Title HYDROCARBONS A-7.pdf ✅

CLASS : XIIth SUBJECT : CHEMISTRY DATE : DPP NO. : 7 1 (a) Lindlar’s catalyst is Pd ― CaCO3deactivated by lead acetate. Cram et. al gave a better catalyst for this purpose as Pd ― BaSO4 poisoned by quinolene. This too is sometimes referred as Lindlar’s catalyst. 2 (b) Remember this value. 3 (a) The aldehydes formed are oxidized by H2O2formed during hydrolysis. 5 (c) The acidic nature is H2O > C2H2 > NH3; thus, conjugate base order will be OH ― > C2H ― > NH― 2 . 8 (d) ‘X’ is a three carbon compound with two halogen atom, so its molecular formula is C3H6Cl2. Only terminal alkynes give red ppt. with ammoniacal Cu2Cl2, so the hydrocarbon produced by the reaction of ‘X’ with alc. KOH, must be a terminal alkyne (i.e., CH3C ≡ CH). C3H6Cl2 Alc.KOH CH3C ≡ CH Amm.Cu2Cl2 CH3C ≡ CCu↓ red ppt. Compound (X) gives an aldehyde when reacts with aqueous KOH. This suggests that both the halogens are present on same terminal carbon atom. Thus, the formula of compound (X) is and the reactions are as follows CH3C ≡ CH Ammoniacal Cu2Cl2 CH3C ≡ CCu↓ red ppt. CH3 CH2 CH Cl Cl (1, 1-dichloropropane) CH3CH2CH Cl Cl Alc.KOH 'X' Topic :-HYDROCARBONS SOLUTION S
9 (a) Removal of H2O is called dehydration. 10 (d) Both vegetable and animal matter are origin of petroleum. 11 (d) All are used in drying alkanes. 12 (b) The stability order is: Staggered>skew>eclipsed 13 (d) Cyclic hydrocarbon, with carbon-carbon bond length between1.34Å and 1.54Å, is benzene in which due to resonance, C ― C, bond length is 1.39Å (ie.,between 1.34Å ― 1.54Å). Benzene is a hexagonal molecule with bond-angle equal to120°. 14 (c) The reaction proceeds via carbocation mechanism. 15 (c) Copper and silver alkylides are obtained by passing to alkynes in the ammoniacal solution of cuprous chloride and silver nitrate respectively. These reactions are used for detecting the presence of acetylenic hydrogen atom. So, alkanes and alkenes remain unaffected. CH3CH2CH Cl Cl Aq.KOH (X) 1, 1-dichloropropane CH3CH2CH OH OH unstable - H2O CH3CH2CHO propanal HC CH + Cu2Cl2 + 2NH4OH acetylene ammoniacal cuprous chloride Cu.C C.Cu + 2NH4Cl + 2H2O copper acetylide (red ppt.) CH3CH2OH H2SO4 -H2O CH2 CH2 ; C6H5 C CH2 +H + C6H5 C CH3 CH3 CH3 + H2O C6H5 C -H + CH3 + OH2 C6H5 C OH CH3 CH3 CH3
16 (b) Benzene reacts with chlorine in presence of sunlight to give gammexane or benzene hexa chloride. C6H6 +3Cl2 Sunlight C6H6Cl6 17 (a) Hydrogenation in presence of Pd and BaSO4 as syn addition and with Na and liquid NH3 at 200 K is anti addition (trans compounds are formed.) 18 (c) In benzene all the six carbon atoms are sp 2 hybridised. Out of these three sp 2 hybrid orbitals of each C-atom, two orbitals overlap with sp 2 hybrid orbitals of adjacent C-atoms to form six C ― C single bonds. The remaining sp 2 orbital of each C-atom overlaps with s- orbitals of each H-atom to form six C ― H single sigma bonds. Each C-atom is now left with one unhybridised p-orbital perpendicular to the plane of the ring. 20 (b) Benzophenone (diphenyl ketone) can be prepared by the Friedel-Crafts’ condensation between benzoyl chloride and benzene C6H6 + C6H5COCl AlCl3 C6H5COC6H5 +HCl(80%) H3C C C CH3 2-butyne pd/BaSO4 H3C H C C H CH3 cis 2-butene
ANSWER-KEY Q. 1 2 3 4 5 6 7 8 9 10 A. A B A C C C B D A D Q. 11 12 13 14 15 16 17 18 19 20 A. D B D C C B A C C B