Predict the product from reaction of the following substance (reddish brown=Br) with:

(a)${\mathbf{\text{PBr}}}_{\mathbf{3}}$

(b) Aqueous${\mathbf{\text{H}}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{4}}$

(c)${\mathbf{\text{SOCl}}}_{\mathbf{2}}$

(d)Dess-Martin periodinane

(e)${\mathbf{\text{Br}}}_{\mathbf{2}}\mathbf{ }\mathbf{,}\mathbf{}\mathit{F}\mathit{e}\mathit{B}{\mathit{r}}_{\mathbf{3}}$

(a) When a given substance reacts with ${\text{PBr}}_3$, the secondary alcohol is converted into alkyl bromide. The ${\text{S}}_{N}2$mechanism follows where nucleophile OH attacks the phosphorus of and one Br leaves. The –OH group is replaced by a bromine atom and is present above the plane. Thus, we get an inverted chirality product.

(b) When a given substance reacts with aqueous ${\text{H}}_{2}S{O}_4$, dehydration of alcohol occurs, where ${\text{H}}_{2}S{O}_4$is an acid catalyst. H and OH are removed, and we get alkene as a product. In this reaction, one major and one minor product are formed.

(c) When a given substance reacts with ${\text{SOCl}}_2$, alkyl chloride is formed as a product. The reaction proceeds through a backside attack of the nucleophile on ${\text{SOCl}}_2$by ${\text{S}}_{N}2$mechanism and inversion of configuration at that carbon are seen in the product.

(d) When a given substance is treated with Dess-Martin periodinane, secondary alcohol is oxidized to the ketone. Thus, the product obtained is shown below,

(e) When a given substance is treated with ${\text{Br}}_2 andFeB{r}_3$, bromination occurs. The two products are formed, one is a para isomer, and the other is an ortho isomer. The alcohol group remains the same because the tribromide ion formed during the reaction does not react with alcohol.