Question

Which among the following will have highest density?

• A. \(CCl_4\)
• B. \(n-C_3H_7I\)
• C. \(n-C_3H_7Br\)
• D. \(n-C_3H_7Cl\)

Hint:

For alkyl halides, density generally increases with increase in atomic mass of halogen: \[ RI \gt RBr \gt RCl \gt RF \]

Answer 1

The Correct Option is B

Step 1: Recall what controls density in haloalkanes.
For similar organic halides, density rises as the molecule gets heavier, which mainly depends on the mass of the halogen atom present.
Step 2: Order the halogens by atomic mass.
The halogens increase in mass as $F < Cl < Br < I$, so iodine compounds tend to be the densest.
Step 3: Compare the propyl halides.
Among $n\text{-}C_3H_7Cl$, $n\text{-}C_3H_7Br$ and $n\text{-}C_3H_7I$, the same propyl group carries chlorine, bromine and iodine respectively, so density follows $Cl < Br < I$. The iodide is the densest of these three.
Step 4: Bring in $CCl_4$.
$CCl_4$ carries four chlorine atoms and is fairly dense, about $1.59\;g\;mL^{-1}$, but each chlorine is much lighter than a single iodine.
Step 5: Compare $CCl_4$ with $n\text{-}C_3H_7I$.
$n$-propyl iodide has a density of roughly $1.74\;g\;mL^{-1}$ because the heavy iodine atom packs a large mass into a fairly small molecule, exceeding that of $CCl_4$.
Step 6: Conclude.
The compound with the highest density is therefore
\[ \boxed{n\text{-}C_3H_7I} \]