Question

The compound which is obtained by treating chloropropane with alcoholic $KOH$, when reacts with $BH _{3} / THF$ followed by acetic acid gives

• A. $CH _{3} CH _{2} CH _{2} OH$
• B. $CH _{3} CH _{2} CH _{3}$
• C. $CH _{3} CH ( OH ) CH _{3}$
• D. $CH _{3} CH _{2} CHOHCH _{3}$

Answer 1

Answer: (B)

$\underset{\text { chloropropane }}{ CH _{3} CH _{2} CH _{2} Cl } \xrightarrow[- KCl ;- H _{2} O ]{\text { Alc. } KOH } \underset{\text { propene }}{ CH _{3} CH = CH _{2}}$
$\xrightarrow{ BF _{3} / THF }\left( CH _{3} CH _{2} CH _{2}\right)_{3} B$
$\xrightarrow{ CH _{3} COOH } \underset{\text { propane }}{ CH _{3} CH _{2} CH _{3}}$