon attached adjacent to the $\mathbf{\text{α}}$ carbon is considered as the $\mathbf{\text{β}}$ carbon.

Identification of the base

The base used here is bulky, so it favors an elimination reaction rather than a nucleophilic substitution reaction. It's easier for a bulky base to abstract a proton than attacking the carbon center where they face steric repulsion.

Choosing β hydrogens for undergoing elimination

For (a), only one type of$\text{β}$ hydrogens is present; hence it gives a single product. For both (b) and (c), two types of $\text{β}$ hydrogens are present. Hence, two different products are obtained.

Representation of hydrogen atoms with alpha and beta carbons in the given compounds

The possible products obtained after elimination are as follows:

a.

b.

The two possible products are obtained from the above two elimination reactions.

c.

The mentioned elimination products yield two possible elimination products.