JEE Questions for Chemistry Organic Chemistry Some Basic Principles And Techniques Quiz 9 - MCQExams.com

trans-2-butene (III) > cis-2- butene (II) > I-butene(I)

p - NO₂ - C₆ H₄ - CH^-₂ is the most stable carbanion since electron-with - drawing – NO₂ group stabilizes the carbanion by dispersal of the –ve charge.

IV is stabilized by +I-effect of two CH3 group and by dispersal of +vecharge on the benzene ring by resonance. III is stabilized by + I-effect ofone CH3 group and by dispersal of +ve charge on the benzene ring. I isstabilized by dispersal of the +ve charge on the benzene ring while II is notstabilized by any of three factors. Thus the overall order of stability is :II < I < III < IV i.e., option (a) is correct.

(b) being the 3º carbocation is the most stable.

Aromatic character increases as the resonance energy increases, Nowresonance energies decrease in the order : pyridine (125.5 kJ mol^-1) >thiophene (117 kJ mol^-1) > pyrrole (88 kJ mol^-1) > furan (71 kJ mol^-1).Thus, option (c) is correct.

Only (b) has two completely delocalized π-electrons and hence is aromatic.

In benzyne, there is 

One sp² - sp² σ - bond 

One p - p, π - bond and 

One sp² - sp², π - bond 

The heat of hydrogenation of cyclohexene is 29 kcal mol-1. Sincecyclohexadiene has two double bonds while cyclohexene has one,therefore, heat of hydrogenation of cyclohexadiene is double(58 kcal mol-1)the heat of hydrogenation of cyclohexene (29 kcal mol^-1).

Delocalization of electrons π− always leads to stabilization is an incorrectstatement since both cyclobutadiene and planar cyclooctatetraene aredestabilized by resonance.

Option (a) is incorrect since the – I-effect of halogens follows the order : I<Br <Cl <F.

Both structures( I and IV ) are stabilized by resonance.

In presence of SbCl5, 2-chloro-2-phenylethane forms a carbocation asshown below:

CH₃ - CHCl - C₆H₅ + SbCl₅  [CH₃ - CH - C₆H₅] SbCl₆^-
Since the carbocation is planar species, therefore, it can be attacked by SbCl₆^- either from the top or the bottom face with equal ease. As a result, a 50 : 50 mixture of two enantiomers of 2 - chloro-2-phenylethane are formed, i.e. (+ ) 2-chloro-2-phenylethane undergoes recemization due to the formation of carbocation intermediate.

Nitrogen (when N is converted into NH₃).

All the three (N,S halogens).

N and S .

Nitrogen

Sodium cyanide (Na + C + N → NaCN)

NH₂NH₂.HCl since it does not contain C which is needed for the formation of NaCN.

Hydrazine since it does not contain C.

HNO₃ is added to decompose Na₂S and NaCN otherwise Na₂S will give black ppt. of Ag₂S and NaCN will white ppt. of AgCN which would interfere with the test of halogens.

Victor-Meyer’s method.

Halogens.

Hydrazine (NH₂NH₂) does not contain C and hence on fusion with Na metal, it cannot form NaCN.Therefore, hydrazine does not show Lassaigne’s test.

0.833 mole contain hydrogen =10g
1 molewill∴ contain hydrogen = 10/0.833 =12 g
Thus 1 mole of carbohydrate contains 12 hydrogen atoms.
But E.F. of carbohydrate = CH₂O
M.F. = (CH₂O) × 6 = C₆H₁₂O₆

In estimation of C, organic compound is heated with CuO

2CuO + C → 2Cu + CO₂

Only urea does not sublime while naphthalene, camphor and benzoic acid do.

Five chain isomers - Refer to review of fundamentals.

2-chlorobutane contains a chiral carbon and hence is optically active

CICH = CHCl.

Because of the bigger size of Cl than H, the rotation about carbon-carbon single bond in hexa chloroethane is more hindered than, in ethane on the other hand because of the presence of double bond in ethylene, and triple bond in acetylene, the rotation about carbon-carbon bond is highly hindered, Thus ethane has the least hindered rotation.

Molecular formula.

Cyclohexane (iv) is non-planar and has chair conformation. In this conformation, the ∠CCC bond angle are the normal tetrahedral angles (109º-28’) and thus has no angle strain and hence is the most stable. The rest of the molecules are nearly planar and hence their stability depends upon the angle strain in accordance with Baeyer’s strain theory. Since cyclopropane has higher angle strain (24º– 44’) than cyclopentane (0º – 44’),
therefore, cyclopentane (iii) is more stable than cyclopropane (ii). Further, because of the presence of a double bond in a three-membered ring, cyclopropene (ii

Since all the six carbon atoms of cyclohexatriene (benzene) are sp² - hybridized, therefore, its stablest structure is planar form.

Both are included in stereoisomerism