Question

A.) Identify the one substrate that reacts the fastest when treated with hydrochloric acid at room temperature.
A.) 5-Chloro-2-methylheptan-2-ol
B.) 6-Chloro-6-methylheptan-3-ol
C.) 2-Chloro-2-methylheptan-1-ol
D.) 2-Chloro-2-methylheptan-3-ol

B.) identify the one substrate that reacts the fastest when treated with PBr3.
A.) 4-Methyl-3-penten-1-ol
B.) 3-Methyl-2-buten-1-ol
C.) 6-Methyl-5-hepten-2-ol
D.) 2,4-Dimethyl-2-hepten-4-ol
please help!!!

Answer 1

A. Reaction of alcohols with HCl takes place via loss of OH as H₂O forming a carbocation intermediate followed by addition of a halide to form the substituted chloride product. Thus the substrate forming most stable carbocation will react fastest.

The given alcohols are:

Out of the given substrates, a) forms a tertiary carbocation b) forms a primary carbocation c) forms a primary carbocation and d) a secondary carbocation. Out of Substrates in part b) and c), the one in c) contains a Cl near to the +ve charge which destabilises it by -I effect. Hence the order of stability becomes:

a) > d) > b) > c)

Therefore, 5-Chloro-2-methylheptan-2-ol reacts fastest with HCl at room temperature.

B. Alcohols on reaction with PBr₃ form alkyl bromide. In the first step SN₂ reaction of 3 equivalents of alcohol and PBr₃ occurs to form phosphite ester with 3 P-O bonds. Bromide ion then attacks via SN₂ mechanism forming alkyl bromide. Thus in this case stearically less hindered alcohol will react fastest.

The given alcohols are:

From the above structures it is clear that alcohol in part a) is leat hindered. Hence, 4-Methyl-3-penten-1-ol will react fastest with PBr₃.