Chapter 25

A Complete Resource Book in Chemistry for JEE Main · 124 exercises

Problem 176

\(2-\) chlorobutane is treated with alcoholic \(\mathrm{KOH}\). Number of alkenes formed is

5 step solution

Problem 178

\(0-\) Xylene is methylated to get trimethyl benzene. The number of possible isomers is

5 step solution

Problem 179

Find the total number of monochlorinated products formed from the pure reactant given below: C=CC(C)C \(\left(-14^{14} \mathrm{C}-\right.\) atom \()\)

4 step solution

Problem 180

\(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3} \stackrel{\mathrm{HBr}}{\longrightarrow}\) product (s). Number of products possible in the above reaction is

5 step solution

Problem 183

Of the following reagents the number of them which will react with \(\mathrm{CH}_{3} \mathrm{MgBr}\) to give \(\mathrm{CH}_{4}\) is \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}, \mathrm{CH}_{3} \mathrm{SH}, \mathrm{CH}_{3} \mathrm{NH}_{2}, \mathrm{RC} \equiv \mathrm{CR}, \mathrm{HCHO}\) \(\mathrm{RC} \equiv \mathrm{CH}\)

3 step solution

Problem 184

Total number of monochlorination products that can be obtained on treating 1 - butene with \(\mathrm{Cl}_{2}\) in presence of light is

5 step solution

Problem 185

In an experimental determination of active hydrogens in an organic compound (Z), \(0.450 \mathrm{~g}\) of compound (Z) was treated with excess of \(\mathrm{CH}_{3} \mathrm{MgBr}\) when \(224 \mathrm{ml}\) of \(\mathrm{CH}_{4}\) was evolved at NTP. Compound (Z) has molar mass equal to \(90 \mathrm{~g}\) per mole. How many active hydrogens are present per molecule of the compound (Z)?

5 step solution

Problem 190

On mixing a certain alkane with chlorine and irradiating it with ultraviolet light, it forms only one monochloroalkane. This alkane could be (a) propane (b) pentane (c) isopentane (d) neopentane

4 step solution

Problem 191

Which one of the following has the minimum boiling point? (a) \(\mathrm{n}\)-butane (b) 1-butyne (c) 1 -butene (d) iso-butene

5 step solution

Problem 192

Acid catalysed hydration of alkenes except ethene leads to the formation of (a) primary alcohol (b) secondary or tertiary alcohol (c) mixture of primary and secondary alcohols (d) mixture of secondary and tertiary alcohols

4 step solution

Problem 193

Elimination of bromine from 2-bromobutane results in the formation of (a) equimolar mixture of 1 and 2 -butene (b) predominantly 2 -butene (c) predominantly 1-butene (d) predominantly 2-butyne

5 step solution

Problem 194

Reaction of one molecule of HBr with one molecule of 1,3 -butadiene at \(40^{\circ} \mathrm{C}\) gives predominantly [2005] (a) 3-bromobutene under kinetically controlled conditions (b) 1-bromo-2-butene under thermodynamically controlled conditions(c) 3-bromobutene under thermodynamically controlled conditions (d) 1-bromo-2-butene under kinetically controlled conditions.

4 step solution

Problem 195

HBr reacts with \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{OCH}_{3}\) under anhydrous conditions at room temperature to give (a) \(\mathrm{CH}_{3} \mathrm{CHO}\) and \(\mathrm{CH}_{3} \mathrm{Br}\) (b) \(\mathrm{BrCH}_{2} \mathrm{CHO}\) and \(\mathrm{CH}_{3} \mathrm{OH}\) (c) \(\mathrm{BrCH}_{2}-\mathrm{CH}_{2}-\mathrm{OCH}_{3}\) (d) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{CHBr}-\mathrm{OCH}_{3}\)

5 step solution

Problem 196

The structure of the compound that gives a tribromo derivative on treatment with bromine water is [2006]\\} (a) Cc1cccc(O)c1 (b) OCc1ccccc1 (c) Cc1ccccc1O (d) Cc1ccc(O)cc1

4 step solution

Problem 198

The compound formed as a result of oxidation of ethyl benzene by \(\mathrm{KMnO}_{4}\) is: [2007] (a) acetophenone (b) benzoic acid (c) benzyl alcohol (d) benzophenone

4 step solution

Problem 200

The treatment of \(\mathrm{CH}_{3} \mathrm{MgX}\) with \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{C}-\mathrm{H}\) pro- duces (a) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}\) (b) \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\) (c) CC=CC (d) \(\mathrm{CH}_{4}\)

4 step solution

Problem 202

In the following sequence of reactions, the alkene affords the compound 'B' \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{3} \stackrel{\mathrm{O}_{3}}{\longrightarrow}(\mathrm{A}) \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow}\) (B) \(\mathrm{Zn}\) The compound (B) is [2008] (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}\) (b) \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{3}\) (d) \(\mathrm{CH}_{3} \mathrm{CHO}\)

6 step solution

Problem 205

The alkene that exhibits geometrical isomerism is: [2009] (a) 2-methyl propene (b) 2 -butene (c) 2-methyl-2-butene (d) propene

6 step solution

Problem 206

vOne mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molcular mass of \(44 \mathrm{u}\). The alkene is (a) Propene (b) 1 -butene (c) 2 -butene (d) ethane

5 step solution

Problem 207

The main product of the following reaction is \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2} \stackrel{\text { conc. } \mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow}\) ? (a) CC=CC(C)C (b) CC(C)=CCc1ccccc1 (c) CC(C)C=CCc1ccccc1 (d) CCC(Cl)(Cl)CCl

5 step solution

Problem 208

Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of (a) a vinyl group (b) an isopropyl group (c) an acetylenic triple bond (d) two ethylenic double bonds

4 step solution

Problem 209

2-Hexyne gives trans \(-2\) - Hexene on treatment with [2012] (a) \(\mathrm{Pd} / \mathrm{BaSO}_{4}\) (b) \(\mathrm{Li} / \mathrm{NH}_{3}\) (c) \(\mathrm{Pt} / \mathrm{H}_{2}\) (d) \(\mathrm{LiAlH}_{4}\)

3 step solution

Problem 210

Which branched chain isomer of the hydrocarbon with molecular mass \(72 \mathrm{u}\) gives only one isomer of mono substituted alkyl halide? [2012] (a) Isohexane (b) Neopentane (c) Tertiary butyl chloride

5 step solution

Problem 211

A gaseous hydrocarbon given upon combustion \(0.72 \mathrm{~g}\) of water and \(3.08 \mathrm{~g}\) of \(\mathrm{CO}_{2}\). The empirical formula of the hydrocarbon is (a) \(\mathrm{C}_{6} \mathrm{H}_{5}\) (b) \(\mathrm{C}_{7} \mathrm{H}_{8}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (d) \(\mathrm{C}_{3} \mathrm{H}_{4}\)

6 step solution

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