$Propanal$ on treatment with dilute sodium hydroxide forms:
- A$CH_3CH_2CH_2CH_2CH_2CHO$
- B$CH_3CH_2CH(OH)CH_2CH_2CHO$
- C$CH_3CH_2CH(OH)CH(CH_3)CHO$
- D$CH_3CH_2COONa$
Explore More
Similar Questions
Which of the following will form a stable hydrate?
Medium
View SolutionIdentify the products of the following reaction:
$HCHO + C_6H_5CHO \xrightarrow[ii. \ H_3O^+]{i. \ conc. \ NaOH} ?$
$HCHO + C_6H_5CHO \xrightarrow[ii. \ H_3O^+]{i. \ conc. \ NaOH} ?$
$3$-hydroxybutanal is formed when $(X)$ reacts with $(Y)$ in dilute $(Z)$ solution. What are $X, Y$ and $Z$?
$A$ primary alcohol was reacted with pyridinium chlorochromate $(PCC)$,which resulted in a product $P$. The product $P$ on treatment with ammoniacal silver nitrate solution produces:
$(A)$ and $(B)$ in the following reaction sequence are:
$RR'C=O$ $\xrightarrow{HCN} (A)$ $\xrightarrow{B} RR'C(OH)CH_2NH_2$
$RR'C=O$ $\xrightarrow{HCN} (A)$ $\xrightarrow{B} RR'C(OH)CH_2NH_2$
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo