${\rm{N}}{{\rm{a}}_2}{\rm{S}} + {\rm{N}}{{\rm{a}}_2}\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_5}{\rm{NO}}} \right]$ $ \to $ Purple colour. It is due to

${\rm{N}}{{\rm{a}}_4}\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_5}{\rm{NOS}}} \right]$

${\rm{N}}{{\rm{a}}_4}\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_3}{\rm{NOS}}} \right]$

${\rm{N}}{{\rm{a}}_3}\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_5}{\rm{NOS}}} \right]$

${\rm{N}}{{\rm{a}}_4}\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_5}{\rm{NO}}} \right]$

$\beta - $ hydroxypropanoic acid

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

Organic chemistry-some basic principles and techniques practice · Mixed difficulty

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211 Medium

211 of 392

Application +4 | -1

An enantiomerically pure acid is treated with racemic mixture of an alcohol having one chiral carbon. The ester formed will be

View answer and explanation

Correct option: A

Optically active mixture

Option A (Optically active mixture) is correct because the reaction of an enantiomerically pure acid with a racemic alcohol produces a mixture of diastereomeric esters, which are optically active.

212 Easy

212 of 392

Remembering +4 | -1

Fractional distillation is useful in distillation of

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because fractional distillation is used to separate components in petroleum, coal-tar, and crude alcohol based on their boiling points.

213 Hard

213 of 392

Understanding +4 | -1

How will you separate a solution (miscible) of benzene

+{CHCl}_3

View answer and explanation

Correct option: C

distillation

Option C is correct because benzene and chloroform are miscible liquids with different boiling points, making distillation the appropriate separation technique.

214 Hard

214 of 392

Understanding +4 | -1

Which of the following is ambident nucleophile

View answer and explanation

Correct option: D

All

Option D (All) is correct because ambident nucleophiles are nucleophiles that can attack from two or more different sites, and all listed options (NH2OH, NCO-, NO2-) exhibit this behavior.

215 Easy

215 of 392

Understanding +4 | -1

Geometrical isomerism is possible in case of

View answer and explanation

Correct option: A

Pentene-2

Option A (Pentene-2) is correct because geometrical isomerism requires restricted rotation around a double bond and two different groups on each carbon of the double bond, which pentene-2 fulfills.

216 Medium

216 of 392

Application +4 | -1

C{H_3} - C \equiv CH + HOH\xrightarrow{{H{g^{2 + }}/{H_2}S{O_4}}}A

The number of sigma and pi bonds that are present in the molecule

A

View answer and explanation

Correct option: A

9 sigma and 1 pi

Option A (9 sigma and 1 pi) is correct because the product A is CH3COCH3 (acetone), which has 9 sigma bonds and 1 pi bond.

217 Medium

217 of 392

Remembering +4 | -1

Fischer projection indicates :

View answer and explanation

Correct option: A

Horizontal substituents above the plane

Option A (Horizontal substituents above the plane) is correct because, in Fischer projections, horizontal lines represent bonds projecting out of the plane (towards the viewer).

218 Easy

218 of 392

Understanding +4 | -1

Which of the following sodium compound/compounds are formed when an organic compound containing both nitrogen and sulphur is fused with sodium?

View answer and explanation

Correct option: B

Thiocyanate

Option B (Thiocyanate) is correct because when an organic compound containing both nitrogen and sulfur is fused with sodium, it forms sodium thiocyanate (NaSCN).

219 Medium

219 of 392

Analyzing +4 | -1

Which of the following compound will give white precipitate on heating with

{HNO}_3

followed by addition of silver nitrate?

View answer and explanation

Correct option: C

Both (1) and (2)

Option C (Both (1) and (2)) is correct because both compounds contain chlorine that can be converted to chloride ions, which then react with silver nitrate to form a white precipitate of silver chloride.

220 Medium

220 of 392

Analyzing +4 | -1

An organic compound A is oxidised with

N{a_2}{O_2}

followed by boiling with

HN{O_3}

. The resultant solution is then treated with ammonium molybdate to yield a yellow precipitate. Based on above observation, the element present in the given compound is

View answer and explanation

Correct option: B

Phosphorus

Option B (Phosphorus) is correct because the yellow precipitate with ammonium molybdate indicates the presence of phosphorus.

