Advertisements
Advertisements
Question
Out of CH3CH2 – CO – CH2 – CH3 and CH3CH2 – CH2 – CO – CH3, which gives iodoform test?
Advertisements
Solution 1
CH3CH2 – CH2 –CO –CH3
Solution 2
Pentan-2-one (CH3-CH2-CH2-CO-CH3) give yellow precipitate of CHI3 with NaOI, thats means it gives iodoform test.

Pentan-3-one (CH3-CH2-CO-CH2-CH3) does not give yellow precipitate with CHI3 with NaOI, so Pentan-3-one does not give iodoform test.
APPEARS IN
RELATED QUESTIONS
Propanal and Propanone
Oxidation of ketones involves carbon-carbon bond cleavage. Name the products formed on oxidation of 2, 5-dimethylhexan-3-one.
Which sugar does not reduce Fehling's solution?
Acetone and acetaldehyde are differentiated by
Fehilng's test is positive for
What is the composition of Fehling's reagent?
The correct set of products obtained in the following reactions:
- \[\ce{RCN ->[reduction]}\]
- \[\ce{RCN ->[(i) CH3MgBr][(ii) H2O]}\]
- \[\ce{RNC ->[hydrolysis]}\]
- \[\ce{RNH2 ->[HNO2]}\]
In Tollen's test for aldehyde, the overall number of electrons(s) transferred to the Tollen's reagent formula \[\ce{[Ag(NH3)2]+}\] per aldehyde group to form silver mirror is ______. (Round off to the nearest integer)
An organic compound 'A' with molecular formula C5H8O2 is reduced to n-pentane with hydrazine followed by heating with NaOH and glycol. 'A' forms a dioxime with hydroxylamine and gives a positive iodoform and Tollen's test. Identify 'A' and give its reaction for iodoform and Tollen's test.
An organic compound 'A' with the molecular formula C4H8O2 undergoes acid hydrolysis to form two compounds 'B' and 'C'. Oxidation of 'C' with acidified potassium permanganate also produces 'B'. Sodium salt of 'B' on heating with soda lime gives methane.
- Identify 'A', 'B' and 'C'.
- Out of 'B' and 'C', which will have higher boiling point? Give reason.
