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NEET • CHEMISTRY • ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES AND TECHNIQUES

Organic chemistry-some basic principles and techniques practice · Medium

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61 Medium
61 of 173
Application +4 | -1
Give IUPAC name of the compound given below
CH3−CH∣Cl−CH2−CH2−CH2−CH∣OH−CH3C{H_3} - \mathop {CH}\limits_{\mathop |\limits_{Cl} } - C{H_2} - C{H_2} - C{H_2} - \mathop {CH}\limits_{\mathop |\limits_{OH} } - C{H_3}
View answer and explanation
Correct option: C
6 – Chloro heptan – 2 – ol
Option D (HC≡Cˉ>CH2=CˉH>CH3−CˉH2\mathrm{HC} \equiv \mathrm{\bar C} > C{\mathrm{H}_2} = \mathrm{\bar CH} > C{\mathrm{H}_3} - \mathrm{\bar C}{\mathrm{H}_2}) is correct because the stability of carbanions increases with increasing s-character of the carbon bearing the negative charge (sp > sp2 > sp3).
62 Medium
62 of 173
Remembering +4 | -1
Mesomeric effect involves delocalisation of
View answer and explanation
Correct option: A
Pi-electrons
Option A (Pi-electrons) is correct because the mesomeric effect involves the delocalization of pi-electrons or lone pairs of electrons.
63 Medium
63 of 173
Remembering +4 | -1
Stereoisomers (geometrical or opticals) which are neither superimposable nor mirror image to each other are called :
View answer and explanation
Correct option: D
Diastereomers
Option D (Diastereomers) is correct because diastereomers are stereoisomers that are not mirror images and are non-superimposable.
64 Medium
64 of 173
Analyzing +4 | -1
Percentage of hydrogen is maximum in .
View answer and explanation
Correct option: B
CH4\mathrm{C}{\mathrm{H}_4}
Option B (CH4\mathrm{C}{\mathrm{H}_4}) is correct because methane has the highest hydrogen to carbon ratio among the given options.
65 Medium
65 of 173
Application +4 | -1
Arrange priority CIP sequence of given groups in decreasing order

i) OHOH \quad ii) COOHCOOH \quad iii) CHOHCH3CHOHCH_3 \quad iv) CH2OHCH_2OH
View answer and explanation
Correct option: A
I >> II >> III >> IV
Option A (I > II > III > IV) is correct because the priority in CIP sequence is based on atomic number, with higher atomic number having higher priority. O > C > C > C.
66 Medium
66 of 173
Analyzing +4 | -1
When thiourea is heated with metallic sodium, the compound which can't be formed is
View answer and explanation
Correct option: C
Na2SO4\mathrm{N}{\mathrm{a}_2}\mathrm{S}{\mathrm{O}_4}
Option C (Na2SO4\mathrm{N}{\mathrm{a}_2}\mathrm{S}{\mathrm{O}_4}) is correct because thiourea when heated with metallic sodium does not produce sodium sulfate.
67 Medium
67 of 173
Application +4 | -1
CH3−C∣CH3CH3∣−CH2−Cl→Δalc. KOHP(major)C{H_3} - \mathop C\limits_{\begin{array}{cccccccccccccccccccc} | \\ {C{H_3}} \end{array}}^{\begin{array}{cccccccccccccccccccc} {C{H_3}} \\ | \end{array}} - C{H_2} - Cl\xrightarrow[\Delta ]{{{\text{alc}}{\text{. }}KOH}}{\text{P}}\left( {{\text{major}}} \right)

Major product 'P\mathrm{P}' is
View answer and explanation
Correct option: A
CH3−CCH3∣=CH−CH3C{H_3} - \mathop C\limits^{\begin{array}{cccccccccccccccccccc} {C{H_3}} \\ | \end{array}} = CH - C{H_3}
Option A (CH3−CCH3∣=CH−CH3C{H_3} - \mathop C\limits^{\begin{array}{cccccccccccccccccccc} {C{H_3}} \\ | \\ \end{array}} = CH - C{H_3}) is correct because the major product of E2 elimination is the most stable alkene, which is the most substituted alkene (Zaitsev's rule).
68 Medium
68 of 173
Analyzing +4 | -1
The stability of