221 Hard

221 of 392

Understanding +4 | -1

Assertion (A): Components of a mixture of red and blue inks can be separated by distributing the components between stationary and mobile phases in paper chromatography.

Reason (R): The coloured components of inks migrate at different rates because paper selectively retains different components according to the difference in their partition between two phases.

View answer and explanation

Correct option: A

Both

\mathrm{A}

and

\mathrm{R}

are correct ;

\mathrm{R}

is the correct explanation of

A

Option A is correct because both the assertion and reason are true, and the reason correctly explains why components of a mixture of inks can be separated by paper chromatography.

222 Easy

222 of 392

Understanding +4 | -1

Which of the following intermediate have the complete octet around the carbon atom?

View answer and explanation

Correct option: B

Carbanion

Option B (Carbanion) is correct because a carbanion has 8 electrons around the carbon atom, fulfilling the octet rule.

223 Easy

223 of 392

Understanding +4 | -1

Inductive effect involves

View answer and explanation

Correct option: B

partial displacement of

\sigma

electrons

Option B (partial displacement of

\sigma

electrons) is correct because the inductive effect involves the partial displacement of sigma electrons along a chain of atoms.

224 Easy

224 of 392

Analyzing +4 | -1

Strongest acid among the following

View answer and explanation

Correct option: C

C{F_3}COOH

Option C (

C{F_3}COOH

) is the strongest acid because the three fluorine atoms, being highly electronegative, exert a strong electron-withdrawing inductive effect (-I effect) that significantly stabilizes the conjugate base (carboxylate ion), thus increasing the acidity of the carboxylic acid.

225 Easy

225 of 392

Analyzing +4 | -1

The compound having molecular formula

{\mathrm{C}_4}{\mathrm{H}_{10}}\mathrm{O}

can show :

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because

{\mathrm{C}_4}{\mathrm{H}_{10}}\mathrm{O}

can exhibit metamerism, functional isomerism, and positional isomerism.

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226 Medium

226 of 392

Application +4 | -1

The number of

\sigma - \,\,{\text{and}}\,\,\pi {\text{ - }}

bonds in but-1-ene- 3-yne are:

View answer and explanation

Correct option: B

7\sigma \,\,{\text{and}}\,\,3\pi

Option A (Allyl carbonium ion

(C{H_2} = CH - \mathop {\mathrm{ }C}\limits^ + {H_2})

is more stable then propyl carbonium ion) is correct because the allyl carbocation is stabilized by resonance, while the propyl carbocation is not.

227 Medium

227 of 392

Analyzing +4 | -1

Which of the following is most reactive towards elimination reaction?

View answer and explanation

Correct option: D

R{\mathrm{O}^ - }

Option D (

RO^−

) is the most reactive towards elimination because alkoxides are strong bases and favor elimination reactions.

228 Medium

228 of 392

Understanding +4 | -1

\mathrm{N}{\mathrm{a}_2}\mathrm{S} + \mathrm{N}{\mathrm{a}_2}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_5}\mathrm{NO}} \right]

\to

Purple colour. It is due to

View answer and explanation

Correct option: D

\mathrm{N}{\mathrm{a}_4}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_5}\mathrm{NOS}} \right]

Option D (

\mathrm{N}{\mathrm{a}_4}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_5}\mathrm{NOS}} \right]

) is correct because the purple color in the test for both nitrogen and sulfur is due to the formation of sodium nitroprusside,

Na_4[Fe(CN)_5NOS]

229 Medium

229 of 392

Remembering +4 | -1

IUPAC name of

{C_6}{H_5}CN

is:

View answer and explanation

Correct option: B

Benzene nitrile

Option B (Benzene nitrile) is correct because it accurately reflects the IUPAC nomenclature for a benzene ring directly bonded to a nitrile group.

230 Easy

230 of 392

Application +4 | -1

The correct IUPAC name of

\mathrm{CH}_3 \mathrm{CH}-\mathrm{C} \equiv\mathrm{C}-\mathrm{CH}_3

is:

View answer and explanation

Correct option: C

4-methyl hex-2-yne

Option B (

R{S^ \ominus }

is less basic but more nucleophilic than

R{O^{\, \ominus }}

) is correct because sulfur is larger and more polarizable than oxygen, making the thiolate anion a better nucleophile but a weaker base.

231 Easy

231 of 392

Remembering +4 | -1

Which of the following statements is wrong?

