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Basics of Organic Chemistry (GOC)

NEET Chemistry · 63 questions · Page 4 of 7 · Click an option or "Show Solution" to reveal answer

Q31
The total number of π\pi -bond electrons in the following structure is
A 12
B 16
C 4
D 8
Correct Answer
Option D
Solution

No. of double bonds = 4 No. of π\pi bond electrons = 2 × no. of double bond = 2 × 4 = 8

Q32
In Duma's method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25 mm, the percentage of nitrogen in the compound is
A 16.75
B 15.76
C 17.36
D 18.20
Correct Answer
Option A
Solution

Wt. of organic substance = 0.25 g V 1 = 40 mL, T 1 = 300 K P 1 = 725 – 25 = 700 mm of Hg P 2 = 760 mm of Hg (at STP) T 2 = 273 K

P1V1T1=P2V2T2{{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}{V_2}} \over {{T_2}}}

V 2 =

700×40×273300×760{{700 \times 40 \times 273} \over {300 \times 760}}

= 33.52 mL Percentage of nitrogen =

28×33.52×10022400×0.25{{28 \times 33.52 \times 100} \over {22400 \times 0.25}}

= 16.76%

Q33
Given : Which of the given compounds can exhibit tautomerism?
A II and III
B I, II and III
C I and II
D I and III
Correct Answer
Option B
Solution

In first two cases α\alpha-H participates in tautomerism. And in third case, γ\gamma-H participate in tautomerism.

Q34
In the Kjeldahl's method for estimation of nitrogen present in a soil sample, ammonia evolved from 0.75 g of sample neutralized 10 mL of 1 M H 2 SO 4 . The percentage of nitrogen in the soil is
A 37.33
B 45.33
C 35.33
D 43.33
Correct Answer
Option A
Solution

10 ml, 1 M H 2 SO 4 = 20 ml, 1 M NH 3 ∵\because wt of N in one mole NH 3 = 14 ∴\therefore 20 × 10 –3 mol NH 3 →\to 20 × 10 –3 × 14 nitrogen ∴\therefore 0.75 g of sample contains =

14×20×10−30.75×100{{14 \times 20 \times {{10}^{ - 3}}} \over {0.75}} \times 100

= 37.33%

Q35
Given : I and II are
A identical
B a pair of conformers
C a pair of geometrical isomers
D a pair of optical isomers
Correct Answer
Option B
Solution

Conformers are form of stereoisomers in which isomers can be interconverted by rotation about single bonds.

I and II are staggered and eclipsed conformers respectively.

Q36
Arrange the following in increasing order of stability
A 5 < 4 < 3 < 1< 2
B 4 < 5 < 3 < 1 < 2
C 1 < 5 < 4 < 3 < 2
D 5 < 4 < 3 < 2 < 1
Correct Answer
Option A
Solution

Greater the number of e – donating alkyl groups (+I effect), greater will be the stability of carbocations.

Q37
Homolytic fission of the following alkanes forms free radicals CH 3 −- CH 3 , CH 3 −- CH 2 −- CH 3 , (CH 3 ) 2 CH −- CH 3 , CH 3 −- CH 2 −-CH(CH 3 ) 2 . Increasing order of stability of the radicals is
A (CH3)2C∙ − CH2CH3 <CH3 − C∙H− CH3<CH3 − C∙H2 <(CH3)3C∙\begin{aligned}& {\left( {C{H_3}} \right)_2}\mathop C\limits^ \bullet \, - \,C{H_2}C{H_3}\, < C{H_3}\, - \,\mathop C\limits^ \bullet H - \,C{H_3} \\ & < C{H_3}\, - \,\mathop C\limits^ \bullet {H_2}\, < {\left( {C{H_3}} \right)_3}\mathop C\limits^ \bullet\end{aligned}
B CH3 − C∙H2 < CH3 − C∙HH − CH3 <(CH3)2C∙ − CH2 − CH3 < (CH3)3C∙\begin{aligned}& C{H_3}\, - \,\mathop C\limits^ \bullet {H_2}\, < \,C{H_3}\, - \,\mathop C\limits^ \bullet HH\, - \,C{H_3}\, < \\ & {\left( {C{H_3}} \right)_2}\mathop C\limits^ \bullet \, - \,C{H_2}\, - \,C{H_3}\, < \,{\left( {C{H_3}} \right)_3}\mathop C\limits^ \bullet\end{aligned}
C CH3−C∙H2<CH3−C∙H−CH3<(CH3)3C∙ <(CH3)2C∙ −CH2CH3\begin{aligned}& C{H_3} - \mathop C\limits^ \bullet {H_2} < C{H_3} - \mathop C\limits^ \bullet H - C{H_3} < \\ & {\left( {C{H_3}} \right)_3}\mathop C\limits^ \bullet \, < {\left( {C{H_3}} \right)_2}\mathop C\limits^ \bullet \, - C{H_2}C{H_3}\end{aligned}
D (CH3)3C∙ <(CH3)2C∙ −CH2CH3<CH3−C∙H−CH3<CH3−C∙H2\begin{aligned}& {\left( {C{H_3}} \right)_3}\mathop C\limits^ \bullet \, < {\left( {C{H_3}} \right)_2}\mathop C\limits^ \bullet \, - C{H_2}C{H_3} < \\ & C{H_3} - \mathop C\limits^ \bullet H - C{H_3} < C{H_3} - \mathop C\limits^ \bullet {H_2}\end{aligned}
Correct Answer
Option B
Solution

Stability depends on number of hyperconjugative structure.

More the number of hyperconjugative structure, the greater is the stability.

Q38
Some meta-directing substituents in aromatic substitution are given. Which one is most deactivating ?
A −-COOH
B −- NO 2
C −- C ≡ \equiv N
D −- SO 3 H
Correct Answer
Option B
Solution

−- NO 2 is most deactivating due to -I ans -M effect.

Q39
The radical, is aromatic because it has
A 7 p-orbitals and 7 unpaired electrons
B 6 p-orbitals and 7 unpaired electrons
C 6 p-orbitals and 6 unpaired electrons
D 7 p-orbitals and 6 unpaired electrons.
Correct Answer
Option C
Solution

In benzyl free radical it has 6 p orbitals, each containing one unpaired electron, in a six membered cyclic structure is in accordance with Huckel rule of aromaticity.

Q40
Which of the following acids does not exhibit optical isomerism?
A Maleic acid
B α\alpha -amino acids
C Lactic acid
D Tartaric acid
Correct Answer
Option A
Solution

HOOC – CH = CH – COOH (Maleic acid) almost show optical isomerism.

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