NEET Practice Questions (MCQs) with Answers & Solutions

Propyne (CH≡C-CH3) undergoes ozonolysis, where ozone (O3) cleaves the carbon-carbon triple bond, forming a molozonide intermediate. The molozonide then rearranges and undergoes reductive workup to give methyl glyoxal (CH3COCHO) as the final product.

Propyne (CH≡C-CH3) undergoes oxidation with selenium dioxide (SeO2) via the formation of a selenoacetal intermediate. The selenoacetal then rearranges to give methyl glyoxal (CH3COCHO) as the final product.

$\left(\mathrm{b}\right) {\mathrm{CH}}_2{\mathrm{BrCH}}_2\mathrm{Br}\underset{\mathrm{dust}}{\overset{\mathrm{Zn}}{\to }}{\mathrm{CH}}_2={\mathrm{CH}}_2 + {\mathrm{ZnBr}}_2$

(a) $\mathrm{CH}\equiv \mathrm{CH} + 2\mathrm{HCHO} \stackrel{ }{\to }{\mathrm{HOH}}_2\mathrm{CC}\equiv {\mathrm{CCH}}_2\mathrm{OH}$

CH₃Br $\underset{-\mathrm{KBr}}{\overset{\mathrm{KCN}}{\to }}$ CH₃CN$\stackrel{{\mathrm{H}}_3{\mathrm{O}}^{+}}{\to }$CH₃COOH $\underset{\mathrm{ether}}{\overset{{\mathrm{LiAlH}}_4}{\to }}$CH₃CH₂OH

methyl            ethane         Ethanoic acid         Ethanol

bromide         nitrile 'A'           'B'                      'C'

CH₃CHO and C₆H₅CH₂CHO both being aliphatic aldehydes react with Tollen's reagent, Fehling solution and Benedict solution. So, these reagents cannot be used to distinguish them. CH₃CHO reacts with NaOH and I₂ to give yellow crystals of iodoform while C₆H₅CH₂CHO does not react with it.

$$\begin{gathered} {\mathrm{CH}}_3\mathrm{CHO}+3{\mathrm{I}}_2+4\mathrm{NaOH}\to {\mathrm{CHI}}_3+\mathrm{HCOONa}+3\mathrm{NaI}+3{\mathrm{H}}_2\mathrm{O} \\ {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2\mathrm{CHO}+{\mathrm{I}}_2+\mathrm{NaOH}\to \mathrm{No} \mathrm{reaction} \end{gathered}$$

Thus, CH₃CHO and C₆H₅CH₂CHO