Aldehyde and Ketone MCQ Questions & Answers in Organic Chemistry | Chemistry

In keto-enol tautomerism keto form should have $$\alpha $$ - hydrogen ( structure I and II ).

Here, $$\gamma - H$$  participates in tautomerism.

Recall that, esters react with excess of Grignard reagents to form $${3^ \circ }$$ alcohols having at least two identical alkyl groups corresponding to Grignard reagent.

Since here Grignard reagent is \[C{{H}_{3}}MgBr,\]   the $${3^ \circ }$$ alcohol should have at least two methyl groups
Thus, the choice with at least two methyl groups at the carbon linked with \[-OH\]  group will be the correct choice. Hence (A) is the correct choice.

No explanation is given for this question. Let's discuss the answer together.

When $$\alpha $$ - hydrogen is absent in carbonyl group, those compound gives cannizaro reaction.
This reaction show disproportionation.
The oxidation product is salt of carboxylic acid and reduced product is alcohol.
\[HCHO+HCHO\xrightarrow{KOH\left( conc. \right)}\underset{\text{Methyl alcohol}}{\mathop{C{{H}_{3}}OH}}\,+HCO{{O}^{-}}{{K}^{+}}\]

\[Zn\left( Hg \right),HCl\]    cannot be used when acid sensitive group like \[-OH\]  is present, but \[N{{H}_{2}}N{{H}_{2}},O{{H}^{-}}\]   can be used.

Since $$Y$$ produces silver mirror with ammoniacal silver nitrate, $$Y$$ must be an aldehyde. $$Y$$ is produced by oxidation of $$X$$ with acidified $${K_2}C{r_2}{O_7},$$   which shows that $$X$$ must be a primary alcohol.

Paraldehyde is used as hypnotic and sleep producing drug.

Aldehydes and ketones containing $$\alpha - H$$  atoms undergo aldol condensation in presence of dilute alkali as catalyst and gives $$\alpha ,\beta $$  unsaturated compound with the elimination of $${H_2}O$$  molecule.