CH3−CH=CH−CH3,CH3(I)−C∣CH3=C∣CH3−CH3(II)CH3−C∣CH3=CH2(III)        CH3−C∣CH3=CH−CH3(IV)\begin{array}{l} \mathop {\mathrm{C}{\mathrm{H}_3} - \mathrm{CH} = \mathrm{CH} - \mathrm{C}{\mathrm{H}_3},\mathrm{C}{\mathrm{H}_3}}\limits_{(I)} - \mathop {\mathop {\mathop \mathrm{C}\limits_| }\limits_{C{H_3}} = \mathop {\mathop \mathrm{C}\limits_| }\limits_{C{H_3}} - \mathrm{C}{\mathrm{H}_3}}\limits_{(II)} \\ \mathop {\mathrm{C}{\mathrm{H}_3} - \mathop {\mathop \mathrm{C}\limits_| }\limits_{C{H_3}} = \mathrm{C}{\mathrm{H}_2}}\limits_{(III)} \mathrm{\;\;\;\;}\mathop {\mathrm{C}{\mathrm{H}_3} - \mathop {\mathop \mathrm{C}\limits_| }\limits_{C{H_3}} = \mathrm{CH} - \mathrm{C}{\mathrm{H}_3}}\limits_{(IV)} \end{array}

In the increasing order is

View answer and explanation
Correct option: A
III<I<IV<IIIII < I < IV < II
Option A (III<I<IV<IIIII < I < IV < II) is correct because the stability of alkenes increases with the number of alkyl substituents on the double-bonded carbons (more substituted alkenes are more stable).
69 Medium
69 of 173
Remembering +4 | -1
Lassaigne's test is used for the detection of
View answer and explanation
Correct option: D
Nitrogen, sulphur and halogens
The correct answer is Nitrogen, sulphur and halogens because Lassaigne's test is specifically designed to detect these elements in organic compounds.
70 Medium
70 of 173
Remembering +4 | -1
Rearrangement is a property of_____
View answer and explanation
Correct option: B
Carbocation
Option B is correct because carbocations are electron-deficient and can undergo rearrangements to form more stable carbocations.
71 Medium
71 of 173
Understanding +4 | -1
In the following groups, −OACI−OMeII−OSO2MeIII−OSO2CF3IV\mathop { - \mathrm{OAC}}\limits_\mathrm{I} - \mathop {\mathrm{OMe}}\limits_{\mathrm{II}} \mathop { - \mathrm{OS}{\mathrm{O}_2}\mathrm{Me}}\limits_{\mathrm{III}} \mathop { - \mathrm{OS}{\mathrm{O}_2}\mathrm{C}{\mathrm{F}_3}}\limits_{\mathrm{IV}} the order of leaving group ability is :
View answer and explanation
Correct option: B
IV > III > I > II
Option B (IV > III > I > II) is correct because the leaving group ability depends on the stability of the leaving group after departure, with stronger acids being better leaving groups.
72 Medium
72 of 173
Remembering +4 | -1
Which of the following compound is synthesised by Kolbe and Berthelot respectively that shows that organic compounds could be synthesised from inorganic sources in a laboratory?
View answer and explanation
Correct option: D
Acetic acid, methane
Option D (Acetic acid, methane) is correct because Kolbe synthesized acetic acid and Berthelot synthesized methane from inorganic sources.
73 Medium
73 of 173
Remembering +4 | -1
Racemic mixture is formed by mixing two
View answer and explanation
Correct option: D
Optical isomers
Option D is correct because a racemic mixture is an equimolar mixture of two enantiomers (optical isomers).
74 Medium
74 of 173
Analyzing +4 | -1
The number of sp2s{p^2} hybridized carbon atoms present in ‘Aspartame’ are
View answer and explanation
Correct option: C
9
Option C (9) is correct because there are 9 carbon atoms in Aspartame that are sp2sp^2 hybridized, meaning they are involved in a double bond or are part of an aromatic ring.
75 Medium
75 of 173
Analyzing +4 | -1
Indicate the wrongly named compound.
View answer and explanation
Correct option: D
CH3CH=CH−CH2−CO∣∣−CH3(3−hexen−2−one)\mathop {C{H_3}CH = CH - C{H_2} - \mathop C\limits^{\begin{array}{cccccccccccccccccccc} O \\ {||} \end{array}} - C{H_3}}\limits_{\left( {3 - {\text{hexen}} - 2 - {\text{one}}} \right)}
Option D is the wrongly named compound because the correct IUPAC name should be 4-hexen-2-one.
76 Medium
76 of 173
Remembering +4 | -1
Which of the following techniques is used to separate glycerol from spent-lye in soap industry?
View answer and explanation
Correct option: B
Distillation under reduced pressure
The correct answer is Distillation under reduced pressure because glycerol has a high boiling point and can decompose at high temperatures, so reducing the pressure lowers the boiling point and prevents decomposition.
77 Medium
77 of 173
Understanding +4 | -1
Correct IUPAC name of [Ni(en)Br2(PPh3)2]\left[ {Ni(en)B{r_2}{{(PP{h_3})}_2}} \right] is
View answer and explanation
Correct option: C
Dibromido ethylene diamine bis (triphenyl phosphine) nickel (II)
Option C (Dibromido ethylene diamine bis (triphenyl phosphine) nickel (II)) is correct because it follows the IUPAC naming conventions for coordination complexes, listing ligands alphabetically and indicating the oxidation state of the metal.
78 Medium
78 of 173
Remembering +4 | -1
The structure of tert-Butyl group in an organic compound is
View answer and explanation
Correct option: D
CH3−C∣CH3CH3∣−C{H_3} - \mathop {\mathop C\limits_{\begin{array}{cccccccccccccccccccc} | \\ {C{H_3}} \end{array}} }\limits^{\begin{array}{cccccccccccccccccccc} {C{H_3}} \\ | \end{array}} -
Option D is correct because a tert-butyl group is a butyl group with three methyl groups attached to the same carbon atom, which is connected to the main chain.
79 Medium
79 of 173
Analyzing +4 | -1
Arrange in decreasing order of +R+\mathrm{R} power in the given groups

I) −NH2-\mathrm{NH}_2

II) −OΘ-O^{\Theta}

III) −OH-\mathrm{OH}

IV) −NHCOCH3-\mathrm{NHCOCH}_3
View answer and explanation
Correct option: A
II >> I >> III >> IV
Option A (II > I > III > IV) is correct because the +R effect depends on the availability of electrons, which is highest for O-, followed by NH2, OH, and NHCOCH3.
80 Medium
80 of 173
Remembering +4 | -1
In Kjeldahl’s method of estimation of nitrogen, K2SO4{\mathrm{K}_2}\mathrm{S}{\mathrm{O}_4} acts as:
View answer and explanation
Correct option: D
Boiling point elevator
Option D (Boiling point elevator) is correct because K2SO4{\mathrm{K}_2}\mathrm{S}{\mathrm{O}_4} is added to increase the boiling point of sulfuric acid, facilitating the digestion process.
81 Medium
81 of 173
Understanding +4 | -1
Which conformation of butane will have the minimum energy
View answer and explanation
Correct option: B
Anti/staggered
Option B (Anti/staggered) is correct because the anti or staggered conformation of butane has the lowest energy due to minimal steric hindrance between the methyl groups.
82 Medium
82 of 173
Understanding +4 | -1
The compound formed in the positive test for nitrogen with the Lassaigne solution of an organic compound is
View answer and explanation
Correct option: A
Fe4[Fe(CN)6]3\mathrm{F}{\mathrm{e}_4}{\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]_3}
Option A (Fe4[Fe(CN)6]3\mathrm{F}{\mathrm{e}_4}{\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]_3}) is correct because the Prussian blue color in the Lassaigne's test for nitrogen is due to the formation of ferric ferrocyanide, Fe4[Fe(CN)6]3Fe_4[Fe(CN)_6]_3.
83 Medium
83 of 173
Analyzing +4 | -1
Which of the following statement(s) is/are correct ?

(i) A carbon atom having an sps p hybrid orbital is less electronegative than carbon atoms possessing sp2s p^2 or sp3s p^3 hybridised orbitals.

