`(CH_(3))_(2)CHBr + MeBr overset(Na//"ether")rarr`[P]. Which product is not expected in this reaction?
(A) Ethane
(B) Propene
(C) 2 -methyl propane
(D) Methane

CH_(3)CH_(2)CH_(2)Cl overset(alc. KOH)rarr B overset(HBr)rarr C overset("Na//ether")rarr D In the above reaction, the product D is

CH_(3) NO_(2) overset(573K) rarr A + B + C In this reaction, the products are

CH_(3)CH_(2)COOH overset("Red P/HI")rarr is overset("alc. KOH")rarr "Product" . Product

How will you convert : (a) Ethane to butane (b) Ethyne to methane (c) Ethene to ethyne (d) Methane to ethane (e) Propene to 2, 3-dimethylbutane (f) Ethane to ethyne (g) Ethyne to but-2-yne

An ethereal solution of an alkyl halide preferably the bromide of iodide, is treated with sodium 2R-X+2Na overset("Ether")toR-R+2NaX 2CH_(3)-Br+2Na overset("Ether")toCH_(3)-CH_(3)+2NaBr CH_(3)CH_(2)Br+2Na+CH_(3)Br overset("Ether")toCH_(3)CH_(2)-CH_(3)+2NaBr In this reaction, the product has new (C-C) bond with the same type of alkyl halide, the product has symmetery and this helps in deciding the nature of reacting halide. Intermediates are free radicals Formation of free radical is easiest in

i. A + 2Na overset ("Dryehter") rarr Me_(2)CH - CJMe_(2) ii. A + 2Na overset ("Dryether") rarr Me_(3)C - Cme_(3) iii. One mole of CH_(3)Br are reacted with 2 mole of sodium metal in dry ether. The productus thus obtained are: a. Ethane + 2NaBr b. Ethane + Propane + 2 NaBr c. Ethane + Propane + Butane + 2 NaBr d. Ethane + Propane + Butane + Ethene + 2 NaBr

overset (Mg//ether)rarr overset (D_(2)O)rarr overset (Na//ether)rarr (X) The compound (X) is: