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Hydrocarbons

JEE Chemistry · 35 questions · Page 4 of 4 · Click an option or "Show Solution" to reveal answer

Q31
Toluene is nitrated and the resulting product is reduced with tin and hydrochloric acid. The product so obtained is diazotised and then heated with cuprous bromide. The reaction mixture so formed contains
A mixture of o− and p−bromotoluenes
B mixture of o− and p−dibromobenzenes
C mixture of o− and p−bromoanilines
D mixture of o− and m−bromotoluenes
Correct Answer
Option A
Solution

on reduction with

Sn/HClSn/HCl

they will form corresponding anilines in which

−NO2- N{O_2}

group changes to

−NH2.- N{H_2}.

These anilines when diazotized and then treated with

CuBrCuBr

forms

o−,p−o-, p-

bromotoluenes.

Q32
Of the five isomeric hexanes, the isomer which can give two monochlorinated compounds is
A n-hexane
B 2, 3-dimethylbutane
C 2,2-dimethylbutane
D 2-methylpentane
Correct Answer
Option B
Solution

Since it contains only two types of

HH

-atoms hence it will give only two mono chlorinated compounds viz. and

Q33
Which of the following reactions will yield 2, 2-dibromopropane?
A CH3 - CH = CH2 + HBr →\to
B CH3 - C ≡\equiv CH + 2HBr →\to
C CH3CH = CHBr + HBr →\to
D CH ≡\equiv CH + 2HBr →\to
Correct Answer
Option B
Solution

The reaction follows Markownikoff rule which states that when unsymmetrical reagent adds across unsymmetrical double or triple bond the negative part adds to carbon atom having lesser number of hydrogen atoms.

Q34
The alkane from below having two secondary hydrogens is :
A 2,2,4,5-Tetramethylheptane
B 2,2,4,4-Tetramethylhexane
C 4-Ethyl-3,4-dimethyloctane
D 2,2,3,3-Tetramethylpentane
Correct Answer
Option D
Solution

4-Ethyl-3,4-dimethyloctane

Q35
Cl∙+CH4→A+B \stackrel{\bullet}{\mathrm{Cl}}+\mathrm{CH}_{4} \rightarrow \mathrm{A}+\mathrm{B} A and B in the above atmospheric reaction step are :
A C2H6\mathrm{C}_{2} \mathrm{H}_{6} and Cl2\mathrm{Cl}_{2}
B C∙HCl2\stackrel{\bullet}{\mathrm{C}} \mathrm{HCl}_{2} and H2\mathrm{H}_{2}
C C∙H3\stackrel{\bullet}{\mathrm{C}}\mathrm{H}_{3} and HCl\mathrm{HCl}
D C2H6\mathrm{C}_{2} \mathrm{H}_{6} and HCl\mathrm{HCl}
Correct Answer
Option C
Solution
Cl∙+CH4⟶C∙H3+HCl\stackrel{\bullet}{\mathrm{Cl}}+\mathrm{CH}_4 \longrightarrow {\stackrel{\bullet}{\mathrm{C}}}\mathrm{H}_3+\mathrm{HCl}
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