JEE Questions for Chemistry Aldehydes Ketones And Carboxylic Acids Quiz 11 - MCQExams.com

1, 3, 5-Trimethylbenzene is called mesitylene

 Phorone 

2-Pentanone (CH₃COCH₂CH₂CH₃) contains CH₃CO group and hence gives positive iodoform test.

I₂ and Na₂CO₃ react with acetophenone (C₆H₅COCH₃) to give yellow ppt. of CHI₃ but benzophenone (C₆H₅COC₆H₅) does not and hence can be used to distinguish between them.

2-Methylpropanal (CH₃)₂ CHCHO contains one α− H and hence undergoes aldol condensation.

Acetaldehyde reacts only with nucleophiles.

Aromatic aldehydes (i.e, benzaldehyde etc.) do not reduce Fehling’s solution. Maltose it is a reducing disaccharide since one of its glucose units has a free CHO group and hence reduces Fehling’ solution.

As the size of the alkyl groups around CO group increases, Crowding increases and hence reactivity decreases, i.e,
CH₃CHO > (CH₃)₂CO > C₂H₅COCH₃

Since the compound does not give a precipitate with 2, 4-dinitrophenylhydrazine, it cannot be an aldehyde or a ketone. Further since it does not react with metallic sodium, therefore, it cannot be an alcohol. Thus, the compound C₃H₆O must be an ether, i.e., option (d) is correct.

Wurtz reaction cannot be used to convert RCOR → RCH₂R

In Rosenmund’s reaction, acid chlorides are reduced to aldehydes

Br₂ in CCl₄ neither reacts with hexanal nor with 2-hexanone and hence cannot be used to distinguish between them.

Dihydrogen sodium phosphate (NaH₂PO₄) does not have a lone pair of electrons on the P atom. As such it cannot act as a nucleophile and hence does not react with aldehydes and ketones.

Since the compound with M.F C₃H₆O did not give a silver mirror with Tollen’s reagent but gives an oxime with hydroxylamine, therefore, it must be a ketone, i.e. CH₃ - CO - CH₃.

Benzaldehyde does not contain an α −hdrogenandhencedoesnot undergo aldol condensation

With conc. H₂SO₄, three molecules of acetone condense to form mesitylene.

Only 2,2-dimethylpropanal does not contain an α-hydrogen and hence undergoes Cannizzaro reaction.

Benzophenone (C₆H₅COC₆H₅₎ does not contain the grouping CH₃CO or CH₃CHOH- linked to carbon and hence does not give iodoform test.

Aldehydes which contain a α-hydrogen on a saturated carbon, i.e, CH₃CH₂CHO undergo aldol condensation. 

Although CH₂ = CH − CHO also contains an α-hydrogen atom but cannot be easily abstracted by a base to form a carbanion and hence does not undergo aldol condensation.