View answer and explanation

Correct option: C

Organic compounds cannot be synthesized in the laboratory

Option C (Organic compounds cannot be synthesized in the laboratory) is wrong because Wöhler's synthesis of urea disproved this statement, showing that organic compounds can indeed be synthesized in the lab.

232 Medium

232 of 392

Application +4 | -1

Compound (A) is formed on treating a benzilwith

\mathrm{NaBH}_4

. The number of possible meso isomers for (A) are

View answer and explanation

Correct option: A

Option A (1) is correct because the reduction of benzil with NaBH4 produces a diol, and only one meso isomer is possible due to the symmetry of the molecule.

233 Easy

233 of 392

Understanding +4 | -1

The suitable method for separation of a mixture of ortho and para nitrophenols is

View answer and explanation

Correct option: D

steam distillation

Option D is correct because ortho and para nitrophenols have different volatilities due to intramolecular hydrogen bonding in ortho-nitrophenol, allowing separation by steam distillation.

234 Easy

234 of 392

Remembering +4 | -1

Lactic acid is:

View answer and explanation

Correct option: C

\alpha -

hydroxypropanoic acid

Option C (

\alpha -

hydroxypropanoic acid) is correct because lactic acid has a hydroxyl group attached to the alpha carbon of propanoic acid.

235 Hard

235 of 392

Application +4 | -1

0.5 g of an organic compound containing nitrogen on Kjeldahlising required 29 mL of

N/5{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

for complete neutralization of ammonia. The percentage of nitrogen in the compound is:

View answer and explanation

Correct option: B

16.2

Option B (

16.2

) is correct because the percentage of nitrogen is calculated using the volume and molarity of the sulfuric acid used to neutralize the ammonia.

236 Medium

236 of 392

Application +4 | -1

The conversion of propan-1-ol to n-butylamine involves the sequential addition of reagents. The correct sequential order of reagents is

View answer and explanation

Correct option: A

({\text{i}})\,\,SOC{l_2}\,\,({\text{ii}})\,\,KCN\,\,({\text{iii}}){H_2}/Ni,Na(Hg)/{C_2}{H_5}OH

Option A

(({\text{i}})\,SOC{l_2}\,({\text{ii}})\,KCN\,({\text{iii}}){H_2}/Ni,Na(Hg)/{C_2}{H_5}OH)

is correct because it outlines the correct sequence of reactions: converting the alcohol to an alkyl chloride using SOCl2, followed by nucleophilic substitution with KCN to form a nitrile, and finally reduction of the nitrile to an amine using H2/Ni or Na(Hg)/C2H5OH.

237 Hard

237 of 392

Analyzing +4 | -1

An organic compound contains

46.78 \%

of a halogen

{X}

. When

2.00 \mathrm{~g}

of this compound is heated with fuming

{HNO}_3

in the presence of

{AgNO}_3, 2.21 \mathrm{g} \,\,{AgX}

was formed. The halogen

{X}

is [Given: atomic weight of

{Ag}=108, {F}=19, {Cl}=35.5, {Bi}=80, {I}=127]

View answer and explanation

Correct option: C

Br

Option C (Br) is correct because the molar mass calculated from the given data matches the molar mass of bromine.

238 Easy

238 of 392

Understanding +4 | -1

Which among the following is not electrophile

View answer and explanation

Correct option: A

\mathop C\limits^ \oplus {H_5}

Option A (

\mathop C\limits^ \oplus {H_5}

) is correct because

CH_5^+

is not a typical electrophile due to its structure and bonding; it is a hypercoordinated carbocation.

239 Easy

239 of 392

Analyzing +4 | -1

Assertion (A) :

\mathrm{SO}_3

is a nucleophile.

Reason (R):

\mathrm{SO}_3

has lone-pair of electrons.

View answer and explanation

Correct option: D

\mathbf{A}

and

\mathbf{R}

both are false.

Option D (A and R both are false) is correct because SO3 is an electrophile, not a nucleophile, and it does not have a lone pair of electrons.

240 Easy

240 of 392

Remembering +4 | -1

Mesomeric effect is due to

View answer and explanation

Correct option: B

Delocalisation of

\pi

electrons

Option B (Delocalisation of

\pi

electrons) is correct because the mesomeric effect involves the delocalization of pi electrons or lone pairs of electrons in a conjugated system.

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