Chemistry MCQs for NEET — Practice Questions with Answers

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 

key idea: the reagent which can convert -CONH2 group into -NH2 group is used for this reaction. among the given reagents only NaOH/Br convert -CONH₂ group into -NH₂ group. Thus, it is used for converting acetamide to methyl amine. This reaction is called Hofmann bromamide reaction.

${\mathrm{CH}}_3{\mathrm{CONH}}_2+\mathrm{NaOH}={\mathrm{Br}}_2^{}\to \underset{\mathrm{methyl} \mathrm{amine}}{{\mathrm{CH}}_3{\mathrm{NH}}_2}+\mathrm{NaBr}+{\mathrm{Na}}_2{\mathrm{CO}}_3+{\mathrm{H}}_2\mathrm{O}$

The given reaction sequence involves: 1) Conversion of phenol to benzene using Zn dust (reduction). 2) Friedel-Crafts acylation of benzene with acetyl chloride (CH3COCl) using anhydrous AlCl3 catalyst to give acetophenone. 3) Oxidation of acetophenone with alkaline KMnO4 to give benzoic acid (Z).

The reduction of methylisocyanide (CH3NC) with lithium aluminium hydride (LiAlH4) yields methylamine, which is a secondary amine. Nitroethane and acetamide yield primary amines, while methylcyanide forms a primary amine on reduction with LiAlH4.

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12.2.2 Role of Macro and Micronutrients

Mineral Nutrition

(d) It is called germplasm collection or the collection of variability. Selection of superior recombinants means selection of the best plant from the whole lot by visual examination and collecting their seeds for growing in field. Cross hybridisation is the method of combining  the characters of different plants together. The selection and evolution are the main steps of hybridisation.

(a) a germplasm is a collection of genetic resources for an organisms. For plants, the germplasm may be stored as a seed collection of all the genes of an organism.