Which of the following gives condensation with hydroxylamine but does not undergo self-condensation?

  • A
    Methanal
  • B
    Propanal
  • C
    Acetone
  • D
    Ethanal

Explore More

Similar Questions

The product formed in Aldol condensation is

$2-$pentanone can be distinguished from $3-$pentanone by the reagent:

Difficult
View Solution

$2(CHO-COOH) \xrightarrow{NaOH} CH_2OH-COOH + HOOC-COONa$. This reaction is:

In the following table,Column-$I$ shows the structure of a compound,Column-$II$ shows its $IUPAC$ name,and Column-$III$ shows its common name. Match the $IUPAC$ name and common name for each structure given in Column-$I$.
| Column-$I$ (Structure) | Column-$II$ ($IUPAC$ name) | Column-$III$ (Common name) |
| :--- | :--- | :--- |
| $(A)$ $C_6H_4(CHO)_2$ (ortho) | $(i)$ Benzophenone | $(p)$ Acrolein |
| $(B)$ $(CH_3)_2C=CHCOCH_3$ | $(ii)$ Benzene$-1,2-$dicarbaldehyde | $(q)$ Diphenyl ketone |
| $(C)$ $CH_2=CH-CHO$ | $(iii)$ Prop$-2-$enal | $(r)$ Phthaldehyde |
| $(D)$ $(C_6H_5)_2CO$ | $(iv)$ $4-$methylpent$-3-$en$-2-$one | $(s)$ Mesityl oxide |

$Ph-CH_2-CN \xrightarrow[(1) EtONa]{(2) CH_3-COCl, (3) H_3O^{\oplus}/\Delta} (P)$; Product $(P)$ of the reaction will be

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo