Acidity decreases as the –I effect of the group decreases, F is the most electronegative atom and hence it has highest –I effect among the halogens.
FCH 2 COOH > ClCH 2 COOH > BrCH 2 COOH > CH 3 COOH
Acidity decreases as the –I effect of the group decreases, F is the most electronegative atom and hence it has highest –I effect among the halogens.
FCH 2 COOH > ClCH 2 COOH > BrCH 2 COOH > CH 3 COOH
The aldehydes or ketones containing -H atom in presence of dilute alkali undergo self condensation reaction to form -hydroxyaldehyde or -hydroxyketone.
This reaction is known as Aldol condensation.
Aldehydes are more reactive than ketones due to +I effect of –CH 3 group.
There are two –CH 3 group in acetone which reduces +ve charge density on carbon atom of carbonyl group.
More hindered carbonyl group too becomes less reactive.
So in the given case CH 3 CHO is the right choice.
ROH + PCl 5 RCl + POCl 3 + HCl RCOOH + PCl 5 RCOCl + POCl 3 + HCl
Carbonyl compounds (aldehydes and ketones) are obtained by the oxidation of 1° and 2° alcohols respectively.
Among the given options, only (a) is 2° alcohol hence it can be oxidized to ketone.
They are resonating forms because the position of the atomic nuclei remain the same and only electron redistribution has occurred.
Tollen's reagent is used to detect aldehydic group.
Tollen's reagent is an ammonical solution of silver nitrate.
When aldehyde is added to Tollen's reagent, silver oxide is reduced to metallic silver which deposits as mirror.
RCHO + Ag 2 O RCOOH + 2Ag
Glucose reduces Fehling solution because glucose has free –CHO group which is readily oxidised.
Here H 2 O is continuously removed from reaction so that reaction always proceeds to product formation i.e., Ethyl benzonate.
Statement I is true But it consume 3 moles of CH3MgBr. So Statement II is false.