In the following sequence of reactions ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{OH}\stackrel{\mathrm{HBr}}{\to }\mathrm{A}\stackrel{\mathrm{alc}.\mathrm{KOH}}{\to }\mathrm{B}$$\underset{Peroxide}{\overset{H Br}{\to }}C\stackrel{aq.KOH}{\to }D$ The original compound and D are:

Organic Reaction Sequence Analysis

Starting Compound:CH₃CH₂CH₂OH (1-propanol)

Step 1: CH₃CH₂CH₂OH + HBr → A

Reaction Type: Nucleophilic substitution (S_N2)

Product A:CH₃CH₂CH₂Br (1-bromopropane)

The hydroxyl group is replaced by bromine via nucleophilic substitution.

Step 2: A + alc. KOH → B

Reaction Type: Elimination (E2)

Product B:CH₂=CHCH₃ (propene)

Alcoholic KOH causes elimination of HBr to form an alkene.

Step 3: B + HBr (peroxide) → C

Reaction Type: Anti-Markovnikov addition (free radical)

Product C:CH₃CH₂CH₂Br (1-bromopropane)

In the presence of peroxide, HBr adds via anti-Markovnikov rule.

Step 4: C + aq. KOH → D

Reaction Type: Nucleophilic substitution (S_N2)

Product D:CH₃CH₂CH₂OH (1-propanol)

Aqueous KOH replaces bromine with hydroxyl group.

Final Answer:

The original compound and compound D are both 1-propanol (CH₃CH₂CH₂OH).

This reaction sequence demonstrates a complete cycle that transforms 1-propanol through various intermediates and ultimately regenerates the same compound, showcasing different types of organic reactions including substitution, elimination, and addition reactions.