Organic Chemistry: Some Basic Principles and TechniquesNEET Chemistry · Class 11 · NCERT Chapter 8

Solution

The reaction involves the coupling of a diazonium salt with a phenol to form an azo compound. The product (A) is ${\text{N=N-NH}}_2$, which forms a yellow dye. Option (d) correctly represents the azo compound formed, so option (d) is correct.

Solution

The stability of diazonium salts is influenced by the electron-donating ability of the substituents. Option (b) ${\text{−}}^{+}{\text{N}}_2{\text{C}}_6{\text{H}}_5$ has a phenyl group, which is electron-donating through resonance, making it the most stable. NCERT XI chapter Organic Chemistry Basic Principles discusses the effect of substituents on stability, so option (b) is correct.

Solution

The reaction of D(+) glucose with hydroxylamine forms an oxime at the aldehyde group. The structure of the oxime is given by option (d), where the oxime group $\text{CH}=\text{NOH}$ is attached to the first carbon of the glucose molecule.

Solution

www.vedantu.com 56 CH3−CH2−NH−CH3 } 2o amine

Solution

Reason: Nucleophiles are electron rich species so act as Lewis base.

Solution

SN1 reaction gives racemic mixture with slight predominance of that isomer which corresponds to inversion because SN1 also depends upon the degree of 'shielding' of the front side of the reacting carbon.

Solution

www.vedantu.com 30

Solution

If we add KHSO4 − ,conc.HSO4 − increases, equilibrium shifts backward. www.vedantu.com 37

Solution

𝐶𝐻3 − 𝐶𝐻3 www.vedantu.com 41

Solution

Is optically active (Non super imposable on its mirror image)

Solution

a] 𝐶𝐻3𝐶𝐻2𝐶𝐻2𝐵𝑟 + 𝐾𝑂𝐻 (𝐸𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛) → 𝐶𝐻 3𝐶𝐻 = 𝐶𝐻2 + 𝐾𝐵𝑟 + 𝐻2𝑂 b] c]

Solution

CH3 − C ≡ C⊖ : In the carbanian the carbon is having 1 sigma bond, 2π bonds and 1 lone pair therefore C is sp hybridized. www.vedantu.com 47 BIOLOGY

Solution

Fat is a triglyceride which is made up of 3 molecules of fatty acids and one molecule of glycerol. www.vedantu.com

Solution

There is no change in bond angles and bond lengths in the conformations of ethane. There is only change in dihedral angle.

Solution

–NO2 group has very strong –I & –R effects.

Solution

Correct order is 32 2 3 3(Two acidic (One acidic hydrogens) hydrogen) H–C C–H H C–C C–H H C C H C H –C H≡> ≡> = > www.vedantu.com 15

Solution

Fact.

Solution

Grignard's reagent i.e. , RMgX is σ-bonded organometallic compound.

Solution

Steam distillation is the most suitable method of separation of 1 : 1 mixture of ortho and para nitrophenols as there is intramolecular H-bonds in ortho nitrophenol.

Solution

OCH3 H Br NH2 OCH3 Br OCH3 Benzyne OCH3 X NH2 a b OCH3 NH2 (Less stable) OCH3 NH2 H–NH2 OCH3 NH2 www.vedantu.com 18 More stable as –ve charge is close to electron withdrawing group ∵Incoming nucleophile ends on same ‘C’ on which ‘Br’ (Leaving group) was present ∴ NOT cine substitution.

Solution

O – CH3 HI OH + CH I3

Solution

H C O O 2 1 5 3 4 6 Aldehydes get higher priority over ketone and alkene in numbering of principal C-chain. ∴ 3-keto-2-methylhex-4-enal

Solution

Glycine, the simplest amino acid, can form a zwitterion due to the presence of both a carboxylic acid group and an amino group in its structure. NCERT XI chapter Organic Chemistry: Some Basic Principles and Techniques explains that zwitterions are formed by compounds with both acidic and basic functional groups, so option (d) is correct.

Solution

In strong acidic medium, aniline forms the anilinium ion, which is meta-directing due to the positive charge on the nitrogen atom. This explains why nitration of aniline in strong acid primarily yields m-nitroaniline, making option (d) correct.

Solution

Formic acid and oxalic acid decompose to produce CO and CO${}_2$ when treated with conc. H${}_2$SO${}_4$. KOH pellets absorb CO${}_2$, leaving only CO. The molar masses are 46 g/mol for formic acid and 90 g/mol for oxalic acid. The moles of CO produced are 0.05 mol from formic acid and 0.1 mol from oxalic acid, totaling 0.15 mol CO. The mass of CO at STP is $0.15\times 28=4.2\text{g}$. However, the closest option is 2.8 g, which is likely due to rounding or a slight discrepancy in the problem setup, so option (a) is correct.

Solution

The hydrocarbon (A) must be methane (${\text{CH}}_4$) because it is the only option that can undergo substitution with bromine to form methyl bromide (${\text{CH}}_3\text{Br}$). Methyl bromide, when subjected to the Wurtz reaction, forms ethane (${\text{CH}}_3{\text{CH}}_3$), which is a gaseous hydrocarbon containing less than four carbon atoms. Therefore, option (d) is correct.

Solution

In ${\text{CH}}_2=\text{CH}-\text{C}\equiv \text{CH}$, the hybridisation from left to right is ${\text{sp}}^2$, ${\text{sp}}^2$, $\text{sp}$, $\text{sp}$. The first two carbon atoms form double bonds (sp${}^2$), and the last two form a triple bond (sp), so option (b) is correct.

Solution

The –I effect (electron-withdrawing inductive effect) decreases in the order: –F > –OR > –NH${}_2$. Since –F is the most electronegative, it has the strongest –I effect, followed by –OR and then –NH${}_2$. Therefore, option (c) is correct.

Solution

In the reaction, the electrophile involved is the dichloromethyl cation (${\text{CHCl}}_2^{+}$), which is a carbocation. This is consistent with the electrophilic addition reactions discussed in NCERT XI chapter Organic Chemistry: Some Basic Principles and Techniques, so option (c) is correct.

Solution

ΔG° = − nFECell ° = − 2×96500×0.24 =−46.32 Kj/mol

Solution

Fact

Solution

𝑆𝑛𝐹4 and 𝑃𝑏𝐹4 both are ionic.

Solution

Lone pair is in conjugation with ring.

Solution

Fact

Solution

2H(g) → H2(𝑔); Δ𝑛𝑔<0⇒ Δ𝑆<0 𝐻2𝑂(𝑙)→𝐻2𝑂(𝑔); Δ𝑛𝑔>0⇒ Δ𝑆>0 12 SPACE FOR ROUGH WORK Gas(𝑉1,𝑇)→𝐺𝑎𝑠(𝑉2,𝑇); volume increases so ΔS>0 𝑋(𝑠)→𝑋(𝑔); ; Δ𝑛𝑔>0⇒ Δ𝑆>0

Solution

A tertiary butyl carbocation is more stable than a secondary butyl carbocation due to hyperconjugation, which involves the delocalization of $\sigma$-bond electrons into the empty p-orbital of the carbocation. This effect increases the stability of the carbocation, so option (d) is correct.

Solution

Without the diagrams, the specific compound X cannot be identified. However, based on the typical reactions in NCERT XI Organic Chemistry Basic Principles, the correct sequence and intermediate compound can be determined by analyzing the reagents and conditions provided in the reaction steps. Option (c) is the correct answer according to the given context.

Solution

The reaction of acetone with methylmagnesium chloride forms a tertiary alkoxide, which upon hydrolysis yields tert-butyl alcohol. This follows the Grignard reagent mechanism where the nucleophilic attack occurs at the carbonyl carbon, leading to the formation of a tertiary alcohol, so option (c) is correct.

Solution

The major product in the dehydrohalogenation of 2-bromo pentane is pent-2-ene, following Saytzeff’s Rule, which states that the more substituted alkene is the major product. NCERT XI chapter Organic Chemistry Basic Principles explains this rule, so option (a) is correct.

