(A) Pyrolysis (or cracking) is the process of breaking down higher alkanes into lower alkanes and alkenes by heating them in the absence of air. For e

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Pyrolysis of higher carbon number of hydrocarbons

(A) Pyrolysis (or cracking) is the process of breaking down higher alkanes into lower alkanes and alkenes by heating them in the absence of air. For example: $C_6H_{14} \xrightarrow[\Delta]{\text{Pyrolysis}} C_2H_4 + C_4H_{10}$ This is the most effective industrial method for obtaining hydrocarbons with a lower carbon number from higher ones.

Class 11 Chemistry Hydrocarbons Alkane

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The most important method of preparation of hydrocarbons of lower carbon number is

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AIPMT 1989 All AIPMT

A
Pyrolysis of higher carbon number of hydrocarbons
B
Electrolysis of salts of fatty acids
C
Sabatier and Senderens reaction
D
Direct synthesis

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Class 11

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Hydrocarbons

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Alkane

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The general formula of an alkane is:

Easy

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In alkane compounds, the $C - H$ bond length and $H - C - H$ bond angle are respectively:

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Arrange the following alkanes in decreasing order of their heats of combustion:
$(i)$ $CH_3-C(CH_3)_2-CH_3$ (Neopentane)
$(ii)$ $CH_3-CH(CH_3)-CH_2-CH_3$ (Isopentane)
$(iii)$ $CH_3-CH_2-CH_2-CH_2-CH_3$ ($n$-pentane)

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When a mixture of methane and oxygen is passed through heated molybdenum oxide,the main product formed is

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Which branched chain isomer of the hydrocarbon with molecular mass $72 \ u$ gives only one isomer of mono-substituted alkyl halide?

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The most important method of preparation of hydrocarbons of lower carbon number is

Similar Questions

The general formula of an alkane is:

In alkane compounds, the $C - H$ bond length and $H - C - H$ bond angle are respectively:

Arrange the following alkanes in decreasing order of their heats of combustion:$(i)$ $CH_3-C(CH_3)_2-CH_3$ (Neopentane)$(ii)$ $CH_3-CH(CH_3)-CH_2-CH_3$ (Isopentane)$(iii)$ $CH_3-CH_2-CH_2-CH_2-CH_3$ ($n$-pentane)

When a mixture of methane and oxygen is passed through heated molybdenum oxide,the main product formed is

Which branched chain isomer of the hydrocarbon with molecular mass $72 \ u$ gives only one isomer of mono-substituted alkyl halide?

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Arrange the following alkanes in decreasing order of their heats of combustion:
$(i)$ $CH_3-C(CH_3)_2-CH_3$ (Neopentane)
$(ii)$ $CH_3-CH(CH_3)-CH_2-CH_3$ (Isopentane)
$(iii)$ $CH_3-CH_2-CH_2-CH_2-CH_3$ ($n$-pentane)