Acetylenic hydrogens are acidic because 
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.
Which is the most suitable reagent among the following to distinguish compound (3) from rest of the compounds ? 
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).
Select the true statement about benzene amongst the following 
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.
The reagent is 
On heating ethylene chloride (1, 1 dichloro ethane) with alcoholic potash followed by sodamide alkyne is obtained
Reduction of 2-butyne with sodium in liquid ammonia gives predominantly : 
Reduction of alkynes with Na/liq. NH3 gives trans-alkenes. This reaction is called Birch reduction
A compound is treated with NaNH2 to give sodium salt. Identify the compound 
Only C2H2 (acetylene) has acidic H-atoms and hence reacts with NaNH2 to form sodium salt, i.e.,
Reactivity of hydrogen atoms attached to different carbon atoms in alkanes has the order 
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.
When hydrochloric acid gas is treated with propene in presence of benzoyl peroxide, it gives 
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 :
Which of the following compounds has the lowest boiling point ? 
Of all the options listed CH3CH2CH2CH3 has the least number of C-atoms and hence has the lowest b.p.
When CH3Cl and AlCl3 are used in Friedel-Crafts reaction, the electrophile is 
In the free radical chlorination of methane, the chain initiating step involves the formation of
The alkene R – CH = CH2 reacts readily with B2H6 and formed the product B which on oxidation with alkaline hydrogen peroxides produces 
When 3, 3-dimethyl 2-butanol is heated with H2SO4, the major product obtained is 
When 3, 3 dimethyl 2-butanol is heated with H2SO4 the major product obtained is 2, 3 dimethyl 2-butene.
In the presence of platinum catalyst, hydrocarbon A adds hydrogen to form n-hexane. When hydrogen bromide is added to A instead of hydrogen, only a single bromo compound is formed. Which of the following is A? 
In reaction sequence 
molecule 'M' and reagent 'R' respectively are
We know that
Which one of the following reactions is expected to readily give a hydrocarbon product in good yields ? 
Electrolysis of a concentrated a queous solution of either sodium or potassium salts of saturated mon-carboxylic acids yields higher alkane at anode.
In commercial gas olines the type of hydrocarbons which are more desirable, is
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.
A hydrocarbon A on chlorination gives B which on heating with alcoholic potassium hydroxide changes into another hydrocarbon C. The latter decolourises Baeyer's reagent and on ozonolysis forms formaldehyde only. A is 
Since hydrocarbon C give only CH2O, on ozonolysis, C should be CH2 = CH2 hence going backward A should be ethane. Thus the reactions are
The reaction of ethyl magnesium bromide with water would give 
Which of the following reagents convert propene to 1-propanol? 
We know that
For the formation of toluene by Friedal Craft reaction, reactants used in presence of anhydrous AlCl3 are 
In preparation of alkene from alcohol using Al2O3 which is effective factor? 
While at 220º – 250ºC it forms ether
Which alkene on ozonolysis gives CH3CH2CHO and
The compound 
on reaction with NaIO4 in the presence of KMnO4 gives
Reaction of HBr with propene in the presence of peroxide gives 
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.
Using anhydrous AlCl3 as catalyst, which one of the following reactions produces ethylbenzene (PhEt)?
Which one of the following alkenes will react faster with H2 under catalytic hydrogenation conditions?  (R = Alkyl Substituent)
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.
Products of the following reaction : 
The glyoxal formed as an intermediate is oxidised by H2O2 to give the acids.`
Predict the product C obtained in the following reaction of butyne-1. 
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.
Which of the compounds with molecular formula C5H10 yields acetone on ozonolysis? 
A(predominantly) is : 
We know that in case of an unsymmetrical alkene there is the possibility of forming two products. In such cases the formation of major product is decided on the basis of Markownikoffs rule which is rationalized in terms of stability of the intermediate carbocation. Also remember that 3° carbocation is more stable than 2° carbocation and 2° carbocation is more stable than 1° carbocation.
of the two possibilities 2° carbocation is more stable so the product of the reaction expected was predominantly one formed by 2° carbocation i.e.
i.e. 2– Bromo-3-Methylbutane
However some electrophilic addition reaction form products that are clearly not the result of the addition of electrophile to the sp2 carbon bonded to the most hydrogens and the addition of a nucleophile to the other sp2 carbon.
In the above cases the addition of HBr to 3-methyl-1-butene the two products formed are shown below.
In this case the major product formed is 2- Bromo-2-methylbutane i.e. option (b) is correct answer. (Note: The unexpected product results from a rearrangement of carbocation intermediate. Please note that all carbocation do not rearrange.
The IUPAC name of the compound having the formula CH ≡ C – CH = CH2 is : 
If both the double and triple bonds are present the compound is regarded as derivative of alkyne.
Further if double and triple bonds are at equidistance from either side, the preference is given to double bond.
Liquid hydrocarbons can be converted to a mixture of gaseous hydrocarbons by : 
During cracking higher hydrocarbons (liquid) are converted to lower gaseous hydrocarbons.
In the following reaction, C6H5CH2Br
In the following reaction, C6H5CH2Br
The IUPAC name of the compound is
If a molecule contains both carbon-carbon double or triple bonds, the two are treated as per in seeking the lowest number combination. However, if the sum of numbers turns out to be the same starting from either of the carbon chain, then lowest number is given to the C = C double bond.
In the following reactions, 
the major products (A) and (C) are respectively :
In this case dehydration is governed by Saytezeff’s rule according to which hydrogen is preferentially eliminated from the carbon atom with fewer number of hydrogen atoms i.e., poor becomes poorer. Thus, 2 methyl butene-2 is the major product.
This reaction is governed by Markownikoff’s rule according to which when an unsymmetrical reagent e.g.
HBr adds to an unsymmetrical alkene, then the negative part of the reagent is added to that carbon atom of the double bond which bears the least number of hydrogen atom. Thus, in above case. 2-methyl 2-bromo butane will be the major product.
The reaction of C6H5CH = CHCH3 with HBr produces :
This is the electrophilic addition reaction in which addition takes place via more stable carbocation according to the Markovnikov's rule.
Which of the following reagents will be able to distinguish between 1-butyne and 2-butyne? [2012 M]
1-Butyne and 2-butyne are distinguish by NaNH2 because 1-Butyne react with NaNH2 due to presence of terminal hydrogen.
The radical,is aromatic because ithas :
Presence of 6p orbitals, each containing one unpaired electron, in a six membered cyclic structure is in accordance with Huckel rule of aromaticity.
In the following reaction :
Product ‘P’ will not give [NEET Kar. 2013]
CH3CHO does not give Victor Meyer test.
Which of the following chemical system is non aromatic?
[NEET Kar. 2013]
Huckel rule is not obeyed. It has only four electrons. Further it does not have continous conjugation.