# Tag Info

## New answers tagged carbonyl-compounds

0

B is the correct answer, but not for the reasons you have given. 4-Me-benzaldehyde cannot self-condense, and is a far more reactive electrophile than acetophenone. Acetophenone has a pKa of around 16 so will readily undergo base-catalysed aldol reaction with the aldehyde. This is the only reaction that will occur. All the other sets of reactants shown are ...

0

For comparing the rate of decarboxylation just compare the stability of the carbanion formed. It's an easy jee concept.

4

Consider the resonance structures of the pyranone. MeMgCl attacks the structure shown second. The perchloric acid workup hydrolyses the enol ether to give the diketone shown bottom right. NaOH promotes the internal aldol cyclisation to give 3,5-dimethylphenol.

3

An aldehyde or a ketone with an alpha hydrogen forms a carbanion that resonates to enolate form. This leads to two canonical structures that are in resonance. The example below is a polyphenol compound extracted from turmeric, called curcumin. source: Biomaterials. 2010 May;31(14):4179-85. doi: 10.1016/j.biomaterials.2010.01.142. Epub 2010 Feb 23. (https://...

0

1.The selective reduction of the carboxylic group is explained in this article. Link:https://www.ch.imperial.ac.uk/rzepa/blog/?p=5114. It could reduce esters but is not favourable at all and difficult.So if u only have an ester grp and its treated with bh3, then you can reduce it. However if both carboxylic and ester grps are present, carboxylic acid gets ...

1

Forget the Beckmann, the oxime is still nucleophilic through N (imagine the N-hydroxy enamine tautomer if that helps). Have it attack the second carbonyl intramolecularly (and hence favoured) to give a dihydropyridine. Dihydropyridines are unstable with respect to aromatisation unless substituted with strongly electron withdrawing groups. The N-hydroxy-...

1

It is because of the fact that a double bond is not usually stable at the bridge head position in smaller fused rings. Therefore the hydrogen cannot be considered as acidic.

1

The first step is called Michael addition. Have a look at this wikipedia page. The Michael reaction or Michael addition is the nucleophilic addition of a carbanion or another nucleophile to an α,β-unsaturated carbonyl compound. A newer definition, proposed by Kohler, is the 1,4-addition of a doubly stabilized carbon nucleophile to an α,β-unsaturated ...

5

I suggest two possible approaches. Narasaka, et al. have effected similar cyclizations photochemically. I prefer to start with the allylic alcohol 1 to avoid complications with oxime formation. Formation of the bis acetate 2b is not a problem. However, given the greater acidity of oximes relative to alcohols does not preclude the formation of oxime acetate ...

1

Here is a short simple available answer provided online by Prachi Sawant in 2016 to quote: Formic acid has both aldehydic (-CHO) and carboxylic (-COOH) functional groups. Hence, it gives both Tollen's and Fehling's test positive. Reaction of formic acid with Tollen's reagent: $\ce{HCOOH + 2[Ag(NH3)2]+ +2OH- → 2Ag + CO2 + 2H2O + 4NH3 }$ ...

0

I think the cinnamaldehyde gives $\ce{FeCl3}$ test due the similar reason why acetic acid can give $\ce{FeCl3}$ test as $\ce{Fe^3+}$ can act as oxidizing agent. As in case of acetic acid $\ce{Fe(CH3COO)3}$ (red colour) is formed, a similar complex may be produced.

2

Wolf-Kishner reduction of ketones and aldehydes proceeds by nucleophilic attack of hydrazine on the carbonyl compound followed by elimination of water to form the intermediate hydrazone mechanism here Nucleophilic attack of hydrazine on esters, acyl halides and anhydrides gives the acyl hydrazide which is a stable species. See the second step of the ...

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