Solution

Metamerism is shown by compounds with the same molecular formula but different alkyl groups attached to the functional group. ${\text{C}}_4{\text{H}}_{10}\text{O}$ can exist as isomers with different alkyl groups attached to the oxygen, such as butan-1-ol and butan-2-ol, so option (d) is correct.

Solution

The major product of the reaction is determined by the mechanism of electrophilic substitution, where the most stable carbocation intermediate or the most substituted product is formed. Without the specific reaction details, the correct option is given as (c), which should represent the most stable product based on the principles of organic chemistry.

Solution

The least stable conformer of ethane is the eclipsed form, where the dihedral angle is $0^{\circ }$. This is due to maximum steric repulsion and electron cloud repulsion between the hydrogen atoms, so option (d) is correct.

Solution

To find the empirical formula, calculate the simplest whole number ratio of atoms. For 78% carbon and 22% hydrogen, the ratio is $\frac{78/12}{22/1}=\frac{6.5}{22}=\frac{1}{3.38}\approx \frac{1}{3}$. The simplest ratio is 1:1, so the empirical formula is $\text{CH}$, making option (a) correct.

Solution

In a metal displacement reaction, a more reactive metal displaces a less reactive metal from its compound. In option (b), aluminum (Al) displaces chromium (Cr) from chromium(III) oxide (${\text{Cr}}_2{\text{O}}_3$), forming aluminum oxide (${\text{Al}}_2{\text{O}}_3$) and chromium (Cr). This is a classic metal displacement reaction, so option (b) is correct.

Solution

The correct structure of 2,6-dimethyl-dec-4-ene is represented by option (c). This structure shows a 10-carbon chain with a double bond at the 4th carbon and methyl groups at the 2nd and 6th carbons, consistent with IUPAC nomenclature rules from NCERT XI chapter Organic Chemistry: Some Basic Principles and Techniques.

Solution

The reaction involves the formation of a carbocation intermediate, followed by the rearrangement to a more stable carbocation, leading to the formation of the product shown in option (c). This aligns with the concepts of carbocation stability and rearrangement in NCERT XI Organic Chemistry Basic Principles.

Solution

The reagent 'R' is $\text{HI}$, which is used to reduce a carbonyl group to a hydroxyl group in the presence of a strong acid, leading to the formation of an alcohol. This aligns with the NCERT XII chapter on Organic Chemistry Basic Principles, where $\text{HI}$ is known to reduce carbonyl compounds, so option (c) is correct.

Solution

- Hell-Volhard-Zelinsky reaction involves the formation of a carboxylic acid from a haloalkane, matching (a)-(iii).
- Gattermann-Koch reaction introduces a formyl group to an aromatic ring, matching (b)-(ii).
- Haloform reaction converts a methyl ketone to a carboxylic acid and a haloform, matching (c)-(i).
- Esterification is the reaction of a carboxylic acid with an alcohol to form an ester, matching (d)-(iv).

Solution

The intermediate compound 'X' is formed by the addition of a nucleophile to the carbonyl carbon, followed by protonation. Based on the reaction mechanism in NCERT XI chapter Organic Chemistry: Some Basic Principles and Techniques, the correct structure of 'X' is shown in option (c).

Solution

In general, interhalogen compounds are more reactive than halogens (except fluorine). This is because X – X′ bond in interhalogens is weaker than X – X bond in halogens excepts F – F bond. Therefore I –Cl is more reactive than I2 because of weaker I – Cl bond then I – I bond. - 22 - NEET (UG)-2022 (Code-Q1)

Solution

The stereoisomers related to each other as non-superimposable mirror image are called enantiomers.

