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Explain : (a) An aq. Solution to tropl...

Explain :
(a) An aq. Solution to troplylium bormide `(C_7 H_7Br)` on treatment with `AgNO_3` gives a pale yellow precipitate.
(b) Cycloheptatrienyl cation has a low `pi` -electron energy than its open-chain counterparts.
(c) Cyclo-octatetraene reacts with 2 mol of potassium to yield a stable compound.

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`(a),(b)`, and `(c)` are all examples of aromatic compounds.

(b) It is aromatic compound that is why it has low energy.
(c) .
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An optically active compound A (assume dextrorotatory) has the molecular formula C_(7)H_(11)Br . A reacts with hydrogen bromide, in the absence of peroxides to yield isomeric products, B and C, with the molecular formula C_(7)H_(12)Br_(2) . Compound B is optically active, C is not. Treating B with 1 mol of potassium butoxide yields (+)-A . Treating C with 1 mol of potassium tert-butoxide yields (+-)-A . Treating A with potassium tert-butoxide yields D(C_(7)H_(10)) . Subjecting 1 mol of D to ozonolysis followed by treatement with zinc and acetic acid yields 2 moles of formaldehyde and 1 mole of 1,3-cyclopentanedione. The compound 'B' is:

An optically active compound A (assume dextrorotatory) has the molecular formula C_(7)H_(11)Br . A reacts with hydrogen bromide, in the absence of peroxides to yield isomeric products, B and C, with the molecular formula C_(7)H_(12)Br_(2) . Compound B is optically active, C is not. Treating B with 1 mol of potassium butoxide yields (+)-A . Treating C with 1 mol of potassium tert-butoxide yields (+-)-A . Treating A with potassium tert-butoxide yields D(C_(7)H_(10)) . Subjecting 1 mol of D to ozonolysis followed by treatement with zinc and acetic acid yields 2 moles of formaldehyde and 1 mole of 1,3-cyclopentanedione. The optically active compound 'A' is

An aromatic compound 'A' (Molecular formula C_(8)H_(8)O) ) gives positive 2, 4-DNP test. It gives a yellow precipitate of compound 'B' on treatment with iodine and sodium hydroxide solution. Compound 'A' does not give Tollen's or Fehling's test. On drastic oxidation with potassium permanganate, it forms a carboxylic acid 'C' (Molecular formula C_(7)H_(6)O_(2) ), which is also formed along with the yellow compound in the above reaction. Identify A, B and C and write all the reactions involved.

An organic compound A, C_(8)H_(6) on reacting with dil. H_(2)SO_(4) and HgSO_(4) gives a compound 'B'-which can also be obtained from reaction of benzene with acid chloride in the presents of anh. AlCl_(3) Compound B when reacted with iodine and aq. NaOH yields C and a yellow compound D. Identify A to D with proper justification.

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