Question

How are the following conversions brought?
i.benzyl chloride to benzyl alcohol
ii. benzyl alcohol to benzoic acid

Answer 1

Hint: The concept of nucleophilic substitution reaction as well as the oxidation of organic compounds to form aldehydes/ketones or carboxylic acids is to be used in this question.

Complete Solution :
- In order to answer this question, we need to learn about the nucleophilic reaction as well as the oxidation reaction, to form acids. Nucleophilic substitution bimolecular or ${{S}_{N}}2$ is a single step bimolecular reaction in which the incoming nucleophile attacks the C-atom of substrate in a direction opposite to the outgoing nucleophile. The reaction passes through a transition state in which both the incoming and outgoing nucleophiles are bonded to the same C-atom. In the transition state, the C-atom is $s{{p}^{2}}$ hybridised with a p-orbital whose one lobe overlaps with an orbital of incoming nucleophile and the other lobe overlaps with an orbital of outgoing nucleophile. The three non-reacting atoms or groups attached to the C-atom are nearly coplanar at an angle of ${{120}^{0}}$. The reaction is completed when the outgoing nucleophile leaves with the bond pair of electrons and simultaneously the incoming nucleophile binds to the C-atom. As the reaction progresses, the configuration of the C-atom under attack is inverted. An ${{S}_{N}}2$ reaction is always accompanied by inversion of configuration. The inversion in configuration implies change in configuration from R to S or S to R (provided the incoming nucleophile and outgoing nucleophile have same priority) and not from (+) to (-) or (-) to (+).
- Primary alcohols are easily oxidised to carboxylic acids with oxidants such as $KMn{{O}_{4}}$, in neutral, acidic or alkaline media or by $Cr{{O}_{3}}/{{K}_{2}}C{{r}_{2}}{{O}_{7}}$ in acidic media. Now, let us come to the question and see the reactions. In (i), an ${{S}_{N}}2$ reaction is taking place.

Hence with the above reagents, we obtain the desired products.

Note: Steric hindrance plays a very vital role in an ${{S}_{N}}2$ reaction. As steric hindrance increases the rate of ${{S}_{N}}2$ reaction decreases. Thus for the same halogen the reactivity order of alkyl halides towards ${{S}_{N}}2$ reaction is as under.
- Methyl halide > primary Alkyl halide > secondary Alkyl halide > tertiary Alkyl halide