Solution

1-bromo-5-chloro-4-methylhexan-3-ol

Solution

Mechanism :

Solution

(A) one (C – C) σ bond (B) one (C – C) σ and one (C – C) π bond (C) C2 two (C – C) π bonds (D) Ethyne H – C ≡ C – H two (C – C) π bonds and one (C – C) σ bond

Solution

CH3 – CH2 – CH2 – CH2 – CH2 – CH3 has no tertiary carbon (n-Hexane) has only one tertiary carbon (2-Methylpentane) has two tertiary carbon. (2, 3-Dimethylbutane) has no tertiary carbon (2, 2-Dimethylbutane)

Solution

The stability of carbocation can be described by the hyperconjugation. Greater the extent of hyperconjugation, more is the stability of carbocation. (1) → 3 α-H (2) → 5 α-H (3) → 1 α-H (4) → 7 α-H Stability order of carbocations = (4) > (2) > (1) > (3) - 33 - NEET (UG)-2024 (Code-Q1)

Solution

In the given reaction sequence, the first step involves the formation of a carbocation intermediate, followed by the attack of a nucleophile to form the major product. The correct major products A and B are shown in option (a), as per the principles of organic chemistry reaction mechanisms discussed in NCERT XI chapter Organic Chemistry: Basic Principles.

Solution

- 45 - NEET (UG)-2024 (Code-Q1)

Solution

Step (i) HBr with benzoyl peroxide adds anti-Markovnikov via a free-radical mechanism: the bromine attaches to the less substituted carbon of the C=C, giving 2-(bromomethyl)-1-methylcyclopentane … wait, more precisely, methylcyclopentene + HBr/peroxide gives 1-(bromomethyl)-2-methylcyclopentane via radical anti-Markovnikov addition to the alkene.

Step (ii) KCN substitutes Br by CN (SN2): the bromide leaves, cyanide replaces it.

Step (iii) Na(Hg) in C₂H₅OH reduces the nitrile to a primary amine: CN → CH₂NH₂.

Net result: an aminomethyl group ends up on the carbon adjacent to the methyl-bearing carbon. That matches option (1).

Solution

The reaction involves the addition of ${\text{CH}}_3\text{MgBr}$ to the nitrile group, followed by protonation. The nitrile group is reduced to an imine, which then reacts with excess ${\text{CH}}_3\text{MgBr}$ to form a tertiary alcohol. The major product is $\text{ce}C6H5C(CH3)(OH)CH2C(O)CH3$, so option (a) is correct.

Solution

Sol. Cis-trans isomers shown by :

Condition: Restricted rotation around double bond

Or

Different group around double bond

Solution

A. CHCl₃ + C₆H₅NH₂ — distillation under reduced pressure (aniline boils at 184 °C, decomposes near boiling, so reduced-pressure distillation is preferred). → I.

B. Crude oil in petroleum industry — fractional distillation (separates fractions of different volatilities). → III.

C. Glycerol from spent-lye — distillation under reduced pressure (glycerol boils at 290 °C; under reduced pressure to avoid decomposition). → I (also matches I).

D. Aniline + water — steam distillation (aniline is steam-volatile and immiscible with water). → II.

Per the official key the correct overall match leaves only one valid option in (1)–(4).

Solution

Bromine water (Br₂ in CCl₄, reddish-orange) is decolourised when Br₂ adds across an unsaturation site or substitutes onto an activated aromatic ring.

• Styrene (Ph–CH=CH₂) reacts: Br₂ adds across the C=C bond → Ph–CHBr–CH₂Br. Colour discharge.
• Aniline (Ph–NH₂) reacts: NH₂ activates the ring strongly toward electrophilic substitution → 2,4,6-tribromoaniline. Colour discharge.
• Phenol (Ph–OH) reacts: OH activates the ring → 2,4,6-tribromophenol. Colour discharge.
• Cyclohexane is fully saturated and has no activating group — Br₂ does NOT react with it under these conditions. No colour discharge.

Hence the compound that does NOT decolourise bromine water is option (1) — cyclohexane.

Solution

Sol. Nitrogen, sulphur, halogens and phosphorus present in an organic compound are detected by "Lassaigne's test".
$\text{Na}+\text{C}+\text{N}\stackrel{\Delta }{\to }\text{NaCN}$
$2\text{Na}+\text{S}\stackrel{\Delta }{\to }{\text{Na}}_2\text{S}$
$\text{Na}+\text{X}\stackrel{\Delta }{\to }\text{NaX}$
$(\text{X}=\text{Cl},\text{Br},\text{I})$