Explanation
Reactivity decreases as the magnitude of +ve charge on the carbonyl carbon decreases or the steric hindrance in the intermediate increases, i.e., H2C=O > RCHO > ArCHO > R2C=O > Ar2C=O
Since compound Y on treatment with I2 and Na2CO3 gives CHI3, therefore, it must contain the grouping CH3CO-,Since compound Y is obtained from X by oxidation with K2Cr2O7,therefore,X must contain the grouping CH3CHOH–. Thus option (d) is correct.
HCHO does not contain α-hydrogen and hence does not undergo aldol condensation
Luca’s test is characteristic of alcohols. Therefore acetaldehyde does not give this test.
Optimum pH for oxime formation is around 3.5.Therefore, option (b) with a pH value of 4.5 is correct.
HCHO is most reactive towards nucleophilic addition reactions.
As the + I - effect of the CH3 group increases and steric hindrance in the tetrahedral intermediate increases, the reactivity decreases,
i.e., HCHO (I) > CH3CHO (II) > CH3COCH3 (III)
2HCHO + NaOH → HCOONa + CH3OH
CH3CHO contains α −hydrogens and hence does not undergo Cannizzaro
Formaldehyde which does not contain α − hydrogen does not undergo aldol condensation
Acetaldehyde does not undergo Cannizzaro reaction since it contains Hα− atoms while formaldehyde, trimethylacetaldehyde and benzaldehyde undergo Cannizzaro reaction since they do not contain α− hydrogen atoms
I (Etard reaction)-C-(chromyl chloride ), II (hydroxylation )-D (dilute alkaline KMnO4), III.(dehydro-halogenation)–A (alcoholic KOH ), IV. (Friedel-Crafts reaction)-B (anhydrous AlCl3) .Thus, option (c) is correct.
I - C (Clemmensen reduction), II - D (Cannizzaro reaction), III-A (Friedel-Crafts reaction). IV-B (Kolbe’s reaction).Thus, option (c) is correct.
With Fehling’s solution, red ppt. of Cu2O are produced
Wolff-Kishner reduction.
Due to electron-withdrawing nature of NO2 group, the partial +ve charge on the carbon atom of the C = O group in p-nitrobenzaldehyde increases and hence becomes more susceptible to nucleophilic attack by the CN- ion.
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