Question

Of the following, the oxime of which compound shows geometrical isomerism?

• A. Acetone
• B. Diethyl ketone
• C. Formaldehyde
• D. Benzaldehyde

Answer 1

The correct option is D Benzaldehyde

For benzaldehyde,

$C_6{H}_{5}CHO+N{H}_{2}OH⟶C_6{H}_{5}CN\left(H\right)\left(OH\right)\left(anti\right)+C_6{H}_{5}CN\left(H\right)\left(OH\right)\left(syn\right)$

gives syn and anti benzaldehyde oximes.

For other cases, only one product is formed.

For acetone,

${\left(C{H}_3\right)}_{2}CO+H_{2}NOH⟶{\left(C{H}_3\right)}_{2}CNOH+H_{2}O$

For formaldehyde,

$HCHO+N{H}_{2}OH⟶H_{2}CNOH$

For diethyl ketone,

$C{H}_{3}C{H}_{2}COC{H}_{2}C{H}_3+N{H}_{2}OH⟶C{H}_{3}C{H}_{2}CN\left(OH\right)C{H}_3$