Arrange the compounds I,II,and III in decreas | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The heat of combustion of a hydrocarbon is directly proportional to the number of carbon atoms present in the molecule.Compound $I$ is ethylcyclop

Vedclass · Accepted Answer

$III > II > I$

Vedclass · Answer

(C) The heat of combustion of a hydrocarbon is directly proportional to the number of carbon atoms present in the molecule.Compound $I$ is ethylcyclopentane $(C_7H_{14})$,which has $7$ carbon atoms.Compound $II$ is ethylcyclohexane $(C_8H_{16})$,which has $8$ carbon atoms.Compound $III$ is ethylcycloheptane $(C_9H_{18})$,which has $9$ carbon atoms.Since the number of carbon atoms follows the order $III > II > I$,the heat of combustion also follows the same order: $III > II > I$.

Arrange the compounds $I$,$II$,and $III$ in decreasing order of their heats of combustion.

Similar Questions

Write the main product of the given reactions:
$(1)$ $C_6H_5CH = CH_2 \xrightarrow{HBr}$
$(2)$ $C_4H_{10} \xrightarrow{\text{air}, \Delta}$

What is the number of moles of $H$ atoms required for the complete reduction of one mole of acetonitrile?

Which of the following hydrocarbons has the lowest dipole moment?

The product $(A)$ is

$A$ sample of petrol is a mixture of $30\%$ $n$-heptane and $70\%$ iso-octane. The sample has an octane number of:

Vedclass Test Series

Exam Paper Generator

Online Exam Module