31 Year NEET Previous Questions: Hydrocarbons Free MCQ Practice Test [Updated]

The acidity of acetylene or 1--alkynes can be explained on the basis of molecular orbital concept according to which formation of C—H bond in acetylene involves sphybridised carbon atom. Now since s electrons are closer to the nucleus than p electrons, the electrons present in a bond having more s character will be correspondingly more closer to the nucleus.
Thus owing to high s character of the C—H bond in alkynes (s = 50%), the electrons constituting this bond are more strongly held by the carbon nucleus (i.e., the acetylenic carbon atom or the sp orbital acts as more electronegative species than the sp² and sp³ with the result the hydrogen  present on such a carbon atom (≡C—H) can be esily removed as a proton.

Br₂ in CCl₄ (a), Br₂ in CH₃ COOH (b) and alk.
KMnO₄ (c) will react with all unsaturated compounds, i.e., 1, 3 and 4 while ammonical AgNO₃ (d) reacts only with terminal alkynes, i.e., 3 and hence compund 3 can be distinguished from 1, 2 and 4 by. ammonical AgNO₃ (d).

Benzene do not show addition reaction like other unsaturated hydrocarbons. However it show substitution reactions. Due to resonance all the C – C bonds have the same nature, which is possible because of the cycli c del oca lisati on of π-electr on s in benzene. Monosubstitution will give only a single product.

On heating ethylene chloride (1, 1 dichloro ethane) with alcoholic potash followed by sodamide alkyne is obtained

Reduction of alkynes with Na/liq. NH₃ gives trans-alkenes. This reaction is called Birch reduction

Only C₂H₂ (acetylene) has acidic H-atoms and hence reacts with NaNH₂ to form sodium salt, i.e.,

The reactivity of H-atoms depends upon the stability of free radicals follows the order : Tertiary > secondary > primary, therefore, reactivity of H-atoms follows the same order, i.e., tertiary > secondary > primary.

Peroxide effect is observed only in case of HBr. Therefore, addition of HCl to propene even in the presence of benzyoyl peroxide occurs according to Markovnikov’s rule :

Of all the options listed CH₃CH₂CH₂CH₃ has the least number of C-atoms and hence has the lowest b.p.

When 3, 3 dimethyl 2-butanol is heated with H₂SO₄ the major product obtained is 2, 3 dimethyl 2-butene.

We know that

Electrolysis of a concentrated a queous solution of either sodium or potassium salts of saturated mon-carboxylic acids yields higher alkane at anode.

Gasoline (petrol) is a mixture of alkanes, alkenes and aromatic hydrocarbons. The quality of a gasoline is determined by the amount of branched chain hydrocarbons (2,2,4-trimethylpentane, commonly known as iso-octane) present in it.

Given

Since hydrocarbon C give only CH₂O, on ozonolysis, C should be CH₂ = CH₂ hence going backward A should be ethane. Thus the reactions are

We know that

Friedel-Crafts Alkylation Reaction:

This reaction is used to introduce an alkyl group (like methyl, ethyl, etc.) onto an aromatic ring, such as benzene.

Reactants:

• Benzene (C₆H₆)

• Alkyl halide (e.g., CH₃Cl — methyl chloride)

• Lewis acid catalyst (AlCl₃ — aluminum chloride)

Step-by-Step Mechanism:

Step 1: Formation of the Electrophile

The alkyl halide (CH₃Cl) reacts with the Lewis acid (AlCl₃):

CH₃Cl + AlCl₃ → CH₃⁺ (methyl carbocation) + AlCl₄⁻

This step generates a carbocation (CH₃⁺), which acts as the electrophile in the reaction.

Step 2: Attack by Benzene

The benzene ring, which is rich in electrons, donates a pair of electrons to the methyl carbocation:

C₆H₆ + CH₃⁺ → C₆H₅CH₃ (toluene)

This forms a new carbon-carbon bond between the benzene ring and the methyl group.

Step 3: Regeneration of Catalyst

A proton (H⁺) is removed from the benzene ring to restore aromaticity. The AlCl4⁻ ion helps in removing this proton:

AlCl⁴⁻ + H⁺ → AlCl₃ + HCl

So, the catalyst AlCl3 is regenerated and can be reused.

Overall Reaction:

C6H6 + CH3Cl → C6H5CH3 (Toluene) + HCl

(in the presence of AlCl3)

Summary:

• The key step is the generation of a carbocation (CH3⁺).

• Benzene attacks this electrophile to form an alkylated product.

• AlCl₃ acts as a catalyst and is regenerated at the end.

Temperature

While at 220º – 250ºC it forms ether

Oxidation of the C=C bond by KMnO₄ gives a vicinal diol, and cleavage of that diol by NaIO₄ breaks the C-C bond to form two carbonyl fragments.

The more substituted double-bond carbon (the carbon without a hydrogen) is converted to a ketone, giving CH₃COCH₃ (acetone).

The less substituted double-bond carbon (the carbon bearing one hydrogen) is converted to an aldehyde, initially giving CH₃CHO (acetaldehyde).

Because KMnO₄ is present under the oxidative conditions, the formed aldehyde is further oxidised to the corresponding carboxylic acid, CH₃COOH (acetic acid). Thus the net products are CH₃COCH₃ and CH₃COOH.

In presence of peroxide, HBr adds on alkenes in anti–markovnikov’s way, thus

Kharasch observed that the addition of HBr to unsymmetrical alkene in the presence of organic peroxides follows an opposite course to that suggested by Markownikoff . This is termed anti-Markownikoff or peroxide effect.

The least stable alkene will react will reacts fastest. So in option d, we have 2 pair of alkyl groups in cis position. SO it will react fastest.

The glyoxal formed as an intermediate is oxidised by H₂O₂ to give the acids.`

This reaction occurs according to Markownikoff’s rule which states that when an unsymmetrical alkene undergo hydrohalogenation, the negative part goes to that C-atom which contain lesser no. of Hatom.