Explanation
Acidity decreases as the electronegativity of the halogen decreases ,i.e.
FCH2COOH > ClCH2COOH > BrCH2COOH
Hell-Volhard Zelinsky reaction.
Koch reaction
BrCH2CH2COOH is the weakest acid since Br has lower – I-effect ascompared to F and is also far removed from –CO2H group.
Due to – I-effect of the COOH group, H-bonds in acids are muchstronger than in alcohols. Aldehydes, however, do not undergo H-bonding Thus option (b) is correct.
CnH2nO2 represents open carboxylic acids as shown below
RCOOH >HOH >ROH> HC ≡ CH
Since C3H6O2 (B) on soda-lime distillation give C2H6, therefore, (B) musta monocarboxylic acid. Since (B) is obtained from (A) by loss a moleculeof CO2 (C4H6O4 – C3H6O2), therefore, A must be 1, 3-dicarboxylic acid inaccordance with Blanc’s rule. Thus (A) must be CH3 — CH(COOH)2 and(B) must be CH3CH2COOH. Therefore, option (a) is correct
As the size of the alcohol increases, the ease of nucleophilic attack ofthe alcohol on the C = O group of COOH group decreases due to sterichindrance and hence ease of esterification decreases. Thus option (b) iscorrect.
As the + I - effect of the alkyl group increases from HCOOH → CH3COOH→ CH3CH2COOH, , the magnitude of the +Vecharge on the carbonyl group decreases and hence the ease with whichCH3OH attacks the C = O decreases and hence the ease of esterificationdecreases from HCOOH → CH3COOH → CH3CH2COOH Thus, option (a) is correct
Hundiecker reaction occurs by a free radical intermediate.
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