Question

Ethyl methyl ketone on reduction with zinc amalgam and Conc. HCl gives:

• A. Sec. alcohol
• B. $1^{\circ }$ alcohol
• C. $3^{\circ }$ alcohol
• D. n -butane

Answer 1

Answer: (D)

$C_2{H}_5-\stackrel{\underset{||}{O}}{C}-C{H}_3\underset{\text{Conc.HCl}}{\overset{\text{Zn-Hg}}{\to }}{C}_2{H}_{5}C{H}_{2}C{H}_3$