(ii) π\pi-bonds provide the most reactive centres in the molecules containing multiple bonds

(iii) The number of σ\sigma and π\pi bonds in compound CH2=C=CHCH3{CH}_2={C}={CHCH}_3 are 7 and 2 respectively.
View answer and explanation
Correct option: C
(ii) only
Option C ((ii) only) is correct because only statement (ii) is correct. Pi-bonds are indeed reactive centers, while (i) is incorrect as sp hybridized carbon is more electronegative, and (iii) has incorrect bond counts.
84 Medium
84 of 173
Understanding +4 | -1
Increasing order of stability of radicals is
View answer and explanation
Correct option: B
CH3−C.H2<CH3−C.H−CH3<(CH3)2C.−CH2−CH3<(CH3)3C.C{H_3} - \mathop C\limits^. {H_2} < C{H_3} - \mathop C\limits^. H - C{H_3} < {\left( {C{H_3}} \right)_2}\mathop C\limits^. - C{H_2} - C{H_3} < {\left( {C{H_3}} \right)_3}\mathop C\limits^.
Option C (4-methyl hex-2-yne) is correct because it follows the IUPAC naming rules by identifying the longest chain containing the triple bond and numbering it to give the triple bond the lowest possible number.
85 Medium
85 of 173
Application +4 | -1
Correct IUPAC structure for the given organic compound is 2,2-Dibromo-1-phenylpentane
View answer and explanation
Correct option: B
Ph−CH2−C∣BrBr∣−CH2−CH2−CH3Ph - C{H_2} - \mathop C\limits_{\begin{array}{cccccccccccccccccccc} | \\ {Br} \end{array}}^{\begin{array}{cccccccccccccccccccc} {Br} \\ | \end{array}} - C{H_2} - C{H_2} - C{H_3}
Option B is correct because it accurately represents the 2,2-Dibromo-1-phenylpentane structure with the correct placement of the phenyl group and bromine atoms.
86 Medium
86 of 173
Application +4 | -1
In Carius method of 0.099 g orgainc compound gave 0.287 g AgCl\mathrm{AgCl}. The percentage of chlorine in the compound will be
View answer and explanation
Correct option: B
71.7
Option B (71.7) is correct because the percentage of chlorine is calculated using the formula: % Chlorine = (Mass of AgCl / Molecular weight of AgCl) * (Atomic weight of Cl / Mass of organic compound) * 100.
87 Medium
87 of 173
Analyzing +4 | -1
Most stable carbocation among the following is
View answer and explanation
Correct option: C
CH3−CH2+C{H_3} - \mathop {C{H_2}}\limits^ +
Option C (CH3−CH2+C{H_3} - \mathop {C{H_2}}\limits^ +) is correct because alkyl groups stabilize carbocations through inductive effect and hyperconjugation.
88 Medium
88 of 173
Remembering +4 | -1
Formulae of phenyl carbinol and chloral are respectively :
View answer and explanation
Correct option: B
C6H5CH2OH  and  CCl3CHO{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{C}{\mathrm{H}_2}\mathrm{OH\;}{\kern 1pt} {\kern 1pt} \mathrm{and}{\kern 1pt} \mathrm{\;CC}{\mathrm{l}_3}\mathrm{CHO}
Option B is correct because phenyl carbinol is benzyl alcohol (C₆H₅CH₂OH), and chloral is trichloroacetaldehyde (CCl₃CHO).
89 Medium
89 of 173
Analyzing +4 | -1
Which of the following species is most stable?
View answer and explanation
Correct option: D
p−CH3O−C6H4−C+H2p - \mathrm{C}{\mathrm{H}_3}\mathrm{O} - {\mathrm{C}_\mathrm{6}}{\mathrm{H}_4} - \mathop \mathrm{C}\limits^ + {\mathrm{H}_2}
Option D (p−CH3O−C6H4−C+H2p - \mathrm{C}{\mathrm{H}_3}\mathrm{O} - {\mathrm{C}_\mathrm{6}}{\mathrm{H}_4} - \mathop \mathrm{C}\limits^ + {\mathrm{H}_2}) is correct because the methoxy group (OCH3OCH_3) is an electron-donating group that stabilizes the carbocation through resonance.
90 Medium
90 of 173
Remembering +4 | -1
3-phenylpropenoic acid is IUPAC name of :
View answer and explanation
Correct option: D
Cinnamic acid
Option D (Cinnamic acid) is correct because 3-phenylpropenoic acid is the IUPAC name for cinnamic acid.
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