Chemistry MCQs for NEET — Practice Questions with Answers

The term pyrolytic elimination literally means an elimination reaction occurring in the organic substrate due to the application of heat (Greek word pyr meaning fire). Pyrolytic eliminations do not require any additional reagent or solvent to occur, and mostly carried out in the gaseous phase though can be performed in inert solvents. Thus, these type of eliminations are different from other types of eliminations (E1,E2 or E1cB) as all other types of elimination reactions require an added base (external base or solvent) to proceed. These thermal elimination reactions of molecules can also be termed as pyrolytic Ei (Elimination internal) reactions. Carboxylic esters, xanthates, amine oxides, sulfoxides and selenoxides are some substrates which commonly undergo pyrolytic eliminations. 

(c) ${\mathrm{CH}}_3\mathrm{CN}\stackrel{\mathrm{Na}/{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}}{\to }$${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{NH}}_2$$\stackrel{{\mathrm{HNO}}_2}{\to }$${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}$$\stackrel{\left[\mathrm{O}\right]}{\to }$${\mathrm{CH}}_3\mathrm{CHO}$$\stackrel{\mathrm{Tollens} \mathrm{reagent}}{\to }$${\mathrm{CH}}_3\mathrm{COOH}$

(a) Molar mass of compound  is 46 which is of ethanol. It also gives iodoform test.                               

(b) Secondary amine is reacted with benzene sulphonyl chloride to form solid compound which is insoluble in alkali

its 2 degree alcohol, which on oxidation gives ketone, and since product will be propanone , so it will give +ve iodoform reaction.

To differentiate between 1,1-dichloroethane and 1,2-dichloroethane, the test involves first treating with aqueous KOH and then with 2,4-dinitrophenylhydrazine (2,4-DNP). 1,2-dichloroethane will give a positive test (orange precipitate) while 1,1-dichloroethane will not react.

Dehydrohalogenation is the process of eliminating hydrogen and halogen atoms from an alkyl halide to form an alkene. The reagents commonly used for this process are alcoholic KOH, sodium amide (NaNH2), and sodium ethoxide (C2H5ONa). Therefore, the correct option is 'all of these'.

The given reaction is an example of addition of hydrogen halide (HX) to an alkene (ethene). According to Markovnikov's rule, the hydrogen atom gets added to the carbon with fewer hydrogen atoms, and the halide (X) gets added to the carbon with more hydrogen atoms. Hence, the product is CH2=CHCHO.